Metamath Proof Explorer


Theorem mof

Description: Version of df-mo with disjoint variable condition replaced by nonfreeness hypothesis. (Contributed by NM, 8-Mar-1995) Extract dfmo from this proof, and prove mof from it (as of 30-Sep-2022, directly from df-mo ). (Revised by Wolf Lammen, 28-May-2019) Avoid ax-13 . (Revised by Wolf Lammen, 16-Oct-2022)

Ref Expression
Hypothesis mof.1 ⊢ Ⅎ y φ
Assertion mof ⊢ ∃* x φ ↔ ∃ y ∀ x φ → x = y

Proof

Step Hyp Ref Expression
1 mof.1 ⊢ Ⅎ y φ
2 dfmo ⊢ ∃* x φ ↔ ∃ z ∀ x φ → x = z
3 nfv ⊢ Ⅎ y x = z
4 1 3 nfim ⊢ 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 imbi2d ⊢ 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