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Theorem ssabf 41372
 Description: Subclass of a class abstraction. (Contributed by Glauco Siliprandi, 26-Jun-2021.)
Hypothesis
Ref Expression
ssabf.1 𝑥𝐴
Assertion
Ref Expression
ssabf (𝐴 ⊆ {𝑥𝜑} ↔ ∀𝑥(𝑥𝐴𝜑))

Proof of Theorem ssabf
StepHypRef Expression
1 ssabf.1 . . . 4 𝑥𝐴
21abid2f 3015 . . 3 {𝑥𝑥𝐴} = 𝐴
32sseq1i 3998 . 2 ({𝑥𝑥𝐴} ⊆ {𝑥𝜑} ↔ 𝐴 ⊆ {𝑥𝜑})
4 ss2ab 4042 . 2 ({𝑥𝑥𝐴} ⊆ {𝑥𝜑} ↔ ∀𝑥(𝑥𝐴𝜑))
53, 4bitr3i 279 1 (𝐴 ⊆ {𝑥𝜑} ↔ ∀𝑥(𝑥𝐴𝜑))
 Colors of variables: wff setvar class Syntax hints:   → wi 4   ↔ wb 208  ∀wal 1534   ∈ wcel 2113  {cab 2802  Ⅎwnfc 2964   ⊆ wss 3939 This theorem was proved from axioms:  ax-mp 5  ax-1 6  ax-2 7  ax-3 8  ax-gen 1795  ax-4 1809  ax-5 1910  ax-6 1969  ax-7 2014  ax-8 2115  ax-9 2123  ax-10 2144  ax-11 2160  ax-12 2176  ax-ext 2796 This theorem depends on definitions:  df-bi 209  df-an 399  df-or 844  df-tru 1539  df-ex 1780  df-nf 1784  df-sb 2069  df-clab 2803  df-cleq 2817  df-clel 2896  df-nfc 2966  df-in 3946  df-ss 3955 This theorem is referenced by:  ssrabf  41387
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