Theorem preqr2 3635

Description: Reverse equality lemma for unordered pairs. If two unordered pairs have the same first element, the second elements are equal. (Contributed by NM, 5-Aug-1993.)

Hypotheses

Ref	Expression
preqr2.1	|- A e. _V
preqr2.2	|- B e. _V

Assertion

Ref	Expression
preqr2	|- ( { C , A } = { C , B } -> A = B )

Proof of Theorem preqr2

Step	Hyp	Ref	Expression
1		prcom 3538	. . 3 |- { C , A } = { A , C }
2		prcom 3538	. . 3 |- { C , B } = { B , C }
3	1, 2	eqeq12i 2108	. 2 |- ( { C , A } = { C , B } <-> { A , C } = { B , C } )
4		preqr2.1	. . 3 |- A e. _V
5		preqr2.2	. . 3 |- B e. _V
6	4, 5	preqr1 3634	. 2 |- ( { A , C } = { B , C } -> A = B )
7	3, 6	sylbi 120	1 |- ( { C , A } = { C , B } -> A = B )

Syntax hints:   -> wi 4   = wceq 1296   e. wcel 1445   _Vcvv 2633   {cpr 3467

This theorem was proved from axioms:  ax-1 5  ax-2 6  ax-mp 7  ax-ia1 105  ax-ia2 106  ax-ia3 107  ax-io 668  ax-5 1388  ax-7 1389  ax-gen 1390  ax-ie1 1434  ax-ie2 1435  ax-8 1447  ax-10 1448  ax-11 1449  ax-i12 1450  ax-bndl 1451  ax-4 1452  ax-17 1471  ax-i9 1475  ax-ial 1479  ax-i5r 1480  ax-ext 2077

This theorem depends on definitions:  df-bi 116  df-tru 1299  df-nf 1402  df-sb 1700  df-clab 2082  df-cleq 2088  df-clel 2091  df-nfc 2224  df-v 2635  df-un 3017  df-sn 3472  df-pr 3473

This theorem is referenced by:  preq12b  3636  opth  4088  opthreg  4400