# Why is $p \land (p \lor q)$ equivalent to $p$?

How can you prove with equivalence laws that $p \land (p\lor q)$ is equivalent to $p$? I know you have to get rid of $q$, but I'm not sure how.

-
To prove using equivalence laws, we need your list of equivalence laws. There are many possible "standard" lists. – user7530 Sep 10 '13 at 19:35
I have to use laws like distributivity, De Morgan, True/False-elimination – Guest001 Sep 10 '13 at 19:39

We have: \begin{align*} p \wedge (p \vee q) &= (p \vee F) \wedge (p \vee q) & \text{identity for } \vee \\ &= p \vee (F \wedge q) & \text{distributivity of } \vee \text{ over } \wedge \\ &= p \vee F & \text{annihilator for } \wedge \\ &= p & \text{identity for } \vee. \end{align*}

-
Note, that $F$ is an arbitrary false statement, such as $p \wedge \neg p$ – AlexR Sep 10 '13 at 19:44
thank you very much! That's what I was looking for – Guest001 Sep 10 '13 at 19:46
@Guest001 If you feel satisfied with this answer, you may accept it. – Doug Spoonwood Sep 11 '13 at 1:30

When proving simple tautologies like this one, I like to break it into cases: if $p$ is $T$ and if $p$ is $F$.

If $p$ is $T$, then we have $T \wedge (T\vee q) = T\wedge T = T$.

If $p$ is $F$, then we have $F \wedge (F\vee q) = F$.

-
This is usually what I do myself. A good approach! – New_to_this Sep 10 '13 at 20:14

Here is a third way, - just pointing out, that there are several ways;

\begin{align*} p\wedge (p\vee q)&\equiv (p\wedge p)\vee(p\wedge q)\ - \ Distributive \ law\\ &\equiv p\vee(p\wedge q) \ - \ Idempotent \ law\\ &\equiv p \ - \ Absorption \ law \end{align*}

-

I have to use laws like distributivity, De Morgan, True/False-elimination

In that case: $$(p ∧ (p ∨ q)) ⇔ (p ∧ (p ∨ q)) ∨ (p ∧ ¬p) ⇔ (p ∧ ((p ∨ q) ∨ ¬p)) ⇔ p$$

-