Theorem axprlem1

Description: Lemma for axpr . There exists a set to which all empty sets belong. (Contributed by Rohan Ridenour, 10-Aug-2023) (Revised by BJ, 13-Aug-2023)

	Ref	Expression
Assertion	axprlem1	⊢ ∃ 𝑥 ∀ 𝑦 ( ∀ 𝑧 ¬ 𝑧 ∈ 𝑦 → 𝑦 ∈ 𝑥 )

Proof

Step	Hyp	Ref	Expression
1		ax-pow	⊢ ∃ 𝑥 ∀ 𝑦 ( ∀ 𝑧 ( 𝑧 ∈ 𝑦 → 𝑧 ∈ 𝑤 ) → 𝑦 ∈ 𝑥 )
2		pm2.21	⊢ ( ¬ 𝑧 ∈ 𝑦 → ( 𝑧 ∈ 𝑦 → 𝑧 ∈ 𝑤 ) )
3	2	alimi	⊢ ( ∀ 𝑧 ¬ 𝑧 ∈ 𝑦 → ∀ 𝑧 ( 𝑧 ∈ 𝑦 → 𝑧 ∈ 𝑤 ) )
4	3	a1i	⊢ ( ∀ 𝑧 ¬ 𝑧 ∈ 𝑤 → ( ∀ 𝑧 ¬ 𝑧 ∈ 𝑦 → ∀ 𝑧 ( 𝑧 ∈ 𝑦 → 𝑧 ∈ 𝑤 ) ) )
5	4	imim1d	⊢ ( ∀ 𝑧 ¬ 𝑧 ∈ 𝑤 → ( ( ∀ 𝑧 ( 𝑧 ∈ 𝑦 → 𝑧 ∈ 𝑤 ) → 𝑦 ∈ 𝑥 ) → ( ∀ 𝑧 ¬ 𝑧 ∈ 𝑦 → 𝑦 ∈ 𝑥 ) ) )
6	5	alimdv	⊢ ( ∀ 𝑧 ¬ 𝑧 ∈ 𝑤 → ( ∀ 𝑦 ( ∀ 𝑧 ( 𝑧 ∈ 𝑦 → 𝑧 ∈ 𝑤 ) → 𝑦 ∈ 𝑥 ) → ∀ 𝑦 ( ∀ 𝑧 ¬ 𝑧 ∈ 𝑦 → 𝑦 ∈ 𝑥 ) ) )
7	6	eximdv	⊢ ( ∀ 𝑧 ¬ 𝑧 ∈ 𝑤 → ( ∃ 𝑥 ∀ 𝑦 ( ∀ 𝑧 ( 𝑧 ∈ 𝑦 → 𝑧 ∈ 𝑤 ) → 𝑦 ∈ 𝑥 ) → ∃ 𝑥 ∀ 𝑦 ( ∀ 𝑧 ¬ 𝑧 ∈ 𝑦 → 𝑦 ∈ 𝑥 ) ) )
8	1 7	mpi	⊢ ( ∀ 𝑧 ¬ 𝑧 ∈ 𝑤 → ∃ 𝑥 ∀ 𝑦 ( ∀ 𝑧 ¬ 𝑧 ∈ 𝑦 → 𝑦 ∈ 𝑥 ) )
9		ax-nul	⊢ ∃ 𝑤 ∀ 𝑧 ¬ 𝑧 ∈ 𝑤
10	8 9	exlimiiv	⊢ ∃ 𝑥 ∀ 𝑦 ( ∀ 𝑧 ¬ 𝑧 ∈ 𝑦 → 𝑦 ∈ 𝑥 )