Question #147069
1.Determine whether ¬(p∨(¬p∧q)) and ¬p∧¬q are equivalent without using truth table.
2.Determine whether the compound proposition ~(p∨q)∨(~p∧q)∨p to tautology.
3.Determine whether (p → q)∧(p → r)≡p → (q ∧r) using a truth table
1
Expert's answer
2020-12-01T06:18:45-0500

Let T denotes True and F denotes False.


1.Let us determine whether ¬(p(¬pq))¬(p∨(¬p∧q)) and ¬p¬q¬p∧¬q are equivalent:


¬(p(¬pq))=¬p¬(¬pq))=¬p(p¬q))=¬(p∨(¬p∧q))=¬p\land\neg(¬p∧q))=¬p\land(p\lor\neg q))=


=¬pp¬p¬q=F¬p¬q=¬p¬q=¬p\land p\lor ¬p\land\neg q=F\lor ¬p\land\neg q= ¬p\land\neg q .


Therefore, the formulas are equivalent.


2.Let us determine whether the compound proposition (pq)(pq)p\sim(p∨q)∨(\sim p∧q)∨p is tautology.


(pq)(pq)p=(pq)(pp)(qp)=(pq)T(qp)=\sim(p∨q)∨(\sim p∧q)∨p=\sim(p∨q)∨(\sim p\lor p)∧(q∨p)=\sim(p∨q)∨T∧(q∨p)=


=(pq)(qp)=(pq)(pq)=T.=\sim(p∨q)∨(q∨p)=\sim(p∨q)∨(p\lor q)=T.


Therefore, the formula (pq)(pq)p\sim(p∨q)∨(\sim p∧q)∨p is tautology.



3.Let us determine whether (p → q)∧(p → r) ≡ p → (q ∧r) using a truth table:


pqrpqpr(pq)(pr)qrp(qr)FFFTTTFTFFTTTTFTFTFTTTFTFTTTTTTTTFFFFFFFTFFFTFFFTTFTFFFFTTTTTTTT\begin{array}{||c|c|c||c|c|c|c|c||} \hline\hline p & q & r & p\to q & p\to r & (p\to q)\land (p\to r) & q\land r & p\to(q\land r)\\ \hline\hline F & F & F & T & T & T & F & T\\ \hline F & F & T & T & T & T & F & T \\ \hline F & T & F & T & T & T & F & T \\ \hline F & T & T & T & T & T & T & T \\ \hline T & F & F & F & F & F & F & F\\ \hline T & F & F & F & T & F & F & F \\ \hline T & T & F & T & F & F & F & F\\ \hline T & T & T & T & T & T & T & T\\ \hline\hline \end{array}


Since the formulas (pq)(pr)(p\to q)\land (p\to r) and p(qr)p\to(q\land r) always have the same truth values, they are logically equivalent.



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