Axiom of Equivalence For test..

Question

We have the axioms:

1. $\vdash x = y \to (A\to A')$ where $A'$ is the formula which is created by replacing some of the free apperances of $x$ in $A$ by $y$

2. $\vdash x=x$ for all $x$

We need to prove that:

1. $\vdash x=y\rightarrow y=x$

2. $\vdash x=y \rightarrow (y=z \rightarrow x=z)$

3. If $F$ is a function of arity 1 then: $\vdash x=y \rightarrow (F(x) \rightarrow F(y))$

I can't really figure out how to solve these and I have a test coming up, it would be a real help guys! I thank anyone who tries and helps me :]

Answer 1

If you want to prove $x= y \implies \phi(x,y)$, apply the axiom scheme with $A : \phi(x,x)$ and $A' : \phi(x,y)$. Then you are left with proving $\phi(x,x)$ which should be straightforward.

For example for $1$, apply the axiom scheme with $A : x = x$ and $A': y = x$ (we only replace the first of the two free occurences of $x$ in $A$). Then, use reflexivity to prove $A$.