Theorem elabgf0 13658

Description: Lemma for elabgf 2868. (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 2229	. 2 ⊢ (𝑥 = 𝐴 → (𝑥 ∈ {𝑥 ∣ 𝜑} ↔ 𝐴 ∈ {𝑥 ∣ 𝜑}))
2		abid 2153	. 2 ⊢ (𝑥 ∈ {𝑥 ∣ 𝜑} ↔ 𝜑)
3	1, 2	bitr3di 194	1 ⊢ (𝑥 = 𝐴 → (𝐴 ∈ {𝑥 ∣ 𝜑} ↔ 𝜑))

Syntax hints:   → wi 4   ↔ wb 104   = wceq 1343   ∈ wcel 2136  {cab 2151

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 1435  ax-gen 1437  ax-ie1 1481  ax-ie2 1482  ax-8 1492  ax-4 1498  ax-17 1514  ax-i9 1518  ax-ial 1522  ax-ext 2147

This theorem depends on definitions:  df-bi 116  df-sb 1751  df-clab 2152  df-cleq 2158  df-clel 2161

This theorem is referenced by:  elabgft1  13659  elabgf2  13661