Theorem elabgf 3628

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 2925	. . . 4 ⊢ Ⅎ𝑥{𝑥 ∣ 𝜑}
3	1, 2	nfel 2937	. . 3 ⊢ Ⅎ𝑥 𝐴 ∈ {𝑥 ∣ 𝜑}
4		elabgf.2	. . 3 ⊢ Ⅎ𝑥𝜓
5	3, 4	nfbi 1936	. 2 ⊢ Ⅎ𝑥(𝐴 ∈ {𝑥 ∣ 𝜑} ↔ 𝜓)
6		eleq1 2849	. . 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 2743	. 2 ⊢ (𝑥 ∈ {𝑥 ∣ 𝜑} ↔ 𝜑)
10	1, 5, 8, 9	vtoclgf 3530	1 ⊢ (𝐴 ∈ 𝐵 → (𝐴 ∈ {𝑥 ∣ 𝜑} ↔ 𝜓))

This proof depends on syntax axioms:   → wi 4   ↔ wb 209   = wceq 1570  Ⅎwnf 1816   ∈ wcel 2145  {cab 2739  Ⅎwnfc 2908

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:  elabf  3629  elab3gf  3638  elrabf  3642  currysetlem  37858  currysetlem1  37860