Theorem eqssd 3289

Description: Equality deduction from two subclass relationships. Compare Theorem 4 of [Suppes] p. 22. (Contributed by NM, 27-Jun-2004.)

Hypotheses

Ref	Expression
eqssd.1	⊢ (φ → A ⊆ B)
eqssd.2	⊢ (φ → B ⊆ A)

Assertion

Ref	Expression
eqssd	⊢ (φ → A = B)

Proof of Theorem eqssd

Step	Hyp	Ref	Expression
1		eqssd.1	. 2 ⊢ (φ → A ⊆ B)
2		eqssd.2	. 2 ⊢ (φ → B ⊆ A)
3		eqss 3287	. 2 ⊢ (A = B ↔ (A ⊆ B ∧ B ⊆ A))
4	1, 2, 3	sylanbrc 645	1 ⊢ (φ → A = B)

Syntax hints:   → wi 4   = wceq 1642   ⊆ wss 3257

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 2478  df-v 2861  df-nin 3211  df-compl 3212  df-in 3213  df-ss 3259

This theorem is referenced by:  uneqdifeq  3638  unissel  3920  intmin  3946  unissint  3950  int0el  3957  vfinspeqtncv  4553  phi11  4596  phi011  4599  dmcosseq  4973  ssreseq  4997  fimacnv  5411  dff3  5420  clos1nrel  5886  enadjlem1  6059  enmap2lem5  6067  enmap1lem5  6073  enprmaplem6  6081  sbthlem1  6203  nchoicelem6  6294  dmfrec  6316