Let us make a truth table for (∼p∧q)∨(p∧∼q)(\sim p∧q) ∨ (p∧\sim q)(∼p∧q)∨(p∧∼q):
pq∼p∼q∼p∧qp∧∼q(∼p∧q)∨(p∧∼q)FFTTFFFFTTFTFTTFFTFTTTTFFFFF\begin{array}{||c|c||c|c|c|c|c||} \hline \hline p & q & \sim p & \sim q & \sim p\land q & p\land \sim q & (\sim p\land q)\lor (p\land \sim q)\\ \hline\hline F & F & T & T & F & F & F\\ \hline F & T & T & F & T & F & T\\ \hline T & F & F & T & F & T & T\\ \hline T & T & F & F & F & F & F\\ \hline\hline \end{array}pFFTTqFTFT∼pTTFF∼qTFTF∼p∧qFTFFp∧∼qFFTF(∼p∧q)∨(p∧∼q)FTTF
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