Theorem eqssi 3200

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 3199	. 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  3488  unv  3489  undifabs  3528  intab  3904  intid  4258  find  4636  limom  4651  dmv  4883  0ima  5030  rnxpid  5105  dftpos4  6330  axaddf  7952  axmulf  7953  dfuzi  9453  unirnioo  10065  0bits  12141  4sqlem19  12603  txuni2  14576  dvef  15047  reeff1o  15093