Theorem elabgf0 10587

Description: Lemma for elabgf 2736. (Contributed by BJ, 21-Nov-2019.)

Assertion

Ref	Expression
elabgf0	⊢ (𝑥 = 𝐴 → (𝐴 ∈ {𝑥 ∣ 𝜑} ↔ 𝜑))

Proof of Theorem elabgf0

Step	Hyp	Ref	Expression
1		abid 2069	. 2 ⊢ (𝑥 ∈ {𝑥 ∣ 𝜑} ↔ 𝜑)
2		eleq1 2141	. 2 ⊢ (𝑥 = 𝐴 → (𝑥 ∈ {𝑥 ∣ 𝜑} ↔ 𝐴 ∈ {𝑥 ∣ 𝜑}))
3	1, 2	syl5rbbr 193	1 ⊢ (𝑥 = 𝐴 → (𝐴 ∈ {𝑥 ∣ 𝜑} ↔ 𝜑))

Syntax hints:   → wi 4   ↔ wb 103   = wceq 1284   ∈ wcel 1433  {cab 2067

This theorem was proved from axioms:  ax-1 5  ax-2 6  ax-mp 7  ax-ia1 104  ax-ia2 105  ax-ia3 106  ax-5 1376  ax-gen 1378  ax-ie1 1422  ax-ie2 1423  ax-8 1435  ax-4 1440  ax-17 1459  ax-i9 1463  ax-ial 1467  ax-ext 2063

This theorem depends on definitions:  df-bi 115  df-sb 1686  df-clab 2068  df-cleq 2074  df-clel 2077

This theorem is referenced by:  elabgft1  10588  elabgf2  10590