MPE Home Metamath Proof Explorer < Previous   Next >
Nearby theorems
Mirrors  >  Home  >  MPE Home  >  Th. List  >  abss Structured version   Visualization version   GIF version

Theorem abss 3868
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 2929 . . 3 {𝑥𝑥𝐴} = 𝐴
21sseq2i 3827 . 2 ({𝑥𝜑} ⊆ {𝑥𝑥𝐴} ↔ {𝑥𝜑} ⊆ 𝐴)
3 ss2ab 3867 . 2 ({𝑥𝜑} ⊆ {𝑥𝑥𝐴} ↔ ∀𝑥(𝜑𝑥𝐴))
42, 3bitr3i 268 1 ({𝑥𝜑} ⊆ 𝐴 ↔ ∀𝑥(𝜑𝑥𝐴))
Colors of variables: wff setvar class
Syntax hints:  wi 4  wb 197  wal 1635  wcel 2156  {cab 2792  wss 3769
This theorem was proved from axioms:  ax-mp 5  ax-1 6  ax-2 7  ax-3 8  ax-gen 1877  ax-4 1894  ax-5 2001  ax-6 2068  ax-7 2104  ax-9 2165  ax-10 2185  ax-11 2201  ax-12 2214  ax-13 2420  ax-ext 2784
This theorem depends on definitions:  df-bi 198  df-an 385  df-or 866  df-tru 1641  df-ex 1860  df-nf 1864  df-sb 2061  df-clab 2793  df-cleq 2799  df-clel 2802  df-nfc 2937  df-in 3776  df-ss 3783
This theorem is referenced by:  abssdv  3873  rabss  3876  uniiunlem  3889  iunss  4753  moabex  5117  reliun  5441  axdc2lem  9555  mptelee  25989  fpwrelmap  29835  ss2iundf  38451  iunssf  39756  hoidmvlelem1  41291
  Copyright terms: Public domain W3C validator