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Theorem abss 4073
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 2877 . . 3 {𝑥𝑥𝐴} = 𝐴
21sseq2i 4025 . 2 ({𝑥𝜑} ⊆ {𝑥𝑥𝐴} ↔ {𝑥𝜑} ⊆ 𝐴)
3 ss2ab 4072 . 2 ({𝑥𝜑} ⊆ {𝑥𝑥𝐴} ↔ ∀𝑥(𝜑𝑥𝐴))
42, 3bitr3i 277 1 ({𝑥𝜑} ⊆ 𝐴 ↔ ∀𝑥(𝜑𝑥𝐴))
Colors of variables: wff setvar class
Syntax hints:  wi 4  wb 206  wal 1535  wcel 2106  {cab 2712  wss 3963
This theorem was proved from axioms:  ax-mp 5  ax-1 6  ax-2 7  ax-3 8  ax-gen 1792  ax-4 1806  ax-5 1908  ax-6 1965  ax-7 2005  ax-8 2108  ax-9 2116  ax-10 2139  ax-11 2155  ax-12 2175  ax-ext 2706
This theorem depends on definitions:  df-bi 207  df-an 396  df-or 848  df-tru 1540  df-ex 1777  df-nf 1781  df-sb 2063  df-clab 2713  df-cleq 2727  df-clel 2814  df-nfc 2890  df-ss 3980
This theorem is referenced by:  abssdvOLD  4079  rabss  4082  uniiunlem  4097  iunssf  5049  iunss  5050  abexd  5331  abex  5332  moabex  5470  reliun  5829  axdc2lem  10486  mptelee  28925  fpwrelmap  32751  hoidmvlelem1  46551
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