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Exercise 12.3 · Q17

Q.The dual of ¬(p∨q)∨[p∨(p∧¬r)]\neg(p\vee q)\vee[p\vee(p\wedge\neg r)] is

(1) ¬(p∧q)∧[p∨(p∧¬r)]\neg(p\wedge q)\wedge[p\vee(p\wedge\neg r)]
(2) (p∧q)∧[p∧(p∨¬r)](p\wedge q)\wedge[p\wedge(p\vee\neg r)]
(3) ¬(p∧q)∧[p∧(p∧r)]\neg(p\wedge q)\wedge[p\wedge(p\wedge r)]
(4) ¬(p∧q)∧[p∧(p∨¬r)]\neg(p\wedge q)\wedge[p\wedge(p\vee\neg r)]
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Forming a dual is a mechanical symbol swap: every ∨\vee becomes ∧\wedge and every ∧\wedge becomes ∨\vee, but ¬\neg is never touched (Definition 12.19, Remark 1).

Step 1. Write the original formula, marking each ∨/∧\vee/\wedge. ¬(p∨‾q)∨‾[p∨‾(p∧‾¬r)]\neg(p\underline\vee q)\underline\vee[p\underline\vee(p\underline\wedge\neg r)].

Step 2. Swap the outermost connective. The outer ¬(p∨q) ∨ […]\neg(p\vee q)\ \vee\ [\ldots] becomes ¬(p∧q) ∧ […]\neg(p\wedge q)\ \wedge\ [\ldots] -- note ¬\neg stays exactly where it is, only the ∨/∧\vee/\wedge inside and around it swap.

Step 3. Swap the connectives inside the bracket. p∨(p∧¬r)p\vee(p\wedge\neg r) becomes p∧(p∨¬r)p\wedge(p\vee\neg r) -- again ¬r\neg r is untouched. …

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