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Metamath Proof Explorer

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 ) ) )