[Math] How to i prove $p\to (q \vee r) \equiv (p \wedge \sim q) \to r$

Question

please Help me in this question

i have tried to solve it like this:

$$p \to (q \vee r) \equiv (p \wedge \sim q)\to r$$
$$p \vee \sim (q \vee r) \equiv \sim(p \wedge q)\vee r$$
$$p \vee \sim q \wedge \sim r \equiv \sim p \vee q \vee r$$
Thank you

Answer 1

Using the rule of implication we have $$p\to (q\vee r) \equiv \sim p \vee (q \vee r).$$ Using associativity we have $$\sim p \vee (q \vee r) \equiv (\sim p \vee q) \vee r.$$ Next is De Morgan $$(\sim p \vee q) \vee r \equiv \sim(p \wedge \sim q) \vee r .$$ And now use implication once again $$\sim(p \wedge \sim q) \vee r \equiv (p \wedge \sim q) \to r .$$

Regarding you answer:

Your first step is what you need to prove. What is the second step?

Edit

The rule of implication is $$ a\to b \equiv \sim a\vee b.$$