Theorem eqss 3287

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 3262	. . 3 ⊢ (A ⊆ B ↔ ∀x(x ∈ A → x ∈ B))
4		dfss2 3262	. . 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 3257

This theorem was proved from axioms:  ax-1 5  ax-2 6  ax-3 7  ax-mp 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:  eqssi  3288  eqssd  3289  sseq1  3292  sseq2  3293  eqimss  3323  dfpss3  3355  uneqin  3506  ss0b  3580  vss  3587  pssdifn0  3611  pwpw0  3855  sssn  3864  ssunsn  3866  pwsnALT  3882  unidif  3923  ssunieq  3924  uniintsn  3963  iuneq1  3982  iuneq2  3985  iunxdif2  4014  ssofeq  4077  dfidk2  4313  sfinltfin  4535  eqrel  4845  eqopr  4847  coeq1  4874  coeq2  4875  cnveq  4886  dmeq  4907  xp11  5056  ssrnres  5059  funeq  5127  fnres  5199  eqfnfv3  5394  fconst4  5458  dfid4  5503  ssetpov  5944