Theorem eqssi 3199

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 3198	. 2 |- ( A = B <-> ( A C_ B /\ B C_ A ) )
4	1, 2, 3	mpbir2an 944	1 |- A = B

Syntax hints:   = wceq 1364   C_ wss 3157

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-5 1461  ax-7 1462  ax-gen 1463  ax-ie1 1507  ax-ie2 1508  ax-8 1518  ax-11 1520  ax-4 1524  ax-17 1540  ax-i9 1544  ax-ial 1548  ax-i5r 1549  ax-ext 2178

This theorem depends on definitions:  df-bi 117  df-nf 1475  df-sb 1777  df-clab 2183  df-cleq 2189  df-clel 2192  df-in 3163  df-ss 3170

This theorem is referenced by:  inv1  3487  unv  3488  undifabs  3527  intab  3903  intid  4257  find  4635  limom  4650  dmv  4882  0ima  5029  rnxpid  5104  dftpos4  6321  axaddf  7935  axmulf  7936  dfuzi  9436  unirnioo  10048  4sqlem19  12578  txuni2  14492  dvef  14963  reeff1o  15009