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GATE | GATE-CS-2006 | Question 27
  • Difficulty Level : Easy
  • Last Updated : 03 Jul, 2014
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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.

 

ABC((A ∧ B) → C))((A → C) ∧ (B → C))((A ∨ B) → C))((A → C) ∨ (B → C))
000TTTT
001TTTT
010TFFT
011TTTT
100TFFT
101TTTT
110FFFF
111TTTT

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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