# Axiom of Equivalence For test..

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 :]

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Hint for 1): Let $A$ be $y=y$ and $A'$ be $y=x$ and use the first axiom. I'm assuming you have the usual rules for first order predicate logic (without equality) already down. – Harald Hanche-Olsen Jan 28 '13 at 15:59
Thaks! I forgot i could actually choose the A and A' – Rachel Bernoulli Jan 28 '13 at 16:02
(sidenote: 1. is not an axiom, but an axiom scheme) – Hagen von Eitzen Jan 28 '13 at 16:04
Axiom scheme? sorry i Just translated that from Hebrew without any knowledge of its real syntax in English so...thanks anyway I managed to solve them :) – Rachel Bernoulli Jan 28 '13 at 16:14

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 $A$.