Theorem eqssi 3195

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 3194	. 2 |- ( A = B <-> ( A C_ B /\ B C_ A ) )
4	1, 2, 3	mpbir2an 944	1 |- A = B

Syntax hints:   = wceq 1364   C_ wss 3153

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 3159  df-ss 3166

This theorem is referenced by:  inv1  3483  unv  3484  undifabs  3523  intab  3899  intid  4253  find  4631  limom  4646  dmv  4878  0ima  5025  rnxpid  5100  dftpos4  6316  axaddf  7928  axmulf  7929  dfuzi  9427  unirnioo  10039  4sqlem19  12547  txuni2  14424  dvef  14873  reeff1o  14908