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)
And also using the axiom but backwards:
(S\implies R) \iff (\lnot S \lor R)
I think the key is in two theorems I already proved.
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