Theorem axprlem3

Description: Lemma for axpr . Eliminate the antecedent of the relevant replacement instance. (Contributed by Rohan Ridenour, 10-Aug-2023) (Proof shortened by Matthew House, 18-Sep-2025)

	Ref	Expression
Assertion	axprlem3	⊢ ∃ 𝑧 ∀ 𝑤 ( 𝑤 ∈ 𝑧 ↔ ∃ 𝑠 ( 𝑠 ∈ 𝑝 ∧ if- ( ∃ 𝑛 𝑛 ∈ 𝑠 , 𝑤 = 𝑥 , 𝑤 = 𝑦 ) ) )

Proof

Step	Hyp	Ref	Expression
1		axrep4v	⊢ ( ∀ 𝑠 ∃ 𝑧 ∀ 𝑤 ( if- ( ∃ 𝑛 𝑛 ∈ 𝑠 , 𝑤 = 𝑥 , 𝑤 = 𝑦 ) → 𝑤 = 𝑧 ) → ∃ 𝑧 ∀ 𝑤 ( 𝑤 ∈ 𝑧 ↔ ∃ 𝑠 ( 𝑠 ∈ 𝑝 ∧ if- ( ∃ 𝑛 𝑛 ∈ 𝑠 , 𝑤 = 𝑥 , 𝑤 = 𝑦 ) ) ) )
2		ifptru	⊢ ( ∃ 𝑛 𝑛 ∈ 𝑠 → ( if- ( ∃ 𝑛 𝑛 ∈ 𝑠 , 𝑤 = 𝑥 , 𝑤 = 𝑦 ) ↔ 𝑤 = 𝑥 ) )
3	2	biimpd	⊢ ( ∃ 𝑛 𝑛 ∈ 𝑠 → ( if- ( ∃ 𝑛 𝑛 ∈ 𝑠 , 𝑤 = 𝑥 , 𝑤 = 𝑦 ) → 𝑤 = 𝑥 ) )
4		equeuclr	⊢ ( 𝑧 = 𝑥 → ( 𝑤 = 𝑥 → 𝑤 = 𝑧 ) )
5	3 4	syl9r	⊢ ( 𝑧 = 𝑥 → ( ∃ 𝑛 𝑛 ∈ 𝑠 → ( if- ( ∃ 𝑛 𝑛 ∈ 𝑠 , 𝑤 = 𝑥 , 𝑤 = 𝑦 ) → 𝑤 = 𝑧 ) ) )
6	5	alrimdv	⊢ ( 𝑧 = 𝑥 → ( ∃ 𝑛 𝑛 ∈ 𝑠 → ∀ 𝑤 ( if- ( ∃ 𝑛 𝑛 ∈ 𝑠 , 𝑤 = 𝑥 , 𝑤 = 𝑦 ) → 𝑤 = 𝑧 ) ) )
7	6	spimevw	⊢ ( ∃ 𝑛 𝑛 ∈ 𝑠 → ∃ 𝑧 ∀ 𝑤 ( if- ( ∃ 𝑛 𝑛 ∈ 𝑠 , 𝑤 = 𝑥 , 𝑤 = 𝑦 ) → 𝑤 = 𝑧 ) )
8		ifpfal	⊢ ( ¬ ∃ 𝑛 𝑛 ∈ 𝑠 → ( if- ( ∃ 𝑛 𝑛 ∈ 𝑠 , 𝑤 = 𝑥 , 𝑤 = 𝑦 ) ↔ 𝑤 = 𝑦 ) )
9	8	biimpd	⊢ ( ¬ ∃ 𝑛 𝑛 ∈ 𝑠 → ( if- ( ∃ 𝑛 𝑛 ∈ 𝑠 , 𝑤 = 𝑥 , 𝑤 = 𝑦 ) → 𝑤 = 𝑦 ) )
10		equeuclr	⊢ ( 𝑧 = 𝑦 → ( 𝑤 = 𝑦 → 𝑤 = 𝑧 ) )
11	9 10	syl9r	⊢ ( 𝑧 = 𝑦 → ( ¬ ∃ 𝑛 𝑛 ∈ 𝑠 → ( if- ( ∃ 𝑛 𝑛 ∈ 𝑠 , 𝑤 = 𝑥 , 𝑤 = 𝑦 ) → 𝑤 = 𝑧 ) ) )
12	11	alrimdv	⊢ ( 𝑧 = 𝑦 → ( ¬ ∃ 𝑛 𝑛 ∈ 𝑠 → ∀ 𝑤 ( if- ( ∃ 𝑛 𝑛 ∈ 𝑠 , 𝑤 = 𝑥 , 𝑤 = 𝑦 ) → 𝑤 = 𝑧 ) ) )
13	12	spimevw	⊢ ( ¬ ∃ 𝑛 𝑛 ∈ 𝑠 → ∃ 𝑧 ∀ 𝑤 ( if- ( ∃ 𝑛 𝑛 ∈ 𝑠 , 𝑤 = 𝑥 , 𝑤 = 𝑦 ) → 𝑤 = 𝑧 ) )
14	7 13	pm2.61i	⊢ ∃ 𝑧 ∀ 𝑤 ( if- ( ∃ 𝑛 𝑛 ∈ 𝑠 , 𝑤 = 𝑥 , 𝑤 = 𝑦 ) → 𝑤 = 𝑧 )
15	1 14	mpg	⊢ ∃ 𝑧 ∀ 𝑤 ( 𝑤 ∈ 𝑧 ↔ ∃ 𝑠 ( 𝑠 ∈ 𝑝 ∧ if- ( ∃ 𝑛 𝑛 ∈ 𝑠 , 𝑤 = 𝑥 , 𝑤 = 𝑦 ) ) )