Theorem eqssi 3186

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 ⊆ 𝐵
eqssi.2	⊢ 𝐵 ⊆ 𝐴

Assertion

Ref	Expression
eqssi	⊢ 𝐴 = 𝐵

Proof of Theorem eqssi

Step	Hyp	Ref	Expression
1		eqssi.1	. 2 ⊢ 𝐴 ⊆ 𝐵
2		eqssi.2	. 2 ⊢ 𝐵 ⊆ 𝐴
3		eqss 3185	. 2 ⊢ (𝐴 = 𝐵 ↔ (𝐴 ⊆ 𝐵 ∧ 𝐵 ⊆ 𝐴))
4	1, 2, 3	mpbir2an 944	1 ⊢ 𝐴 = 𝐵

Syntax hints:   = wceq 1364   ⊆ wss 3144

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 2171

This theorem depends on definitions:  df-bi 117  df-nf 1472  df-sb 1774  df-clab 2176  df-cleq 2182  df-clel 2185  df-in 3150  df-ss 3157

This theorem is referenced by:  inv1  3474  unv  3475  undifabs  3514  intab  3888  intid  4242  find  4616  limom  4631  dmv  4861  0ima  5006  rnxpid  5081  dftpos4  6288  axaddf  7897  axmulf  7898  dfuzi  9393  unirnioo  10003  4sqlem19  12441  txuni2  14213  dvef  14645  reeff1o  14651