Theorem elabgf0 13155

Description: Lemma for elabgf 2830. (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		eleq1 2203	. 2 ⊢ (𝑥 = 𝐴 → (𝑥 ∈ {𝑥 ∣ 𝜑} ↔ 𝐴 ∈ {𝑥 ∣ 𝜑}))
2		abid 2128	. 2 ⊢ (𝑥 ∈ {𝑥 ∣ 𝜑} ↔ 𝜑)
3	1, 2	bitr3di 194	1 ⊢ (𝑥 = 𝐴 → (𝐴 ∈ {𝑥 ∣ 𝜑} ↔ 𝜑))

Syntax hints:   → wi 4   ↔ wb 104   = wceq 1332   ∈ wcel 1481  {cab 2126

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 1424  ax-gen 1426  ax-ie1 1470  ax-ie2 1471  ax-8 1483  ax-4 1488  ax-17 1507  ax-i9 1511  ax-ial 1515  ax-ext 2122

This theorem depends on definitions:  df-bi 116  df-sb 1737  df-clab 2127  df-cleq 2133  df-clel 2136

This theorem is referenced by:  elabgft1  13156  elabgf2  13158