Theorem elabgf0 15390

Description: Lemma for elabgf 2906. (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 2259	. 2 ⊢ (𝑥 = 𝐴 → (𝑥 ∈ {𝑥 ∣ 𝜑} ↔ 𝐴 ∈ {𝑥 ∣ 𝜑}))
2		abid 2184	. 2 ⊢ (𝑥 ∈ {𝑥 ∣ 𝜑} ↔ 𝜑)
3	1, 2	bitr3di 195	1 ⊢ (𝑥 = 𝐴 → (𝐴 ∈ {𝑥 ∣ 𝜑} ↔ 𝜑))

Syntax hints:   → wi 4   ↔ wb 105   = wceq 1364   ∈ wcel 2167  {cab 2182

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 1461  ax-gen 1463  ax-ie1 1507  ax-ie2 1508  ax-8 1518  ax-4 1524  ax-17 1540  ax-i9 1544  ax-ial 1548  ax-ext 2178

This theorem depends on definitions:  df-bi 117  df-sb 1777  df-clab 2183  df-cleq 2189  df-clel 2192

This theorem is referenced by:  elabgft1  15391  elabgf2  15393