Theorem eqssd 3290

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 3288	. 2 ⊢ (A = B ↔ (A ⊆ B ∧ B ⊆ A))
4	1, 2, 3	sylanbrc 645	1 ⊢ (φ → A = B)

Syntax hints:   → wi 4   = 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:  uneqdifeq  3639  unissel  3921  intmin  3947  unissint  3951  int0el  3958  vfinspeqtncv  4554  phi11  4597  phi011  4600  dmcosseq  4974  ssreseq  4998  fimacnv  5412  dff3  5421  clos1nrel  5887  enadjlem1  6060  enmap2lem5  6068  enmap1lem5  6074  enprmaplem6  6082  sbthlem1  6204  nchoicelem6  6295  dmfrec  6317