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I'm really stuck on proving this propositional logic equivalency. I've tried De Morgan's law, and double negation to see if I could get it, but no luck. Any help is greatly appreciated! P.S Please, tell me which laws you use to prove it.

$(p \land q) \lor (q \land r) \lor (r \land p) \equiv (p \lor q) \land (q \lor r) \land (r \lor p)$

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$(p\lor q)\land(p\lor r)\land(q\lor r)$

$\Leftrightarrow (p\lor(q\land r))\land(q\lor r)$ by the distributive law

$\Leftrightarrow ((p\lor(q\land r))\land q)\lor ((p\lor(q\land r))\land r)$ again by the distributive law

$\Leftrightarrow ((p\land q)\lor((q\land r)\land q))\lor ((p\land r)\lor((q\land r)\land r))$ again by the distributive law

$\Leftrightarrow (p\land q)\lor(q\land r\land q)\lor (p\land r)\lor(q\land r\land r)$ dropping parantheses since certain parts are well defined

$\Leftrightarrow (p\land q)\lor(r\land q\land q)\lor (p\land r)\lor(q\land r\land r)$ by commutative law

$\Leftrightarrow (p\land q)\lor(q\land r)\lor (p\land r)\lor(q\land r)$ by idempotent law

$\Leftrightarrow (p\land q)\lor (p\land r)\lor(q\land r)\lor(q\land r)$ by commutative law

$\Leftrightarrow (p\land q)\lor(q\land r)\lor (p\land r)$ by idempotent law

Hence, $(p\lor q)\land(p\lor r)\land(q\lor r) \Leftrightarrow (p\land q)\lor(q\land r)\lor (p\land r)$

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