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Theorem abss 3134
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 2236 . . 3 {𝑥𝑥𝐴} = 𝐴
21sseq2i 3092 . 2 ({𝑥𝜑} ⊆ {𝑥𝑥𝐴} ↔ {𝑥𝜑} ⊆ 𝐴)
3 ss2ab 3133 . 2 ({𝑥𝜑} ⊆ {𝑥𝑥𝐴} ↔ ∀𝑥(𝜑𝑥𝐴))
42, 3bitr3i 185 1 ({𝑥𝜑} ⊆ 𝐴 ↔ ∀𝑥(𝜑𝑥𝐴))
Colors of variables: wff set class
Syntax hints:  wi 4  wb 104  wal 1312  wcel 1463  {cab 2101  wss 3039
This theorem was proved from axioms:  ax-mp 5  ax-1 6  ax-2 7  ax-ia1 105  ax-ia2 106  ax-ia3 107  ax-io 681  ax-5 1406  ax-7 1407  ax-gen 1408  ax-ie1 1452  ax-ie2 1453  ax-8 1465  ax-10 1466  ax-11 1467  ax-i12 1468  ax-bndl 1469  ax-4 1470  ax-17 1489  ax-i9 1493  ax-ial 1497  ax-i5r 1498  ax-ext 2097
This theorem depends on definitions:  df-bi 116  df-nf 1420  df-sb 1719  df-clab 2102  df-cleq 2108  df-clel 2111  df-nfc 2245  df-in 3045  df-ss 3052
This theorem is referenced by:  abssdv  3139  rabss  3142  uniiunlem  3153  iunss  3822  reliun  4628  funimaexglem  5174
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