Confusing Axiomatic Set Theory Proof

1. Aug 30, 2013

darkchild

This proof makes no sense to me.

The theorem to be proved is

Theorem 44. {x,y} = {u,v} → (x = u & y = v) V (x = v & y = u)

where {x,y} and {u,v} are sets with exactly two members, which can be either sets or individuals. The proof relies on:

Theorem 43. z $\in$ {x,y} z = x V z = y.

The given proof is:

"By virtue of Theorem 43
u $\in$ {u,v},​
and thus by the hypothesis of the theorem
u $\in$ {x,y}.​
Hence, by virtue of Th. 43 again
(1)
u = x V u = y.​
By exactly similar arguments
(2)
v = x V v = y,​
(3)
x = u V x = v,​
(4)
y = u V y = v.​
We may now consider two cases.
Case 1: x = y. Then by virtue of (1), x = u, and by virtue of (2), y = v."

This is where I got lost. Couldn't I just as easily argue that x = v by virtue of (2) and y = u by virtue of (1)? Or by virtue of (3) and (4)? What's the rationale behind the assumed values of x and y, and couldn't any of the four propositions support it? And if x = y, how could I justify the argument that x and y were equal to two ostensibly different variables without showing that u = v? On one, hand I can sort of see that the assumption x = y and the conditions (1) - (4) would necessarily make it true that u = v...but, given that then either x or y could be said to be equal to either v or u, would there be any need for this part of the proof at all?

In the interest of completeness, the rest of the proof is:

Case 2: x ≠ y. In view of (1), either x = u or y = u. Suppose x ≠ u. Then y = u and by (3) x = v. On the other hand, suppose y ≠ u. Then x = u and by (4), y = v.

2. Sep 1, 2013

CompuChip

Yes, if x = y then basically the four propositions read (replace x and y by (x = y) everywhere):
(1)' u = x = y V u = x = y
(2)' v = x = y V v = x = y
(3)' x = y = u V x = y = v
(4)' x = y = u V x = y = v

Which reduces to
(1)'' u = x = y
(2)'' v = x = y
(3)'' x = y = u V x = y = v

Which reduces to
(1,2) u = x = y = v
(3)'' x = y = u V x = y = v

Which reduces to
(1-3) x = y = u = v

So indeed, you can use any of them to draw that conclusion.

Actually, x = y is a pretty boring case, but you have to handle it :-)