Theorem abss 2120

Description: Class abstraction in a subclass relationship.

Assertion

Ref	Expression
abss	|- ({x | ph} (_ A <-> A.x(ph -> x e. A))

Distinct variable group:   x,A

Proof of Theorem abss

Step	Hyp	Ref	Expression
1		abid2 1583	. . 3 |- {x | x e. A} = A
2	1	sseq2i 2089	. 2 |- ({x | ph} (_ {x | x e. A} <-> {x | ph} (_ A)
3		ss2ab 2119	. 2 |- ({x | ph} (_ {x | x e. A} <-> A.x(ph -> x e. A))
4	2, 3	bitr3 175	1 |- ({x | ph} (_ A <-> A.x(ph -> x e. A))

Syntax hints:   -> wi 3   <-> wb 146  A.wal 956   e. wcel 960  {cab 1466   (_ wss 2050

This theorem is referenced by:  abssdv 2124  rabss 2127  uniiunlem 2135  homcard 10525  qusp 10541

This theorem was proved from axioms:  ax-1 4  ax-2 5  ax-3 6  ax-mp 7  ax-7 964  ax-gen 965  ax-8 966  ax-10 968  ax-12 970  ax-17 973  ax-4 975  ax-5o 977  ax-6o 980  ax-9o 1125  ax-10o 1142  ax-16 1212  ax-11o 1220  ax-ext 1462

This theorem depends on definitions:  df-bi 147  df-an 225  df-ex 983  df-sb 1174  df-clab 1467  df-cleq 1472  df-clel 1475  df-in 2054  df-ss 2056

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