Metamath Proof Explorer


Theorem 1fv

Description: A function on a singleton. (Contributed by Alexander van der Vekens, 3-Dec-2017) (Proof shortened by AV, 18-Apr-2021)

Ref Expression
Assertion 1fv ⊢ N ∈ V ∧ P = 0 N → P : 0 … 0 ⟶ V ∧ P ⁡ 0 = N

Proof

Step Hyp Ref Expression
1 0z ⊢ 0 ∈ ℤ
2 1 a1i ⊢ N ∈ V → 0 ∈ ℤ
3 id ⊢ N ∈ V → N ∈ V
4 2 3 fsnd ⊢ N ∈ V → 0 N : 0 ⟶ V
5 fvsng ⊢ 0 ∈ ℤ ∧ N ∈ V → 0 N ⁡ 0 = N
6 1 5 mpan ⊢ N ∈ V → 0 N ⁡ 0 = N
7 4 6 jca ⊢ N ∈ V → 0 N : 0 ⟶ V ∧ 0 N ⁡ 0 = N
8 7 adantr ⊢ N ∈ V ∧ P = 0 N → 0 N : 0 ⟶ V ∧ 0 N ⁡ 0 = N
9 id ⊢ P = 0 N → P = 0 N
10 fz0sn ⊢ 0 … 0 = 0
11 10 a1i ⊢ P = 0 N → 0 … 0 = 0
12 9 11 feq12d ⊢ P = 0 N → P : 0 … 0 ⟶ V ↔ 0 N : 0 ⟶ V
13 fveq1 ⊢ P = 0 N → P ⁡ 0 = 0 N ⁡ 0
14 13 eqeq1d ⊢ P = 0 N → P ⁡ 0 = N ↔ 0 N ⁡ 0 = N
15 12 14 anbi12d ⊢ P = 0 N → P : 0 … 0 ⟶ V ∧ P ⁡ 0 = N ↔ 0 N : 0 ⟶ V ∧ 0 N ⁡ 0 = N
16 15 adantl ⊢ N ∈ V ∧ P = 0 N → P : 0 … 0 ⟶ V ∧ P ⁡ 0 = N ↔ 0 N : 0 ⟶ V ∧ 0 N ⁡ 0 = N
17 8 16 mpbird ⊢ N ∈ V ∧ P = 0 N → P : 0 … 0 ⟶ V ∧ P ⁡ 0 = N