Theorem preqr1 3568

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 3506	. . . 4 |- A e. { A , C }
3		eleq2 2143	. . . 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 3427	. . 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 3506	. . . 4 |- B e. { B , C }
9		eleq2 2143	. . . 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 3427	. . 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 2084	. 2 |- ( A = B <-> B = A )
14		eqeq2 2091	. 2 |- ( A = C -> ( B = A <-> B = C ) )
15	6, 12, 13, 14	oplem1 917	1 |- ( { A , C } = { B , C } -> A = B )

Syntax hints:   -> wi 4   \/ wo 662   = wceq 1285   e. wcel 1434   _Vcvv 2602   {cpr 3407

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 663  ax-5 1377  ax-7 1378  ax-gen 1379  ax-ie1 1423  ax-ie2 1424  ax-8 1436  ax-10 1437  ax-11 1438  ax-i12 1439  ax-bndl 1440  ax-4 1441  ax-17 1460  ax-i9 1464  ax-ial 1468  ax-i5r 1469  ax-ext 2064

This theorem depends on definitions:  df-bi 115  df-tru 1288  df-nf 1391  df-sb 1687  df-clab 2069  df-cleq 2075  df-clel 2078  df-nfc 2209  df-v 2604  df-un 2978  df-sn 3412  df-pr 3413

This theorem is referenced by:  preqr2  3569