Ordered pairs, property, proof

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enter image description here Can some please make an exposition of this proof. I certainly am able to understand the trivial direction. But on the interesting direction, the first case is that (b) and (c) must hold and the second is (a) and (d) must hold and produce the desired conclusion. It is clear to me that if the first case holds then the conclusion is reached. On the otherhand, if the second hold, i cannot quite grasp why we can get either $u=y$ or $v=y$ part. Can anyone help me. Can someone also draw the sketch of this proof, why was it done this way. Thanks.

This is from the elements of set theory by enderton.

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Assume (x,y) = (u,v). Thus
{x} = $\cap$(x,y) = $\cap$(u,v) = {u}. x = u.
So {x,y} = $\cup${{x}, {x,y}} = $\cup${{x}, {x,v}} = {x,v}.
This reduces the tedium of separate cases to two.
Namely, x = y and not x = y.