Theorem elabd3 3630

Description: Membership in a class abstraction, using implicit substitution. Deduction version of elab 3638. (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 2762	. 2 ⊢ (𝜑 → {𝑥 ∣ 𝜓} = {𝑥 ∣ 𝜓})
3		elabd3.is	. 2 ⊢ ((𝜑 ∧ 𝑥 = 𝐴) → (𝜓 ↔ 𝜒))
4	1, 2, 3	elabd2 3629	1 ⊢ (𝜑 → (𝐴 ∈ {𝑥 ∣ 𝜓} ↔ 𝜒))

Syntax hints:   → wi 4   ↔ wb 208   ∧ wa 399   = wceq 1559   ∈ wcel 2141  {cab 2739

This theorem was proved from axioms:  ax-mp 5  ax-1 6  ax-2 7  ax-3 8  ax-gen 1814  ax-4 1828  ax-5 1929  ax-6 1986  ax-7 2027  ax-8 2143  ax-9 2151  ax-ext 2733

This theorem depends on definitions:  df-bi 209  df-an 400  df-tru 1562  df-ex 1799  df-sb 2090  df-clab 2740  df-cleq 2753  df-clel 2836

This theorem is referenced by:  sbcied  3787