English Logic Puzzle












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On this site, I found this:




"If some Smaugs are Thors and some Thors are Thrains, then some Smaugs are definitely Thrains."




Options: TRUE, FALSE, NEITHER



The answer given is FALSE. However, I need convincing that the answer isn't 'NEITHER'. How can I see this?



(I got all the rest right btw)










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  • $begingroup$
    Can you explain why it would be "neither"?
    $endgroup$
    – jafe
    yesterday










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    I am using $Acap Bneemptyset, Bcap Cneemptyset implies Acap Cneemptyset$, and I have a counter-example, but I also have an example where it is true, so it is both true and false.
    $endgroup$
    – JonMark Perry
    yesterday








  • 7




    $begingroup$
    If we have a counterexample, the claim should be false, no? I.e. it's not true that some Smaugs are definitely Thrains.
    $endgroup$
    – jafe
    yesterday






  • 2




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    Not seeing the puzzle, just an english/logic question.
    $endgroup$
    – deep thought
    yesterday






  • 1




    $begingroup$
    Yeah off-topic though, isn't it? Why is this a puzzle, or about puzzles.
    $endgroup$
    – deep thought
    yesterday
















3












$begingroup$


On this site, I found this:




"If some Smaugs are Thors and some Thors are Thrains, then some Smaugs are definitely Thrains."




Options: TRUE, FALSE, NEITHER



The answer given is FALSE. However, I need convincing that the answer isn't 'NEITHER'. How can I see this?



(I got all the rest right btw)










share|improve this question









$endgroup$












  • $begingroup$
    Can you explain why it would be "neither"?
    $endgroup$
    – jafe
    yesterday










  • $begingroup$
    I am using $Acap Bneemptyset, Bcap Cneemptyset implies Acap Cneemptyset$, and I have a counter-example, but I also have an example where it is true, so it is both true and false.
    $endgroup$
    – JonMark Perry
    yesterday








  • 7




    $begingroup$
    If we have a counterexample, the claim should be false, no? I.e. it's not true that some Smaugs are definitely Thrains.
    $endgroup$
    – jafe
    yesterday






  • 2




    $begingroup$
    Not seeing the puzzle, just an english/logic question.
    $endgroup$
    – deep thought
    yesterday






  • 1




    $begingroup$
    Yeah off-topic though, isn't it? Why is this a puzzle, or about puzzles.
    $endgroup$
    – deep thought
    yesterday














3












3








3


1



$begingroup$


On this site, I found this:




"If some Smaugs are Thors and some Thors are Thrains, then some Smaugs are definitely Thrains."




Options: TRUE, FALSE, NEITHER



The answer given is FALSE. However, I need convincing that the answer isn't 'NEITHER'. How can I see this?



(I got all the rest right btw)










share|improve this question









$endgroup$




On this site, I found this:




"If some Smaugs are Thors and some Thors are Thrains, then some Smaugs are definitely Thrains."




Options: TRUE, FALSE, NEITHER



The answer given is FALSE. However, I need convincing that the answer isn't 'NEITHER'. How can I see this?



(I got all the rest right btw)







english logic-theory






share|improve this question













share|improve this question











share|improve this question




share|improve this question










asked yesterday









JonMark PerryJonMark Perry

18.9k63891




18.9k63891












  • $begingroup$
    Can you explain why it would be "neither"?
    $endgroup$
    – jafe
    yesterday










  • $begingroup$
    I am using $Acap Bneemptyset, Bcap Cneemptyset implies Acap Cneemptyset$, and I have a counter-example, but I also have an example where it is true, so it is both true and false.
    $endgroup$
    – JonMark Perry
    yesterday








  • 7




    $begingroup$
    If we have a counterexample, the claim should be false, no? I.e. it's not true that some Smaugs are definitely Thrains.
    $endgroup$
    – jafe
    yesterday






  • 2




    $begingroup$
    Not seeing the puzzle, just an english/logic question.
    $endgroup$
    – deep thought
    yesterday






  • 1




    $begingroup$
    Yeah off-topic though, isn't it? Why is this a puzzle, or about puzzles.
    $endgroup$
    – deep thought
    yesterday


















  • $begingroup$
    Can you explain why it would be "neither"?
    $endgroup$
    – jafe
    yesterday










  • $begingroup$
    I am using $Acap Bneemptyset, Bcap Cneemptyset implies Acap Cneemptyset$, and I have a counter-example, but I also have an example where it is true, so it is both true and false.
    $endgroup$
    – JonMark Perry
    yesterday








  • 7




    $begingroup$
    If we have a counterexample, the claim should be false, no? I.e. it's not true that some Smaugs are definitely Thrains.
    $endgroup$
    – jafe
    yesterday






  • 2




    $begingroup$
    Not seeing the puzzle, just an english/logic question.
    $endgroup$
    – deep thought
    yesterday






  • 1




    $begingroup$
    Yeah off-topic though, isn't it? Why is this a puzzle, or about puzzles.
    $endgroup$
    – deep thought
    yesterday
















$begingroup$
Can you explain why it would be "neither"?
$endgroup$
– jafe
yesterday




$begingroup$
Can you explain why it would be "neither"?
$endgroup$
– jafe
yesterday












$begingroup$
I am using $Acap Bneemptyset, Bcap Cneemptyset implies Acap Cneemptyset$, and I have a counter-example, but I also have an example where it is true, so it is both true and false.
$endgroup$
– JonMark Perry
yesterday






$begingroup$
I am using $Acap Bneemptyset, Bcap Cneemptyset implies Acap Cneemptyset$, and I have a counter-example, but I also have an example where it is true, so it is both true and false.
$endgroup$
– JonMark Perry
yesterday






7




7




$begingroup$
If we have a counterexample, the claim should be false, no? I.e. it's not true that some Smaugs are definitely Thrains.
$endgroup$
– jafe
yesterday




$begingroup$
If we have a counterexample, the claim should be false, no? I.e. it's not true that some Smaugs are definitely Thrains.
$endgroup$
– jafe
yesterday




2




2




$begingroup$
Not seeing the puzzle, just an english/logic question.
$endgroup$
– deep thought
yesterday




$begingroup$
Not seeing the puzzle, just an english/logic question.
$endgroup$
– deep thought
yesterday




1




1




$begingroup$
Yeah off-topic though, isn't it? Why is this a puzzle, or about puzzles.
$endgroup$
– deep thought
yesterday




$begingroup$
Yeah off-topic though, isn't it? Why is this a puzzle, or about puzzles.
$endgroup$
– deep thought
yesterday










12 Answers
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13












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In more an English sense than a Puzzling one...



In the question there is the keyword:




definitely.




That means if 100% true, then the statement is true. else it is false, even it is true 99%.




In a sentence with "definitely", neither is the first choice that should go out of your mind. then you choose from true or false. hope this helps!




In other words:




this is a question of definitely true or not definitely true (maybe true), not definitely true, definitely false, or maybe true




Hope this helps!






share|improve this answer









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  • 1




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    Even without "definitely", the answer is still "false"...
    $endgroup$
    – BlueRaja - Danny Pflughoeft
    yesterday










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    @BlueRaja-DannyPflughoeft No, it's not. Some Smaugs being Thrains is consistent with some Smaugs being Thors and some Thors being Thrains.
    $endgroup$
    – Acccumulation
    yesterday










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    I agree with this answer. But I have a question: what is an example question with NEITHER as its answer?
    $endgroup$
    – athin
    yesterday






  • 1




    $begingroup$
    Good question. For any question with the word definite, there is never the option NEITHER. As definite is absolute, its is either true or false @athin
    $endgroup$
    – Omega Krypton
    yesterday






  • 1




    $begingroup$
    @athin: From a mathematical-logic standpoint, there is none. It's called the law of the excluded middle.
    $endgroup$
    – BlueRaja - Danny Pflughoeft
    21 hours ago





















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I think this is just a matter of understanding the language used in logic.



In the implication




If A, then B




you seem to be arguing that, since there are cases where A is true but B can be either true or false, we should say "the implication is neither true nor false".



However, every mathematician I know would say that the implication is false. In order for it to be true, it should always hold. If there are cases where it doesn't hold, we simply say it's false.





For example, take the claim




If the ground is wet, then it is raining




There are certainly cases where both sides of the implication are true; however, since it's not always true, we say it's a false claim. I don't know of any logician who would call this "neither a true nor a false statement".






share|improve this answer









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  • 1




    $begingroup$
    But if the implication is sometimes always true, and sometimes always false, then it is neither true nor false.
    $endgroup$
    – Acccumulation
    yesterday



















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You can draw a Venn diagram where the Smaug set overlaps with the Thor set, and the latter with the Thrain set, but not the first with the last. On the other hand, you can include the Thrain set in the Thor one and the latter in the Smaug one, but the word "definitely" implies ALL the possible cases, so it's false.







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    1












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    I'm inclined to say that you are wrong because for example :

    In the case of "Some dogs are animals and some animals are chicken" then the statement is false.

    On the other hand "Some dogs are animals and some animals are dogs" for which it is true.

    So it could be EITHER, but not NEITHER. Also we have definitely which the above two examples contradict with, so I would select FALSE.







    share|improve this answer











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      i believe that the word "definitely" rot13(vf veeryrinag. Rira jvgubhg vg, gur nafjre jbhyq or snyfr orpnhfr gur pbapyhfvba pna abg or ybtvpnyyl bognvarq sebz gur gjb cerzvfr fgngrzragf. Vg pbhyq or gehr be vg pbhyq or snyfr nf n fgnaq nybar fgngrzrag, ohg gur VS GURA fgngrzrag vf nyjnlf SNYFR va guvf pnfr.)
      $endgroup$
      – SteveV
      yesterday



















    1












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    A lot depends on exactly how we move from (ambiguous) English to (unambiguous) logic.




    If some Smaugs are Thors and some Thors are Thrains, then some Smaugs are definitely Thrains




    The first way we can do it is to treat it as a syllogism.



    Some Smaugs are Thors.
    Some Thors are Thrains.
    Therefore some Smaugs are definitely Thrains.


    Syllogisms aren't true or false; they just contain valid or invalid reasoning. In this case, it's invalid: the last line does not follow from the first two. But in colloquial English, it's perfectly reasonable to say that invalid reasoning is false reasoning.





    Alternately, we can translate it to a symbolic logical statement:



    "Some Smaugs are Thors" would be "∃ a : Smaug(a) ∧ Thor(a)" - There exists at least one a such that a is both a Smaug and a Thor.

    Similarly, "Some Thors are Thrains" is "∃ a : Thor(a) ∧ Thrain(a)" and "some Smaugs are definitely Thrains" is "∃ a : Smaug(a) ∧ Thrain(a)"; "definitely" doesn't add anything except emphasis.



    So the full statement becomes (∃ a : Smaug(a) ∧ Thor(a)) ∧ (∃ a : Thor(a) ∧ Thrain(a)) ⇒ (∃ a : Smaug(a) ∧ Thrain(a)).



    Again, the last clause doesn't follow from the first two, however the statement as a whole can be trivially true in the degenerate case where either (∃ a : Smaug(a) ∧ Thor(a)) ∧ (∃ a : Thor(a) ∧ Thrain(a)) is always false or (∃ a : Smaug(a) ∧ Thrain(a)) is always true.



    Effectively, the statement P ⇒ Q is equivalent to saying that Q is true in all situations where P is true, so as long as P is never true, or as long as Q is always true, no counterexamples can exist.



    So the truth value of (∃ a : Smaug(a) ∧ Thor(a)) ∧ (∃ a : Thor(a) ∧ Thrain(a)) ⇒ (∃ a : Smaug(a) ∧ Thrain(a)) is unknown without further information, but we can state the circumstances under which it is true or false. (A specific case in which it would be true is if there exist no Thrains.)



    (Which is to say, just in case this hasn't been unreadable enough yet, that ((∃ a : Smaug(a) ∧ Thor(a)) ∧ (∃ a : Thor(a) ∧ Thrain(a)) ⇒ (∃ a : Smaug(a) ∧ Thrain(a))) ⇔ (~[(∃ a : Smaug(a) ∧ Thor(a)) ∧ (∃ a : Thor(a) ∧ Thrain(a))] ∨ [∀ a : Smaug(a) ∧ Thrain(a)]) is necessarily true.)






    share|improve this answer









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      1












      $begingroup$

      The word "definitely" is ambiguous as to what it modifies. Taken literally, it modifies "Thrains"; according to the standard rules of English grammar, the default is that modifiers modify the next word. Under this interpretation, the statement is saying that there exists a nonempty set (and if we take the plural literally, the set needs to not only be nonempty, but have at least two members) S such that all members of S are Smaugs, and all members of S are definitely Thrains. But there is not enough information to decide whether this statement is true or not, and so "neither" is a reasonable choice.



      Now, if we take "definitely" to be modifying the entire statement ("If some Smaugs are Thors and some Thors are Thrains, then it is definitely true that some Smaugs are Thrains."), then that's a different matter. But that is not what the statement says. And even in that case, it depends on whether it's a statement about all possible values of "Smaugs", "Thors", and "Thrains". If it's taken to be a general statement about any possible Smaugs, Thors, and Thrains, then it is false. But if it refer to specific values of those words, then it could be true. For instance, if "Smaugs" means "triangles", "Thors" means "isosceles triangles", and "Thrains" means "right triangles", then we have "If some triangles are isosceles triangles, and some isosceles triangles are right triangles, then some triangles are right triangles". And according to the standard rules of logic, that is true, because any statement of the form "If X, then Y" is true if Y is true.



      So the writers of this test clearly failed the writing a clear test test.






      share|improve this answer











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        1












        $begingroup$

        The rules leaves it open ended whether Smaugs are Thrains.



        So it could describe a relationship like :-




        Some Humans are Female, Some Females are Mothers



        in this case, some human females are mothers.






        Or it could be




        some Humans are Female, some Females are Kangaroos



        in this case, No Human is a Kangaroo




        The last case clearly shows that Smaugs can't definitely be Thrains without having more information






        share|improve this answer








        New contributor




        Keith Nicholas is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
        Check out our Code of Conduct.






        $endgroup$





















          0












          $begingroup$

          Consider this scenario:




          A is a Smaug and a Thor.

          B is Thor and a Thrain.

          In this case, while the first two conditions are met, there are no Smaugs that are Thrains. Thus, we cannot be certain (note 'definitely' in the the question) that there are some Smaugs that are Thrains, so FALSE is the correct answer.







          share|improve this answer









          $endgroup$





















            0












            $begingroup$

            The statement could only be true if ALL of the Thors were also Smaugs or all of the Thors were Thrains



            The statement is FALSE because you cannot guarantee that some Smaugs are DEFINITELY Thrains (perhaps only the Thors that are not Smaugs are the ones that are Thrains)



            The only way it could be NEITHER is if there is sufficient ambiguity in the facts or the statement was written in a way to be unable to say with certainty if the statement was TRUE or FALSE and cannot be proven to be either.



            For instance, If the statement was 'Some Thors might be Smaugs and Thrains' then you could not argue that it was unquestionably TRUE because with the facts, it's possible that none may be all three. You also couldn't state it was FALSE, because it is possible that some might be all three. It's only in this case that you would say NEITHER, as it would be impossible to say which it is without further information.






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              0












              $begingroup$

              I already agree with some answers presented. Let me provide a graphical one. This is the 3-set venn diagram of all posible sets an element can be a member of:



              enter image description here



              I assume "some" means "1 or more". Thus, let's interpret the question:




              some Smaugs are Thors




              means elements of Green + Red are 1 or more




              some Thors are Thrains




              means elements of Blue + Red are 1 or more
              These two are facts.



              Now we are asked to deduce whether:




              some Smaugs are definitely Thrains




              First of all, I find "definitely" useless. It should be obvious this is about boolean logic; both true and false are always "definitely". Maybe it's put there as a user-friendly term? Disregarding that, this one would mean elements of Red are 1 or more.



              You can clearly see that if green >= 1, blue >=1 and red=0 then the initial facts are satisfied, but the deducted one is not. Thus, "FALSE".






              share|improve this answer









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                0












                $begingroup$

                The 3 over-laying circles don't always make sense.



                Smaugs may be touching into Thrain territory, or it may not.



                50% chance it's the 3 over-laying rings theory above; 50% chance it looks more like 3 serial links- as only the Smaugs and Thors must intersect, and only the Thors and Thrain must intersect. There is no data to intersect Smaugs with Thrains (or not to intersect)…



                So by adding in "definitely", it knocks off the True and Neither logic paths; so it's false.






                share|improve this answer








                New contributor




                tlr is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
                Check out our Code of Conduct.






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                  0












                  $begingroup$

                  Consider the 1st part:





                  • If some Smaugs are Thors




                  Part 1:The following diagram shows how some Smaugs are Thors. The violet shaded part is the one that represents the few Smaugs that are Thors.





                  • some Thors are Thrains




                  This can have 2 possibilities:



                  1.enter image description here In the above case, as seen by the red shaded part, some Smaugs are Thrains.



                  2.But then, the other possibility of the second part:enter image description here Here, you can see that the red shaded part is outside the scope of the Smaugs, which means that there is a possibility that NONE of the Smaugs are Thrains. Hence, either some Smaugs are Thrains or none are Thrains. Which would further deduce that the statement




                  some Smaugs are definitely Thrains




                  is not true. Since, all Smaugs are not definitely Thrains, as there is a possibility that they're not Thrains.






                  share|improve this answer











                  $endgroup$













                  • $begingroup$
                    there is a third possibility, some smaugs are thrains (but not thors)
                    $endgroup$
                    – JonMark Perry
                    21 hours ago










                  • $begingroup$
                    Thanks for pointing it out @JonMarkPerry. I overlooked it. But the 2nd case shown would still keep the outcome as not definitely Thrains. Right? Will update nevertheless.
                    $endgroup$
                    – Rai
                    21 hours ago











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                  12 Answers
                  12






                  active

                  oldest

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                  12 Answers
                  12






                  active

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                  active

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                  active

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                  13












                  $begingroup$

                  In more an English sense than a Puzzling one...



                  In the question there is the keyword:




                  definitely.




                  That means if 100% true, then the statement is true. else it is false, even it is true 99%.




                  In a sentence with "definitely", neither is the first choice that should go out of your mind. then you choose from true or false. hope this helps!




                  In other words:




                  this is a question of definitely true or not definitely true (maybe true), not definitely true, definitely false, or maybe true




                  Hope this helps!






                  share|improve this answer









                  $endgroup$









                  • 1




                    $begingroup$
                    Even without "definitely", the answer is still "false"...
                    $endgroup$
                    – BlueRaja - Danny Pflughoeft
                    yesterday










                  • $begingroup$
                    @BlueRaja-DannyPflughoeft No, it's not. Some Smaugs being Thrains is consistent with some Smaugs being Thors and some Thors being Thrains.
                    $endgroup$
                    – Acccumulation
                    yesterday










                  • $begingroup$
                    I agree with this answer. But I have a question: what is an example question with NEITHER as its answer?
                    $endgroup$
                    – athin
                    yesterday






                  • 1




                    $begingroup$
                    Good question. For any question with the word definite, there is never the option NEITHER. As definite is absolute, its is either true or false @athin
                    $endgroup$
                    – Omega Krypton
                    yesterday






                  • 1




                    $begingroup$
                    @athin: From a mathematical-logic standpoint, there is none. It's called the law of the excluded middle.
                    $endgroup$
                    – BlueRaja - Danny Pflughoeft
                    21 hours ago


















                  13












                  $begingroup$

                  In more an English sense than a Puzzling one...



                  In the question there is the keyword:




                  definitely.




                  That means if 100% true, then the statement is true. else it is false, even it is true 99%.




                  In a sentence with "definitely", neither is the first choice that should go out of your mind. then you choose from true or false. hope this helps!




                  In other words:




                  this is a question of definitely true or not definitely true (maybe true), not definitely true, definitely false, or maybe true




                  Hope this helps!






                  share|improve this answer









                  $endgroup$









                  • 1




                    $begingroup$
                    Even without "definitely", the answer is still "false"...
                    $endgroup$
                    – BlueRaja - Danny Pflughoeft
                    yesterday










                  • $begingroup$
                    @BlueRaja-DannyPflughoeft No, it's not. Some Smaugs being Thrains is consistent with some Smaugs being Thors and some Thors being Thrains.
                    $endgroup$
                    – Acccumulation
                    yesterday










                  • $begingroup$
                    I agree with this answer. But I have a question: what is an example question with NEITHER as its answer?
                    $endgroup$
                    – athin
                    yesterday






                  • 1




                    $begingroup$
                    Good question. For any question with the word definite, there is never the option NEITHER. As definite is absolute, its is either true or false @athin
                    $endgroup$
                    – Omega Krypton
                    yesterday






                  • 1




                    $begingroup$
                    @athin: From a mathematical-logic standpoint, there is none. It's called the law of the excluded middle.
                    $endgroup$
                    – BlueRaja - Danny Pflughoeft
                    21 hours ago
















                  13












                  13








                  13





                  $begingroup$

                  In more an English sense than a Puzzling one...



                  In the question there is the keyword:




                  definitely.




                  That means if 100% true, then the statement is true. else it is false, even it is true 99%.




                  In a sentence with "definitely", neither is the first choice that should go out of your mind. then you choose from true or false. hope this helps!




                  In other words:




                  this is a question of definitely true or not definitely true (maybe true), not definitely true, definitely false, or maybe true




                  Hope this helps!






                  share|improve this answer









                  $endgroup$



                  In more an English sense than a Puzzling one...



                  In the question there is the keyword:




                  definitely.




                  That means if 100% true, then the statement is true. else it is false, even it is true 99%.




                  In a sentence with "definitely", neither is the first choice that should go out of your mind. then you choose from true or false. hope this helps!




                  In other words:




                  this is a question of definitely true or not definitely true (maybe true), not definitely true, definitely false, or maybe true




                  Hope this helps!







                  share|improve this answer












                  share|improve this answer



                  share|improve this answer










                  answered yesterday









                  Omega KryptonOmega Krypton

                  3,5061336




                  3,5061336








                  • 1




                    $begingroup$
                    Even without "definitely", the answer is still "false"...
                    $endgroup$
                    – BlueRaja - Danny Pflughoeft
                    yesterday










                  • $begingroup$
                    @BlueRaja-DannyPflughoeft No, it's not. Some Smaugs being Thrains is consistent with some Smaugs being Thors and some Thors being Thrains.
                    $endgroup$
                    – Acccumulation
                    yesterday










                  • $begingroup$
                    I agree with this answer. But I have a question: what is an example question with NEITHER as its answer?
                    $endgroup$
                    – athin
                    yesterday






                  • 1




                    $begingroup$
                    Good question. For any question with the word definite, there is never the option NEITHER. As definite is absolute, its is either true or false @athin
                    $endgroup$
                    – Omega Krypton
                    yesterday






                  • 1




                    $begingroup$
                    @athin: From a mathematical-logic standpoint, there is none. It's called the law of the excluded middle.
                    $endgroup$
                    – BlueRaja - Danny Pflughoeft
                    21 hours ago
















                  • 1




                    $begingroup$
                    Even without "definitely", the answer is still "false"...
                    $endgroup$
                    – BlueRaja - Danny Pflughoeft
                    yesterday










                  • $begingroup$
                    @BlueRaja-DannyPflughoeft No, it's not. Some Smaugs being Thrains is consistent with some Smaugs being Thors and some Thors being Thrains.
                    $endgroup$
                    – Acccumulation
                    yesterday










                  • $begingroup$
                    I agree with this answer. But I have a question: what is an example question with NEITHER as its answer?
                    $endgroup$
                    – athin
                    yesterday






                  • 1




                    $begingroup$
                    Good question. For any question with the word definite, there is never the option NEITHER. As definite is absolute, its is either true or false @athin
                    $endgroup$
                    – Omega Krypton
                    yesterday






                  • 1




                    $begingroup$
                    @athin: From a mathematical-logic standpoint, there is none. It's called the law of the excluded middle.
                    $endgroup$
                    – BlueRaja - Danny Pflughoeft
                    21 hours ago










                  1




                  1




                  $begingroup$
                  Even without "definitely", the answer is still "false"...
                  $endgroup$
                  – BlueRaja - Danny Pflughoeft
                  yesterday




                  $begingroup$
                  Even without "definitely", the answer is still "false"...
                  $endgroup$
                  – BlueRaja - Danny Pflughoeft
                  yesterday












                  $begingroup$
                  @BlueRaja-DannyPflughoeft No, it's not. Some Smaugs being Thrains is consistent with some Smaugs being Thors and some Thors being Thrains.
                  $endgroup$
                  – Acccumulation
                  yesterday




                  $begingroup$
                  @BlueRaja-DannyPflughoeft No, it's not. Some Smaugs being Thrains is consistent with some Smaugs being Thors and some Thors being Thrains.
                  $endgroup$
                  – Acccumulation
                  yesterday












                  $begingroup$
                  I agree with this answer. But I have a question: what is an example question with NEITHER as its answer?
                  $endgroup$
                  – athin
                  yesterday




                  $begingroup$
                  I agree with this answer. But I have a question: what is an example question with NEITHER as its answer?
                  $endgroup$
                  – athin
                  yesterday




                  1




                  1




                  $begingroup$
                  Good question. For any question with the word definite, there is never the option NEITHER. As definite is absolute, its is either true or false @athin
                  $endgroup$
                  – Omega Krypton
                  yesterday




                  $begingroup$
                  Good question. For any question with the word definite, there is never the option NEITHER. As definite is absolute, its is either true or false @athin
                  $endgroup$
                  – Omega Krypton
                  yesterday




                  1




                  1




                  $begingroup$
                  @athin: From a mathematical-logic standpoint, there is none. It's called the law of the excluded middle.
                  $endgroup$
                  – BlueRaja - Danny Pflughoeft
                  21 hours ago






                  $begingroup$
                  @athin: From a mathematical-logic standpoint, there is none. It's called the law of the excluded middle.
                  $endgroup$
                  – BlueRaja - Danny Pflughoeft
                  21 hours ago













                  5












                  $begingroup$

                  I think this is just a matter of understanding the language used in logic.



                  In the implication




                  If A, then B




                  you seem to be arguing that, since there are cases where A is true but B can be either true or false, we should say "the implication is neither true nor false".



                  However, every mathematician I know would say that the implication is false. In order for it to be true, it should always hold. If there are cases where it doesn't hold, we simply say it's false.





                  For example, take the claim




                  If the ground is wet, then it is raining




                  There are certainly cases where both sides of the implication are true; however, since it's not always true, we say it's a false claim. I don't know of any logician who would call this "neither a true nor a false statement".






                  share|improve this answer









                  $endgroup$









                  • 1




                    $begingroup$
                    But if the implication is sometimes always true, and sometimes always false, then it is neither true nor false.
                    $endgroup$
                    – Acccumulation
                    yesterday
















                  5












                  $begingroup$

                  I think this is just a matter of understanding the language used in logic.



                  In the implication




                  If A, then B




                  you seem to be arguing that, since there are cases where A is true but B can be either true or false, we should say "the implication is neither true nor false".



                  However, every mathematician I know would say that the implication is false. In order for it to be true, it should always hold. If there are cases where it doesn't hold, we simply say it's false.





                  For example, take the claim




                  If the ground is wet, then it is raining




                  There are certainly cases where both sides of the implication are true; however, since it's not always true, we say it's a false claim. I don't know of any logician who would call this "neither a true nor a false statement".






                  share|improve this answer









                  $endgroup$









                  • 1




                    $begingroup$
                    But if the implication is sometimes always true, and sometimes always false, then it is neither true nor false.
                    $endgroup$
                    – Acccumulation
                    yesterday














                  5












                  5








                  5





                  $begingroup$

                  I think this is just a matter of understanding the language used in logic.



                  In the implication




                  If A, then B




                  you seem to be arguing that, since there are cases where A is true but B can be either true or false, we should say "the implication is neither true nor false".



                  However, every mathematician I know would say that the implication is false. In order for it to be true, it should always hold. If there are cases where it doesn't hold, we simply say it's false.





                  For example, take the claim




                  If the ground is wet, then it is raining




                  There are certainly cases where both sides of the implication are true; however, since it's not always true, we say it's a false claim. I don't know of any logician who would call this "neither a true nor a false statement".






                  share|improve this answer









                  $endgroup$



                  I think this is just a matter of understanding the language used in logic.



                  In the implication




                  If A, then B




                  you seem to be arguing that, since there are cases where A is true but B can be either true or false, we should say "the implication is neither true nor false".



                  However, every mathematician I know would say that the implication is false. In order for it to be true, it should always hold. If there are cases where it doesn't hold, we simply say it's false.





                  For example, take the claim




                  If the ground is wet, then it is raining




                  There are certainly cases where both sides of the implication are true; however, since it's not always true, we say it's a false claim. I don't know of any logician who would call this "neither a true nor a false statement".







                  share|improve this answer












                  share|improve this answer



                  share|improve this answer










                  answered yesterday









                  BlueRaja - Danny PflughoeftBlueRaja - Danny Pflughoeft

                  264110




                  264110








                  • 1




                    $begingroup$
                    But if the implication is sometimes always true, and sometimes always false, then it is neither true nor false.
                    $endgroup$
                    – Acccumulation
                    yesterday














                  • 1




                    $begingroup$
                    But if the implication is sometimes always true, and sometimes always false, then it is neither true nor false.
                    $endgroup$
                    – Acccumulation
                    yesterday








                  1




                  1




                  $begingroup$
                  But if the implication is sometimes always true, and sometimes always false, then it is neither true nor false.
                  $endgroup$
                  – Acccumulation
                  yesterday




                  $begingroup$
                  But if the implication is sometimes always true, and sometimes always false, then it is neither true nor false.
                  $endgroup$
                  – Acccumulation
                  yesterday











                  2












                  $begingroup$


                  You can draw a Venn diagram where the Smaug set overlaps with the Thor set, and the latter with the Thrain set, but not the first with the last. On the other hand, you can include the Thrain set in the Thor one and the latter in the Smaug one, but the word "definitely" implies ALL the possible cases, so it's false.







                  share|improve this answer









                  $endgroup$


















                    2












                    $begingroup$


                    You can draw a Venn diagram where the Smaug set overlaps with the Thor set, and the latter with the Thrain set, but not the first with the last. On the other hand, you can include the Thrain set in the Thor one and the latter in the Smaug one, but the word "definitely" implies ALL the possible cases, so it's false.







                    share|improve this answer









                    $endgroup$
















                      2












                      2








                      2





                      $begingroup$


                      You can draw a Venn diagram where the Smaug set overlaps with the Thor set, and the latter with the Thrain set, but not the first with the last. On the other hand, you can include the Thrain set in the Thor one and the latter in the Smaug one, but the word "definitely" implies ALL the possible cases, so it's false.







                      share|improve this answer









                      $endgroup$




                      You can draw a Venn diagram where the Smaug set overlaps with the Thor set, and the latter with the Thrain set, but not the first with the last. On the other hand, you can include the Thrain set in the Thor one and the latter in the Smaug one, but the word "definitely" implies ALL the possible cases, so it's false.








                      share|improve this answer












                      share|improve this answer



                      share|improve this answer










                      answered yesterday









                      NautilusNautilus

                      3,674523




                      3,674523























                          1












                          $begingroup$


                          I'm inclined to say that you are wrong because for example :

                          In the case of "Some dogs are animals and some animals are chicken" then the statement is false.

                          On the other hand "Some dogs are animals and some animals are dogs" for which it is true.

                          So it could be EITHER, but not NEITHER. Also we have definitely which the above two examples contradict with, so I would select FALSE.







                          share|improve this answer











                          $endgroup$













                          • $begingroup$
                            i believe that the word "definitely" rot13(vf veeryrinag. Rira jvgubhg vg, gur nafjre jbhyq or snyfr orpnhfr gur pbapyhfvba pna abg or ybtvpnyyl bognvarq sebz gur gjb cerzvfr fgngrzragf. Vg pbhyq or gehr be vg pbhyq or snyfr nf n fgnaq nybar fgngrzrag, ohg gur VS GURA fgngrzrag vf nyjnlf SNYFR va guvf pnfr.)
                            $endgroup$
                            – SteveV
                            yesterday
















                          1












                          $begingroup$


                          I'm inclined to say that you are wrong because for example :

                          In the case of "Some dogs are animals and some animals are chicken" then the statement is false.

                          On the other hand "Some dogs are animals and some animals are dogs" for which it is true.

                          So it could be EITHER, but not NEITHER. Also we have definitely which the above two examples contradict with, so I would select FALSE.







                          share|improve this answer











                          $endgroup$













                          • $begingroup$
                            i believe that the word "definitely" rot13(vf veeryrinag. Rira jvgubhg vg, gur nafjre jbhyq or snyfr orpnhfr gur pbapyhfvba pna abg or ybtvpnyyl bognvarq sebz gur gjb cerzvfr fgngrzragf. Vg pbhyq or gehr be vg pbhyq or snyfr nf n fgnaq nybar fgngrzrag, ohg gur VS GURA fgngrzrag vf nyjnlf SNYFR va guvf pnfr.)
                            $endgroup$
                            – SteveV
                            yesterday














                          1












                          1








                          1





                          $begingroup$


                          I'm inclined to say that you are wrong because for example :

                          In the case of "Some dogs are animals and some animals are chicken" then the statement is false.

                          On the other hand "Some dogs are animals and some animals are dogs" for which it is true.

                          So it could be EITHER, but not NEITHER. Also we have definitely which the above two examples contradict with, so I would select FALSE.







                          share|improve this answer











                          $endgroup$




                          I'm inclined to say that you are wrong because for example :

                          In the case of "Some dogs are animals and some animals are chicken" then the statement is false.

                          On the other hand "Some dogs are animals and some animals are dogs" for which it is true.

                          So it could be EITHER, but not NEITHER. Also we have definitely which the above two examples contradict with, so I would select FALSE.








                          share|improve this answer














                          share|improve this answer



                          share|improve this answer








                          edited yesterday

























                          answered yesterday









                          rhsquaredrhsquared

                          8,16021849




                          8,16021849












                          • $begingroup$
                            i believe that the word "definitely" rot13(vf veeryrinag. Rira jvgubhg vg, gur nafjre jbhyq or snyfr orpnhfr gur pbapyhfvba pna abg or ybtvpnyyl bognvarq sebz gur gjb cerzvfr fgngrzragf. Vg pbhyq or gehr be vg pbhyq or snyfr nf n fgnaq nybar fgngrzrag, ohg gur VS GURA fgngrzrag vf nyjnlf SNYFR va guvf pnfr.)
                            $endgroup$
                            – SteveV
                            yesterday


















                          • $begingroup$
                            i believe that the word "definitely" rot13(vf veeryrinag. Rira jvgubhg vg, gur nafjre jbhyq or snyfr orpnhfr gur pbapyhfvba pna abg or ybtvpnyyl bognvarq sebz gur gjb cerzvfr fgngrzragf. Vg pbhyq or gehr be vg pbhyq or snyfr nf n fgnaq nybar fgngrzrag, ohg gur VS GURA fgngrzrag vf nyjnlf SNYFR va guvf pnfr.)
                            $endgroup$
                            – SteveV
                            yesterday
















                          $begingroup$
                          i believe that the word "definitely" rot13(vf veeryrinag. Rira jvgubhg vg, gur nafjre jbhyq or snyfr orpnhfr gur pbapyhfvba pna abg or ybtvpnyyl bognvarq sebz gur gjb cerzvfr fgngrzragf. Vg pbhyq or gehr be vg pbhyq or snyfr nf n fgnaq nybar fgngrzrag, ohg gur VS GURA fgngrzrag vf nyjnlf SNYFR va guvf pnfr.)
                          $endgroup$
                          – SteveV
                          yesterday




                          $begingroup$
                          i believe that the word "definitely" rot13(vf veeryrinag. Rira jvgubhg vg, gur nafjre jbhyq or snyfr orpnhfr gur pbapyhfvba pna abg or ybtvpnyyl bognvarq sebz gur gjb cerzvfr fgngrzragf. Vg pbhyq or gehr be vg pbhyq or snyfr nf n fgnaq nybar fgngrzrag, ohg gur VS GURA fgngrzrag vf nyjnlf SNYFR va guvf pnfr.)
                          $endgroup$
                          – SteveV
                          yesterday











                          1












                          $begingroup$

                          A lot depends on exactly how we move from (ambiguous) English to (unambiguous) logic.




                          If some Smaugs are Thors and some Thors are Thrains, then some Smaugs are definitely Thrains




                          The first way we can do it is to treat it as a syllogism.



                          Some Smaugs are Thors.
                          Some Thors are Thrains.
                          Therefore some Smaugs are definitely Thrains.


                          Syllogisms aren't true or false; they just contain valid or invalid reasoning. In this case, it's invalid: the last line does not follow from the first two. But in colloquial English, it's perfectly reasonable to say that invalid reasoning is false reasoning.





                          Alternately, we can translate it to a symbolic logical statement:



                          "Some Smaugs are Thors" would be "∃ a : Smaug(a) ∧ Thor(a)" - There exists at least one a such that a is both a Smaug and a Thor.

                          Similarly, "Some Thors are Thrains" is "∃ a : Thor(a) ∧ Thrain(a)" and "some Smaugs are definitely Thrains" is "∃ a : Smaug(a) ∧ Thrain(a)"; "definitely" doesn't add anything except emphasis.



                          So the full statement becomes (∃ a : Smaug(a) ∧ Thor(a)) ∧ (∃ a : Thor(a) ∧ Thrain(a)) ⇒ (∃ a : Smaug(a) ∧ Thrain(a)).



                          Again, the last clause doesn't follow from the first two, however the statement as a whole can be trivially true in the degenerate case where either (∃ a : Smaug(a) ∧ Thor(a)) ∧ (∃ a : Thor(a) ∧ Thrain(a)) is always false or (∃ a : Smaug(a) ∧ Thrain(a)) is always true.



                          Effectively, the statement P ⇒ Q is equivalent to saying that Q is true in all situations where P is true, so as long as P is never true, or as long as Q is always true, no counterexamples can exist.



                          So the truth value of (∃ a : Smaug(a) ∧ Thor(a)) ∧ (∃ a : Thor(a) ∧ Thrain(a)) ⇒ (∃ a : Smaug(a) ∧ Thrain(a)) is unknown without further information, but we can state the circumstances under which it is true or false. (A specific case in which it would be true is if there exist no Thrains.)



                          (Which is to say, just in case this hasn't been unreadable enough yet, that ((∃ a : Smaug(a) ∧ Thor(a)) ∧ (∃ a : Thor(a) ∧ Thrain(a)) ⇒ (∃ a : Smaug(a) ∧ Thrain(a))) ⇔ (~[(∃ a : Smaug(a) ∧ Thor(a)) ∧ (∃ a : Thor(a) ∧ Thrain(a))] ∨ [∀ a : Smaug(a) ∧ Thrain(a)]) is necessarily true.)






                          share|improve this answer









                          $endgroup$


















                            1












                            $begingroup$

                            A lot depends on exactly how we move from (ambiguous) English to (unambiguous) logic.




                            If some Smaugs are Thors and some Thors are Thrains, then some Smaugs are definitely Thrains




                            The first way we can do it is to treat it as a syllogism.



                            Some Smaugs are Thors.
                            Some Thors are Thrains.
                            Therefore some Smaugs are definitely Thrains.


                            Syllogisms aren't true or false; they just contain valid or invalid reasoning. In this case, it's invalid: the last line does not follow from the first two. But in colloquial English, it's perfectly reasonable to say that invalid reasoning is false reasoning.





                            Alternately, we can translate it to a symbolic logical statement:



                            "Some Smaugs are Thors" would be "∃ a : Smaug(a) ∧ Thor(a)" - There exists at least one a such that a is both a Smaug and a Thor.

                            Similarly, "Some Thors are Thrains" is "∃ a : Thor(a) ∧ Thrain(a)" and "some Smaugs are definitely Thrains" is "∃ a : Smaug(a) ∧ Thrain(a)"; "definitely" doesn't add anything except emphasis.



                            So the full statement becomes (∃ a : Smaug(a) ∧ Thor(a)) ∧ (∃ a : Thor(a) ∧ Thrain(a)) ⇒ (∃ a : Smaug(a) ∧ Thrain(a)).



                            Again, the last clause doesn't follow from the first two, however the statement as a whole can be trivially true in the degenerate case where either (∃ a : Smaug(a) ∧ Thor(a)) ∧ (∃ a : Thor(a) ∧ Thrain(a)) is always false or (∃ a : Smaug(a) ∧ Thrain(a)) is always true.



                            Effectively, the statement P ⇒ Q is equivalent to saying that Q is true in all situations where P is true, so as long as P is never true, or as long as Q is always true, no counterexamples can exist.



                            So the truth value of (∃ a : Smaug(a) ∧ Thor(a)) ∧ (∃ a : Thor(a) ∧ Thrain(a)) ⇒ (∃ a : Smaug(a) ∧ Thrain(a)) is unknown without further information, but we can state the circumstances under which it is true or false. (A specific case in which it would be true is if there exist no Thrains.)



                            (Which is to say, just in case this hasn't been unreadable enough yet, that ((∃ a : Smaug(a) ∧ Thor(a)) ∧ (∃ a : Thor(a) ∧ Thrain(a)) ⇒ (∃ a : Smaug(a) ∧ Thrain(a))) ⇔ (~[(∃ a : Smaug(a) ∧ Thor(a)) ∧ (∃ a : Thor(a) ∧ Thrain(a))] ∨ [∀ a : Smaug(a) ∧ Thrain(a)]) is necessarily true.)






                            share|improve this answer









                            $endgroup$
















                              1












                              1








                              1





                              $begingroup$

                              A lot depends on exactly how we move from (ambiguous) English to (unambiguous) logic.




                              If some Smaugs are Thors and some Thors are Thrains, then some Smaugs are definitely Thrains




                              The first way we can do it is to treat it as a syllogism.



                              Some Smaugs are Thors.
                              Some Thors are Thrains.
                              Therefore some Smaugs are definitely Thrains.


                              Syllogisms aren't true or false; they just contain valid or invalid reasoning. In this case, it's invalid: the last line does not follow from the first two. But in colloquial English, it's perfectly reasonable to say that invalid reasoning is false reasoning.





                              Alternately, we can translate it to a symbolic logical statement:



                              "Some Smaugs are Thors" would be "∃ a : Smaug(a) ∧ Thor(a)" - There exists at least one a such that a is both a Smaug and a Thor.

                              Similarly, "Some Thors are Thrains" is "∃ a : Thor(a) ∧ Thrain(a)" and "some Smaugs are definitely Thrains" is "∃ a : Smaug(a) ∧ Thrain(a)"; "definitely" doesn't add anything except emphasis.



                              So the full statement becomes (∃ a : Smaug(a) ∧ Thor(a)) ∧ (∃ a : Thor(a) ∧ Thrain(a)) ⇒ (∃ a : Smaug(a) ∧ Thrain(a)).



                              Again, the last clause doesn't follow from the first two, however the statement as a whole can be trivially true in the degenerate case where either (∃ a : Smaug(a) ∧ Thor(a)) ∧ (∃ a : Thor(a) ∧ Thrain(a)) is always false or (∃ a : Smaug(a) ∧ Thrain(a)) is always true.



                              Effectively, the statement P ⇒ Q is equivalent to saying that Q is true in all situations where P is true, so as long as P is never true, or as long as Q is always true, no counterexamples can exist.



                              So the truth value of (∃ a : Smaug(a) ∧ Thor(a)) ∧ (∃ a : Thor(a) ∧ Thrain(a)) ⇒ (∃ a : Smaug(a) ∧ Thrain(a)) is unknown without further information, but we can state the circumstances under which it is true or false. (A specific case in which it would be true is if there exist no Thrains.)



                              (Which is to say, just in case this hasn't been unreadable enough yet, that ((∃ a : Smaug(a) ∧ Thor(a)) ∧ (∃ a : Thor(a) ∧ Thrain(a)) ⇒ (∃ a : Smaug(a) ∧ Thrain(a))) ⇔ (~[(∃ a : Smaug(a) ∧ Thor(a)) ∧ (∃ a : Thor(a) ∧ Thrain(a))] ∨ [∀ a : Smaug(a) ∧ Thrain(a)]) is necessarily true.)






                              share|improve this answer









                              $endgroup$



                              A lot depends on exactly how we move from (ambiguous) English to (unambiguous) logic.




                              If some Smaugs are Thors and some Thors are Thrains, then some Smaugs are definitely Thrains




                              The first way we can do it is to treat it as a syllogism.



                              Some Smaugs are Thors.
                              Some Thors are Thrains.
                              Therefore some Smaugs are definitely Thrains.


                              Syllogisms aren't true or false; they just contain valid or invalid reasoning. In this case, it's invalid: the last line does not follow from the first two. But in colloquial English, it's perfectly reasonable to say that invalid reasoning is false reasoning.





                              Alternately, we can translate it to a symbolic logical statement:



                              "Some Smaugs are Thors" would be "∃ a : Smaug(a) ∧ Thor(a)" - There exists at least one a such that a is both a Smaug and a Thor.

                              Similarly, "Some Thors are Thrains" is "∃ a : Thor(a) ∧ Thrain(a)" and "some Smaugs are definitely Thrains" is "∃ a : Smaug(a) ∧ Thrain(a)"; "definitely" doesn't add anything except emphasis.



                              So the full statement becomes (∃ a : Smaug(a) ∧ Thor(a)) ∧ (∃ a : Thor(a) ∧ Thrain(a)) ⇒ (∃ a : Smaug(a) ∧ Thrain(a)).



                              Again, the last clause doesn't follow from the first two, however the statement as a whole can be trivially true in the degenerate case where either (∃ a : Smaug(a) ∧ Thor(a)) ∧ (∃ a : Thor(a) ∧ Thrain(a)) is always false or (∃ a : Smaug(a) ∧ Thrain(a)) is always true.



                              Effectively, the statement P ⇒ Q is equivalent to saying that Q is true in all situations where P is true, so as long as P is never true, or as long as Q is always true, no counterexamples can exist.



                              So the truth value of (∃ a : Smaug(a) ∧ Thor(a)) ∧ (∃ a : Thor(a) ∧ Thrain(a)) ⇒ (∃ a : Smaug(a) ∧ Thrain(a)) is unknown without further information, but we can state the circumstances under which it is true or false. (A specific case in which it would be true is if there exist no Thrains.)



                              (Which is to say, just in case this hasn't been unreadable enough yet, that ((∃ a : Smaug(a) ∧ Thor(a)) ∧ (∃ a : Thor(a) ∧ Thrain(a)) ⇒ (∃ a : Smaug(a) ∧ Thrain(a))) ⇔ (~[(∃ a : Smaug(a) ∧ Thor(a)) ∧ (∃ a : Thor(a) ∧ Thrain(a))] ∨ [∀ a : Smaug(a) ∧ Thrain(a)]) is necessarily true.)







                              share|improve this answer












                              share|improve this answer



                              share|improve this answer










                              answered yesterday









                              RayRay

                              28514




                              28514























                                  1












                                  $begingroup$

                                  The word "definitely" is ambiguous as to what it modifies. Taken literally, it modifies "Thrains"; according to the standard rules of English grammar, the default is that modifiers modify the next word. Under this interpretation, the statement is saying that there exists a nonempty set (and if we take the plural literally, the set needs to not only be nonempty, but have at least two members) S such that all members of S are Smaugs, and all members of S are definitely Thrains. But there is not enough information to decide whether this statement is true or not, and so "neither" is a reasonable choice.



                                  Now, if we take "definitely" to be modifying the entire statement ("If some Smaugs are Thors and some Thors are Thrains, then it is definitely true that some Smaugs are Thrains."), then that's a different matter. But that is not what the statement says. And even in that case, it depends on whether it's a statement about all possible values of "Smaugs", "Thors", and "Thrains". If it's taken to be a general statement about any possible Smaugs, Thors, and Thrains, then it is false. But if it refer to specific values of those words, then it could be true. For instance, if "Smaugs" means "triangles", "Thors" means "isosceles triangles", and "Thrains" means "right triangles", then we have "If some triangles are isosceles triangles, and some isosceles triangles are right triangles, then some triangles are right triangles". And according to the standard rules of logic, that is true, because any statement of the form "If X, then Y" is true if Y is true.



                                  So the writers of this test clearly failed the writing a clear test test.






                                  share|improve this answer











                                  $endgroup$


















                                    1












                                    $begingroup$

                                    The word "definitely" is ambiguous as to what it modifies. Taken literally, it modifies "Thrains"; according to the standard rules of English grammar, the default is that modifiers modify the next word. Under this interpretation, the statement is saying that there exists a nonempty set (and if we take the plural literally, the set needs to not only be nonempty, but have at least two members) S such that all members of S are Smaugs, and all members of S are definitely Thrains. But there is not enough information to decide whether this statement is true or not, and so "neither" is a reasonable choice.



                                    Now, if we take "definitely" to be modifying the entire statement ("If some Smaugs are Thors and some Thors are Thrains, then it is definitely true that some Smaugs are Thrains."), then that's a different matter. But that is not what the statement says. And even in that case, it depends on whether it's a statement about all possible values of "Smaugs", "Thors", and "Thrains". If it's taken to be a general statement about any possible Smaugs, Thors, and Thrains, then it is false. But if it refer to specific values of those words, then it could be true. For instance, if "Smaugs" means "triangles", "Thors" means "isosceles triangles", and "Thrains" means "right triangles", then we have "If some triangles are isosceles triangles, and some isosceles triangles are right triangles, then some triangles are right triangles". And according to the standard rules of logic, that is true, because any statement of the form "If X, then Y" is true if Y is true.



                                    So the writers of this test clearly failed the writing a clear test test.






                                    share|improve this answer











                                    $endgroup$
















                                      1












                                      1








                                      1





                                      $begingroup$

                                      The word "definitely" is ambiguous as to what it modifies. Taken literally, it modifies "Thrains"; according to the standard rules of English grammar, the default is that modifiers modify the next word. Under this interpretation, the statement is saying that there exists a nonempty set (and if we take the plural literally, the set needs to not only be nonempty, but have at least two members) S such that all members of S are Smaugs, and all members of S are definitely Thrains. But there is not enough information to decide whether this statement is true or not, and so "neither" is a reasonable choice.



                                      Now, if we take "definitely" to be modifying the entire statement ("If some Smaugs are Thors and some Thors are Thrains, then it is definitely true that some Smaugs are Thrains."), then that's a different matter. But that is not what the statement says. And even in that case, it depends on whether it's a statement about all possible values of "Smaugs", "Thors", and "Thrains". If it's taken to be a general statement about any possible Smaugs, Thors, and Thrains, then it is false. But if it refer to specific values of those words, then it could be true. For instance, if "Smaugs" means "triangles", "Thors" means "isosceles triangles", and "Thrains" means "right triangles", then we have "If some triangles are isosceles triangles, and some isosceles triangles are right triangles, then some triangles are right triangles". And according to the standard rules of logic, that is true, because any statement of the form "If X, then Y" is true if Y is true.



                                      So the writers of this test clearly failed the writing a clear test test.






                                      share|improve this answer











                                      $endgroup$



                                      The word "definitely" is ambiguous as to what it modifies. Taken literally, it modifies "Thrains"; according to the standard rules of English grammar, the default is that modifiers modify the next word. Under this interpretation, the statement is saying that there exists a nonempty set (and if we take the plural literally, the set needs to not only be nonempty, but have at least two members) S such that all members of S are Smaugs, and all members of S are definitely Thrains. But there is not enough information to decide whether this statement is true or not, and so "neither" is a reasonable choice.



                                      Now, if we take "definitely" to be modifying the entire statement ("If some Smaugs are Thors and some Thors are Thrains, then it is definitely true that some Smaugs are Thrains."), then that's a different matter. But that is not what the statement says. And even in that case, it depends on whether it's a statement about all possible values of "Smaugs", "Thors", and "Thrains". If it's taken to be a general statement about any possible Smaugs, Thors, and Thrains, then it is false. But if it refer to specific values of those words, then it could be true. For instance, if "Smaugs" means "triangles", "Thors" means "isosceles triangles", and "Thrains" means "right triangles", then we have "If some triangles are isosceles triangles, and some isosceles triangles are right triangles, then some triangles are right triangles". And according to the standard rules of logic, that is true, because any statement of the form "If X, then Y" is true if Y is true.



                                      So the writers of this test clearly failed the writing a clear test test.







                                      share|improve this answer














                                      share|improve this answer



                                      share|improve this answer








                                      edited yesterday

























                                      answered yesterday









                                      AcccumulationAcccumulation

                                      504111




                                      504111























                                          1












                                          $begingroup$

                                          The rules leaves it open ended whether Smaugs are Thrains.



                                          So it could describe a relationship like :-




                                          Some Humans are Female, Some Females are Mothers



                                          in this case, some human females are mothers.






                                          Or it could be




                                          some Humans are Female, some Females are Kangaroos



                                          in this case, No Human is a Kangaroo




                                          The last case clearly shows that Smaugs can't definitely be Thrains without having more information






                                          share|improve this answer








                                          New contributor




                                          Keith Nicholas is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
                                          Check out our Code of Conduct.






                                          $endgroup$


















                                            1












                                            $begingroup$

                                            The rules leaves it open ended whether Smaugs are Thrains.



                                            So it could describe a relationship like :-




                                            Some Humans are Female, Some Females are Mothers



                                            in this case, some human females are mothers.






                                            Or it could be




                                            some Humans are Female, some Females are Kangaroos



                                            in this case, No Human is a Kangaroo




                                            The last case clearly shows that Smaugs can't definitely be Thrains without having more information






                                            share|improve this answer








                                            New contributor




                                            Keith Nicholas is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
                                            Check out our Code of Conduct.






                                            $endgroup$
















                                              1












                                              1








                                              1





                                              $begingroup$

                                              The rules leaves it open ended whether Smaugs are Thrains.



                                              So it could describe a relationship like :-




                                              Some Humans are Female, Some Females are Mothers



                                              in this case, some human females are mothers.






                                              Or it could be




                                              some Humans are Female, some Females are Kangaroos



                                              in this case, No Human is a Kangaroo




                                              The last case clearly shows that Smaugs can't definitely be Thrains without having more information






                                              share|improve this answer








                                              New contributor




                                              Keith Nicholas is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
                                              Check out our Code of Conduct.






                                              $endgroup$



                                              The rules leaves it open ended whether Smaugs are Thrains.



                                              So it could describe a relationship like :-




                                              Some Humans are Female, Some Females are Mothers



                                              in this case, some human females are mothers.






                                              Or it could be




                                              some Humans are Female, some Females are Kangaroos



                                              in this case, No Human is a Kangaroo




                                              The last case clearly shows that Smaugs can't definitely be Thrains without having more information







                                              share|improve this answer








                                              New contributor




                                              Keith Nicholas is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
                                              Check out our Code of Conduct.









                                              share|improve this answer



                                              share|improve this answer






                                              New contributor




                                              Keith Nicholas is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
                                              Check out our Code of Conduct.









                                              answered yesterday









                                              Keith NicholasKeith Nicholas

                                              1112




                                              1112




                                              New contributor




                                              Keith Nicholas is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
                                              Check out our Code of Conduct.





                                              New contributor





                                              Keith Nicholas is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
                                              Check out our Code of Conduct.






                                              Keith Nicholas is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
                                              Check out our Code of Conduct.























                                                  0












                                                  $begingroup$

                                                  Consider this scenario:




                                                  A is a Smaug and a Thor.

                                                  B is Thor and a Thrain.

                                                  In this case, while the first two conditions are met, there are no Smaugs that are Thrains. Thus, we cannot be certain (note 'definitely' in the the question) that there are some Smaugs that are Thrains, so FALSE is the correct answer.







                                                  share|improve this answer









                                                  $endgroup$


















                                                    0












                                                    $begingroup$

                                                    Consider this scenario:




                                                    A is a Smaug and a Thor.

                                                    B is Thor and a Thrain.

                                                    In this case, while the first two conditions are met, there are no Smaugs that are Thrains. Thus, we cannot be certain (note 'definitely' in the the question) that there are some Smaugs that are Thrains, so FALSE is the correct answer.







                                                    share|improve this answer









                                                    $endgroup$
















                                                      0












                                                      0








                                                      0





                                                      $begingroup$

                                                      Consider this scenario:




                                                      A is a Smaug and a Thor.

                                                      B is Thor and a Thrain.

                                                      In this case, while the first two conditions are met, there are no Smaugs that are Thrains. Thus, we cannot be certain (note 'definitely' in the the question) that there are some Smaugs that are Thrains, so FALSE is the correct answer.







                                                      share|improve this answer









                                                      $endgroup$



                                                      Consider this scenario:




                                                      A is a Smaug and a Thor.

                                                      B is Thor and a Thrain.

                                                      In this case, while the first two conditions are met, there are no Smaugs that are Thrains. Thus, we cannot be certain (note 'definitely' in the the question) that there are some Smaugs that are Thrains, so FALSE is the correct answer.








                                                      share|improve this answer












                                                      share|improve this answer



                                                      share|improve this answer










                                                      answered yesterday









                                                      ZanyGZanyG

                                                      388114




                                                      388114























                                                          0












                                                          $begingroup$

                                                          The statement could only be true if ALL of the Thors were also Smaugs or all of the Thors were Thrains



                                                          The statement is FALSE because you cannot guarantee that some Smaugs are DEFINITELY Thrains (perhaps only the Thors that are not Smaugs are the ones that are Thrains)



                                                          The only way it could be NEITHER is if there is sufficient ambiguity in the facts or the statement was written in a way to be unable to say with certainty if the statement was TRUE or FALSE and cannot be proven to be either.



                                                          For instance, If the statement was 'Some Thors might be Smaugs and Thrains' then you could not argue that it was unquestionably TRUE because with the facts, it's possible that none may be all three. You also couldn't state it was FALSE, because it is possible that some might be all three. It's only in this case that you would say NEITHER, as it would be impossible to say which it is without further information.






                                                          share|improve this answer











                                                          $endgroup$


















                                                            0












                                                            $begingroup$

                                                            The statement could only be true if ALL of the Thors were also Smaugs or all of the Thors were Thrains



                                                            The statement is FALSE because you cannot guarantee that some Smaugs are DEFINITELY Thrains (perhaps only the Thors that are not Smaugs are the ones that are Thrains)



                                                            The only way it could be NEITHER is if there is sufficient ambiguity in the facts or the statement was written in a way to be unable to say with certainty if the statement was TRUE or FALSE and cannot be proven to be either.



                                                            For instance, If the statement was 'Some Thors might be Smaugs and Thrains' then you could not argue that it was unquestionably TRUE because with the facts, it's possible that none may be all three. You also couldn't state it was FALSE, because it is possible that some might be all three. It's only in this case that you would say NEITHER, as it would be impossible to say which it is without further information.






                                                            share|improve this answer











                                                            $endgroup$
















                                                              0












                                                              0








                                                              0





                                                              $begingroup$

                                                              The statement could only be true if ALL of the Thors were also Smaugs or all of the Thors were Thrains



                                                              The statement is FALSE because you cannot guarantee that some Smaugs are DEFINITELY Thrains (perhaps only the Thors that are not Smaugs are the ones that are Thrains)



                                                              The only way it could be NEITHER is if there is sufficient ambiguity in the facts or the statement was written in a way to be unable to say with certainty if the statement was TRUE or FALSE and cannot be proven to be either.



                                                              For instance, If the statement was 'Some Thors might be Smaugs and Thrains' then you could not argue that it was unquestionably TRUE because with the facts, it's possible that none may be all three. You also couldn't state it was FALSE, because it is possible that some might be all three. It's only in this case that you would say NEITHER, as it would be impossible to say which it is without further information.






                                                              share|improve this answer











                                                              $endgroup$



                                                              The statement could only be true if ALL of the Thors were also Smaugs or all of the Thors were Thrains



                                                              The statement is FALSE because you cannot guarantee that some Smaugs are DEFINITELY Thrains (perhaps only the Thors that are not Smaugs are the ones that are Thrains)



                                                              The only way it could be NEITHER is if there is sufficient ambiguity in the facts or the statement was written in a way to be unable to say with certainty if the statement was TRUE or FALSE and cannot be proven to be either.



                                                              For instance, If the statement was 'Some Thors might be Smaugs and Thrains' then you could not argue that it was unquestionably TRUE because with the facts, it's possible that none may be all three. You also couldn't state it was FALSE, because it is possible that some might be all three. It's only in this case that you would say NEITHER, as it would be impossible to say which it is without further information.







                                                              share|improve this answer














                                                              share|improve this answer



                                                              share|improve this answer








                                                              edited yesterday

























                                                              answered yesterday









                                                              SmockSmock

                                                              112




                                                              112























                                                                  0












                                                                  $begingroup$

                                                                  I already agree with some answers presented. Let me provide a graphical one. This is the 3-set venn diagram of all posible sets an element can be a member of:



                                                                  enter image description here



                                                                  I assume "some" means "1 or more". Thus, let's interpret the question:




                                                                  some Smaugs are Thors




                                                                  means elements of Green + Red are 1 or more




                                                                  some Thors are Thrains




                                                                  means elements of Blue + Red are 1 or more
                                                                  These two are facts.



                                                                  Now we are asked to deduce whether:




                                                                  some Smaugs are definitely Thrains




                                                                  First of all, I find "definitely" useless. It should be obvious this is about boolean logic; both true and false are always "definitely". Maybe it's put there as a user-friendly term? Disregarding that, this one would mean elements of Red are 1 or more.



                                                                  You can clearly see that if green >= 1, blue >=1 and red=0 then the initial facts are satisfied, but the deducted one is not. Thus, "FALSE".






                                                                  share|improve this answer









                                                                  $endgroup$


















                                                                    0












                                                                    $begingroup$

                                                                    I already agree with some answers presented. Let me provide a graphical one. This is the 3-set venn diagram of all posible sets an element can be a member of:



                                                                    enter image description here



                                                                    I assume "some" means "1 or more". Thus, let's interpret the question:




                                                                    some Smaugs are Thors




                                                                    means elements of Green + Red are 1 or more




                                                                    some Thors are Thrains




                                                                    means elements of Blue + Red are 1 or more
                                                                    These two are facts.



                                                                    Now we are asked to deduce whether:




                                                                    some Smaugs are definitely Thrains




                                                                    First of all, I find "definitely" useless. It should be obvious this is about boolean logic; both true and false are always "definitely". Maybe it's put there as a user-friendly term? Disregarding that, this one would mean elements of Red are 1 or more.



                                                                    You can clearly see that if green >= 1, blue >=1 and red=0 then the initial facts are satisfied, but the deducted one is not. Thus, "FALSE".






                                                                    share|improve this answer









                                                                    $endgroup$
















                                                                      0












                                                                      0








                                                                      0





                                                                      $begingroup$

                                                                      I already agree with some answers presented. Let me provide a graphical one. This is the 3-set venn diagram of all posible sets an element can be a member of:



                                                                      enter image description here



                                                                      I assume "some" means "1 or more". Thus, let's interpret the question:




                                                                      some Smaugs are Thors




                                                                      means elements of Green + Red are 1 or more




                                                                      some Thors are Thrains




                                                                      means elements of Blue + Red are 1 or more
                                                                      These two are facts.



                                                                      Now we are asked to deduce whether:




                                                                      some Smaugs are definitely Thrains




                                                                      First of all, I find "definitely" useless. It should be obvious this is about boolean logic; both true and false are always "definitely". Maybe it's put there as a user-friendly term? Disregarding that, this one would mean elements of Red are 1 or more.



                                                                      You can clearly see that if green >= 1, blue >=1 and red=0 then the initial facts are satisfied, but the deducted one is not. Thus, "FALSE".






                                                                      share|improve this answer









                                                                      $endgroup$



                                                                      I already agree with some answers presented. Let me provide a graphical one. This is the 3-set venn diagram of all posible sets an element can be a member of:



                                                                      enter image description here



                                                                      I assume "some" means "1 or more". Thus, let's interpret the question:




                                                                      some Smaugs are Thors




                                                                      means elements of Green + Red are 1 or more




                                                                      some Thors are Thrains




                                                                      means elements of Blue + Red are 1 or more
                                                                      These two are facts.



                                                                      Now we are asked to deduce whether:




                                                                      some Smaugs are definitely Thrains




                                                                      First of all, I find "definitely" useless. It should be obvious this is about boolean logic; both true and false are always "definitely". Maybe it's put there as a user-friendly term? Disregarding that, this one would mean elements of Red are 1 or more.



                                                                      You can clearly see that if green >= 1, blue >=1 and red=0 then the initial facts are satisfied, but the deducted one is not. Thus, "FALSE".







                                                                      share|improve this answer












                                                                      share|improve this answer



                                                                      share|improve this answer










                                                                      answered yesterday









                                                                      George MenoutisGeorge Menoutis

                                                                      950212




                                                                      950212























                                                                          0












                                                                          $begingroup$

                                                                          The 3 over-laying circles don't always make sense.



                                                                          Smaugs may be touching into Thrain territory, or it may not.



                                                                          50% chance it's the 3 over-laying rings theory above; 50% chance it looks more like 3 serial links- as only the Smaugs and Thors must intersect, and only the Thors and Thrain must intersect. There is no data to intersect Smaugs with Thrains (or not to intersect)…



                                                                          So by adding in "definitely", it knocks off the True and Neither logic paths; so it's false.






                                                                          share|improve this answer








                                                                          New contributor




                                                                          tlr is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
                                                                          Check out our Code of Conduct.






                                                                          $endgroup$


















                                                                            0












                                                                            $begingroup$

                                                                            The 3 over-laying circles don't always make sense.



                                                                            Smaugs may be touching into Thrain territory, or it may not.



                                                                            50% chance it's the 3 over-laying rings theory above; 50% chance it looks more like 3 serial links- as only the Smaugs and Thors must intersect, and only the Thors and Thrain must intersect. There is no data to intersect Smaugs with Thrains (or not to intersect)…



                                                                            So by adding in "definitely", it knocks off the True and Neither logic paths; so it's false.






                                                                            share|improve this answer








                                                                            New contributor




                                                                            tlr is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
                                                                            Check out our Code of Conduct.






                                                                            $endgroup$
















                                                                              0












                                                                              0








                                                                              0





                                                                              $begingroup$

                                                                              The 3 over-laying circles don't always make sense.



                                                                              Smaugs may be touching into Thrain territory, or it may not.



                                                                              50% chance it's the 3 over-laying rings theory above; 50% chance it looks more like 3 serial links- as only the Smaugs and Thors must intersect, and only the Thors and Thrain must intersect. There is no data to intersect Smaugs with Thrains (or not to intersect)…



                                                                              So by adding in "definitely", it knocks off the True and Neither logic paths; so it's false.






                                                                              share|improve this answer








                                                                              New contributor




                                                                              tlr is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
                                                                              Check out our Code of Conduct.






                                                                              $endgroup$



                                                                              The 3 over-laying circles don't always make sense.



                                                                              Smaugs may be touching into Thrain territory, or it may not.



                                                                              50% chance it's the 3 over-laying rings theory above; 50% chance it looks more like 3 serial links- as only the Smaugs and Thors must intersect, and only the Thors and Thrain must intersect. There is no data to intersect Smaugs with Thrains (or not to intersect)…



                                                                              So by adding in "definitely", it knocks off the True and Neither logic paths; so it's false.







                                                                              share|improve this answer








                                                                              New contributor




                                                                              tlr is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
                                                                              Check out our Code of Conduct.









                                                                              share|improve this answer



                                                                              share|improve this answer






                                                                              New contributor




                                                                              tlr is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
                                                                              Check out our Code of Conduct.









                                                                              answered yesterday









                                                                              tlrtlr

                                                                              1




                                                                              1




                                                                              New contributor




                                                                              tlr is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
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                                                                              New contributor





                                                                              tlr is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
                                                                              Check out our Code of Conduct.






                                                                              tlr is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
                                                                              Check out our Code of Conduct.























                                                                                  0












                                                                                  $begingroup$

                                                                                  Consider the 1st part:





                                                                                  • If some Smaugs are Thors




                                                                                  Part 1:The following diagram shows how some Smaugs are Thors. The violet shaded part is the one that represents the few Smaugs that are Thors.





                                                                                  • some Thors are Thrains




                                                                                  This can have 2 possibilities:



                                                                                  1.enter image description here In the above case, as seen by the red shaded part, some Smaugs are Thrains.



                                                                                  2.But then, the other possibility of the second part:enter image description here Here, you can see that the red shaded part is outside the scope of the Smaugs, which means that there is a possibility that NONE of the Smaugs are Thrains. Hence, either some Smaugs are Thrains or none are Thrains. Which would further deduce that the statement




                                                                                  some Smaugs are definitely Thrains




                                                                                  is not true. Since, all Smaugs are not definitely Thrains, as there is a possibility that they're not Thrains.






                                                                                  share|improve this answer











                                                                                  $endgroup$













                                                                                  • $begingroup$
                                                                                    there is a third possibility, some smaugs are thrains (but not thors)
                                                                                    $endgroup$
                                                                                    – JonMark Perry
                                                                                    21 hours ago










                                                                                  • $begingroup$
                                                                                    Thanks for pointing it out @JonMarkPerry. I overlooked it. But the 2nd case shown would still keep the outcome as not definitely Thrains. Right? Will update nevertheless.
                                                                                    $endgroup$
                                                                                    – Rai
                                                                                    21 hours ago
















                                                                                  0












                                                                                  $begingroup$

                                                                                  Consider the 1st part:





                                                                                  • If some Smaugs are Thors




                                                                                  Part 1:The following diagram shows how some Smaugs are Thors. The violet shaded part is the one that represents the few Smaugs that are Thors.





                                                                                  • some Thors are Thrains




                                                                                  This can have 2 possibilities:



                                                                                  1.enter image description here In the above case, as seen by the red shaded part, some Smaugs are Thrains.



                                                                                  2.But then, the other possibility of the second part:enter image description here Here, you can see that the red shaded part is outside the scope of the Smaugs, which means that there is a possibility that NONE of the Smaugs are Thrains. Hence, either some Smaugs are Thrains or none are Thrains. Which would further deduce that the statement




                                                                                  some Smaugs are definitely Thrains




                                                                                  is not true. Since, all Smaugs are not definitely Thrains, as there is a possibility that they're not Thrains.






                                                                                  share|improve this answer











                                                                                  $endgroup$













                                                                                  • $begingroup$
                                                                                    there is a third possibility, some smaugs are thrains (but not thors)
                                                                                    $endgroup$
                                                                                    – JonMark Perry
                                                                                    21 hours ago










                                                                                  • $begingroup$
                                                                                    Thanks for pointing it out @JonMarkPerry. I overlooked it. But the 2nd case shown would still keep the outcome as not definitely Thrains. Right? Will update nevertheless.
                                                                                    $endgroup$
                                                                                    – Rai
                                                                                    21 hours ago














                                                                                  0












                                                                                  0








                                                                                  0





                                                                                  $begingroup$

                                                                                  Consider the 1st part:





                                                                                  • If some Smaugs are Thors




                                                                                  Part 1:The following diagram shows how some Smaugs are Thors. The violet shaded part is the one that represents the few Smaugs that are Thors.





                                                                                  • some Thors are Thrains




                                                                                  This can have 2 possibilities:



                                                                                  1.enter image description here In the above case, as seen by the red shaded part, some Smaugs are Thrains.



                                                                                  2.But then, the other possibility of the second part:enter image description here Here, you can see that the red shaded part is outside the scope of the Smaugs, which means that there is a possibility that NONE of the Smaugs are Thrains. Hence, either some Smaugs are Thrains or none are Thrains. Which would further deduce that the statement




                                                                                  some Smaugs are definitely Thrains




                                                                                  is not true. Since, all Smaugs are not definitely Thrains, as there is a possibility that they're not Thrains.






                                                                                  share|improve this answer











                                                                                  $endgroup$



                                                                                  Consider the 1st part:





                                                                                  • If some Smaugs are Thors




                                                                                  Part 1:The following diagram shows how some Smaugs are Thors. The violet shaded part is the one that represents the few Smaugs that are Thors.





                                                                                  • some Thors are Thrains




                                                                                  This can have 2 possibilities:



                                                                                  1.enter image description here In the above case, as seen by the red shaded part, some Smaugs are Thrains.



                                                                                  2.But then, the other possibility of the second part:enter image description here Here, you can see that the red shaded part is outside the scope of the Smaugs, which means that there is a possibility that NONE of the Smaugs are Thrains. Hence, either some Smaugs are Thrains or none are Thrains. Which would further deduce that the statement




                                                                                  some Smaugs are definitely Thrains




                                                                                  is not true. Since, all Smaugs are not definitely Thrains, as there is a possibility that they're not Thrains.







                                                                                  share|improve this answer














                                                                                  share|improve this answer



                                                                                  share|improve this answer








                                                                                  edited 21 hours ago

























                                                                                  answered 21 hours ago









                                                                                  RaiRai

                                                                                  966111




                                                                                  966111












                                                                                  • $begingroup$
                                                                                    there is a third possibility, some smaugs are thrains (but not thors)
                                                                                    $endgroup$
                                                                                    – JonMark Perry
                                                                                    21 hours ago










                                                                                  • $begingroup$
                                                                                    Thanks for pointing it out @JonMarkPerry. I overlooked it. But the 2nd case shown would still keep the outcome as not definitely Thrains. Right? Will update nevertheless.
                                                                                    $endgroup$
                                                                                    – Rai
                                                                                    21 hours ago


















                                                                                  • $begingroup$
                                                                                    there is a third possibility, some smaugs are thrains (but not thors)
                                                                                    $endgroup$
                                                                                    – JonMark Perry
                                                                                    21 hours ago










                                                                                  • $begingroup$
                                                                                    Thanks for pointing it out @JonMarkPerry. I overlooked it. But the 2nd case shown would still keep the outcome as not definitely Thrains. Right? Will update nevertheless.
                                                                                    $endgroup$
                                                                                    – Rai
                                                                                    21 hours ago
















                                                                                  $begingroup$
                                                                                  there is a third possibility, some smaugs are thrains (but not thors)
                                                                                  $endgroup$
                                                                                  – JonMark Perry
                                                                                  21 hours ago




                                                                                  $begingroup$
                                                                                  there is a third possibility, some smaugs are thrains (but not thors)
                                                                                  $endgroup$
                                                                                  – JonMark Perry
                                                                                  21 hours ago












                                                                                  $begingroup$
                                                                                  Thanks for pointing it out @JonMarkPerry. I overlooked it. But the 2nd case shown would still keep the outcome as not definitely Thrains. Right? Will update nevertheless.
                                                                                  $endgroup$
                                                                                  – Rai
                                                                                  21 hours ago




                                                                                  $begingroup$
                                                                                  Thanks for pointing it out @JonMarkPerry. I overlooked it. But the 2nd case shown would still keep the outcome as not definitely Thrains. Right? Will update nevertheless.
                                                                                  $endgroup$
                                                                                  – Rai
                                                                                  21 hours ago


















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