Equality and its axioms I'm reading a Wikipedia article about equality and its axioms and I'm wondering how we can prove (using equality axioms) that:


*

*If $x = y$ then $y = x$ (symmetry) 

*If $x = y$ and $y = z$ then $x = z$ (transitivity)


Equality axioms:


*

*For each variable $x$: $x = x$

*For all variables x and y, and any function symbol f:   $x=y
    \rightarrow f(...,x,...)=f(...,y...,)$ 

*$x=y \rightarrow (\varphi \rightarrow \varphi')$,  where and $y$ is
free for $x$ in $\varphi$ and $\varphi'$ is replacing some free
variable $x$ by $y$ in $\varphi$
The Wikipedia article:
https://en.wikipedia.org/wiki/First-order_logic#Equality_and_its_axioms

For symmetry :
1) $x = y → (x = x → y = x)$ --- equality axiom 3, with $x=x$ as $\varphi$
2) $x = x → (x = y → y = x)$ --- from 1) with the tautological equivalence : $[p \to (q \to r)] \leftrightarrow [q \to (p \to r)]$ and modus ponens
3) $x = x$ --- equality axiom 1

4) $x = y → y = x$ --- from 2) and 3) by mp.


Similar for transitivity; from axiom 3 we have : $y=x \to (y=z \to x=z)$, with $y=z$ as $\varphi$.
Thus, with symmetry : $x=y \to y=x$ and tautological equivalence, we get :

$(x=y \land y=z) \to x=z$.

