Theorem elabgf0 15713

Description: Lemma for elabgf 2915. (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		eleq1 2268	. 2 |- ( x = A -> ( x e. { x | ph } <-> A e. { x | ph } ) )
2		abid 2193	. 2 |- ( x e. { x | ph } <-> ph )
3	1, 2	bitr3di 195	1 |- ( x = A -> ( A e. { x | ph } <-> ph ) )

Syntax hints:   -> wi 4   <-> wb 105   = wceq 1373   e. wcel 2176   {cab 2191

This theorem was proved from axioms:  ax-mp 5  ax-1 6  ax-2 7  ax-ia1 106  ax-ia2 107  ax-ia3 108  ax-5 1470  ax-gen 1472  ax-ie1 1516  ax-ie2 1517  ax-8 1527  ax-4 1533  ax-17 1549  ax-i9 1553  ax-ial 1557  ax-ext 2187

This theorem depends on definitions:  df-bi 117  df-sb 1786  df-clab 2192  df-cleq 2198  df-clel 2201

This theorem is referenced by:  elabgft1  15714  elabgf2  15716