Theorem abvALT 3501

Description: Alternate proof of abv 3500, shorter but using more axioms. (Contributed by BJ, 19-Mar-2021.) (Proof modification is discouraged.) (New usage is discouraged.)

Assertion

Ref	Expression
abvALT	⊢ ({𝑥 ∣ 𝜑} = V ↔ ∀𝑥𝜑)

Proof of Theorem abvALT

Dummy variable 𝑦 is distinct from all other variables.

Step	Hyp	Ref	Expression
1		df-clab 2718	. . 3 ⊢ (𝑦 ∈ {𝑥 ∣ 𝜑} ↔ [𝑦 / 𝑥]𝜑)
2	1	albii 1817	. 2 ⊢ (∀𝑦 𝑦 ∈ {𝑥 ∣ 𝜑} ↔ ∀𝑦[𝑦 / 𝑥]𝜑)
3		eqv 3498	. 2 ⊢ ({𝑥 ∣ 𝜑} = V ↔ ∀𝑦 𝑦 ∈ {𝑥 ∣ 𝜑})
4		sb8v 2358	. 2 ⊢ (∀𝑥𝜑 ↔ ∀𝑦[𝑦 / 𝑥]𝜑)
5	2, 3, 4	3bitr4i 303	1 ⊢ ({𝑥 ∣ 𝜑} = V ↔ ∀𝑥𝜑)

Syntax hints:   ↔ wb 206  ∀wal 1535   = wceq 1537  [wsb 2064   ∈ wcel 2108  {cab 2717  Vcvv 3488

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-11 2158  ax-ext 2711

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

This theorem is referenced by: (None)