Theorem axrep4v 5237

Description: Version of axrep4 5238 with a disjoint variable condition, requiring fewer axioms. (Contributed by Matthew House, 18-Sep-2025.)

Assertion

Ref	Expression
axrep4v	⊢ (∀𝑥∃𝑧∀𝑦(𝜑 → 𝑦 = 𝑧) → ∃𝑧∀𝑦(𝑦 ∈ 𝑧 ↔ ∃𝑥(𝑥 ∈ 𝑤 ∧ 𝜑)))

Distinct variable groups:   𝑥,𝑤,𝑦,𝑧   𝜑,𝑧

Allowed substitution hints:   𝜑(𝑥, 𝑦, 𝑤)

Proof of Theorem axrep4v

Step	Hyp	Ref	Expression
1		ax-rep 5232	. 2 ⊢ (∀𝑥∃𝑧∀𝑦(∀𝑧𝜑 → 𝑦 = 𝑧) → ∃𝑧∀𝑦(𝑦 ∈ 𝑧 ↔ ∃𝑥(𝑥 ∈ 𝑤 ∧ ∀𝑧𝜑)))
2		19.3v 2015	. . . . . 6 ⊢ (∀𝑧𝜑 ↔ 𝜑)
3	2	imbi1i 352	. . . . 5 ⊢ ((∀𝑧𝜑 → 𝑦 = 𝑧) ↔ (𝜑 → 𝑦 = 𝑧))
4	3	albii 1852	. . . 4 ⊢ (∀𝑦(∀𝑧𝜑 → 𝑦 = 𝑧) ↔ ∀𝑦(𝜑 → 𝑦 = 𝑧))
5	4	exbii 1881	. . 3 ⊢ (∃𝑧∀𝑦(∀𝑧𝜑 → 𝑦 = 𝑧) ↔ ∃𝑧∀𝑦(𝜑 → 𝑦 = 𝑧))
6	5	albii 1852	. 2 ⊢ (∀𝑥∃𝑧∀𝑦(∀𝑧𝜑 → 𝑦 = 𝑧) ↔ ∀𝑥∃𝑧∀𝑦(𝜑 → 𝑦 = 𝑧))
7	2	anbi2i 635	. . . . . 6 ⊢ ((𝑥 ∈ 𝑤 ∧ ∀𝑧𝜑) ↔ (𝑥 ∈ 𝑤 ∧ 𝜑))
8	7	exbii 1881	. . . . 5 ⊢ (∃𝑥(𝑥 ∈ 𝑤 ∧ ∀𝑧𝜑) ↔ ∃𝑥(𝑥 ∈ 𝑤 ∧ 𝜑))
9	8	bibi2i 340	. . . 4 ⊢ ((𝑦 ∈ 𝑧 ↔ ∃𝑥(𝑥 ∈ 𝑤 ∧ ∀𝑧𝜑)) ↔ (𝑦 ∈ 𝑧 ↔ ∃𝑥(𝑥 ∈ 𝑤 ∧ 𝜑)))
10	9	albii 1852	. . 3 ⊢ (∀𝑦(𝑦 ∈ 𝑧 ↔ ∃𝑥(𝑥 ∈ 𝑤 ∧ ∀𝑧𝜑)) ↔ ∀𝑦(𝑦 ∈ 𝑧 ↔ ∃𝑥(𝑥 ∈ 𝑤 ∧ 𝜑)))
11	10	exbii 1881	. 2 ⊢ (∃𝑧∀𝑦(𝑦 ∈ 𝑧 ↔ ∃𝑥(𝑥 ∈ 𝑤 ∧ ∀𝑧𝜑)) ↔ ∃𝑧∀𝑦(𝑦 ∈ 𝑧 ↔ ∃𝑥(𝑥 ∈ 𝑤 ∧ 𝜑)))
12	1, 6, 11	3imtr3i 294	1 ⊢ (∀𝑥∃𝑧∀𝑦(𝜑 → 𝑦 = 𝑧) → ∃𝑧∀𝑦(𝑦 ∈ 𝑧 ↔ ∃𝑥(𝑥 ∈ 𝑤 ∧ 𝜑)))

This proof depends on syntax axioms:   → wi 4   ↔ wb 209   ∧ wa 401  ∀wal 1568  ∃wex 1812

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-6 2000  ax-rep 5232

This proof depends on definitions:  df-bi 210  df-an 402  df-ex 1813

This theorem is used by:  axrep6  5240  axprlem3  5387