Proving formulae from Theory - problem

Question

I want to prove this formulae $(f(a,f(y,x)) = f(a,f(x,y)))$ from theory T ($a$ and $b$ are functions without arguments):

$$T:$$

$$\alpha: f(f(x,y),z) = f(x,f(y,z))$$

$$\beta: f(x,y) = f(y,x)$$ $$\gamma: P(f(f(a,a),b))$$

So I'm trying to proove that $T |- f(a,f(y,x)) = f(a,f(x,y))$ using rules.

The problem is that my solution is too simple - it is probably incorrect.

So what I need is $\beta$ formulae and reflexivity rule ( $x=x$ )

1. Since $\beta$ says: $f(x,y) = f(y,x)$, from $f(a,f(x,y))$ I can conclude $f(a,(f(y,x))$
2. So now I have $f(a,f(y,x)) = f(a,f(y,x))$
3. The formula from 2. is in fact reflexivity axiom so it's proven.

Is this proof correct? I don't think so but I can't figure out where is the problem.

Answer 1

It's got the right idea, but it's backwards: you've gone from the thing you're trying to prove, to a true statement, and you need to go the other way.

Start with $f(a, f(x, y))=f(a, f(x, y))$ - true by reflexivity - and apply $\beta$ to get $f(a, f(x, y))=f(a, f(y, x))$.