How to prove the following propositional logic equivalence?

Question

The question asked me to prove the following equivalence. $$（r \rightarrow p)\rightarrow (p\land q)\equiv (\lnot r \rightarrow p)\land(\lnot r \rightarrow q)\land (p \rightarrow q)$$

And I did the followings:

For L.H.S., $（r\rightarrow p)\rightarrow (p\land q)\\\equiv \lnot (\lnot r\lor p)\lor (p\land q)\\ \equiv (r \land \lnot p) \lor (p \land q)$

For R.H.S., $(\lnot r \rightarrow p)\land(\lnot r \rightarrow q)\land (p \rightarrow q)\\ \equiv (r\lor p) \land (r \lor q) \land (\lnot p \lor q)\\ \equiv (r\lor (p\land q))\land (\lnot p\lor q)\\ $

Now, I'm a bit stuck as to how to show that LHS = RHS.

Answer 1

I’ve done the proof and here is the answer.
L.H.S. = $ （r\rightarrow p)\rightarrow (p\land q)\\ \equiv \lnot (\lnot r\lor p)\lor (p\land q)\\ \equiv (r \land \lnot p) \lor (p \land q)\\ \equiv (r\lor p)\land (\lnot p \lor p)\land (r\lor q)\land (\lnot p \lor q)\\ \equiv(r\lor p)\land T\land (r\lor q)\land(\lnot p\lor q)\\ \equiv(\lnot r\rightarrow p)\land (\lnot r\rightarrow q)\land (p\rightarrow q)\\ \equiv R.H.S. $

Thx for all your helps !

Answer 2

After writing out the left hand side using $\;\phi \rightarrow \psi \;\equiv\; \lnot \phi \lor \psi\;$ and DeMorgan, it is an $\;\lor\;$ of $\;\land\;$'s. Similarly, the right hand side becomes a $\;\land\;$ of $\;\lor\;$'s.

How do you get from one to the other? Through distribution.

So you choose one of the two sides, and distribute $\;\land\;$ over $\;\lor\;$, or $\;\lor\;$ over $\;\land\;$, and work towards the other side.

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After you completed the proof, look up conjunctive normal form and disjunctive normal form and see how those terms relate to your proof.