Metamath Proof Explorer


Theorem rneqdmfinf1o

Description: Any function from a finite set onto the same set must be a bijection. (Contributed by AV, 5-Jul-2021)

Ref Expression
Assertion rneqdmfinf1o ⊢ A ∈ Fin ∧ F Fn A ∧ ran ⁡ F = A → F : A ⟶ 1-1 onto A

Proof

Step Hyp Ref Expression
1 dffn4 ⊢ F Fn A ↔ F : A ⟶ onto ran ⁡ F
2 1 biimpi ⊢ F Fn A → F : A ⟶ onto ran ⁡ F
3 2 3ad2ant2 ⊢ A ∈ Fin ∧ F Fn A ∧ ran ⁡ F = A → F : A ⟶ onto ran ⁡ F
4 foeq3 ⊢ ran ⁡ F = A → F : A ⟶ onto ran ⁡ F ↔ F : A ⟶ onto A
5 4 3ad2ant3 ⊢ A ∈ Fin ∧ F Fn A ∧ ran ⁡ F = A → F : A ⟶ onto ran ⁡ F ↔ F : A ⟶ onto A
6 3 5 mpbid ⊢ A ∈ Fin ∧ F Fn A ∧ ran ⁡ F = A → F : A ⟶ onto A
7 enrefg ⊢ A ∈ Fin → A ≈ A
8 7 3ad2ant1 ⊢ A ∈ Fin ∧ F Fn A ∧ ran ⁡ F = A → A ≈ A
9 simp1 ⊢ A ∈ Fin ∧ F Fn A ∧ ran ⁡ F = A → A ∈ Fin
10 fofinf1o ⊢ F : A ⟶ onto A ∧ A ≈ A ∧ A ∈ Fin → F : A ⟶ 1-1 onto A
11 6 8 9 10 syl3anc ⊢ A ∈ Fin ∧ F Fn A ∧ ran ⁡ F = A → F : A ⟶ 1-1 onto A