Question #156785

Simplify the following expressions using laws of logic:

  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. ~[p --> ~(p ^ q)]




Expert's answer

Let us simplify the following expressions using laws of logic:


1.p∨∼(∼p→q)=p∨∼(p∨q)=p∨(∼p∧∼q)=(p∨∼p)∧(p∨∼q)=T∧(p∨∼q)=p∨∼qp \lor \sim(\sim p \to q)=p \lor \sim( p \lor q)=p \lor (\sim p \land\sim q)= (p \lor \sim p) \land (p \lor \sim q)=T \land (p \lor \sim q)=p \lor \sim q


2.[(p→q)∧∼q]→∼p=∼[(∼p∨q)∧∼q]∨∼p=∼(∼p∨q)∨q∨∼p=(p∧∼q)∨q∨∼p=(p∨q∨∼p)∧(∼q∨q∨∼p)=(q∨T)∧(T∨∼p)=T∧T=T[(p \to q)\land \sim q] \to \sim p= \sim[(\sim p \lor q)\land \sim q] \lor \sim p= \sim(\sim p \lor q)\lor q \lor \sim p= ( p \land\sim q)\lor q \lor \sim p= ( p \lor q \lor \sim p) \land(\sim q \lor q \lor \sim p)= ( q \lor T) \land(T \lor \sim p)=T\land T=T


3. [(p∨q)∧(p→∼r)∧r]→q=[(p∨q)∧(∼p∨∼r)∧r]→q=[(p∨q)∧((∼p∧r)∨(∼r∧r))]→q=[(p∨q)∧((∼p∧r)∨F)]→q=[(p∨q)∧(∼p∧r)]→q=[(p∧∼p∧r)∨(q∧∼p∧r)]→q=[(F∧r)∨(q∧∼p∧r)]→q=[F∨(q∧∼p∧r)]→q=q∧∼p∧r→q=∼(q∧∼p∧r)∨q=∼q∨p∨∼r∨q=T∨p∨∼r=T[(p \lor q) \land (p \to \sim r) \land r ]\to q= [(p \lor q) \land (\sim p \lor \sim r) \land r ]\to q= [(p \lor q) \land ((\sim p\land r) \lor (\sim r \land r) )]\to q= [(p \lor q) \land( (\sim p\land r) \lor F) ]\to q= [(p \lor q) \land (\sim p\land r) ]\to q= [(p \land \sim p\land r)\lor (q\land \sim p\land r) ]\to q= [(F \land r)\lor (q\land \sim p\land r) ]\to q= [F \lor (q\land \sim p\land r) ]\to q= q\land \sim p\land r\to q= \sim(q\land \sim p\land r)\lor q= \sim q\lor p\lor \sim r\lor q=T\lor p\lor \sim r=T


4. (p∨∼q)∧(p∨q)=p∨(∼q∧q)=p∨F=p(p \lor \sim q) \land (p \lor q)= p \lor (\sim q\land q)=p\lor F=p


 5. ∼[p→∼(p∧q)]=∼[∼p∨∼(p∧q)]=p∧(p∧q)=(p∧p)∧q=p∧q\sim[p\to \sim(p \land q)]= \sim[\sim p\lor \sim(p \land q)]= p\land(p \land q)=(p\land p) \land q=p\land q





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