Theorem eqssi 3196

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 3195	. 2 |- ( A = B <-> ( A C_ B /\ B C_ A ) )
4	1, 2, 3	mpbir2an 944	1 |- A = B

Syntax hints:   = wceq 1364   C_ wss 3154

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 1458  ax-7 1459  ax-gen 1460  ax-ie1 1504  ax-ie2 1505  ax-8 1515  ax-11 1517  ax-4 1521  ax-17 1537  ax-i9 1541  ax-ial 1545  ax-i5r 1546  ax-ext 2175

This theorem depends on definitions:  df-bi 117  df-nf 1472  df-sb 1774  df-clab 2180  df-cleq 2186  df-clel 2189  df-in 3160  df-ss 3167

This theorem is referenced by:  inv1  3484  unv  3485  undifabs  3524  intab  3900  intid  4254  find  4632  limom  4647  dmv  4879  0ima  5026  rnxpid  5101  dftpos4  6318  axaddf  7930  axmulf  7931  dfuzi  9430  unirnioo  10042  4sqlem19  12550  txuni2  14435  dvef  14906  reeff1o  14949