Theorem sbc5ALT 3833

Description: Alternate proof of sbc5 3832. This proof helps show how clelab 2890 works, since it is equivalent but shorter thanks to now-available library theorems like vtoclbg 3569 and isset 3502. (Contributed by NM, 23-Aug-1993.) (Revised by Mario Carneiro, 12-Oct-2016.) (New usage is discouraged.) (Proof modification is discouraged.)

Assertion

Ref	Expression
sbc5ALT	⊢ ([𝐴 / 𝑥]𝜑 ↔ ∃𝑥(𝑥 = 𝐴 ∧ 𝜑))

Distinct variable group:   𝑥,𝐴

Allowed substitution hint:   𝜑(𝑥)

Proof of Theorem sbc5ALT

Dummy variable 𝑦 is distinct from all other variables.

Step	Hyp	Ref	Expression
1		sbcex 3814	. 2 ⊢ ([𝐴 / 𝑥]𝜑 → 𝐴 ∈ V)
2		exsimpl 1867	. . 3 ⊢ (∃𝑥(𝑥 = 𝐴 ∧ 𝜑) → ∃𝑥 𝑥 = 𝐴)
3		isset 3502	. . 3 ⊢ (𝐴 ∈ V ↔ ∃𝑥 𝑥 = 𝐴)
4	2, 3	sylibr 234	. 2 ⊢ (∃𝑥(𝑥 = 𝐴 ∧ 𝜑) → 𝐴 ∈ V)
5		dfsbcq2 3807	. . 3 ⊢ (𝑦 = 𝐴 → ([𝑦 / 𝑥]𝜑 ↔ [𝐴 / 𝑥]𝜑))
6		eqeq2 2752	. . . . 5 ⊢ (𝑦 = 𝐴 → (𝑥 = 𝑦 ↔ 𝑥 = 𝐴))
7	6	anbi1d 630	. . . 4 ⊢ (𝑦 = 𝐴 → ((𝑥 = 𝑦 ∧ 𝜑) ↔ (𝑥 = 𝐴 ∧ 𝜑)))
8	7	exbidv 1920	. . 3 ⊢ (𝑦 = 𝐴 → (∃𝑥(𝑥 = 𝑦 ∧ 𝜑) ↔ ∃𝑥(𝑥 = 𝐴 ∧ 𝜑)))
9		sb5 2277	. . 3 ⊢ ([𝑦 / 𝑥]𝜑 ↔ ∃𝑥(𝑥 = 𝑦 ∧ 𝜑))
10	5, 8, 9	vtoclbg 3569	. 2 ⊢ (𝐴 ∈ V → ([𝐴 / 𝑥]𝜑 ↔ ∃𝑥(𝑥 = 𝐴 ∧ 𝜑)))
11	1, 4, 10	pm5.21nii 378	1 ⊢ ([𝐴 / 𝑥]𝜑 ↔ ∃𝑥(𝑥 = 𝐴 ∧ 𝜑))

Syntax hints:   ↔ wb 206   ∧ wa 395   = wceq 1537  ∃wex 1777  [wsb 2064   ∈ wcel 2108  Vcvv 3488  [wsbc 3804

This theorem was proved from axioms:  ax-mp 5  ax-1 6  ax-2 7  ax-3 8  ax-gen 1793  ax-4 1807  ax-5 1909  ax-6 1967  ax-7 2007  ax-8 2110  ax-9 2118  ax-10 2141  ax-12 2178  ax-ext 2711

This theorem depends on definitions:  df-bi 207  df-an 396  df-tru 1540  df-ex 1778  df-nf 1782  df-sb 2065  df-clab 2718  df-cleq 2732  df-clel 2819  df-v 3490  df-sbc 3805

This theorem is referenced by: (None)