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

Question

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.

Answer 1

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,

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

Answer 2

The statement to be proved is that $$p \equiv p \land (p \lor q),$$ which from the definition of the biconditional can be broken into 2 statements \begin{equation} \label{1} \tag{1} p \rightarrow p \land (p \lor q) \end{equation} and \begin{equation} \label{2} \tag{2} p \land (p \lor q) \rightarrow p. \end{equation}

$\ref{2} \ $ follows very simply, we just use the rule of simplification or conjunction elimination to conclude $p$ immediately.

For $ \ \ref{1} \ $ we start with $p$ and use addition or disjunction introduction to write $p \lor q$, and then combine the 2 expressions using conjunction introduction to conclude $ p \land (p \lor q)$