Theorem ssab 1555

Description: Restriction of class abstraction creates subclass.

Assertion

Ref	Expression
ssab	|- {x | (x e. A /\ ph)} (_ A

Distinct variable group(s):   x,A

Proof of Theorem ssab

Step	Hyp	Ref	Expression
1		pm3.26 256	. . 3 |- ((x e. A /\ ph) -> x e. A)
2	1	ss2abi 1552	. 2 |- {x | (x e. A /\ ph)} (_ {x | x e. A}
3		abid2 1186	. 2 |- {x | x e. A} = A
4	2, 3	sseqtr 1532	1 |- {x | (x e. A /\ ph)} (_ A

Syntax hints:   /\ wa 196  {cab 1090   e. wcel 1092   (_ wss 1487

This theorem is referenced by:  ssrab 1556  zfausab 1704  exss 1881  onminex 2275  dmopabss 2540  chsssh 5129

This theorem was proved from axioms:  ax-1 3  ax-2 4  ax-3 5  ax-mp 6  ax-4 673  ax-5 674  ax-6 675  ax-7 676  ax-gen 677  ax-8 798  ax-9 799  ax-10 800  ax-11 801  ax-12 802  ax-16 922  ax-17 925  ax-ext 1074

This theorem depends on definitions:  df-bi 128  df-an 198  df-ex 679  df-sb 853  df-clab 1093  df-cleq 1097  df-clel 1099  df-in 1491  df-ss 1492

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