Using two different properties to prove a relation is an equivalence relation

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I'm currently taking Introduction to Pure Mathematics, and we've been working on equivalence relations for the last week. I had a problem with one question on our last tutorial which asks us to consider a relation ~ on a set X which satisfies the following two conditions:

a) $\forall$ $x$ $\in X,$ $x$ ~ $x$

b) $\forall$ $x,$ $y,$ $z$ $\in X,$ if $x$ ~ $y$ and $y$ ~ $z$, then $z$ ~ $x$

We were asked to prove that these two conditions alone are enough to prove that ~ is an equivalence relation.

I was unable to do this, and the answer sheet was posted today, stating that prove symmetry:

"Suppose that x ~ y. Then by reflexivity y ~ y. Hence using the second condition (with x = x, y, z = y) we conclude y ~ x."

Assuming we can prove symmetry, the transitivity proof is straight forward, but I don't understand the explanation given for the proof of symmetry. Namely, why can is be assumed that z = y? If someone could straighten this out for me it would be much appreciated.

Edit: All sorted - thanks, I'll leave this up for the time being though.

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Hint: for symmetry, if $x\sim y$ and since $y\sim y$ then by $(b)$ we have $y\sim x$,

for the transitive proprety: try $(b)$ and the symmetry.