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Theorem abss 3951
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 2875 . . 3 {𝑥𝑥𝐴} = 𝐴
21sseq2i 3907 . 2 ({𝑥𝜑} ⊆ {𝑥𝑥𝐴} ↔ {𝑥𝜑} ⊆ 𝐴)
3 ss2ab 3950 . 2 ({𝑥𝜑} ⊆ {𝑥𝑥𝐴} ↔ ∀𝑥(𝜑𝑥𝐴))
42, 3bitr3i 280 1 ({𝑥𝜑} ⊆ 𝐴 ↔ ∀𝑥(𝜑𝑥𝐴))
Colors of variables: wff setvar class
Syntax hints:  wi 4  wb 209  wal 1540  wcel 2114  {cab 2717  wss 3844
This theorem was proved from axioms:  ax-mp 5  ax-1 6  ax-2 7  ax-3 8  ax-gen 1802  ax-4 1816  ax-5 1917  ax-6 1975  ax-7 2020  ax-8 2116  ax-9 2124  ax-10 2145  ax-11 2162  ax-12 2179  ax-ext 2711
This theorem depends on definitions:  df-bi 210  df-an 400  df-or 847  df-tru 1545  df-ex 1787  df-nf 1791  df-sb 2075  df-clab 2718  df-cleq 2731  df-clel 2812  df-nfc 2882  df-v 3401  df-in 3851  df-ss 3861
This theorem is referenced by:  abssdv  3959  rabss  3962  uniiunlem  3976  iunssf  4931  iunss  4932  moabex  5318  reliun  5661  axdc2lem  9951  mptelee  26844  fpwrelmap  30646  ss2iundf  40836  hoidmvlelem1  43698
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