Theorem zfauscl 5250

Description: Separation Scheme (Aussonderung) using a class variable. To derive this from ax-sep 5248, we invoke the Axiom of Extensionality (indirectly via vtocl 3511), 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 5310 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 2826	. . . . . 6 ⊢ (𝑧 = 𝐴 → (𝑥 ∈ 𝑧 ↔ 𝑥 ∈ 𝐴))
3	2	anbi1d 631	. . . . 5 ⊢ (𝑧 = 𝐴 → ((𝑥 ∈ 𝑧 ∧ 𝜑) ↔ (𝑥 ∈ 𝐴 ∧ 𝜑)))
4	3	bibi2d 343	. . . 4 ⊢ (𝑧 = 𝐴 → ((𝑥 ∈ 𝑦 ↔ (𝑥 ∈ 𝑧 ∧ 𝜑)) ↔ (𝑥 ∈ 𝑦 ↔ (𝑥 ∈ 𝐴 ∧ 𝜑))))
5	4	albidv 1923	. . 3 ⊢ (𝑧 = 𝐴 → (∀𝑥(𝑥 ∈ 𝑦 ↔ (𝑥 ∈ 𝑧 ∧ 𝜑)) ↔ ∀𝑥(𝑥 ∈ 𝑦 ↔ (𝑥 ∈ 𝐴 ∧ 𝜑))))
6	5	exbidv 1924	. 2 ⊢ (𝑧 = 𝐴 → (∃𝑦∀𝑥(𝑥 ∈ 𝑦 ↔ (𝑥 ∈ 𝑧 ∧ 𝜑)) ↔ ∃𝑦∀𝑥(𝑥 ∈ 𝑦 ↔ (𝑥 ∈ 𝐴 ∧ 𝜑))))
7		ax-sep 5248	. 2 ⊢ ∃𝑦∀𝑥(𝑥 ∈ 𝑦 ↔ (𝑥 ∈ 𝑧 ∧ 𝜑))
8	1, 6, 7	vtocl 3511	1 ⊢ ∃𝑦∀𝑥(𝑥 ∈ 𝑦 ↔ (𝑥 ∈ 𝐴 ∧ 𝜑))

Syntax hints:   ↔ wb 205   ∧ wa 397  ∀wal 1539   = wceq 1541  ∃wex 1781   ∈ wcel 2106  Vcvv 3442

This theorem was proved from axioms:  ax-mp 5  ax-1 6  ax-2 7  ax-3 8  ax-gen 1797  ax-4 1811  ax-5 1913  ax-6 1971  ax-7 2011  ax-8 2108  ax-9 2116  ax-ext 2708  ax-sep 5248

This theorem depends on definitions:  df-bi 206  df-an 398  df-ex 1782  df-cleq 2729  df-clel 2815

This theorem is referenced by:  inex1  5266