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## Here's the Solution to this Question

(1) $p \vee \sim (\sim p\to q)$

$=p \vee \sim(p\vee q)$ Implication

$=p\vee (\sim p \wedge \sim q)$ De- Morgan's law

$=(p \vee \sim p)\wedge(p \vee \sim q)$ Distributive law

$=1 \wedge(p\vee \sim q)$ Known tautology

$=(p \vee \sim q)$ Dominance

$=(\sim q \vee p)$ Commutative

$=q\to p$ Implication

(2)$[(p \to q) \wedge \sim q]\to \sim p$

$=\sim[(p \to q) \wedge \sim q] \vee \sim p$ Implication

$=\sim[(\sim p \vee q) \wedge \sim q] \vee \sim p$ Implication

$=\sim[(\sim p \wedge \sim q) \vee (q \wedge \sim q)]\vee \sim p$ Distributive

$=$ $\sim [(\sim p \wedge \sim q)\vee 0]\vee \sim p$ Known contradiction

$=\sim [(\sim p \wedge \sim q)] \vee \sim p$ Dominance

$=(p\vee q) \vee \sim p$ De Morgan's Law

$=(p\vee \sim p) \vee q$ Associativity

$= 1\vee q$ Known tautology

$=1$ Dominance

(3)$[(p \vee q)\wedge (p \to \sim r) \wedge r] \to q$

$=[(p\vee q)\wedge ( \sim p \vee \sim r)\wedge r]\to q$ Implication

$=[(p \vee q) \wedge (\sim p \wedge r) \vee ( \sim r \wedge r)]\to q$ Distributive

$=[(p \vee q) \wedge(\sim p \wedge r) \vee 0]\to q$ Known contradiction

$=[(p \vee q] \wedge (\sim p \wedge r)]\to q$ Dominance

$=\sim[(p \vee q) \wedge (\sim p \wedge r)] \to q$ Implication

$=\sim(p \vee q) \vee \sim (\sim p \wedge r) \vee q$ De Morgan

$=\sim (p \vee q) \vee (p \vee \sim r) \vee q$ De Morgan

$=\sim(p \vee q) \vee (p\vee q)\vee \sim r$ Associativity

$=1 \vee \sim r$ Known tautology

$=1$ Dominance

(4)$(p \vee \sim q)\wedge (p \vee q)$

$=p \vee (\sim q \wedge q)$ Distributive law

$=p \vee 0$ Known contradiction

$=p$ Dominance

(5)$\sim[p \to \sim (p \wedge q)]$

$=\sim[\sim p\vee \sim(p \wedge q)]$ Implication

$=p \wedge(p \wedge q)$ De Morgan's Law

$=(p \wedge p) \wedge q$ Associative

$=p \wedge q$ Idempotent