Theorem zfauscl 5198

Description: Separation Scheme (Aussonderung) using a class variable. To derive this from ax-sep 5196, we invoke the Axiom of Extensionality (indirectly via vtocl 3560), which is needed for the justification of class variable notation.

If we omit the requirement that 𝑦 not occur in 𝜑, we can derive a contradiction, as notzfaus 5255 shows. (Contributed by NM, 21-Jun-1993.)

Hypothesis

Ref	Expression
zfauscl.1	⊢ 𝐴 ∈ V

Assertion

Ref	Expression
zfauscl	⊢ ∃𝑦∀𝑥(𝑥 ∈ 𝑦 ↔ (𝑥 ∈ 𝐴 ∧ 𝜑))

Distinct variable groups:   𝑥,𝑦,𝐴   𝜑,𝑦

Allowed substitution hint:   𝜑(𝑥)

Proof of Theorem zfauscl

Dummy variable 𝑧 is distinct from all other variables.

Step	Hyp	Ref	Expression
1		zfauscl.1	. 2 ⊢ 𝐴 ∈ V
2		eleq2 2901	. . . . . 6 ⊢ (𝑧 = 𝐴 → (𝑥 ∈ 𝑧 ↔ 𝑥 ∈ 𝐴))
3	2	anbi1d 631	. . . . 5 ⊢ (𝑧 = 𝐴 → ((𝑥 ∈ 𝑧 ∧ 𝜑) ↔ (𝑥 ∈ 𝐴 ∧ 𝜑)))
4	3	bibi2d 345	. . . 4 ⊢ (𝑧 = 𝐴 → ((𝑥 ∈ 𝑦 ↔ (𝑥 ∈ 𝑧 ∧ 𝜑)) ↔ (𝑥 ∈ 𝑦 ↔ (𝑥 ∈ 𝐴 ∧ 𝜑))))
5	4	albidv 1917	. . 3 ⊢ (𝑧 = 𝐴 → (∀𝑥(𝑥 ∈ 𝑦 ↔ (𝑥 ∈ 𝑧 ∧ 𝜑)) ↔ ∀𝑥(𝑥 ∈ 𝑦 ↔ (𝑥 ∈ 𝐴 ∧ 𝜑))))
6	5	exbidv 1918	. 2 ⊢ (𝑧 = 𝐴 → (∃𝑦∀𝑥(𝑥 ∈ 𝑦 ↔ (𝑥 ∈ 𝑧 ∧ 𝜑)) ↔ ∃𝑦∀𝑥(𝑥 ∈ 𝑦 ↔ (𝑥 ∈ 𝐴 ∧ 𝜑))))
7		ax-sep 5196	. 2 ⊢ ∃𝑦∀𝑥(𝑥 ∈ 𝑦 ↔ (𝑥 ∈ 𝑧 ∧ 𝜑))
8	1, 6, 7	vtocl 3560	1 ⊢ ∃𝑦∀𝑥(𝑥 ∈ 𝑦 ↔ (𝑥 ∈ 𝐴 ∧ 𝜑))

Syntax hints:   ↔ wb 208   ∧ wa 398  ∀wal 1531   = wceq 1533  ∃wex 1776   ∈ wcel 2110  Vcvv 3495

This theorem was proved from axioms:  ax-mp 5  ax-1 6  ax-2 7  ax-3 8  ax-gen 1792  ax-4 1806  ax-5 1907  ax-6 1966  ax-7 2011  ax-8 2112  ax-9 2120  ax-ext 2793  ax-sep 5196

This theorem depends on definitions:  df-bi 209  df-an 399  df-ex 1777  df-cleq 2814  df-clel 2893

This theorem is referenced by:  inex1  5214