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Theorem abss 3671
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 2745 . . 3 {𝑥𝑥𝐴} = 𝐴
21sseq2i 3630 . 2 ({𝑥𝜑} ⊆ {𝑥𝑥𝐴} ↔ {𝑥𝜑} ⊆ 𝐴)
3 ss2ab 3670 . 2 ({𝑥𝜑} ⊆ {𝑥𝑥𝐴} ↔ ∀𝑥(𝜑𝑥𝐴))
42, 3bitr3i 266 1 ({𝑥𝜑} ⊆ 𝐴 ↔ ∀𝑥(𝜑𝑥𝐴))
Colors of variables: wff setvar class
Syntax hints:  wi 4  wb 196  wal 1481  wcel 1990  {cab 2608  wss 3574
This theorem was proved from axioms:  ax-mp 5  ax-1 6  ax-2 7  ax-3 8  ax-gen 1722  ax-4 1737  ax-5 1839  ax-6 1888  ax-7 1935  ax-9 1999  ax-10 2019  ax-11 2034  ax-12 2047  ax-13 2246  ax-ext 2602
This theorem depends on definitions:  df-bi 197  df-or 385  df-an 386  df-tru 1486  df-ex 1705  df-nf 1710  df-sb 1881  df-clab 2609  df-cleq 2615  df-clel 2618  df-nfc 2753  df-in 3581  df-ss 3588
This theorem is referenced by:  abssdv  3676  rabss  3679  uniiunlem  3691  iunss  4561  moabex  4927  reliun  5239  axdc2lem  9270  mptelee  25775  fpwrelmap  29508  ss2iundf  37951  iunssf  39263  hoidmvlelem1  40809
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