Theorem axprlem2 5386

Description: Lemma for axpr 5389. There exists a set to which all sets whose only members are empty sets belong. (Contributed by Rohan Ridenour, 9-Aug-2023.) (Revised by BJ, 13-Aug-2023.)

Assertion

Ref	Expression
axprlem2	⊢ ∃𝑥∀𝑦(∀𝑧 ∈ 𝑦 ∀𝑤 ¬ 𝑤 ∈ 𝑧 → 𝑦 ∈ 𝑥)

Distinct variable group:   𝑥,𝑦,𝑧,𝑤

Proof of Theorem axprlem2

Dummy variable 𝑣 is distinct from all other variables.

Step	Hyp	Ref	Expression
1		ax-pow 5327	. . 3 ⊢ ∃𝑥∀𝑦(∀𝑧(𝑧 ∈ 𝑦 → 𝑧 ∈ 𝑣) → 𝑦 ∈ 𝑥)
2		df-ral 3078	. . . . . . 7 ⊢ (∀𝑧 ∈ 𝑦 ∀𝑤 ¬ 𝑤 ∈ 𝑧 ↔ ∀𝑧(𝑧 ∈ 𝑦 → ∀𝑤 ¬ 𝑤 ∈ 𝑧))
3		imim2 59	. . . . . . . 8 ⊢ ((∀𝑤 ¬ 𝑤 ∈ 𝑧 → 𝑧 ∈ 𝑣) → ((𝑧 ∈ 𝑦 → ∀𝑤 ¬ 𝑤 ∈ 𝑧) → (𝑧 ∈ 𝑦 → 𝑧 ∈ 𝑣)))
4	3	al2imi 1848	. . . . . . 7 ⊢ (∀𝑧(∀𝑤 ¬ 𝑤 ∈ 𝑧 → 𝑧 ∈ 𝑣) → (∀𝑧(𝑧 ∈ 𝑦 → ∀𝑤 ¬ 𝑤 ∈ 𝑧) → ∀𝑧(𝑧 ∈ 𝑦 → 𝑧 ∈ 𝑣)))
5	2, 4	biimtrid 245	. . . . . 6 ⊢ (∀𝑧(∀𝑤 ¬ 𝑤 ∈ 𝑧 → 𝑧 ∈ 𝑣) → (∀𝑧 ∈ 𝑦 ∀𝑤 ¬ 𝑤 ∈ 𝑧 → ∀𝑧(𝑧 ∈ 𝑦 → 𝑧 ∈ 𝑣)))
6	5	imim1d 83	. . . . 5 ⊢ (∀𝑧(∀𝑤 ¬ 𝑤 ∈ 𝑧 → 𝑧 ∈ 𝑣) → ((∀𝑧(𝑧 ∈ 𝑦 → 𝑧 ∈ 𝑣) → 𝑦 ∈ 𝑥) → (∀𝑧 ∈ 𝑦 ∀𝑤 ¬ 𝑤 ∈ 𝑧 → 𝑦 ∈ 𝑥)))
7	6	alimdv 1949	. . . 4 ⊢ (∀𝑧(∀𝑤 ¬ 𝑤 ∈ 𝑧 → 𝑧 ∈ 𝑣) → (∀𝑦(∀𝑧(𝑧 ∈ 𝑦 → 𝑧 ∈ 𝑣) → 𝑦 ∈ 𝑥) → ∀𝑦(∀𝑧 ∈ 𝑦 ∀𝑤 ¬ 𝑤 ∈ 𝑧 → 𝑦 ∈ 𝑥)))
8	7	eximdv 1950	. . 3 ⊢ (∀𝑧(∀𝑤 ¬ 𝑤 ∈ 𝑧 → 𝑧 ∈ 𝑣) → (∃𝑥∀𝑦(∀𝑧(𝑧 ∈ 𝑦 → 𝑧 ∈ 𝑣) → 𝑦 ∈ 𝑥) → ∃𝑥∀𝑦(∀𝑧 ∈ 𝑦 ∀𝑤 ¬ 𝑤 ∈ 𝑧 → 𝑦 ∈ 𝑥)))
9	1, 8	mpi 21	. 2 ⊢ (∀𝑧(∀𝑤 ¬ 𝑤 ∈ 𝑧 → 𝑧 ∈ 𝑣) → ∃𝑥∀𝑦(∀𝑧 ∈ 𝑦 ∀𝑤 ¬ 𝑤 ∈ 𝑧 → 𝑦 ∈ 𝑥))
10		axprlem1 5385	. 2 ⊢ ∃𝑣∀𝑧(∀𝑤 ¬ 𝑤 ∈ 𝑧 → 𝑧 ∈ 𝑣)
11	9, 10	exlimiiv 1964	1 ⊢ ∃𝑥∀𝑦(∀𝑧 ∈ 𝑦 ∀𝑤 ¬ 𝑤 ∈ 𝑧 → 𝑦 ∈ 𝑥)

This proof depends on syntax axioms:  ¬ wn 3   → wi 4  ∀wal 1568  ∃wex 1812  ∀wral 3077

This proof depends on axioms:  ax-mp 5  ax-1 6  ax-2 7  ax-3 8  ax-gen 1828  ax-4 1842  ax-5 1943  ax-pow 5327

This proof depends on definitions:  df-bi 210  df-ex 1813  df-ral 3078

This theorem is used by:  axpr  5389