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Question 112
For any two statements p, q the statement $$\left(p \Rightarrow q \right) \vee \left(\sim \left(\left(\sim p\right) \Leftrightarrow q \right)\right)$$ is false only when
Solution
| p | $$\rightsquigarrow$$q | $$\rightsquigarrow$$p$$\Rightarrow$$q | p⇒∼p | (p$$\rightsquigarrow$$q |
| T | F | T | F | F |
| F | T | F | T | F |
From the above table, the truth value of the expression (∼p⇒p)∧(p⇒∼p) is always F. Hence it's a contradiction. so the p is always true and q is false Answer
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