Theorem preqr1 3690

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 3624	. . . 4 |- A e. { A , C }
3		eleq2 2201	. . . 4 |- ( { A , C } = { B , C } -> ( A e. { A , C } <-> A e. { B , C } ) )
4	2, 3	mpbii 147	. . 3 |- ( { A , C } = { B , C } -> A e. { B , C } )
5	1	elpr 3543	. . 3 |- ( A e. { B , C } <-> ( A = B \/ A = C ) )
6	4, 5	sylib 121	. 2 |- ( { A , C } = { B , C } -> ( A = B \/ A = C ) )
7		preqr1.2	. . . . 5 |- B e. _V
8	7	prid1 3624	. . . 4 |- B e. { B , C }
9		eleq2 2201	. . . 4 |- ( { A , C } = { B , C } -> ( B e. { A , C } <-> B e. { B , C } ) )
10	8, 9	mpbiri 167	. . 3 |- ( { A , C } = { B , C } -> B e. { A , C } )
11	7	elpr 3543	. . 3 |- ( B e. { A , C } <-> ( B = A \/ B = C ) )
12	10, 11	sylib 121	. 2 |- ( { A , C } = { B , C } -> ( B = A \/ B = C ) )
13		eqcom 2139	. 2 |- ( A = B <-> B = A )
14		eqeq2 2147	. 2 |- ( A = C -> ( B = A <-> B = C ) )
15	6, 12, 13, 14	oplem1 959	1 |- ( { A , C } = { B , C } -> A = B )

Syntax hints:   -> wi 4   \/ wo 697   = wceq 1331   e. wcel 1480   _Vcvv 2681   {cpr 3523

This theorem was proved from axioms:  ax-mp 5  ax-1 6  ax-2 7  ax-ia1 105  ax-ia2 106  ax-ia3 107  ax-io 698  ax-5 1423  ax-7 1424  ax-gen 1425  ax-ie1 1469  ax-ie2 1470  ax-8 1482  ax-10 1483  ax-11 1484  ax-i12 1485  ax-bndl 1486  ax-4 1487  ax-17 1506  ax-i9 1510  ax-ial 1514  ax-i5r 1515  ax-ext 2119

This theorem depends on definitions:  df-bi 116  df-tru 1334  df-nf 1437  df-sb 1736  df-clab 2124  df-cleq 2130  df-clel 2133  df-nfc 2268  df-v 2683  df-un 3070  df-sn 3528  df-pr 3529

This theorem is referenced by:  preqr2  3691