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Theorem abssf 41363
 Description: Class abstraction in a subclass relationship. (Contributed by Glauco Siliprandi, 26-Jun-2021.)
Hypothesis
Ref Expression
abssf.1 𝑥𝐴
Assertion
Ref Expression
abssf ({𝑥𝜑} ⊆ 𝐴 ↔ ∀𝑥(𝜑𝑥𝐴))

Proof of Theorem abssf
StepHypRef Expression
1 abssf.1 . . . 4 𝑥𝐴
21abid2f 3010 . . 3 {𝑥𝑥𝐴} = 𝐴
32sseq2i 3994 . 2 ({𝑥𝜑} ⊆ {𝑥𝑥𝐴} ↔ {𝑥𝜑} ⊆ 𝐴)
4 ss2ab 4037 . 2 ({𝑥𝜑} ⊆ {𝑥𝑥𝐴} ↔ ∀𝑥(𝜑𝑥𝐴))
53, 4bitr3i 279 1 ({𝑥𝜑} ⊆ 𝐴 ↔ ∀𝑥(𝜑𝑥𝐴))
 Colors of variables: wff setvar class Syntax hints:   → wi 4   ↔ wb 208  ∀wal 1528   ∈ wcel 2107  {cab 2797  Ⅎwnfc 2959   ⊆ wss 3934 This theorem was proved from axioms:  ax-mp 5  ax-1 6  ax-2 7  ax-3 8  ax-gen 1789  ax-4 1803  ax-5 1904  ax-6 1963  ax-7 2008  ax-8 2109  ax-9 2117  ax-10 2138  ax-11 2153  ax-12 2169  ax-ext 2791 This theorem depends on definitions:  df-bi 209  df-an 399  df-or 844  df-tru 1533  df-ex 1774  df-nf 1778  df-sb 2063  df-clab 2798  df-cleq 2812  df-clel 2891  df-nfc 2961  df-in 3941  df-ss 3950 This theorem is referenced by:  rabssf  41370
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