Theorem bj-zfauscl 36260

Description: General version of zfauscl 5291.

Remark: the comment in zfauscl 5291 is misleading: the essential use of ax-ext 2695 is the one via eleq2 2814 and not the one via vtocl 3538, since the latter can be proved without ax-ext 2695 (see bj-vtoclg 36256).

(Contributed by BJ, 2-Jul-2022.) (Proof modification is discouraged.)

Assertion

Ref	Expression
bj-zfauscl	⊢ (𝐴 ∈ 𝑉 → ∃𝑦∀𝑥(𝑥 ∈ 𝑦 ↔ (𝑥 ∈ 𝐴 ∧ 𝜑)))

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

Allowed substitution hints:   𝜑(𝑥)   𝑉(𝑥,𝑦)

Proof of Theorem bj-zfauscl

Dummy variable 𝑧 is distinct from all other variables.

Step	Hyp	Ref	Expression
1		eleq2 2814	. . . . . . 7 ⊢ (𝑧 = 𝐴 → (𝑥 ∈ 𝑧 ↔ 𝑥 ∈ 𝐴))
2	1	anbi1d 629	. . . . . 6 ⊢ (𝑧 = 𝐴 → ((𝑥 ∈ 𝑧 ∧ 𝜑) ↔ (𝑥 ∈ 𝐴 ∧ 𝜑)))
3	2	bibi2d 342	. . . . 5 ⊢ (𝑧 = 𝐴 → ((𝑥 ∈ 𝑦 ↔ (𝑥 ∈ 𝑧 ∧ 𝜑)) ↔ (𝑥 ∈ 𝑦 ↔ (𝑥 ∈ 𝐴 ∧ 𝜑))))
4	3	biimpd 228	. . . 4 ⊢ (𝑧 = 𝐴 → ((𝑥 ∈ 𝑦 ↔ (𝑥 ∈ 𝑧 ∧ 𝜑)) → (𝑥 ∈ 𝑦 ↔ (𝑥 ∈ 𝐴 ∧ 𝜑))))
5	4	alimdv 1911	. . 3 ⊢ (𝑧 = 𝐴 → (∀𝑥(𝑥 ∈ 𝑦 ↔ (𝑥 ∈ 𝑧 ∧ 𝜑)) → ∀𝑥(𝑥 ∈ 𝑦 ↔ (𝑥 ∈ 𝐴 ∧ 𝜑))))
6	5	eximdv 1912	. 2 ⊢ (𝑧 = 𝐴 → (∃𝑦∀𝑥(𝑥 ∈ 𝑦 ↔ (𝑥 ∈ 𝑧 ∧ 𝜑)) → ∃𝑦∀𝑥(𝑥 ∈ 𝑦 ↔ (𝑥 ∈ 𝐴 ∧ 𝜑))))
7		ax-sep 5289	. 2 ⊢ ∃𝑦∀𝑥(𝑥 ∈ 𝑦 ↔ (𝑥 ∈ 𝑧 ∧ 𝜑))
8	6, 7	bj-vtoclg 36256	1 ⊢ (𝐴 ∈ 𝑉 → ∃𝑦∀𝑥(𝑥 ∈ 𝑦 ↔ (𝑥 ∈ 𝐴 ∧ 𝜑)))

Syntax hints:   → wi 4   ↔ wb 205   ∧ wa 395  ∀wal 1531   = wceq 1533  ∃wex 1773   ∈ wcel 2098

This theorem was proved from axioms:  ax-mp 5  ax-1 6  ax-2 7  ax-3 8  ax-gen 1789  ax-4 1803  ax-5 1905  ax-6 1963  ax-7 2003  ax-8 2100  ax-9 2108  ax-ext 2695  ax-sep 5289

This theorem depends on definitions:  df-bi 206  df-an 396  df-ex 1774  df-cleq 2716  df-clel 2802

This theorem is referenced by: (None)