Dear Student (p∧~q)∨(~p∧q)∨(~p∧~q)Apply distributive property=[p∨(~p∧q) ∧ ~q∨(~p∧q)]∨(~p∧~q)=[T∧(p∨q) ∧ (~p∨~q)∧T]∨(~p∧~q)=(p∨q)∨(~p∧~q)∧(~p∨~q)=(p∨~q)∨(q∨~p)∧(~p∨~q)=(q→p)∨(p→q)∧(~p∨~q)=T∧(~p∨~q)=(~p∨~q)hence proved Regards

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Dear Student

(p \land ~ q) \lor (~ p \land q) \lor (~ p \land ~ q) Apply distributive property = [p \lor (~ p \land q) \land ~ q \lor (~ p \land q)] \lor (~ p \land ~ q) = [T \land (p \lor q) \land (~ p \lor ~ q) \land T] \lor (~ p \land ~ q) = (p \lor q) \lor (~ p \land ~ q) \land (~ p \lor ~ q) = (p \lor ~ q) \lor (q \lor ~ p) \land (~ p \lor ~ q) = (q \to p) \lor (p \to q) \land (~ p \lor ~ q) = T \land (~ p \lor ~ q) = (~ p \lor ~ q) hence proved

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