Theorem eqssi 3264

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 3263	. 2 ⊢ (𝐴 = 𝐵 ↔ (𝐴 ⊆ 𝐵 ∧ 𝐵 ⊆ 𝐴))
4	1, 2, 3	mpbir2an 955	1 ⊢ 𝐴 = 𝐵

This proof depends on syntax axioms:   = wceq 1402   ⊆ wss 3220

This proof depends on axioms:  ax-mp 5  ax-1 6  ax-2 7  ax-ia1 106  ax-ia2 107  ax-ia3 108  ax-5 1500  ax-7 1501  ax-gen 1502  ax-ie1 1546  ax-ie2 1547  ax-8 1557  ax-11 1559  ax-4 1563  ax-17 1579  ax-i9 1583  ax-ial 1587  ax-i5r 1588  ax-ext 2220

This proof depends on definitions:  df-bi 117  df-nf 1514  df-sb 1816  df-clab 2225  df-cleq 2231  df-clel 2234  df-in 3226  df-ss 3233

This theorem is used by:  inv1  3559  unv  3560  undifabs  3604  intab  3999  intid  4366  find  4748  limom  4763  dmv  4999  0ima  5150  rnxpid  5225  rinvf1o  6038  dftpos4  6539  axaddf  8241  axmulf  8242  dfuzi  9766  unirnioo  10391  0bits  12750  4sqlem19  13216  ballotfilemth  13338  txuni2  15499  dvef  15970  reeff1o  16016