Theorem elabgf 3631

Description: Membership in a class abstraction, using implicit substitution. Compare Theorem 6.13 of [Quine] p. 44. This version has bound-variable hypotheses in place of distinct variable restrictions. (Contributed by NM, 21-Sep-2003.) (Revised by Mario Carneiro, 12-Oct-2016.)

Hypotheses

Ref	Expression
elabgf.1	⊢ Ⅎ𝑥𝐴
elabgf.2	⊢ Ⅎ𝑥𝜓
elabgf.3	⊢ (𝑥 = 𝐴 → (𝜑 ↔ 𝜓))

Assertion

Ref	Expression
elabgf	⊢ (𝐴 ∈ 𝐵 → (𝐴 ∈ {𝑥 ∣ 𝜑} ↔ 𝜓))

Proof of Theorem elabgf

Step	Hyp	Ref	Expression
1		elabgf.1	. 2 ⊢ Ⅎ𝑥𝐴
2		nfab1 2928	. . . 4 ⊢ Ⅎ𝑥{𝑥 ∣ 𝜑}
3	1, 2	nfel 2940	. . 3 ⊢ Ⅎ𝑥 𝐴 ∈ {𝑥 ∣ 𝜑}
4		elabgf.2	. . 3 ⊢ Ⅎ𝑥𝜓
5	3, 4	nfbi 1937	. 2 ⊢ Ⅎ𝑥(𝐴 ∈ {𝑥 ∣ 𝜑} ↔ 𝜓)
6		eleq1 2852	. . 3 ⊢ (𝑥 = 𝐴 → (𝑥 ∈ {𝑥 ∣ 𝜑} ↔ 𝐴 ∈ {𝑥 ∣ 𝜑}))
7		elabgf.3	. . 3 ⊢ (𝑥 = 𝐴 → (𝜑 ↔ 𝜓))
8	6, 7	bibi12d 348	. 2 ⊢ (𝑥 = 𝐴 → ((𝑥 ∈ {𝑥 ∣ 𝜑} ↔ 𝜑) ↔ (𝐴 ∈ {𝑥 ∣ 𝜑} ↔ 𝜓)))
9		abid 2745	. 2 ⊢ (𝑥 ∈ {𝑥 ∣ 𝜑} ↔ 𝜑)
10	1, 5, 8, 9	vtoclgf 3533	1 ⊢ (𝐴 ∈ 𝐵 → (𝐴 ∈ {𝑥 ∣ 𝜑} ↔ 𝜓))

This proof depends on syntax axioms:   → wi 4   ↔ wb 209   = wceq 1569  Ⅎwnf 1815   ∈ wcel 2146  {cab 2741  Ⅎwnfc 2911

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 1944  ax-6 2001  ax-7 2042  ax-8 2148  ax-9 2156  ax-10 2179  ax-11 2196  ax-12 2215  ax-ext 2735

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 2101  df-clab 2742  df-cleq 2755  df-clel 2839  df-nfc 2913  df-v 3456

This theorem is used by:  elabf  3632  elab3gf  3641  elrabf  3645  currysetlem  37991  currysetlem1  37993