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Theorem abss 3994
Description: Class abstraction in a subclass relationship. (Contributed by NM, 16-Aug-2006.)
Assertion
Ref Expression
abss ({𝑥𝜑} ⊆ 𝐴 ↔ ∀𝑥(𝜑𝑥𝐴))
Distinct variable group:   𝑥,𝐴
Allowed substitution hint:   𝜑(𝑥)

Proof of Theorem abss
StepHypRef Expression
1 abid2 2882 . . 3 {𝑥𝑥𝐴} = 𝐴
21sseq2i 3950 . 2 ({𝑥𝜑} ⊆ {𝑥𝑥𝐴} ↔ {𝑥𝜑} ⊆ 𝐴)
3 ss2ab 3993 . 2 ({𝑥𝜑} ⊆ {𝑥𝑥𝐴} ↔ ∀𝑥(𝜑𝑥𝐴))
42, 3bitr3i 276 1 ({𝑥𝜑} ⊆ 𝐴 ↔ ∀𝑥(𝜑𝑥𝐴))
Colors of variables: wff setvar class
Syntax hints:  wi 4  wb 205  wal 1537  wcel 2106  {cab 2715  wss 3887
This theorem was proved from axioms:  ax-mp 5  ax-1 6  ax-2 7  ax-3 8  ax-gen 1798  ax-4 1812  ax-5 1913  ax-6 1971  ax-7 2011  ax-8 2108  ax-9 2116  ax-10 2137  ax-11 2154  ax-12 2171  ax-ext 2709
This theorem depends on definitions:  df-bi 206  df-an 397  df-or 845  df-tru 1542  df-ex 1783  df-nf 1787  df-sb 2068  df-clab 2716  df-cleq 2730  df-clel 2816  df-nfc 2889  df-v 3434  df-in 3894  df-ss 3904
This theorem is referenced by:  abssdv  4002  rabss  4005  uniiunlem  4019  iunssf  4974  iunss  4975  moabex  5374  reliun  5726  axdc2lem  10204  mptelee  27263  fpwrelmap  31068  ss2iundf  41267  hoidmvlelem1  44133
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