Metamath Proof Explorer


Theorem mof0

Description: There is at most one function into the empty set. (Contributed by Zhi Wang, 19-Sep-2024)

Ref Expression
Assertion mof0 ⊢ ∃* f f : A ⟶ ∅

Proof

Step Hyp Ref Expression
1 0ex ⊢ ∅ ∈ V
2 eqeq2 ⊢ g = ∅ → f = g ↔ f = ∅
3 2 imbi2d ⊢ g = ∅ → f : A ⟶ ∅ → f = g ↔ f : A ⟶ ∅ → f = ∅
4 3 albidv ⊢ g = ∅ → ∀ f f : A ⟶ ∅ → f = g ↔ ∀ f f : A ⟶ ∅ → f = ∅
5 1 4 spcev ⊢ ∀ f f : A ⟶ ∅ → f = ∅ → ∃ g ∀ f f : A ⟶ ∅ → f = g
6 f00 ⊢ f : A ⟶ ∅ ↔ f = ∅ ∧ A = ∅
7 6 simplbi ⊢ f : A ⟶ ∅ → f = ∅
8 5 7 mpg ⊢ ∃ g ∀ f f : A ⟶ ∅ → f = g
9 dfmo ⊢ ∃* f f : A ⟶ ∅ ↔ ∃ g ∀ f f : A ⟶ ∅ → f = g
10 8 9 mpbir ⊢ ∃* f f : A ⟶ ∅