Theorem elabd3 3640

Description: Membership in a class abstraction, using implicit substitution. Deduction version of elab 3649. (Contributed by GG, 12-Oct-2024.)

Hypotheses

Ref	Expression
elabd3.ex	⊢ (𝜑 → 𝐴 ∈ 𝑉)
elabd3.is	⊢ ((𝜑 ∧ 𝑥 = 𝐴) → (𝜓 ↔ 𝜒))

Assertion

Ref	Expression
elabd3	⊢ (𝜑 → (𝐴 ∈ {𝑥 ∣ 𝜓} ↔ 𝜒))

Distinct variable groups:   𝜑,𝑥   𝜒,𝑥   𝑥,𝐴

Allowed substitution hints:   𝜓(𝑥)   𝑉(𝑥)

Proof of Theorem elabd3

Step	Hyp	Ref	Expression
1		elabd3.ex	. 2 ⊢ (𝜑 → 𝐴 ∈ 𝑉)
2		eqidd 2731	. 2 ⊢ (𝜑 → {𝑥 ∣ 𝜓} = {𝑥 ∣ 𝜓})
3		elabd3.is	. 2 ⊢ ((𝜑 ∧ 𝑥 = 𝐴) → (𝜓 ↔ 𝜒))
4	1, 2, 3	elabd2 3639	1 ⊢ (𝜑 → (𝐴 ∈ {𝑥 ∣ 𝜓} ↔ 𝜒))

Syntax hints:   → wi 4   ↔ wb 206   ∧ wa 395   = wceq 1540   ∈ wcel 2109  {cab 2708

This theorem was proved from axioms:  ax-mp 5  ax-1 6  ax-2 7  ax-3 8  ax-gen 1795  ax-4 1809  ax-5 1910  ax-6 1967  ax-7 2008  ax-8 2111  ax-9 2119  ax-ext 2702

This theorem depends on definitions:  df-bi 207  df-an 396  df-tru 1543  df-ex 1780  df-sb 2066  df-clab 2709  df-cleq 2722  df-clel 2804

This theorem is referenced by:  sbcied  3800