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Theorem elabgf0 12984
Description: Lemma for elabgf 2826. (Contributed by BJ, 21-Nov-2019.)
Assertion
Ref Expression
elabgf0 (𝑥 = 𝐴 → (𝐴 ∈ {𝑥𝜑} ↔ 𝜑))

Proof of Theorem elabgf0
StepHypRef Expression
1 abid 2127 . 2 (𝑥 ∈ {𝑥𝜑} ↔ 𝜑)
2 eleq1 2202 . 2 (𝑥 = 𝐴 → (𝑥 ∈ {𝑥𝜑} ↔ 𝐴 ∈ {𝑥𝜑}))
31, 2syl5rbbr 194 1 (𝑥 = 𝐴 → (𝐴 ∈ {𝑥𝜑} ↔ 𝜑))
Colors of variables: wff set class
Syntax hints:  wi 4  wb 104   = wceq 1331  wcel 1480  {cab 2125
This theorem was proved from axioms:  ax-mp 5  ax-1 6  ax-2 7  ax-ia1 105  ax-ia2 106  ax-ia3 107  ax-5 1423  ax-gen 1425  ax-ie1 1469  ax-ie2 1470  ax-8 1482  ax-4 1487  ax-17 1506  ax-i9 1510  ax-ial 1514  ax-ext 2121
This theorem depends on definitions:  df-bi 116  df-sb 1736  df-clab 2126  df-cleq 2132  df-clel 2135
This theorem is referenced by:  elabgft1  12985  elabgf2  12987
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