Theorem sbc8g 3410

Description: This is the closest we can get to df-sbc 3403 if we start from dfsbcq 3404 (see its comments) and dfsbcq2 3405. (Contributed by NM, 18-Nov-2008.) (Proof shortened by Andrew Salmon, 29-Jun-2011.) (Proof modification is discouraged.)

Assertion

Ref	Expression
sbc8g	⊢ (𝐴 ∈ 𝑉 → ([𝐴 / 𝑥]𝜑 ↔ 𝐴 ∈ {𝑥 ∣ 𝜑}))

Proof of Theorem sbc8g

Dummy variable 𝑦 is distinct from all other variables.

Step	Hyp	Ref	Expression
1		dfsbcq 3404	. 2 ⊢ (𝑦 = 𝐴 → ([𝑦 / 𝑥]𝜑 ↔ [𝐴 / 𝑥]𝜑))
2		eleq1 2676	. 2 ⊢ (𝑦 = 𝐴 → (𝑦 ∈ {𝑥 ∣ 𝜑} ↔ 𝐴 ∈ {𝑥 ∣ 𝜑}))
3		df-clab 2597	. . 3 ⊢ (𝑦 ∈ {𝑥 ∣ 𝜑} ↔ [𝑦 / 𝑥]𝜑)
4		equid 1926	. . . 4 ⊢ 𝑦 = 𝑦
5		dfsbcq2 3405	. . . 4 ⊢ (𝑦 = 𝑦 → ([𝑦 / 𝑥]𝜑 ↔ [𝑦 / 𝑥]𝜑))
6	4, 5	ax-mp 5	. . 3 ⊢ ([𝑦 / 𝑥]𝜑 ↔ [𝑦 / 𝑥]𝜑)
7	3, 6	bitr2i 264	. 2 ⊢ ([𝑦 / 𝑥]𝜑 ↔ 𝑦 ∈ {𝑥 ∣ 𝜑})
8	1, 2, 7	vtoclbg 3240	1 ⊢ (𝐴 ∈ 𝑉 → ([𝐴 / 𝑥]𝜑 ↔ 𝐴 ∈ {𝑥 ∣ 𝜑}))

Syntax hints:   → wi 4   ↔ wb 195  [wsb 1867   ∈ wcel 1977  {cab 2596  [wsbc 3402

This theorem was proved from axioms:  ax-mp 5  ax-1 6  ax-2 7  ax-3 8  ax-gen 1713  ax-4 1728  ax-5 1827  ax-6 1875  ax-7 1922  ax-12 2034  ax-ext 2590

This theorem depends on definitions:  df-bi 196  df-or 384  df-an 385  df-tru 1478  df-ex 1696  df-nf 1701  df-sb 1868  df-clab 2597  df-cleq 2603  df-clel 2606  df-v 3175  df-sbc 3403

This theorem is referenced by:  bnj984  30276  rusbcALT  37662