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We know that $p\leftrightarrow q\equiv (p\rightarrow q)\wedge (q\rightarrow p)$

                       $p\leftrightarrow q\equiv(\sim p\vee q)\wedge (\sim q\vee p)\equiv(\sim p\wedge\sim q)\vee(p\wedge q)$

                       $p\leftrightarrow q\equiv(p\wedge q)\vee (\sim p\wedge\sim q)$

lets prove using truth table

$p$ $q$ $\sim p$ $\sim q$ $ (p\wedge q)$ $(\sim p\wedge\sim q)$ $(p\wedge q)\vee(\sim p\wedge\sim q)$ $p\leftrightarrow q$
T T F F T F T T
T F F T F F F F
F T T F F F F F
F F T T F T T T
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.........

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As given in question: "either show that both sides are true, or that both sides are false, for exactly the same combinations of truth values of the propositional variables in these expressions (whichever is easier).”

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