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Theorem elabgf0 7023
Description: Lemma for elabgf 2662. (Contributed by BJ, 21-Nov-2019.)
Assertion
Ref Expression
elabgf0 (x = A → (A {xφ} ↔ φ))

Proof of Theorem elabgf0
StepHypRef Expression
1 abid 2010 . 2 (x {xφ} ↔ φ)
2 eleq1 2082 . 2 (x = A → (x {xφ} ↔ A {xφ}))
31, 2syl5rbbr 184 1 (x = A → (A {xφ} ↔ φ))
Colors of variables: wff set class
Syntax hints:  wi 4  wb 98   = wceq 1228   wcel 1374  {cab 2008
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 1316  ax-gen 1318  ax-ie1 1363  ax-ie2 1364  ax-8 1376  ax-4 1381  ax-17 1400  ax-i9 1404  ax-ial 1409  ax-ext 2004
This theorem depends on definitions:  df-bi 110  df-sb 1628  df-clab 2009  df-cleq 2015  df-clel 2018
This theorem is referenced by:  elabgft1  7024  elabgf2  7026
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