Theorem elabd3 3625

Description: Membership in a class abstraction, using implicit substitution. Deduction version of elab 3633. (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 3624	1 ⊢ (𝜑 → (𝐴 ∈ {𝑥 ∣ 𝜓} ↔ 𝜒))

This proof depends on syntax axioms:   → wi 4   ↔ wb 209   ∧ wa 401   = wceq 1570   ∈ wcel 2145  {cab 2739

This proof depends on axioms:  ax-mp 5  ax-1 6  ax-2 7  ax-3 8  ax-gen 1828  ax-4 1842  ax-5 1943  ax-6 2000  ax-7 2041  ax-8 2147  ax-9 2155  ax-ext 2733

This proof depends on definitions:  df-bi 210  df-an 402  df-tru 1573  df-ex 1813  df-sb 2100  df-clab 2740  df-cleq 2753  df-clel 2836

This theorem is used by:  sbcied  3782