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Theorem ssab 4064
Description: Subclass of a class abstraction. (Contributed by NM, 16-Aug-2006.)
Assertion
Ref Expression
ssab (𝐴 ⊆ {𝑥𝜑} ↔ ∀𝑥(𝑥𝐴𝜑))
Distinct variable group:   𝑥,𝐴
Allowed substitution hint:   𝜑(𝑥)

Proof of Theorem ssab
StepHypRef Expression
1 abid2 2879 . . 3 {𝑥𝑥𝐴} = 𝐴
21sseq1i 4012 . 2 ({𝑥𝑥𝐴} ⊆ {𝑥𝜑} ↔ 𝐴 ⊆ {𝑥𝜑})
3 ss2ab 4062 . 2 ({𝑥𝑥𝐴} ⊆ {𝑥𝜑} ↔ ∀𝑥(𝑥𝐴𝜑))
42, 3bitr3i 277 1 (𝐴 ⊆ {𝑥𝜑} ↔ ∀𝑥(𝑥𝐴𝜑))
Colors of variables: wff setvar class
Syntax hints:  wi 4  wb 206  wal 1538  wcel 2108  {cab 2714  wss 3951
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 1967  ax-7 2007  ax-8 2110  ax-9 2118  ax-10 2141  ax-11 2157  ax-12 2177  ax-ext 2708
This theorem depends on definitions:  df-bi 207  df-an 396  df-or 849  df-tru 1543  df-ex 1780  df-nf 1784  df-sb 2065  df-clab 2715  df-cleq 2729  df-clel 2816  df-nfc 2892  df-ss 3968
This theorem is referenced by:  ssabral  4065  ssrab  4073  wdomd  9621  ixpiunwdom  9630  lidldvgen  21344  prdsxmslem2  24542  ballotlem2  34491
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