Consider the following propositional statements:
P1 : ((A ∧ B) → C)) ≡ ((A → C) ∧ (B → C))
P2 : ((A ∨ B) → C)) ≡ ((A → C) ∨ (B → C))
Which one of the following is true?
(A) P1 is a tautology, but not P2
(B) P2 is a tautology, but not P1
(C) P1 and P2 are both tautologies
(D) Both P1 and P2 are not tautologies
Answer: (D)
Explanation: The easiest way to solve this question by creating truth tables for the expressions given. Note that P1 will be a tautology if truth table for left expression is exactly same as truth table for right expression. Same holds for P2 also.
| A | B | C | ((A ∧ B) → C)) | ((A → C) ∧ (B → C)) | ((A ∨ B) → C)) | ((A → C) ∨ (B → C)) |
|---|---|---|---|---|---|---|
| 0 | 0 | 0 | T | T | T | T |
| 0 | 0 | 1 | T | T | T | T |
| 0 | 1 | 0 | T | F | F | T |
| 0 | 1 | 1 | T | T | T | T |
| 1 | 0 | 0 | T | F | F | T |
| 1 | 0 | 1 | T | T | T | T |
| 1 | 1 | 0 | F | F | F | F |
| 1 | 1 | 1 | T | T | T | T |
So as we see from table, none of the P1 or P2 are tautologies, so option (D) is correct.
Source: www.cse.iitd.ac.in/~mittal/gate/gate_math_2006.html
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