Question #171387

Find simpler statement forms that are logically equivalent to p ⊕ p and (p ⊕ p) ⊕ p.

b) Is (p ⊕ q) ⊕ r ≡ p ⊕ (q ⊕ r)? Justify your answer.

c) Is (p ⊕ q) ∧ r ≡ (p ∧ r) ⊕ (q ∧ r)? Justify your answer.


1
Expert's answer
2021-03-17T12:09:17-0400

a) Let us find the truth table for ppp ⊕ p and (pp)p:(p ⊕ p) ⊕ p:


ppp(pp)p000101\begin{array}{|c|c|c|c|c|c|c|} \hline p & p\oplus p & (p\oplus p)\oplus p \\ \hline 0 & 0 & 0 \\ \hline 1 & 0 & 1 \\ \hline \end{array}


It follows that pp0p ⊕ p\equiv 0 and (pp)pp(p ⊕ p) ⊕ p\equiv p.



b) Let us find the truth table for (pq)r(p ⊕ q) ⊕ r and p(qr)p ⊕ (q ⊕ r):


pqrpq(pq)rqrp(qr)00000000010111010111101110001001101101101011000101110101\begin{array}{|c|c|c|c|c|c|c|} \hline p & q & r & p\oplus q & (p\oplus q)\oplus r & q\oplus r & p\oplus (q\oplus r)\\ \hline 0 & 0 & 0 & 0 & 0 & 0 & 0\\ \hline 0 & 0 & 1 & 0 & 1 & 1 & 1\\ \hline 0 & 1 & 0 & 1 & 1 & 1 & 1\\ \hline 0 & 1 & 1 & 1 & 0 & 0 & 0\\ \hline 1 & 0 & 0 & 1 & 1 & 0 & 1\\ \hline 1 & 0 & 1 & 1 & 0 & 1 & 0\\ \hline 1 & 1 & 0 & 0 & 0 & 1 & 0\\ \hline 1 & 1 & 1 & 0 & 1 & 0 & 1\\ \hline \end{array}


Since the formulas (pq)r(p ⊕ q) ⊕ r and p(qr)p ⊕ (q ⊕ r) have the same truth value in all cases, they are logically equivalent. It follows that (pq)rp(qr).(p ⊕ q) ⊕ r ≡ p ⊕ (q ⊕ r).


c) Let us find the truth table for (pq)r(p ⊕ q) ∧ r and (pr)(qr)(p ∧ r) ⊕ (q ∧ r):


pqrpq(pq)rprqrprqr0000000000100000010100000111101110010000101111011100000011100110\begin{array}{|c|c|c|c|c|c|c|c|} \hline p & q & r & p\oplus q & (p\oplus q)\land r & p\land r & q\land r &p\land r\oplus q\land r\\ \hline 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0\\ \hline 0 & 0 & 1 & 0 & 0 & 0 & 0 & 0\\ \hline 0 & 1 & 0 & 1 & 0 & 0 & 0 & 0\\ \hline 0 & 1 & 1 & 1 & 1 & 0 & 1 & 1\\ \hline 1 & 0 & 0 & 1 & 0 & 0 & 0 & 0\\ \hline 1 & 0 & 1 & 1 & 1 & 1 & 0 & 1\\ \hline 1 & 1 & 0 & 0 & 0 & 0 & 0 & 0\\ \hline 1 & 1 & 1 & 0 & 0 & 1 & 1 & 0\\ \hline \end{array}


Since the formulas (pq)r(p ⊕ q) ∧ r and (pr)(qr)(p ∧ r) ⊕ (q ∧ r) have the same truth value in all cases, they are logically equivalent. It follows that (pq)r(pr)(qr).(p ⊕ q) ∧ r ≡ (p ∧ r) ⊕ (q ∧ r).


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