Theorem eqss 3288

Description: The subclass relationship is antisymmetric. Compare Theorem 4 of [Suppes] p. 22. (Contributed by NM, 5-Aug-1993.)

Assertion

Ref	Expression
eqss	⊢ (A = B ↔ (A ⊆ B ∧ B ⊆ A))

Proof of Theorem eqss

Dummy variable x is distinct from all other variables.

Step	Hyp	Ref	Expression
1		albiim 1611	. 2 ⊢ (∀x(x ∈ A ↔ x ∈ B) ↔ (∀x(x ∈ A → x ∈ B) ∧ ∀x(x ∈ B → x ∈ A)))
2		dfcleq 2347	. 2 ⊢ (A = B ↔ ∀x(x ∈ A ↔ x ∈ B))
3		dfss2 3263	. . 3 ⊢ (A ⊆ B ↔ ∀x(x ∈ A → x ∈ B))
4		dfss2 3263	. . 3 ⊢ (B ⊆ A ↔ ∀x(x ∈ B → x ∈ A))
5	3, 4	anbi12i 678	. 2 ⊢ ((A ⊆ B ∧ B ⊆ A) ↔ (∀x(x ∈ A → x ∈ B) ∧ ∀x(x ∈ B → x ∈ A)))
6	1, 2, 5	3bitr4i 268	1 ⊢ (A = B ↔ (A ⊆ B ∧ B ⊆ A))

Syntax hints:   → wi 4   ↔ wb 176   ∧ wa 358  ∀wal 1540   = wceq 1642   ∈ wcel 1710   ⊆ 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:  eqssi  3289  eqssd  3290  sseq1  3293  sseq2  3294  eqimss  3324  dfpss3  3356  uneqin  3507  ss0b  3581  vss  3588  pssdifn0  3612  pwpw0  3856  sssn  3865  ssunsn  3867  pwsnALT  3883  unidif  3924  ssunieq  3925  uniintsn  3964  iuneq1  3983  iuneq2  3986  iunxdif2  4015  ssofeq  4078  dfidk2  4314  sfinltfin  4536  eqrel  4846  eqopr  4848  coeq1  4875  coeq2  4876  cnveq  4887  dmeq  4908  xp11  5057  ssrnres  5060  funeq  5128  fnres  5200  eqfnfv3  5395  fconst4  5459  dfid4  5504  ssetpov  5945