# Proving Logic statement

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?

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?

• I was thinking of using the distributive law but I thought that could only be used for 3 or more variables? – sanch May 26 '16 at 15:15
• @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 $\neg y$. – Brian M. Scott May 26 '16 at 15:16
• Ah I see.. Thanks! – sanch May 26 '16 at 15:19
• @Matt: You’re welcome! – Brian M. Scott May 26 '16 at 15:20