Theorem elabf 3628

Description: Membership in a class abstraction, using implicit substitution. (Contributed by NM, 1-Aug-1994.) (Revised by Mario Carneiro, 12-Oct-2016.)

Hypotheses

Ref	Expression
elabf.1	⊢ Ⅎ𝑥𝜓
elabf.2	⊢ 𝐴 ∈ V
elabf.3	⊢ (𝑥 = 𝐴 → (𝜑 ↔ 𝜓))

Assertion

Ref	Expression
elabf	⊢ (𝐴 ∈ {𝑥 ∣ 𝜑} ↔ 𝜓)

Distinct variable group:   𝑥,𝐴

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

Proof of Theorem elabf

Step	Hyp	Ref	Expression
1		elabf.2	. 2 ⊢ 𝐴 ∈ V
2		nfcv 2922	. . 3 ⊢ Ⅎ𝑥𝐴
3		elabf.1	. . 3 ⊢ Ⅎ𝑥𝜓
4		elabf.3	. . 3 ⊢ (𝑥 = 𝐴 → (𝜑 ↔ 𝜓))
5	2, 3, 4	elabgf 3627	. 2 ⊢ (𝐴 ∈ V → (𝐴 ∈ {𝑥 ∣ 𝜑} ↔ 𝜓))
6	1, 5	ax-mp 5	1 ⊢ (𝐴 ∈ {𝑥 ∣ 𝜑} ↔ 𝜓)

This proof depends on syntax axioms:   → wi 4   ↔ wb 209   = wceq 1569  Ⅎwnf 1815   ∈ wcel 2144  {cab 2738  Vcvv 3450

This proof depends on axioms:  ax-mp 5  ax-1 6  ax-2 7  ax-3 8  ax-gen 1827  ax-4 1841  ax-5 1942  ax-6 1999  ax-7 2040  ax-8 2146  ax-9 2154  ax-10 2177  ax-11 2193  ax-12 2212  ax-ext 2732

This proof depends on definitions:  df-bi 210  df-an 402  df-or 862  df-tru 1572  df-ex 1812  df-nf 1816  df-sb 2099  df-clab 2739  df-cleq 2752  df-clel 2835  df-nfc 2909  df-v 3452

This theorem is used by:  scottabf  9943  sdclem2  38770  sdclem1  38771