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)
english logic-theory
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show 4 more comments
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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)
english logic-theory
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Can you explain why it would be "neither"?
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– 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.
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– JonMark Perry
yesterday
7
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If we have a counterexample, the claim should be false, no? I.e. it's not true that some Smaugs are definitely Thrains.
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– jafe
yesterday
2
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Not seeing the puzzle, just an english/logic question.
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– deep thought
yesterday
1
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Yeah off-topic though, isn't it? Why is this a puzzle, or about puzzles.
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– deep thought
yesterday
|
show 4 more comments
$begingroup$
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)
english logic-theory
$endgroup$
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
english logic-theory
asked yesterday
JonMark PerryJonMark Perry
18.9k63891
18.9k63891
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Can you explain why it would be "neither"?
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– 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.
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– JonMark Perry
yesterday
7
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If we have a counterexample, the claim should be false, no? I.e. it's not true that some Smaugs are definitely Thrains.
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– jafe
yesterday
2
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Not seeing the puzzle, just an english/logic question.
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– deep thought
yesterday
1
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Yeah off-topic though, isn't it? Why is this a puzzle, or about puzzles.
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– deep thought
yesterday
|
show 4 more comments
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Can you explain why it would be "neither"?
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– 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.
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– deep thought
yesterday
1
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Yeah off-topic though, isn't it? Why is this a puzzle, or about puzzles.
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– deep thought
yesterday
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Can you explain why it would be "neither"?
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– jafe
yesterday
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Can you explain why it would be "neither"?
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– 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
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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
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Not seeing the puzzle, just an english/logic question.
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– deep thought
yesterday
$begingroup$
Not seeing the puzzle, just an english/logic question.
$endgroup$
– deep thought
yesterday
1
1
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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
|
show 4 more comments
12 Answers
12
active
oldest
votes
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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!
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1
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Even without "definitely", the answer is still "false"...
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– 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.
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– 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?
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– athin
yesterday
1
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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
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– Omega Krypton
yesterday
1
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@athin: From a mathematical-logic standpoint, there is none. It's called the law of the excluded middle.
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– BlueRaja - Danny Pflughoeft
21 hours ago
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show 1 more comment
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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".
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But if the implication is sometimes always true, and sometimes always false, then it is neither true nor false.
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– Acccumulation
yesterday
add a comment |
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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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add a comment |
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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.
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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.)
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– SteveV
yesterday
add a comment |
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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.)
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add a comment |
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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.
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add a comment |
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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
New contributor
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add a comment |
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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.
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add a comment |
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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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add a comment |
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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:
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".
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add a comment |
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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.
New contributor
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add a comment |
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Consider the 1st part:
If some Smaugs are Thors
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. 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: 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.
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there is a third possibility, some smaugs are thrains (but not thors)
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– JonMark Perry
21 hours ago
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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.
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– Rai
21 hours ago
add a comment |
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12 Answers
12
active
oldest
votes
12 Answers
12
active
oldest
votes
active
oldest
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active
oldest
votes
$begingroup$
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!
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1
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Even without "definitely", the answer is still "false"...
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– 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.
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– 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?
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– athin
yesterday
1
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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
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– Omega Krypton
yesterday
1
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@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
|
show 1 more comment
$begingroup$
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!
$endgroup$
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?
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– 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
|
show 1 more comment
$begingroup$
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!
$endgroup$
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!
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?
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– 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
|
show 1 more comment
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
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I agree with this answer. But I have a question: what is an example question with NEITHER as its answer?
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– athin
yesterday
1
1
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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
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– Omega Krypton
yesterday
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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
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– Omega Krypton
yesterday
1
1
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@athin: From a mathematical-logic standpoint, there is none. It's called the law of the excluded middle.
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– BlueRaja - Danny Pflughoeft
21 hours ago
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@athin: From a mathematical-logic standpoint, there is none. It's called the law of the excluded middle.
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– BlueRaja - Danny Pflughoeft
21 hours ago
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show 1 more comment
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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".
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1
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But if the implication is sometimes always true, and sometimes always false, then it is neither true nor false.
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– Acccumulation
yesterday
add a comment |
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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".
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1
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But if the implication is sometimes always true, and sometimes always false, then it is neither true nor false.
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– Acccumulation
yesterday
add a comment |
$begingroup$
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".
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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".
answered yesterday
BlueRaja - Danny PflughoeftBlueRaja - Danny Pflughoeft
264110
264110
1
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But if the implication is sometimes always true, and sometimes always false, then it is neither true nor false.
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– Acccumulation
yesterday
add a comment |
1
$begingroup$
But if the implication is sometimes always true, and sometimes always false, then it is neither true nor false.
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– Acccumulation
yesterday
1
1
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But if the implication is sometimes always true, and sometimes always false, then it is neither true nor false.
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– Acccumulation
yesterday
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But if the implication is sometimes always true, and sometimes always false, then it is neither true nor false.
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– Acccumulation
yesterday
add a comment |
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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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add a comment |
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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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add a comment |
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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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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.
answered yesterday
NautilusNautilus
3,674523
3,674523
add a comment |
add a comment |
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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.
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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.)
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– SteveV
yesterday
add a comment |
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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.
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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.)
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– SteveV
yesterday
add a comment |
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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.
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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.
edited yesterday
answered yesterday
rhsquaredrhsquared
8,16021849
8,16021849
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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.)
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– SteveV
yesterday
add a comment |
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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.)
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– SteveV
yesterday
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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.)
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– SteveV
yesterday
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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.)
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– SteveV
yesterday
add a comment |
$begingroup$
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.)
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add a comment |
$begingroup$
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.)
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add a comment |
$begingroup$
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.)
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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.)
answered yesterday
RayRay
28514
28514
add a comment |
add a comment |
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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.
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add a comment |
$begingroup$
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.
$endgroup$
add a comment |
$begingroup$
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.
$endgroup$
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.
edited yesterday
answered yesterday
AcccumulationAcccumulation
504111
504111
add a comment |
add a comment |
$begingroup$
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
New contributor
$endgroup$
add a comment |
$begingroup$
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
New contributor
$endgroup$
add a comment |
$begingroup$
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
New contributor
$endgroup$
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
New contributor
New contributor
answered yesterday
Keith NicholasKeith Nicholas
1112
1112
New contributor
New contributor
add a comment |
add a comment |
$begingroup$
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.
$endgroup$
add a comment |
$begingroup$
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.
$endgroup$
add a comment |
$begingroup$
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.
$endgroup$
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.
answered yesterday
ZanyGZanyG
388114
388114
add a comment |
add a comment |
$begingroup$
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.
$endgroup$
add a comment |
$begingroup$
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.
$endgroup$
add a comment |
$begingroup$
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.
$endgroup$
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.
edited yesterday
answered yesterday
SmockSmock
112
112
add a comment |
add a comment |
$begingroup$
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:
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".
$endgroup$
add a comment |
$begingroup$
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:
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".
$endgroup$
add a comment |
$begingroup$
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:
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".
$endgroup$
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:
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".
answered yesterday
George MenoutisGeorge Menoutis
950212
950212
add a comment |
add a comment |
$begingroup$
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.
New contributor
$endgroup$
add a comment |
$begingroup$
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.
New contributor
$endgroup$
add a comment |
$begingroup$
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.
New contributor
$endgroup$
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.
New contributor
New contributor
answered yesterday
tlrtlr
1
1
New contributor
New contributor
add a comment |
add a comment |
$begingroup$
Consider the 1st part:
If some Smaugs are Thors
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. 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: 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.
$endgroup$
$begingroup$
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
add a comment |
$begingroup$
Consider the 1st part:
If some Smaugs are Thors
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. 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: 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.
$endgroup$
$begingroup$
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
add a comment |
$begingroup$
Consider the 1st part:
If some Smaugs are Thors
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. 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: 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.
$endgroup$
Consider the 1st part:
If some Smaugs are Thors
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. 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: 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.
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
add a comment |
$begingroup$
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
add a comment |
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$begingroup$
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