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Exercise 12.3 · Q11

Q.Which one is the inverse of the statement (p∨q)→(p∧q)(p\vee q)\to(p\wedge q)?

(1) (p∧q)→(p∨q)(p\wedge q)\to(p\vee q)
(2) ¬(p∨q)→(p∧q)\neg(p\vee q)\to(p\wedge q)
(3) (¬p∨¬q)→(¬p∧¬q)(\neg p\vee \neg q)\to(\neg p\wedge\neg q)
(4) (¬p∧¬q)→(¬p∨¬q)(\neg p\wedge\neg q)\to(\neg p\vee\neg q)
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The inverse of a conditional P→QP\to Q is ¬P→¬Q\neg P\to\neg Q; here P=(p∨q)P=(p\vee q) and Q=(p∧q)Q=(p\wedge q), so we need ¬(p∨q)\neg(p\vee q) and ¬(p∧q)\neg(p\wedge q), each simplified with De Morgan's Law.

Step 1. Identify P,QP,Q. P=p∨qP=p\vee q, Q=p∧qQ=p\wedge q.

Step 2. Negate PP using De Morgan's Law. ¬P=¬(p∨q)≡¬p∧¬q\neg P=\neg(p\vee q)\equiv\neg p\wedge\neg q.

Step 3. Negate QQ using De Morgan's Law. ¬Q=¬(p∧q)≡¬p∨¬q\neg Q=\neg(p\wedge q)\equiv\neg p\vee\neg q. …

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