Theorem preqr1 3780

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 3710	. . . 4 |- A e. { A , C }
3		eleq2 2251	. . . 4 |- ( { A , C } = { B , C } -> ( A e. { A , C } <-> A e. { B , C } ) )
4	2, 3	mpbii 148	. . 3 |- ( { A , C } = { B , C } -> A e. { B , C } )
5	1	elpr 3625	. . 3 |- ( A e. { B , C } <-> ( A = B \/ A = C ) )
6	4, 5	sylib 122	. 2 |- ( { A , C } = { B , C } -> ( A = B \/ A = C ) )
7		preqr1.2	. . . . 5 |- B e. _V
8	7	prid1 3710	. . . 4 |- B e. { B , C }
9		eleq2 2251	. . . 4 |- ( { A , C } = { B , C } -> ( B e. { A , C } <-> B e. { B , C } ) )
10	8, 9	mpbiri 168	. . 3 |- ( { A , C } = { B , C } -> B e. { A , C } )
11	7	elpr 3625	. . 3 |- ( B e. { A , C } <-> ( B = A \/ B = C ) )
12	10, 11	sylib 122	. 2 |- ( { A , C } = { B , C } -> ( B = A \/ B = C ) )
13		eqcom 2189	. 2 |- ( A = B <-> B = A )
14		eqeq2 2197	. 2 |- ( A = C -> ( B = A <-> B = C ) )
15	6, 12, 13, 14	oplem1 976	1 |- ( { A , C } = { B , C } -> A = B )

Syntax hints:   -> wi 4   \/ wo 709   = wceq 1363   e. wcel 2158   _Vcvv 2749   {cpr 3605

This theorem was proved from axioms:  ax-mp 5  ax-1 6  ax-2 7  ax-ia1 106  ax-ia2 107  ax-ia3 108  ax-io 710  ax-5 1457  ax-7 1458  ax-gen 1459  ax-ie1 1503  ax-ie2 1504  ax-8 1514  ax-10 1515  ax-11 1516  ax-i12 1517  ax-bndl 1519  ax-4 1520  ax-17 1536  ax-i9 1540  ax-ial 1544  ax-i5r 1545  ax-ext 2169

This theorem depends on definitions:  df-bi 117  df-tru 1366  df-nf 1471  df-sb 1773  df-clab 2174  df-cleq 2180  df-clel 2183  df-nfc 2318  df-v 2751  df-un 3145  df-sn 3610  df-pr 3611

This theorem is referenced by:  preqr2  3781