Equality and its axioms

Question

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

1. If $x = y$ then $y = x$ (symmetry)
2. If $x = y$ and $y = z$ then $x = z$ (transitivity)

Equality axioms:

1. For each variable $x$: $x = x$

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

3. $x=y \rightarrow (\varphi \rightarrow \varphi')$, 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

Answer 1

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.

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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 :

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