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Theorem dfss 3130
Description: Variant of subclass definition df-ss 3129. (Contributed by NM, 3-Sep-2004.)
Assertion
Ref Expression
dfss (𝐴𝐵𝐴 = (𝐴𝐵))

Proof of Theorem dfss
StepHypRef Expression
1 df-ss 3129 . 2 (𝐴𝐵 ↔ (𝐴𝐵) = 𝐴)
2 eqcom 2167 . 2 ((𝐴𝐵) = 𝐴𝐴 = (𝐴𝐵))
31, 2bitri 183 1 (𝐴𝐵𝐴 = (𝐴𝐵))
Colors of variables: wff set class
Syntax hints:  wb 104   = wceq 1343  cin 3115  wss 3116
This theorem was proved from axioms:  ax-mp 5  ax-1 6  ax-2 7  ax-ia1 105  ax-ia2 106  ax-ia3 107  ax-5 1435  ax-gen 1437  ax-ext 2147
This theorem depends on definitions:  df-bi 116  df-cleq 2158  df-ss 3129
This theorem is referenced by:  dfss2  3131  onelini  4408  cnvcnv  5056  funimass1  5265  sbthlemi5  6926  dmaddpi  7266  dmmulpi  7267  tgioo  13186
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