Metamath Proof Explorer


Theorem f102g

Description: A function that maps the empty set to a class is injective. (Contributed by Zhi Wang, 1-Oct-2024)

Ref Expression
Assertion f102g ⊢ A = ∅ ∧ F : A ⟶ B → F : A ⟶ 1-1 B

Proof

Step Hyp Ref Expression
1 feq2 ⊢ A = ∅ → F : A ⟶ B ↔ F : ∅ ⟶ B
2 1 biimpa ⊢ A = ∅ ∧ F : A ⟶ B → F : ∅ ⟶ B
3 f0bi ⊢ F : ∅ ⟶ B ↔ F = ∅
4 f10 ⊢ ∅ : ∅ ⟶ 1-1 B
5 f1eq1 ⊢ F = ∅ → F : ∅ ⟶ 1-1 B ↔ ∅ : ∅ ⟶ 1-1 B
6 4 5 mpbiri ⊢ F = ∅ → F : ∅ ⟶ 1-1 B
7 3 6 sylbi ⊢ F : ∅ ⟶ B → F : ∅ ⟶ 1-1 B
8 2 7 syl ⊢ A = ∅ ∧ F : A ⟶ B → F : ∅ ⟶ 1-1 B
9 f1eq2 ⊢ A = ∅ → F : A ⟶ 1-1 B ↔ F : ∅ ⟶ 1-1 B
10 9 adantr ⊢ A = ∅ ∧ F : A ⟶ B → F : A ⟶ 1-1 B ↔ F : ∅ ⟶ 1-1 B
11 8 10 mpbird ⊢ A = ∅ ∧ F : A ⟶ B → F : A ⟶ 1-1 B