Theorem preqr2 3561

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 3468	. . 3 |- { C , A } = { A , C }
2		prcom 3468	. . 3 |- { C , B } = { B , C }
3	1, 2	eqeq12i 2094	. 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 3560	. 2 |- ( { A , C } = { B , C } -> A = B )
7	3, 6	sylbi 119	1 |- ( { C , A } = { C , B } -> A = B )

Syntax hints:   -> wi 4   = wceq 1284   e. wcel 1433   _Vcvv 2601   {cpr 3399

This theorem was proved from axioms:  ax-1 5  ax-2 6  ax-mp 7  ax-ia1 104  ax-ia2 105  ax-ia3 106  ax-io 662  ax-5 1376  ax-7 1377  ax-gen 1378  ax-ie1 1422  ax-ie2 1423  ax-8 1435  ax-10 1436  ax-11 1437  ax-i12 1438  ax-bndl 1439  ax-4 1440  ax-17 1459  ax-i9 1463  ax-ial 1467  ax-i5r 1468  ax-ext 2063

This theorem depends on definitions:  df-bi 115  df-tru 1287  df-nf 1390  df-sb 1686  df-clab 2068  df-cleq 2074  df-clel 2077  df-nfc 2208  df-v 2603  df-un 2977  df-sn 3404  df-pr 3405

This theorem is referenced by:  preq12b  3562  opth  3992  opthreg  4299