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Consider the following propositional statements:

  • $P_1: ((A ∧ B) → C)) ≡ ((A → C) ∧ (B → C))$
  • $P_2: ((A ∨ B) → C)) ≡ ((A → C) ∨ (B → C))$


Which one of the following is true?

  1. $P_1$ is a tautology, but not $P_2$
  2. $P_2$ is a tautology, but not $P_1$
  3. $P_1$ and $P_2$ are both tautologies
  4. Both $P_1$ and $P_2$ are not tautologies
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(D) Both $P_1$ and $P_2$ are not tautologies.

$P_1:$ If $A$ is true and $B$ is false, LHS of $P_1$ is true but RHS becomes false. Hence not tautology.

$P_2:$ Forward side is true. But reverse side is not true. When $A$ is false and $B$ is true and C is false, RHS is true but LHS is false.

LHS of $P_2$ can be simplified as follows:

$((A∨B) → C) \equiv (~(A∨B) ∨ C)$
$\quad \quad \equiv (~A ∧~B) ∨C)$
$\quad \quad \equiv (~A ∨C) ∧ (~B ∨C)$
$\quad \quad \equiv (A→C) ∧ (B→C)$
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Answer is "D" for this question

but what if we exchange in right side of P1 & P2 let's see what happen

P1 : ((A ∧ B) → C)) ≡ ((A → C) ∨ (B → C))       
P2 : ((A ∨ B) → C)) ≡ ((A → C) ∧ (B → C)) 

P1: ~A +~B +C  ≡ ~A +~B+C   

P2:  ~A~B + C ≡(~A+C)(~B+C)     // Here we know that C+~A~B = (C+~A)(C+~B)

 P1 and P2 are both not tautologies here

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