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Metamath Proof Explorer

Theorem axrep4v

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

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

Proof

Step Hyp Ref Expression
1 ax-rep ⊢ ( ∀ 𝑥 ∃ 𝑧 ∀ 𝑦 ( ∀ 𝑧 𝜑 → 𝑦 = 𝑧 ) → ∃ 𝑧 ∀ 𝑦 ( 𝑦 ∈ 𝑧 ↔ ∃ 𝑥 ( 𝑥 ∈ 𝑤 ∧ ∀ 𝑧 𝜑 ) ) )
2 19.3v ⊢ ( ∀ 𝑧 𝜑 ↔ 𝜑 )
3 2 imbi1i ⊢ ( ( ∀ 𝑧 𝜑 → 𝑦 = 𝑧 ) ↔ ( 𝜑 → 𝑦 = 𝑧 ) )
4 3 albii ⊢ ( ∀ 𝑦 ( ∀ 𝑧 𝜑 → 𝑦 = 𝑧 ) ↔ ∀ 𝑦 ( 𝜑 → 𝑦 = 𝑧 ) )
5 4 exbii ⊢ ( ∃ 𝑧 ∀ 𝑦 ( ∀ 𝑧 𝜑 → 𝑦 = 𝑧 ) ↔ ∃ 𝑧 ∀ 𝑦 ( 𝜑 → 𝑦 = 𝑧 ) )
6 5 albii ⊢ ( ∀ 𝑥 ∃ 𝑧 ∀ 𝑦 ( ∀ 𝑧 𝜑 → 𝑦 = 𝑧 ) ↔ ∀ 𝑥 ∃ 𝑧 ∀ 𝑦 ( 𝜑 → 𝑦 = 𝑧 ) )
7 2 anbi2i ⊢ ( ( 𝑥 ∈ 𝑤 ∧ ∀ 𝑧 𝜑 ) ↔ ( 𝑥 ∈ 𝑤 ∧ 𝜑 ) )
8 7 exbii ⊢ ( ∃ 𝑥 ( 𝑥 ∈ 𝑤 ∧ ∀ 𝑧 𝜑 ) ↔ ∃ 𝑥 ( 𝑥 ∈ 𝑤 ∧ 𝜑 ) )
9 8 bibi2i ⊢ ( ( 𝑦 ∈ 𝑧 ↔ ∃ 𝑥 ( 𝑥 ∈ 𝑤 ∧ ∀ 𝑧 𝜑 ) ) ↔ ( 𝑦 ∈ 𝑧 ↔ ∃ 𝑥 ( 𝑥 ∈ 𝑤 ∧ 𝜑 ) ) )
10 9 albii ⊢ ( ∀ 𝑦 ( 𝑦 ∈ 𝑧 ↔ ∃ 𝑥 ( 𝑥 ∈ 𝑤 ∧ ∀ 𝑧 𝜑 ) ) ↔ ∀ 𝑦 ( 𝑦 ∈ 𝑧 ↔ ∃ 𝑥 ( 𝑥 ∈ 𝑤 ∧ 𝜑 ) ) )
11 10 exbii ⊢ ( ∃ 𝑧 ∀ 𝑦 ( 𝑦 ∈ 𝑧 ↔ ∃ 𝑥 ( 𝑥 ∈ 𝑤 ∧ ∀ 𝑧 𝜑 ) ) ↔ ∃ 𝑧 ∀ 𝑦 ( 𝑦 ∈ 𝑧 ↔ ∃ 𝑥 ( 𝑥 ∈ 𝑤 ∧ 𝜑 ) ) )
12 1 6 11 3imtr3i ⊢ ( ∀ 𝑥 ∃ 𝑧 ∀ 𝑦 ( 𝜑 → 𝑦 = 𝑧 ) → ∃ 𝑧 ∀ 𝑦 ( 𝑦 ∈ 𝑧 ↔ ∃ 𝑥 ( 𝑥 ∈ 𝑤 ∧ 𝜑 ) ) )