At the moment I'm going through a book which treats logic in a very rigorous axiomatic way. But I just got stuck in this theorem that I can't seem to be able to solve (I'm still trying hard). The thing is, that I already went through all the theorems before, and all the theorems after this particular theorem, but I still can't solve it.

This is the theorem I have to prove:

The symbol 'y' is equal to 'and', because the book is in Spanish. At the moment, I'm going this way, and I think I'm 'very near' to prove it.

I start by the **axiom 2**, which it is the material conditional:

$

(R\implies S) \iff (\lnot R \lor S)

$ (1)

And also using the axiom but backwards:

$

(S\implies R) \iff (\lnot S \lor R)

$ (2)

I think the key is in two theorems I already proved.

**First theorem:**

If $A \implies B$ and $ C \implies D $ are both true, then $ (A \land C) \implies (B \land D) $ is also true

**Similarly, second theorem:**

If $A \implies B$ and $ C \implies D $ are both true, then $ (A \lor C) \implies (B \lor D) $ is also true

So by using this, I go this way. By using (1) and (2) and **theorem 1** I get:

$

(S \implies R) \land (R \implies S) \iff [(\lnot R \lor S) \land (\lnot S \lor R)]

$ is true

This is equivalent to

$

(R \iff S) \iff [(\lnot R \lor S) \land (\lnot S \lor R)]

$

But I haven't been able to match the other side through already proved theorems or axioms. Some help will be greatly appreciated.

**I'm editing to add the theorems of distribution:**

Let A, B and C be statements. Then:

$ (A \lor B) \land C \implies [ (A \land C) \lor (B \land C)] $ is true

$ (A \land B) \lor C \iff [ (A \lor C) \land (B \lor C)] $ is true

## Best Answer

I'm using natural deduction:

Theorem:$(R\leftrightarrow S) \leftrightarrow (R\land S)\lor(\neg R \land \neg S)$Can you continue from here?