Theorem axprlem3 5398

Description: Lemma for axpr 5400. Eliminate the antecedent of the relevant replacement instance. (Contributed by Rohan Ridenour, 10-Aug-2023.) (Proof shortened by Matthew House, 18-Sep-2025.)

Assertion

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

Distinct variable groups:   𝑥,𝑧,𝑤   𝑦,𝑧,𝑤   𝑧,𝑛,𝑤   𝑧,𝑠,𝑝,𝑤

Proof of Theorem axprlem3

Step	Hyp	Ref	Expression
1		axrep4v 5244	. 2 ⊢ (∀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 1091	. . . . . . 7 ⊢ (∃𝑛 𝑛 ∈ 𝑠 → (if-(∃𝑛 𝑛 ∈ 𝑠, 𝑤 = 𝑥, 𝑤 = 𝑦) ↔ 𝑤 = 𝑥))
3	2	biimpd 232	. . . . . 6 ⊢ (∃𝑛 𝑛 ∈ 𝑠 → (if-(∃𝑛 𝑛 ∈ 𝑠, 𝑤 = 𝑥, 𝑤 = 𝑦) → 𝑤 = 𝑥))
4		equeuclr 2053	. . . . . 6 ⊢ (𝑧 = 𝑥 → (𝑤 = 𝑥 → 𝑤 = 𝑧))
5	3, 4	syl9r 79	. . . . 5 ⊢ (𝑧 = 𝑥 → (∃𝑛 𝑛 ∈ 𝑠 → (if-(∃𝑛 𝑛 ∈ 𝑠, 𝑤 = 𝑥, 𝑤 = 𝑦) → 𝑤 = 𝑧)))
6	5	alrimdv 1959	. . . 4 ⊢ (𝑧 = 𝑥 → (∃𝑛 𝑛 ∈ 𝑠 → ∀𝑤(if-(∃𝑛 𝑛 ∈ 𝑠, 𝑤 = 𝑥, 𝑤 = 𝑦) → 𝑤 = 𝑧)))
7	6	spimevw 2015	. . 3 ⊢ (∃𝑛 𝑛 ∈ 𝑠 → ∃𝑧∀𝑤(if-(∃𝑛 𝑛 ∈ 𝑠, 𝑤 = 𝑥, 𝑤 = 𝑦) → 𝑤 = 𝑧))
8		ifpfal 1092	. . . . . . 7 ⊢ (¬ ∃𝑛 𝑛 ∈ 𝑠 → (if-(∃𝑛 𝑛 ∈ 𝑠, 𝑤 = 𝑥, 𝑤 = 𝑦) ↔ 𝑤 = 𝑦))
9	8	biimpd 232	. . . . . 6 ⊢ (¬ ∃𝑛 𝑛 ∈ 𝑠 → (if-(∃𝑛 𝑛 ∈ 𝑠, 𝑤 = 𝑥, 𝑤 = 𝑦) → 𝑤 = 𝑦))
10		equeuclr 2053	. . . . . 6 ⊢ (𝑧 = 𝑦 → (𝑤 = 𝑦 → 𝑤 = 𝑧))
11	9, 10	syl9r 79	. . . . 5 ⊢ (𝑧 = 𝑦 → (¬ ∃𝑛 𝑛 ∈ 𝑠 → (if-(∃𝑛 𝑛 ∈ 𝑠, 𝑤 = 𝑥, 𝑤 = 𝑦) → 𝑤 = 𝑧)))
12	11	alrimdv 1959	. . . 4 ⊢ (𝑧 = 𝑦 → (¬ ∃𝑛 𝑛 ∈ 𝑠 → ∀𝑤(if-(∃𝑛 𝑛 ∈ 𝑠, 𝑤 = 𝑥, 𝑤 = 𝑦) → 𝑤 = 𝑧)))
13	12	spimevw 2015	. . 3 ⊢ (¬ ∃𝑛 𝑛 ∈ 𝑠 → ∃𝑧∀𝑤(if-(∃𝑛 𝑛 ∈ 𝑠, 𝑤 = 𝑥, 𝑤 = 𝑦) → 𝑤 = 𝑧))
14	7, 13	pm2.61i 184	. 2 ⊢ ∃𝑧∀𝑤(if-(∃𝑛 𝑛 ∈ 𝑠, 𝑤 = 𝑥, 𝑤 = 𝑦) → 𝑤 = 𝑧)
15	1, 14	mpg 1827	1 ⊢ ∃𝑧∀𝑤(𝑤 ∈ 𝑧 ↔ ∃𝑠(𝑠 ∈ 𝑝 ∧ if-(∃𝑛 𝑛 ∈ 𝑠, 𝑤 = 𝑥, 𝑤 = 𝑦)))

Syntax hints:  ¬ wn 3   → wi 4   ↔ wb 209   ∧ wa 400  if-wif 1078  ∀wal 1568  ∃wex 1809

This theorem was proved from axioms:  ax-mp 5  ax-1 6  ax-2 7  ax-3 8  ax-gen 1825  ax-4 1839  ax-5 1940  ax-6 1997  ax-7 2038  ax-rep 5239

This theorem depends on definitions:  df-bi 210  df-an 401  df-or 861  df-ifp 1079  df-ex 1810

This theorem is referenced by:  axpr  5400  axprALT2  35484