Theorem dfss 1493

Description: A frequently-used variant of subclass definition df-ss 1492.

Assertion

Ref	Expression
dfss	⊢ (A ⊆ B ↔ A = (A ∩ B))

Proof of Theorem dfss

Step	Hyp	Ref	Expression
1		df-ss 1492	. 2 ⊢ (A ⊆ B ↔ (A ∩ B) = A)
2		cleqcom 1103	. 2 ⊢ ((A ∩ B) = A ↔ A = (A ∩ B))
3	1, 2	bitr 151	1 ⊢ (A ⊆ B ↔ A = (A ∩ B))

Syntax hints:   ↔ wb 127   = wceq 1091   ∩ cin 1486   ⊆ wss 1487

This theorem is referenced by:  dfss2 1497  wefrc 2195  onelin 2351  resabs2 2593  cnvcnv 2661  funimass1 2712  tz7.44-2 2967  tz7.44-3 2968  frfnom 2989  sbthlem5 3353  dmaddpi 3812  dmmulpi 3813  mdbr3 5729  mdbr4 5730  ssmd1 5734

This theorem was proved from axioms:  ax-1 3  ax-2 4  ax-3 5  ax-mp 6  ax-4 673  ax-5 674  ax-gen 677  ax-ext 1074

This theorem depends on definitions:  df-bi 128  df-an 198  df-cleq 1097  df-ss 1492

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