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Theorem elabgf0 14465
Description: Lemma for elabgf 2879. (Contributed by BJ, 21-Nov-2019.)
Assertion
Ref Expression
elabgf0 (𝑥 = 𝐴 → (𝐴 ∈ {𝑥𝜑} ↔ 𝜑))

Proof of Theorem elabgf0
StepHypRef Expression
1 eleq1 2240 . 2 (𝑥 = 𝐴 → (𝑥 ∈ {𝑥𝜑} ↔ 𝐴 ∈ {𝑥𝜑}))
2 abid 2165 . 2 (𝑥 ∈ {𝑥𝜑} ↔ 𝜑)
31, 2bitr3di 195 1 (𝑥 = 𝐴 → (𝐴 ∈ {𝑥𝜑} ↔ 𝜑))
Colors of variables: wff set class
Syntax hints:  wi 4  wb 105   = wceq 1353  wcel 2148  {cab 2163
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 1447  ax-gen 1449  ax-ie1 1493  ax-ie2 1494  ax-8 1504  ax-4 1510  ax-17 1526  ax-i9 1530  ax-ial 1534  ax-ext 2159
This theorem depends on definitions:  df-bi 117  df-sb 1763  df-clab 2164  df-cleq 2170  df-clel 2173
This theorem is referenced by:  elabgft1  14466  elabgf2  14468
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