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Theorem elabgf0 9916
Description: Lemma for elabgf 2685. (Contributed by BJ, 21-Nov-2019.)
Assertion
Ref Expression
elabgf0  |-  ( x  =  A  ->  ( A  e.  { x  |  ph }  <->  ph ) )

Proof of Theorem elabgf0
StepHypRef Expression
1 abid 2028 . 2  |-  ( x  e.  { x  | 
ph }  <->  ph )
2 eleq1 2100 . 2  |-  ( x  =  A  ->  (
x  e.  { x  |  ph }  <->  A  e.  { x  |  ph }
) )
31, 2syl5rbbr 184 1  |-  ( x  =  A  ->  ( A  e.  { x  |  ph }  <->  ph ) )
Colors of variables: wff set class
Syntax hints:    -> wi 4    <-> wb 98    = wceq 1243    e. wcel 1393   {cab 2026
This theorem was proved from axioms:  ax-1 5  ax-2 6  ax-mp 7  ax-ia1 99  ax-ia2 100  ax-ia3 101  ax-5 1336  ax-gen 1338  ax-ie1 1382  ax-ie2 1383  ax-8 1395  ax-4 1400  ax-17 1419  ax-i9 1423  ax-ial 1427  ax-ext 2022
This theorem depends on definitions:  df-bi 110  df-sb 1646  df-clab 2027  df-cleq 2033  df-clel 2036
This theorem is referenced by:  elabgft1  9917  elabgf2  9919
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