Theorem elabf 3629

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 2923	. . 3 ⊢ Ⅎ𝑥𝐴
3		elabf.1	. . 3 ⊢ Ⅎ𝑥𝜓
4		elabf.3	. . 3 ⊢ (𝑥 = 𝐴 → (𝜑 ↔ 𝜓))
5	2, 3, 4	elabgf 3628	. 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 1570  Ⅎwnf 1816   ∈ wcel 2145  {cab 2739  Vcvv 3451

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-10 2178  ax-11 2194  ax-12 2213  ax-ext 2733

This proof depends on definitions:  df-bi 210  df-an 402  df-or 862  df-tru 1573  df-ex 1813  df-nf 1817  df-sb 2100  df-clab 2740  df-cleq 2753  df-clel 2836  df-nfc 2910  df-v 3453

This theorem is used by:  scottabf  9920  dfon2lem1  36515  sdclem2  38644  sdclem1  38645