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Theorem ssab 3089
Description: Subclass of a class abstraction. (Contributed by NM, 16-Aug-2006.)
Assertion
Ref Expression
ssab (𝐴 ⊆ {𝑥𝜑} ↔ ∀𝑥(𝑥𝐴𝜑))
Distinct variable group:   𝑥,𝐴
Allowed substitution hint:   𝜑(𝑥)

Proof of Theorem ssab
StepHypRef Expression
1 abid2 2208 . . 3 {𝑥𝑥𝐴} = 𝐴
21sseq1i 3048 . 2 ({𝑥𝑥𝐴} ⊆ {𝑥𝜑} ↔ 𝐴 ⊆ {𝑥𝜑})
3 ss2ab 3087 . 2 ({𝑥𝑥𝐴} ⊆ {𝑥𝜑} ↔ ∀𝑥(𝑥𝐴𝜑))
42, 3bitr3i 184 1 (𝐴 ⊆ {𝑥𝜑} ↔ ∀𝑥(𝑥𝐴𝜑))
Colors of variables: wff set class
Syntax hints:  wi 4  wb 103  wal 1287  wcel 1438  {cab 2074  wss 2997
This theorem was proved from axioms:  ax-1 5  ax-2 6  ax-mp 7  ax-ia1 104  ax-ia2 105  ax-ia3 106  ax-io 665  ax-5 1381  ax-7 1382  ax-gen 1383  ax-ie1 1427  ax-ie2 1428  ax-8 1440  ax-10 1441  ax-11 1442  ax-i12 1443  ax-bndl 1444  ax-4 1445  ax-17 1464  ax-i9 1468  ax-ial 1472  ax-i5r 1473  ax-ext 2070
This theorem depends on definitions:  df-bi 115  df-nf 1395  df-sb 1693  df-clab 2075  df-cleq 2081  df-clel 2084  df-nfc 2217  df-in 3003  df-ss 3010
This theorem is referenced by:  ssabral  3090  ssrab  3097
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