Theorem elabgf0 9916

Description: Lemma for elabgf 2685. (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 2028	. 2 ⊢ (𝑥 ∈ {𝑥 ∣ 𝜑} ↔ 𝜑)
2		eleq1 2100	. 2 ⊢ (𝑥 = 𝐴 → (𝑥 ∈ {𝑥 ∣ 𝜑} ↔ 𝐴 ∈ {𝑥 ∣ 𝜑}))
3	1, 2	syl5rbbr 184	1 ⊢ (𝑥 = 𝐴 → (𝐴 ∈ {𝑥 ∣ 𝜑} ↔ 𝜑))

Syntax hints:   → wi 4   ↔ wb 98   = wceq 1243   ∈ 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