Kenneth Rosen Edition 7 Exercise 1.3 Question 51 (Page No. 36)

Question

Find a compound proposition logically equivalent to p->q using only the logical operator NOR.

Answer 1

Logic equivalent to p->q using NOR gate only.

Comments

¬p≡p↓p

p→q≡((p↓p)↓q)↓((p↓p)↓q)

Answer 2

p$\rightarrow$q

=$\neg$p$\vee$q

=$\neg$($\neg$p$\vee$q) by applying Demorgan's Law,

=($\neg$p$\downarrow$q)

=($\neg$(p$\vee$p)$\downarrow$q) by applying Idempotent Law,

=((p$\downarrow$p)$\downarrow$q) .

edit: as identified by @ankitgupta.1729, this answer is wrong.

So right answer is, 

$p\rightarrow$q

=$\neg$p$\vee$q

=$\neg$($\neg$($\neg$p$\vee$q)) by applying Double negation Law,

=$\neg$($\neg$p$\downarrow$q)

=$\neg$($\neg$(p$\vee$p)$\downarrow$q)) by applying Idempotent Law,

=$\neg$((p$\downarrow$p)$\downarrow$q) 

=$\neg$[$\color{Red}($(p$\downarrow$p)$\downarrow$q$\color{Red})$$\vee$$\color{Red}($(p$\downarrow$p)$\downarrow$q$\color{Red})$] by applying Idempotent law

=$\color{Red}($(p$\downarrow$p)$\downarrow$q$\color{Red})$$\downarrow$$\color{Red}($(p$\downarrow$p)$\downarrow$q$\color{Red})$

Comments

@ankitgupta.1729

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you have done mistake while applying De-Morgan’s law.

$p \rightarrow q \equiv \neg p \vee q \equiv \neg \neg (\neg p \vee q) \equiv \neg (p \wedge \neg q) $

$\equiv \neg ((p \wedge \neg q ) \vee (p \wedge \neg q)) \equiv \neg (\neg \neg(p \wedge \neg q ) \vee \neg \neg(p \wedge \neg q))$

$ \equiv \neg (\neg (\neg p \vee q ) \vee \neg (\neg p \vee q))$

$ \equiv \neg (\neg (\neg (p \vee p) \vee q  ) \vee \neg (\neg (p \vee p) \vee q))$

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@ankitgupta.1729 Thanks for identifying my mistake. I have edited my answer.