Simplify a proposition of logic: $p ∨ (p ∧ (\lnot p ∧ q ∨ r ∧ (p ∧ r)))$ I'm trying to come up with some concrete simplification for the following proposition:

$$p ∨ (p ∧ (\lnot p ∧ q ∨ r ∧ (p ∧ r)))$$

Any ideas on what is the resulting form?
 A: I'll touch on a comment left below another answer:

I can manipulate $p∨(p∧\varphi)$ a lot using distributivity, but I ran in circles when I tried to drop $\varphi$ using simple rules.

Let's take a look at a truth-table for $p \lor(p \land \varphi)$, and compare those truth values, with the truth value assignment column for $p$:
$\qquad\qquad\qquad$
(Truth-table image compliments to WolframAlpha). We see that $p \equiv p \lor (p \land \varphi)$, and can justify the equivalence by merely considering each of the possible $4$ truth value assignments for $p$ and $\phi$, i.e., we have a proof-by-cases.
Now, if we let $\varphi = \lnot p ∧ q ∨ r ∧ (p ∧ r)$, then we have $$p ∨ (p ∧ (\lnot p ∧ q ∨ r ∧ (p ∧ r))) \equiv p \lor (p \land \varphi) \equiv p$$
A: If $p$ is true, the statement is true. If $p$ is false the statement is false. From this we conclude the statement$\iff$$p$, so the statement is equivalent to $p$.
A: Hint:


*

*Let $f_{\alpha}(p) = p \lor (p \land \alpha)$, show that $f_\alpha$ does not depend on $\alpha$, that is, $f_\alpha(p) = p$.

*This behaves much like $g_\beta(x) = \max(x,\min(x,\beta))$, i.e. $g_\beta(x) \leq x$ because of minimum and $g_\beta(x) \geq x$ because of maximum, hence $g_\beta(x) = x$.


I hope this helps $\ddot\smile$
