Metamath Proof Explorer


Theorem euf

Description: Version of eu6 with disjoint variable condition replaced by nonfreeness hypothesis. (Contributed by NM, 12-Aug-1993) (Proof shortened by Wolf Lammen, 30-Oct-2018) Avoid ax-13 . (Revised by Wolf Lammen, 16-Oct-2022)

Ref Expression
Hypothesis euf.1 ⊢ Ⅎ 𝑦 𝜑
Assertion euf ( ∃! 𝑥 𝜑 ↔ ∃ 𝑦 ∀ 𝑥 ( 𝜑 ↔ 𝑥 = 𝑦 ) )

Proof

Step Hyp Ref Expression
1 euf.1 ⊢ Ⅎ 𝑦 𝜑
2 eu6 ⊢ ( ∃! 𝑥 𝜑 ↔ ∃ 𝑧 ∀ 𝑥 ( 𝜑 ↔ 𝑥 = 𝑧 ) )
3 nfv ⊢ Ⅎ 𝑦 𝑥 = 𝑧
4 1 3 nfbi ⊢ Ⅎ 𝑦 ( 𝜑 ↔ 𝑥 = 𝑧 )
5 4 nfal ⊢ Ⅎ 𝑦 ∀ 𝑥 ( 𝜑 ↔ 𝑥 = 𝑧 )
6 nfv ⊢ Ⅎ 𝑧 ∀ 𝑥 ( 𝜑 ↔ 𝑥 = 𝑦 )
7 equequ2 ⊢ ( 𝑧 = 𝑦 → ( 𝑥 = 𝑧 ↔ 𝑥 = 𝑦 ) )
8 7 bibi2d ⊢ ( 𝑧 = 𝑦 → ( ( 𝜑 ↔ 𝑥 = 𝑧 ) ↔ ( 𝜑 ↔ 𝑥 = 𝑦 ) ) )
9 8 albidv ⊢ ( 𝑧 = 𝑦 → ( ∀ 𝑥 ( 𝜑 ↔ 𝑥 = 𝑧 ) ↔ ∀ 𝑥 ( 𝜑 ↔ 𝑥 = 𝑦 ) ) )
10 5 6 9 cbvexv1 ⊢ ( ∃ 𝑧 ∀ 𝑥 ( 𝜑 ↔ 𝑥 = 𝑧 ) ↔ ∃ 𝑦 ∀ 𝑥 ( 𝜑 ↔ 𝑥 = 𝑦 ) )
11 2 10 bitri ⊢ ( ∃! 𝑥 𝜑 ↔ ∃ 𝑦 ∀ 𝑥 ( 𝜑 ↔ 𝑥 = 𝑦 ) )