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Exercise 12.2 · Q8

Q.Show that

(i) ¬(p∧q)≡¬p∨¬q\neg(p\wedge q)\equiv \neg p\vee\neg q
(ii) ¬(p→q)≡p∧¬q\neg(p\to q)\equiv p\wedge\neg q.
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For each part we build both sides' truth-table columns side by side across all four rows and confirm they match exactly, which is the definition of logical equivalence.

Step 1. (i) ¬(p∧q)\neg(p\wedge q) vs ¬p∨¬q\neg p\vee\neg q.

ppqqp∧qp\wedge q¬(p∧q)\neg(p\wedge q)¬p\neg p¬q\neg q¬p∨¬q\neg p\vee\neg q
TTTFFFF
TFFTFTT
FTFTTFT
FFFTTTT

The columns for ¬(p∧q)\neg(p\wedge q) and ¬p∨¬q\neg p\vee\neg q both read F,T,T,TF,T,T,T -- identical. So ¬(p∧q)≡¬p∨¬q\neg(p\wedge q)\equiv\neg p\vee\neg q (De Morgan's Law).

Step 2. (ii) ¬(p→q)\neg(p\to q) vs p∧¬qp\wedge\neg q.

| pp | qq | p→qp\to q | ¬(p→q)\neg(p\to q) | ¬q\neg q | p∧¬qp\wedge\neg q |

|---|---|---|---|---|---| …

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