- #1

darkchild

- 155

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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 [itex]\in[/itex] {x,y} z = x V z = y.

The given proof is:

"By virtue of Theorem 43

u [itex]\in[/itex] {u,v},

and thus by the hypothesis of the theorem u [itex]\in[/itex] {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.