Proving equivalence of statements without using a truth table [closed]

How can I prove these statements without using a truth table?

$1.\quad (p \leftrightarrow q) \equiv (p \land q) \lor (\neg p \land \neg q)$

$2. \quad (p \land q \land r) \equiv \neg p \lor \neg q \lor \neg r$

$3.\quad p \land (p \lor r) \equiv p$

closed as off-topic by Pragabhava, Davide Giraudo, Jean-Claude Arbaut, Daniel W. Farlow, ShaileshApr 15 '17 at 0:02

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• You're looking for the symbol ⇔. Also, please use LaTeX. – DHMO Apr 14 '17 at 14:40
• @DHMO Actually, he is using LaTeX (or at least MathJax), but only on single symbols, not whole expressions. – Arthur Apr 14 '17 at 14:43
• @Arthur I only saw that after I clicked edit, so whatever. – DHMO Apr 14 '17 at 14:44
• Work with that on your own. If you were to narrow/edit your question post so that, indeed, only one question is asked, perhaps you'll get more help, – Namaste Apr 14 '17 at 14:51
• @DHMO You're right, of course. – Arthur Apr 14 '17 at 14:51

You've posted three separate questions in one question field.

So I will not answer them all. I will give hints for each:

$(1)\;$ Note that by definition, \begin{align} p\leftrightarrow q &\equiv (p\to q) \land (q \to p)\tag{Definition of biconditional}\\ \\ &\equiv (\lnot p \lor q) \land (\lnot q \lor p)\tag{implication} \end{align}

Use the distributive property (a couple of times, if needed) we can work to show it is equivalent to $$(p\land q) \lor (\lnot p \land \lnot q)$$ I'll let you take it from here.

$(2)\;$ Use DeMorgan's twice (and make use of the associative property), on $\neg (p \land q \land r),\,$ to show that is equivalent to $\;\neg p \lor \neg q \lor \neg r.$

\begin{align} \neg (p \land q \land r) &\equiv \lnot ((p\land q) \land r)\tag{associativity}\\ \\ &\equiv (\lnot (p\land q) \lor \lnot r) \tag{DeMorgan's}\\ \\ &\equiv (\lnot p \lor \lnot q)\lor \lnot r \tag{DeMorgan's again}\\ \\ &\equiv \lnot p \lor \lnot q \lor \lnot r\tag{associativity} \end{align}

$(3)\;$ use the distributive property on $\;p \land (p \lor r)\;$ to show it is equivalent to $\;p.$ \begin{align} p \land (p \lor r)&\equiv (p\land p) \lor (p \land r)\tag{distributive property}\\ \\ &\equiv p \lor (p\land r) \tag{Why is p\land p \equiv p?} \end{align}

I'll let you finish the proof for $(3)$.

• HJY: If you'd like me to check your work upon your completion of the proofs of $(1), (3),\,$ feel free to post your work, and I'll be happy to review it. – Namaste Apr 14 '17 at 15:46