Some Symbolic Logic

Logic Level 3

If the following statements are true:

A B A \equiv B

B C \sim B \vee C

D ( A & C ) D \supset (A \text{ } \& \sim C)

Then which of the following cannot be true?

A A B B C C D D

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4 solutions

Varsha Dani
Oct 3, 2018

Hi, I believe the problem is incorrect, and that you meant to write ( B ) ( C ) (\sim B) \vee (\sim C) which is equivalent to C B C \implies \sim B .

As it stands, the last proposition means that if D D is true then A A and C C are both true, and by the first statement, so is B B , and this is a consistent assignment of truth values, in that it does not contradict the second statement B C \sim B \vee C which is true since C C is.

On the other hand if the second statement said C B C \implies \sim B then the truth of D D would contradict this, so D D would have to be false.

The third statement should have been D ( A & C ) D \supset (A \text{ } \& \sim C) , not D ( A & C ) D \supset (A \text{ } \& C) . I had some LaTeX problems (LaTeX didn't like the "~" symbol that I had, but I was able to change it with "\sim"). Anyway, it should be fixed now. Sorry for the confusion!

David Vreken - 2 years, 8 months ago
Johanan Paul
Oct 14, 2018

Using the rules of inference from classical logic:

1) A B A \equiv B

2) A B A \supset B (Biconditional elimination from (1))

3) B C \sim B \lor C

4) B C B \supset C (Material implication from (3))

5) A C A \supset C (Hypothetical syllogism from (2) and (4))

6) D ( A & C ) D \supset (A \& \sim C)

7) ( D A ) & ( D C ) (D \supset A) \& ( D \supset \sim C) (Distribution of (6))

8) D C D \supset \sim C (Conjunctive elimination of (7))

9) D A D \supset A (Conjunctive elimination of (7))

10) D C D \supset C (Hypothetical syllogism of (9) and (5))

Assume indirect proof: Assume D to be true

11) C \sim C (Modus ponens of (8))

12) C C (Modus ponens of (10))

13) C & C C \& \sim C (Conjunctive introduction of (11) and (12))

(13) is a contradiction, hence the assumption that D is true is false.

C) D cannot be true

Fabricio Kolberg
Oct 7, 2018

Since A B A \Leftrightarrow B , we can infer, from the second premise, that ¬ A C \neg A \vee C , which is the opposite of A ¬ C A \wedge \neg C , which is therefore false.

The third premise then implies that D implies a falsehood, therefore D is false.

Affan Morshed
Oct 4, 2018

I think you made a second (besides the one pointed by Varsha Dani) mistake while defining D; you used set symbols (and that to wrongly as otherwise we would have no idea of contradicting D as it could include any possibility, besides having to include (A&C) in addition to something else) instead of logic symbols. There is technically no solution to this problem as I pointed out, but I assumed your mistake (luckily correctly) and got the right answer.

The third statement should have been D ( A & C ) D \supset (A \text{ } \& \sim C) , not D ( A & C ) D \supset (A \text{ } \& C) . I had some LaTeX problems (LaTeX didn't like the "~" symbol that I had, but I was able to change it with "\sim"). Anyway, it should be fixed now. Sorry for the confusion!

David Vreken - 2 years, 8 months ago

I was confused by the choice of symbols as well, but according to Wikipedia's list of logic symbols, \supset is an alternative symbol for \implies . I believe the three statements translate to A B A \iff B , ¬ B C \neg B \vee C , and D ( A ¬ C ) D \implies (A \wedge \neg C) , respectively.

William Allbritain - 2 years, 8 months ago

1 pending report

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