Is there a mistake in this logical problem?

Question

I've found an oldish book with logical puzzles.

In one of the puzzles I was not able to find a single correct answer - IMO at least two solutions fit - either I miss something or there is a mistake.

The puzzle goes like this:

there are tree paths - you need to choose one - only one should be right;
each path has a sign - at least one sign is false.

Now, here are the signs:

A: The correct path is either B or C.
B: If (and only if) this is the wrong path, then the correct path is C.
C: The correct path is either this one or B.

Essentially A and C say the same, so they are either both true or both false.

It is easy to see that path A may be the answer (then all 3 signs are wrong) and path C is wrong (this would make all 3 signs right).

My confusion comes from trying path B:

if path B is the right path:
- sign A is true;
- sign C is true;
- then sign B must be false:
  - as far as I understand the implication on sign B allows this situation - while the answer in the book says that this sign is right as well, hence all three signs are true, hence B cannot be the answer.

Am I right saying that it's not possible to deterministically say that A is a correct path (i.e. it may be A or B)?

When sign B says "... the correct one is C", does it mean that C is the correct path? Or is it perhaps saying that what its says on C's sign is true? — Theo Bendit, Aug 13 '21 at 08:05
@TheoBendit, it says about the path - corrected to make it clear. — Werolik, Aug 13 '21 at 09:17

score 3 · Accepted Answer · answered Aug 13 '21 at 08:05

3

If path B is correct sign B must indeed be false and the sign says

"B is wrong" $\iff$ "C is correct".

But in this situation the truth value of "C is correct" is false and "B is wrong" also has value "false" (as we're in the situation where B is actually correct, remember), so the equivalence is true. So sign B would be true, your book is correct, and we again have a contradiction.

answered Aug 13 '21 at 08:05

Henno Brandsma

242,131

If you interpret the statement on $B$ as a counterfactual (ie "in every possible world where $B$ is wrong $C$ is right, and vice-versa"), then it is false. Because there still exists the logically valid possibility of both $B$ and $C$ being wrong, regardless what their present actual status is. But that's a pretty non-standard interpretation for the implication. – user3257842 Aug 13 '21 at 08:17
Ah, damn. it's the and only if part that my brain was skipping. I was thinking about !B ⟹ C - hence the confusion. – Werolik Aug 13 '21 at 09:24
@Werolik you see that the exact formulation is indeed essential. – Henno Brandsma Aug 13 '21 at 09:25

score 0 · Answer 2 · answered Aug 13 '21 at 08:04

0

Sign B is saying that either both halves are true, or both halves are false.
If B is the correct path, then both halves are false, so sign B is true.

answered Aug 13 '21 at 08:04

Empy2

50,853

score 0 · Answer 3 · answered Aug 13 '21 at 08:44

If path B is correct, then both these statements are vacuously true: $$\text{B is the wrong path } \implies \text{the correct path is C}\\ \text{the correct path is C } \implies \text{B is the wrong path}.$$ In other words: if path B is correct, then sign B is correct.

Is there a mistake in this logical problem?

3 Answers3