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The following propositional statement is  $\left(P \implies \left(Q \vee R\right)\right) \implies \left(\left(P \wedge Q \right)\implies R\right)$

  1.    satisfiable but not valid
  2.    valid
  3.    a contradiction
  4.    None of the above
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5 Answers

Best answer
36 votes
36 votes
Answer a

It is false when P = T, Q = T, R = F and hence cannot be valid.

It is true (satisfiable) when P = T, Q = T, R = T
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Given statement  $(P \Rightarrow (Q \vee R)) \Rightarrow ((P \wedge Q) \Rightarrow R)$ can be simplified as follows

  $ \equiv (\neg P \vee (Q \vee R) ) \Rightarrow (\neg (P \wedge Q) \vee R) \\ \equiv (\neg P \vee R \vee Q ) \Rightarrow (\neg P \vee \neg Q \vee R) \\ \equiv ((\neg P \vee R )\vee Q ) \Rightarrow ((\neg P \vee R) \vee \neg Q)  \\ $

Let $(\neg P \vee R ) = A $, then $(A \vee Q ) \Rightarrow (A \vee \neg Q) $  [simplifying this further]

$ \neg(A \vee Q ) \vee (A \vee \neg Q) $

$ (\neg A \wedge \neg Q)  \vee (A \vee \neg Q) $

$[(\neg A \wedge \neg Q)  \vee A] \vee \neg Q $     [As, $ X \vee (\neg X \wedge Y) = X \vee Y  $ ]

$  (\neg Q \vee A) \vee \neg Q $

$  (\neg Q \vee A) $

$  (\neg Q \vee (\neg P \vee R ) ) $

For $Q=F$, result is True

For $Q=T, P=T, R=F$, result is False

Hence, Satisfiable(true atleast once) but not valid
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Check for validity by making given implication $false$

Option $\large A$


 THE BELOW ANSWER IS WRONG


By  Exportation Law,

$P \implies Q \implies R \equiv (P\land Q ) \implies R$

applying it to given proposition

$\left(P \implies \left(Q \vee R\right)\right) \implies \left(\left(P \wedge Q \right)\implies R\right) \equiv \left( P \land \left( Q \lor R\right) \land \left( P \land Q\right)  \implies R \right)$

checking validity first

 

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Applying by case method,

case: Let P = T,

(Q V R) → (Q → R)

(Q V R) → (Q’ V R)

(Q V R)’ + (Q’ V R)

Q’R’ + Q’ + R

Q’(1+R’) + R = Q’+R

Here, Q’+R is satisfiable but not valid.
Answer:

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