Theorem eqssi 3289

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 3288	. 2 ⊢ (A = B ↔ (A ⊆ B ∧ B ⊆ A))
4	1, 2, 3	mpbir2an 886	1 ⊢ A = B

Syntax hints:   = wceq 1642   ⊆ wss 3258

This theorem was proved from axioms:  ax-mp 5  ax-1 6  ax-2 7  ax-3 8  ax-gen 1546  ax-5 1557  ax-17 1616  ax-9 1654  ax-8 1675  ax-6 1729  ax-7 1734  ax-11 1746  ax-12 1925  ax-ext 2334

This theorem depends on definitions:  df-bi 177  df-or 359  df-an 360  df-nan 1288  df-tru 1319  df-ex 1542  df-nf 1545  df-sb 1649  df-clab 2340  df-cleq 2346  df-clel 2349  df-nfc 2479  df-v 2862  df-nin 3212  df-compl 3213  df-in 3214  df-ss 3260

This theorem is referenced by:  inv1  3578  unv  3579  intab  3957  evenoddnnnul  4515  dmv  4921  0ima  5015  clos10  5888  cenc  6182