What logic rule is used to show $(p \lor q) \land (p)=p~$? 
What logic rule is used to show: $(p \vee q) \wedge (p)=p~$?

This is obvious, because if p is true the whole expression is true, and if p is false, the whole expression is false, but I'm not sure which logic rule is used to draw this conclusion.

One can show this with Boolean algebra as follows:
$$
(p \vee q)\wedge p = (p \vee q)\wedge (p\vee F) = p \vee (q \wedge F) = p \vee F = p
$$
If we have $\cdot$ for $\wedge$ and $+$ for $\vee$, then here's what this looks like:
$$
(p+q)p = (p+q)(p+0) = p+q0 = p+0 = p
$$
To prove this, I am using the "distributive laws" of and/or.  That is,


*

*$p \wedge(q \vee r) = (p \wedge q) \vee (p \wedge r)$

*$p \vee(q \wedge r) = (p \vee q) \wedge (p \vee r)$

