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So I have an statement that I need to prove using Logical Equivalences: $$(p\land q) \lor [p \land (\lnot( \lnot p \lor q)) ] \equiv p $$ I made it through some steps but I can't seem to make it to the end. Here is my work: $$ \equiv (p\land q) \lor [p \land ( p \land \lnot q)) ] $$ $$ \equiv (p\land q) \lor [(p \land p) \land \lnot q] $$ $$ \equiv (p\land q) \lor (p \land \lnot q) $$ This is about as far as a I get. Can someone show me where I went wrong or point me in the right direction?

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You’re almost there: $(p\land q)\lor(p\land\neg q)\equiv p\land(q\lor\neg q)$ by one of the distributive laws. Can you finish it now?

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  • $\begingroup$ I was thinking of using the distributive law but I thought that could only be used for 3 or more variables? $\endgroup$ – sanch May 26 '16 at 15:15
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    $\begingroup$ @Matt: It just says that $(x\land y)\lor(x\land z)\equiv x\land(y\lor z)$; it doesn’t matter what $x,y$, and $z$ are. They don’t even have to be atomic propositions. In particular, $z$ can be $\neg y$. $\endgroup$ – Brian M. Scott May 26 '16 at 15:16
  • $\begingroup$ Ah I see.. Thanks! $\endgroup$ – sanch May 26 '16 at 15:19
  • $\begingroup$ @Matt: You’re welcome! $\endgroup$ – Brian M. Scott May 26 '16 at 15:20

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