Theorem elabgf0 11677

Description: Lemma for elabgf 2758. (Contributed by BJ, 21-Nov-2019.)

Assertion

Ref	Expression
elabgf0	|- ( x = A -> ( A e. { x | ph } <-> ph ) )

Proof of Theorem elabgf0

Step	Hyp	Ref	Expression
1		abid 2076	. 2 |- ( x e. { x | ph } <-> ph )
2		eleq1 2150	. 2 |- ( x = A -> ( x e. { x | ph } <-> A e. { x | ph } ) )
3	1, 2	syl5rbbr 193	1 |- ( x = A -> ( A e. { x | ph } <-> ph ) )

Syntax hints:   -> wi 4   <-> wb 103   = wceq 1289   e. wcel 1438   {cab 2074

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-5 1381  ax-gen 1383  ax-ie1 1427  ax-ie2 1428  ax-8 1440  ax-4 1445  ax-17 1464  ax-i9 1468  ax-ial 1472  ax-ext 2070

This theorem depends on definitions:  df-bi 115  df-sb 1693  df-clab 2075  df-cleq 2081  df-clel 2084

This theorem is referenced by:  elabgft1  11678  elabgf2  11680