If the following statements are true:
A ≡ B
∼ B ∨ C
D ⊃ ( A & ∼ C )
Then which of the following cannot be true?
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The third statement should have been D ⊃ ( A & ∼ C ) , not D ⊃ ( A & 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!
Using the rules of inference from classical logic:
1) A ≡ B
2) A ⊃ B (Biconditional elimination from (1))
3) ∼ B ∨ C
4) B ⊃ C (Material implication from (3))
5) A ⊃ C (Hypothetical syllogism from (2) and (4))
6) D ⊃ ( A & ∼ C )
7) ( D ⊃ A ) & ( D ⊃ ∼ C ) (Distribution of (6))
8) D ⊃ ∼ C (Conjunctive elimination of (7))
9) D ⊃ A (Conjunctive elimination of (7))
10) D ⊃ C (Hypothetical syllogism of (9) and (5))
Assume indirect proof: Assume D to be true
11) ∼ C (Modus ponens of (8))
12) C (Modus ponens of (10))
13) C & ∼ 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
Since A ⇔ B , we can infer, from the second premise, that ¬ A ∨ C , which is the opposite of A ∧ ¬ C , which is therefore false.
The third premise then implies that D implies a falsehood, therefore D is false.
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 ) , not D ⊃ ( A & 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!
I was confused by the choice of symbols as well, but according to Wikipedia's list of logic symbols, ⊃ is an alternative symbol for ⟹ . I believe the three statements translate to A ⟺ B , ¬ B ∨ C , and D ⟹ ( A ∧ ¬ C ) , respectively.
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Hi, I believe the problem is incorrect, and that you meant to write ( ∼ B ) ∨ ( ∼ C ) which is equivalent to C ⟹ ∼ B .
As it stands, the last proposition means that if D is true then A and C are both true, and by the first statement, so is B , and this is a consistent assignment of truth values, in that it does not contradict the second statement ∼ B ∨ C which is true since C is.
On the other hand if the second statement said C ⟹ ∼ B then the truth of D would contradict this, so D would have to be false.