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Theorem abss 3633
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 2731 . . 3 {𝑥𝑥𝐴} = 𝐴
21sseq2i 3592 . 2 ({𝑥𝜑} ⊆ {𝑥𝑥𝐴} ↔ {𝑥𝜑} ⊆ 𝐴)
3 ss2ab 3632 . 2 ({𝑥𝜑} ⊆ {𝑥𝑥𝐴} ↔ ∀𝑥(𝜑𝑥𝐴))
42, 3bitr3i 264 1 ({𝑥𝜑} ⊆ 𝐴 ↔ ∀𝑥(𝜑𝑥𝐴))
Colors of variables: wff setvar class
Syntax hints:  wi 4  wb 194  wal 1472  wcel 1976  {cab 2595  wss 3539
This theorem was proved from axioms:  ax-mp 5  ax-1 6  ax-2 7  ax-3 8  ax-gen 1712  ax-4 1727  ax-5 1826  ax-6 1874  ax-7 1921  ax-10 2005  ax-11 2020  ax-12 2033  ax-13 2233  ax-ext 2589
This theorem depends on definitions:  df-bi 195  df-or 383  df-an 384  df-tru 1477  df-ex 1695  df-nf 1700  df-sb 1867  df-clab 2596  df-cleq 2602  df-clel 2605  df-nfc 2739  df-in 3546  df-ss 3553
This theorem is referenced by:  abssdv  3638  rabss  3641  uniiunlem  3652  iunss  4491  moabex  4847  reliun  5150  axdc2lem  9130  mptelee  25520  fpwrelmap  28689  ss2iundf  36753  iunssf  38073  hoidmvlelem1  39268
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