Answer to Question #156318 in Discrete Mathematics for Jason

Question #156318

Simplify the following expressions using laws of logic and put what law of logic did you use or apply.

  1. p v ~(~p --> q)
  2. [(p --> q)^ ~q] --> ~p
  3. [(p v q) ^ (p --> ~r) ^ r ] --> q
  4. (p v ~q) ^ (p v q)
  5. 5. ~[p --> ~(p ^ q)]
1
Expert's answer
2021-01-25T01:43:40-0500

(1) p(pq)p \vee \sim (\sim p\to q)

=p(pq)=p \vee \sim(p\vee q) Implication

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

=(pp)(pq)=(p \vee \sim p)\wedge(p \vee \sim q) Distributive law

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

=(pq)=(p \vee \sim q) Dominance

=(qp)=(\sim q \vee p) Commutative

=qp=q\to p Implication


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

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

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

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

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

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

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

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

=1q= 1\vee q Known tautology

=1=1 Dominance


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

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

=[(pq)(pr)(rr)]q=[(p \vee q) \wedge (\sim p \wedge r) \vee ( \sim r \wedge r)]\to q Distributive

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

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

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

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

=(pq)(pr)q=\sim (p \vee q) \vee (p \vee \sim r) \vee q De Morgan

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

=1r=1 \vee \sim r Known tautology

=1=1 Dominance


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

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

=p0=p \vee 0 Known contradiction

=p=p Dominance


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

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

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

=(pp)q=(p \wedge p) \wedge q Associative

=pq=p \wedge q Idempotent


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