Show that ¬[q∧ ( p → ¬ q ) ] → p ≡ ( p ⋁ q ) using equivalence laws.

Question

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Answer 1

¬[q∧ ( p → ¬ q ) ] → p ≡ ( p ⋁ q )

Note that your precise way of writing some of this will depend on your textbook/professor's choices.

1 | ¬[q∧ ( p → ¬ q ) ] → p....................ACP (want p v q)

2 | [¬q v ¬(p → ¬ q)] → p.....................DM 1

3 | [¬q v ¬(¬p v ¬ q)] → p.....................MI 2

4 | [¬q v (¬¬p ∧ ¬¬ q)] → p................DM 3

5 | [¬q v (¬¬p ∧ ¬¬ q)] → p.................DM 3

6 | [¬q v (p ∧ q)] → p...........................DN 6

7 || ¬q....................................................ACP (want p)

8 || ¬q v (p ∧ q)....................................Add 7

9 || p.......................................................MP 6, 8

10| ¬q → p.............................................CP 7-9

11| ¬¬q v p.............................................MI 10

12| q v p...................................................DN 11

13| p v q...................................................Comm 12

14 [¬[q∧ ( p → ¬ q ) ] → p]→( p ⋁ q )....CP 1-13

There's half the answer -- we've proven that one implies the other. If you'd like to complete the biconditional and prove that one is equivalent to the other, please contact me via Wyzant messenger. Thanks!

Answer 2

statement reason

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not [ q and ( p -> not q) ] --> p given

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[ q and ( p -> not q) ] or p not A or B = A --> B

A is the stuff inside braces;

B is p

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p or q and p or (p -> not q) distributive

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p or q and p or ( not p or not q) not A or B = A --> B

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p or q and (p or not p or p or not q) distributive

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p or q and ( true or p or not q) bichotomy law: A or not A must be true

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p or q and (true) true or anything is true

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p or q the truth value depends on p or q