Theorem elabgf0 15847

Description: Lemma for elabgf 2919. (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 2269	. 2 ⊢ (𝑥 = 𝐴 → (𝑥 ∈ {𝑥 ∣ 𝜑} ↔ 𝐴 ∈ {𝑥 ∣ 𝜑}))
2		abid 2194	. 2 ⊢ (𝑥 ∈ {𝑥 ∣ 𝜑} ↔ 𝜑)
3	1, 2	bitr3di 195	1 ⊢ (𝑥 = 𝐴 → (𝐴 ∈ {𝑥 ∣ 𝜑} ↔ 𝜑))

Syntax hints:   → wi 4   ↔ wb 105   = wceq 1373   ∈ wcel 2177  {cab 2192

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 1471  ax-gen 1473  ax-ie1 1517  ax-ie2 1518  ax-8 1528  ax-4 1534  ax-17 1550  ax-i9 1554  ax-ial 1558  ax-ext 2188

This theorem depends on definitions:  df-bi 117  df-sb 1787  df-clab 2193  df-cleq 2199  df-clel 2202

This theorem is referenced by:  elabgft1  15848  elabgf2  15850