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Exercise 12.2 · Q12

Q.Check whether the statement p→(q→p)p\to(q\to p) is a tautology or a contradiction without using the truth table.

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We convert every conditional to its ¬∨\neg\vee form (from Logical Equivalence, A→B≡¬A∨BA\to B\equiv\neg A\vee B), then simplify using the standard laws until the formula collapses to T\mathbb T.

Step 1. Rewrite the inner conditional q→pq\to p using A→B≡¬A∨BA\to B\equiv\neg A\vee B. q→p≡¬q∨pq\to p\equiv\neg q\vee p.

Step 2. Substitute back. p→(q→p)≡p→(¬q∨p)p\to(q\to p)\equiv p\to(\neg q\vee p).

Step 3. Rewrite the outer conditional the same way. p→(¬q∨p)≡¬p∨(¬q∨p)p\to(\neg q\vee p)\equiv\neg p\vee(\neg q\vee p).

Step 4. Regroup using the Associative and Commutative Laws. ¬p∨(¬q∨p)≡(¬p∨p)∨¬q\neg p\vee(\neg q\vee p)\equiv(\neg p\vee p)\vee\neg q. …

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