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Theorem abss 4018
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 2872 . . 3 {𝑥𝑥𝐴} = 𝐴
21sseq2i 3974 . 2 ({𝑥𝜑} ⊆ {𝑥𝑥𝐴} ↔ {𝑥𝜑} ⊆ 𝐴)
3 ss2ab 4017 . 2 ({𝑥𝜑} ⊆ {𝑥𝑥𝐴} ↔ ∀𝑥(𝜑𝑥𝐴))
42, 3bitr3i 277 1 ({𝑥𝜑} ⊆ 𝐴 ↔ ∀𝑥(𝜑𝑥𝐴))
Colors of variables: wff setvar class
Syntax hints:  wi 4  wb 205  wal 1540  wcel 2107  {cab 2710  wss 3911
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 1914  ax-6 1972  ax-7 2012  ax-8 2109  ax-9 2117  ax-10 2138  ax-11 2155  ax-12 2172  ax-ext 2704
This theorem depends on definitions:  df-bi 206  df-an 398  df-or 847  df-tru 1545  df-ex 1783  df-nf 1787  df-sb 2069  df-clab 2711  df-cleq 2725  df-clel 2811  df-nfc 2886  df-v 3446  df-in 3918  df-ss 3928
This theorem is referenced by:  abssdvOLD  4027  rabss  4030  uniiunlem  4045  iunssf  5005  iunss  5006  moabex  5417  reliun  5773  axdc2lem  10389  mptelee  27886  fpwrelmap  31697  hoidmvlelem1  44922
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