Theorem eqssi 3015

Description: Infer equality from two subclass relationships. Compare Theorem 4 of [Suppes] p. 22. (Contributed by NM, 9-Sep-1993.)

Hypotheses

Ref	Expression
eqssi.1	|- A C_ B
eqssi.2	|- B C_ A

Assertion

Ref	Expression
eqssi	|- A = B

Proof of Theorem eqssi

Step	Hyp	Ref	Expression
1		eqssi.1	. 2 |- A C_ B
2		eqssi.2	. 2 |- B C_ A
3		eqss 3014	. 2 |- ( A = B <-> ( A C_ B /\ B C_ A ) )
4	1, 2, 3	mpbir2an 883	1 |- A = B

Syntax hints:   = wceq 1284   C_ wss 2973

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-5 1376  ax-7 1377  ax-gen 1378  ax-ie1 1422  ax-ie2 1423  ax-8 1435  ax-11 1437  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-nf 1390  df-sb 1686  df-clab 2068  df-cleq 2074  df-clel 2077  df-in 2979  df-ss 2986

This theorem is referenced by:  inv1  3280  unv  3281  undifabs  3320  intab  3665  intid  3979  find  4340  limom  4354  dmv  4569  0ima  4705  rnxpid  4775  dftpos4  5901  dfuzi  8457  unirnioo  8996