Theorem preqr1 3560

Description: Reverse equality lemma for unordered pairs. If two unordered pairs have the same second element, the first elements are equal. (Contributed by NM, 18-Oct-1995.)

Hypotheses

Ref	Expression
preqr1.1	|- A e. _V
preqr1.2	|- B e. _V

Assertion

Ref	Expression
preqr1	|- ( { A , C } = { B , C } -> A = B )

Proof of Theorem preqr1

Step	Hyp	Ref	Expression
1		preqr1.1	. . . . 5 |- A e. _V
2	1	prid1 3498	. . . 4 |- A e. { A , C }
3		eleq2 2142	. . . 4 |- ( { A , C } = { B , C } -> ( A e. { A , C } <-> A e. { B , C } ) )
4	2, 3	mpbii 146	. . 3 |- ( { A , C } = { B , C } -> A e. { B , C } )
5	1	elpr 3419	. . 3 |- ( A e. { B , C } <-> ( A = B \/ A = C ) )
6	4, 5	sylib 120	. 2 |- ( { A , C } = { B , C } -> ( A = B \/ A = C ) )
7		preqr1.2	. . . . 5 |- B e. _V
8	7	prid1 3498	. . . 4 |- B e. { B , C }
9		eleq2 2142	. . . 4 |- ( { A , C } = { B , C } -> ( B e. { A , C } <-> B e. { B , C } ) )
10	8, 9	mpbiri 166	. . 3 |- ( { A , C } = { B , C } -> B e. { A , C } )
11	7	elpr 3419	. . 3 |- ( B e. { A , C } <-> ( B = A \/ B = C ) )
12	10, 11	sylib 120	. 2 |- ( { A , C } = { B , C } -> ( B = A \/ B = C ) )
13		eqcom 2083	. 2 |- ( A = B <-> B = A )
14		eqeq2 2090	. 2 |- ( A = C -> ( B = A <-> B = C ) )
15	6, 12, 13, 14	oplem1 916	1 |- ( { A , C } = { B , C } -> A = B )

Syntax hints:   -> wi 4   \/ wo 661   = 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:  preqr2  3561