# Propositional logic simplification using laws

I've been given the propisition below and the task to simplify it to the simplest equal proposition.

$$(p \rightarrow (q \vee r)) \rightarrow (p \wedge (q \vee r))$$

I've been trying to do this for a bit now and the only steps I can possibly think of are these.

Distribution $$( p \rightarrow (q \vee r)) \rightarrow ((p \wedge r) \vee (p \wedge r))$$

or (not sure if this one is correct)

Implication $$(\neg p \vee (q \vee r)) \rightarrow ((p \wedge r) \vee (p \wedge r))$$

Because I had no idea where to start, I attempted to create a truth table (I left out some of the steps below, it's a pain to create this in markdown). This shows that the proposition is equal to p. This shows me where I should end up.

p q r $$(p \rightarrow (q \vee r)) \rightarrow (p \wedge (q \vee r))$$
0 0 0 0
0 0 1 0
0 1 0 0
0 1 1 0
1 0 0 1
1 0 1 1
1 1 0 1
1 1 1 1

If someone could point me in the right direction and especially tell me which steps they took and more importantly why, that would be very helpful. I really want to understand this but I am struggeling.

Hint: Let $$q \vee r = s$$. The proposition would be $$(p \rightarrow s) \rightarrow (p \wedge s)$$
• So, if I am correct, that leaves me with: $\neg(\neg p \vee s) \vee (p \vee s)$. Can I then get rid of the $\vee s$ on both sides? Since they are the same on both sides? Commented Sep 17, 2021 at 17:15
• @user214910485 it should be $\neg(\neg p \vee s) \vee(p \wedge s)$ which equivalent with $(p \wedge \neg s)\vee (p \wedge s)$ Commented Sep 17, 2021 at 17:17
• Yes my bad, i messed up the latex thing and can't edit it anymore. So when I simplify the negation it basically says p and not s OR p and s. Which makes the value of s irrelevant (is that the negation law?). That leaves me with $p \vee p$ which is p, right? Commented Sep 17, 2021 at 17:37
• @user214910485 It is also distribution law: $$(p \wedge \neg s) \vee (p \wedge s) \equiv p \vee (\neg s \wedge s).$$ Commented Sep 17, 2021 at 17:40