Simplify the following expressions using laws of logic: p v ~(~p --> q) [(p --> q)^ ~q] --> ~p [(p v q) ^ (p --> ~r) ^ r ] --> q (p v ~q) ^ (p v q) 5. ~[p --> ~(p ^ q)]

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Let us simplify the following expressions using laws of logic:

1. $p \lor \sim(\sim p \to q)=p \lor \sim( p \lor q)=p \lor (\sim p \land\sim q)= (p \lor \sim p) \land (p \lor \sim q)=T \land (p \lor \sim q)=p \lor \sim q$

2. $[(p \to q)\land \sim q] \to \sim p= \sim[(\sim p \lor q)\land \sim q] \lor \sim p= \sim(\sim p \lor q)\lor q \lor \sim p= ( p \land\sim q)\lor q \lor \sim p= ( p \lor q \lor \sim p) \land(\sim q \lor q \lor \sim p)= ( q \lor T) \land(T \lor \sim p)=T\land T=T$

3. $[(p \lor q) \land (p \to \sim r) \land r ]\to q= [(p \lor q) \land (\sim p \lor \sim r) \land r ]\to q= [(p \lor q) \land ((\sim p\land r) \lor (\sim r \land r) )]\to q= [(p \lor q) \land( (\sim p\land r) \lor F) ]\to q= [(p \lor q) \land (\sim p\land r) ]\to q= [(p \land \sim p\land r)\lor (q\land \sim p\land r) ]\to q= [(F \land r)\lor (q\land \sim p\land r) ]\to q= [F \lor (q\land \sim p\land r) ]\to q= q\land \sim p\land r\to q= \sim(q\land \sim p\land r)\lor q= \sim q\lor p\lor \sim r\lor q=T\lor p\lor \sim r=T$

4. $(p \lor \sim q) \land (p \lor q)= p \lor (\sim q\land q)=p\lor F=p$

5. $\sim[p\to \sim(p \land q)]= \sim[\sim p\lor \sim(p \land q)]= p\land(p \land q)=(p\land p) \land q=p\land q$

Let us simplify the following expressions using laws of logic:

1. $p \lor \sim(\sim p \to q)=p \lor \sim( p \lor q)=p \lor (\sim p \land\sim q)= (p \lor \sim p) \land (p \lor \sim q)=T \land (p \lor \sim q)=p \lor \sim q$

2. $[(p \to q)\land \sim q] \to \sim p= \sim[(\sim p \lor q)\land \sim q] \lor \sim p= \sim(\sim p \lor q)\lor q \lor \sim p= ( p \land\sim q)\lor q \lor \sim p= ( p \lor q \lor \sim p) \land(\sim q \lor q \lor \sim p)= ( q \lor T) \land(T \lor \sim p)=T\land T=T$

3. $[(p \lor q) \land (p \to \sim r) \land r ]\to q= [(p \lor q) \land (\sim p \lor \sim r) \land r ]\to q= [(p \lor q) \land ((\sim p\land r) \lor (\sim r \land r) )]\to q= [(p \lor q) \land( (\sim p\land r) \lor F) ]\to q= [(p \lor q) \land (\sim p\land r) ]\to q= [(p \land \sim p\land r)\lor (q\land \sim p\land r) ]\to q= [(F \land r)\lor (q\land \sim p\land r) ]\to q= [F \lor (q\land \sim p\land r) ]\to q= q\land \sim p\land r\to q= \sim(q\land \sim p\land r)\lor q= \sim q\lor p\lor \sim r\lor q=T\lor p\lor \sim r=T$

4. $(p \lor \sim q) \land (p \lor q)= p \lor (\sim q\land q)=p\lor F=p$

5. $\sim[p\to \sim(p \land q)]= \sim[\sim p\lor \sim(p \land q)]= p\land(p \land q)=(p\land p) \land q=p\land q$

Let us simplify the following expressions using laws of logic:

1. $p \lor \sim(\sim p \to q)=p \lor \sim( p \lor q)=p \lor (\sim p \land\sim q)= (p \lor \sim p) \land (p \lor \sim q)=T \land (p \lor \sim q)=p \lor \sim q$

2. $[(p \to q)\land \sim q] \to \sim p= \sim[(\sim p \lor q)\land \sim q] \lor \sim p= \sim(\sim p \lor q)\lor q \lor \sim p= ( p \land\sim q)\lor q \lor \sim p= ( p \lor q \lor \sim p) \land(\sim q \lor q \lor \sim p)= ( q \lor T) \land(T \lor \sim p)=T\land T=T$

3. $[(p \lor q) \land (p \to \sim r) \land r ]\to q= [(p \lor q) \land (\sim p \lor \sim r) \land r ]\to q= [(p \lor q) \land ((\sim p\land r) \lor (\sim r \land r) )]\to q= [(p \lor q) \land( (\sim p\land r) \lor F) ]\to q= [(p \lor q) \land (\sim p\land r) ]\to q= [(p \land \sim p\land r)\lor (q\land \sim p\land r) ]\to q= [(F \land r)\lor (q\land \sim p\land r) ]\to q= [F \lor (q\land \sim p\land r) ]\to q= q\land \sim p\land r\to q= \sim(q\land \sim p\land r)\lor q= \sim q\lor p\lor \sim r\lor q=T\lor p\lor \sim r=T$

4. $(p \lor \sim q) \land (p \lor q)= p \lor (\sim q\land q)=p\lor F=p$

5. $\sim[p\to \sim(p \land q)]= \sim[\sim p\lor \sim(p \land q)]= p\land(p \land q)=(p\land p) \land q=p\land q$

Let us simplify the following expressions using laws of logic:

1. $p \lor \sim(\sim p \to q)=p \lor \sim( p \lor q)=p \lor (\sim p \land\sim q)= (p \lor \sim p) \land (p \lor \sim q)=T \land (p \lor \sim q)=p \lor \sim q$

2. $[(p \to q)\land \sim q] \to \sim p= \sim[(\sim p \lor q)\land \sim q] \lor \sim p= \sim(\sim p \lor q)\lor q \lor \sim p= ( p \land\sim q)\lor q \lor \sim p= ( p \lor q \lor \sim p) \land(\sim q \lor q \lor \sim p)= ( q \lor T) \land(T \lor \sim p)=T\land T=T$

3. $[(p \lor q) \land (p \to \sim r) \land r ]\to q= [(p \lor q) \land (\sim p \lor \sim r) \land r ]\to q= [(p \lor q) \land ((\sim p\land r) \lor (\sim r \land r) )]\to q= [(p \lor q) \land( (\sim p\land r) \lor F) ]\to q= [(p \lor q) \land (\sim p\land r) ]\to q= [(p \land \sim p\land r)\lor (q\land \sim p\land r) ]\to q= [(F \land r)\lor (q\land \sim p\land r) ]\to q= [F \lor (q\land \sim p\land r) ]\to q= q\land \sim p\land r\to q= \sim(q\land \sim p\land r)\lor q= \sim q\lor p\lor \sim r\lor q=T\lor p\lor \sim r=T$

4. $(p \lor \sim q) \land (p \lor q)= p \lor (\sim q\land q)=p\lor F=p$

5. $\sim[p\to \sim(p \land q)]= \sim[\sim p\lor \sim(p \land q)]= p\land(p \land q)=(p\land p) \land q=p\land q$

Let us simplify the following expressions using laws of logic:

1. $p \lor \sim(\sim p \to q)=p \lor \sim( p \lor q)=p \lor (\sim p \land\sim q)= (p \lor \sim p) \land (p \lor \sim q)=T \land (p \lor \sim q)=p \lor \sim q$

2. $[(p \to q)\land \sim q] \to \sim p= \sim[(\sim p \lor q)\land \sim q] \lor \sim p= \sim(\sim p \lor q)\lor q \lor \sim p= ( p \land\sim q)\lor q \lor \sim p= ( p \lor q \lor \sim p) \land(\sim q \lor q \lor \sim p)= ( q \lor T) \land(T \lor \sim p)=T\land T=T$

3. $[(p \lor q) \land (p \to \sim r) \land r ]\to q= [(p \lor q) \land (\sim p \lor \sim r) \land r ]\to q= [(p \lor q) \land ((\sim p\land r) \lor (\sim r \land r) )]\to q= [(p \lor q) \land( (\sim p\land r) \lor F) ]\to q= [(p \lor q) \land (\sim p\land r) ]\to q= [(p \land \sim p\land r)\lor (q\land \sim p\land r) ]\to q= [(F \land r)\lor (q\land \sim p\land r) ]\to q= [F \lor (q\land \sim p\land r) ]\to q= q\land \sim p\land r\to q= \sim(q\land \sim p\land r)\lor q= \sim q\lor p\lor \sim r\lor q=T\lor p\lor \sim r=T$

4. $(p \lor \sim q) \land (p \lor q)= p \lor (\sim q\land q)=p\lor F=p$

5. $\sim[p\to \sim(p \land q)]= \sim[\sim p\lor \sim(p \land q)]= p\land(p \land q)=(p\land p) \land q=p\land q$

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Question ID: mtid-5-stid-8-sqid-3401-qpid-2100