Theorem preqr1OLD 4380

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.) Obsolete version of preqr1 4379 as of 18-Dec-2020. (New usage is discouraged.) (Proof modification is discouraged.)

Hypotheses

Ref	Expression
preqr1.a	|- A e. _V
preqr1.b	|- B e. _V

Assertion

Ref	Expression
preqr1OLD	|- ( { A , C } = { B , C } -> A = B )

Proof of Theorem preqr1OLD

Step	Hyp	Ref	Expression
1		preqr1.a	. . . . 5 |- A e. _V
2	1	prid1 4297	. . . 4 |- A e. { A , C }
3		eleq2 2690	. . . 4 |- ( { A , C } = { B , C } -> ( A e. { A , C } <-> A e. { B , C } ) )
4	2, 3	mpbii 223	. . 3 |- ( { A , C } = { B , C } -> A e. { B , C } )
5	1	elpr 4198	. . 3 |- ( A e. { B , C } <-> ( A = B \/ A = C ) )
6	4, 5	sylib 208	. 2 |- ( { A , C } = { B , C } -> ( A = B \/ A = C ) )
7		preqr1.b	. . . . 5 |- B e. _V
8	7	prid1 4297	. . . 4 |- B e. { B , C }
9		eleq2 2690	. . . 4 |- ( { A , C } = { B , C } -> ( B e. { A , C } <-> B e. { B , C } ) )
10	8, 9	mpbiri 248	. . 3 |- ( { A , C } = { B , C } -> B e. { A , C } )
11	7	elpr 4198	. . 3 |- ( B e. { A , C } <-> ( B = A \/ B = C ) )
12	10, 11	sylib 208	. 2 |- ( { A , C } = { B , C } -> ( B = A \/ B = C ) )
13		eqcom 2629	. 2 |- ( A = B <-> B = A )
14		eqeq2 2633	. 2 |- ( A = C -> ( B = A <-> B = C ) )
15	6, 12, 13, 14	oplem1 1007	1 |- ( { A , C } = { B , C } -> A = B )

Syntax hints:   -> wi 4   \/ wo 383   = wceq 1483   e. wcel 1990   _Vcvv 3200   {cpr 4179

This theorem was proved from axioms:  ax-mp 5  ax-1 6  ax-2 7  ax-3 8  ax-gen 1722  ax-4 1737  ax-5 1839  ax-6 1888  ax-7 1935  ax-9 1999  ax-10 2019  ax-11 2034  ax-12 2047  ax-13 2246  ax-ext 2602

This theorem depends on definitions:  df-bi 197  df-or 385  df-an 386  df-tru 1486  df-ex 1705  df-nf 1710  df-sb 1881  df-clab 2609  df-cleq 2615  df-clel 2618  df-nfc 2753  df-v 3202  df-un 3579  df-sn 4178  df-pr 4180

This theorem is referenced by: (None)