Show that (~𝒑∨𝒒)∧(𝒑∧~𝒒) is a contradiction
(p˜∨q)∧(p∧q˜)=(p˜∧p∧q˜)∨(q∧p∧q˜)=((p˜∧p)∧q˜)∨((q∧q˜)∧p)=(0∧q˜)∨(0∧p)=0∨0=0\left( {\~p \vee q} \right) \wedge \left( {p \wedge \~q} \right) = \left( {\~p \wedge p \wedge \~q} \right) \vee \left( {q \wedge p \wedge \~q} \right) = \left( {\left( {\~p \wedge p} \right) \wedge \~q} \right) \vee \left( {\left( {q \wedge \~q} \right) \wedge p} \right) = \left( {0 \wedge \~q} \right) \vee \left( {0 \wedge p} \right) = 0 \vee 0 = 0(p˜∨q)∧(p∧q˜)=(p˜∧p∧q˜)∨(q∧p∧q˜)=((p˜∧p)∧q˜)∨((q∧q˜)∧p)=(0∧q˜)∨(0∧p)=0∨0=0
Q.E.D.
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