Metamath Proof Explorer


Theorem dig0

Description: All digits of 0 are 0. (Contributed by AV, 24-May-2020)

Ref Expression
Assertion dig0 ⊢ B ∈ ℕ ∧ K ∈ ℤ → K digit ⁡ B 0 = 0

Proof

Step Hyp Ref Expression
1 0e0icopnf ⊢ 0 ∈ 0 +∞
2 digval ⊢ B ∈ ℕ ∧ K ∈ ℤ ∧ 0 ∈ 0 +∞ → K digit ⁡ B 0 = B − K ⋅ 0 mod B
3 1 2 mp3an3 ⊢ B ∈ ℕ ∧ K ∈ ℤ → K digit ⁡ B 0 = B − K ⋅ 0 mod B
4 nncn ⊢ B ∈ ℕ → B ∈ ℂ
5 4 adantr ⊢ B ∈ ℕ ∧ K ∈ ℤ → B ∈ ℂ
6 nnne0 ⊢ B ∈ ℕ → B ≠ 0
7 6 adantr ⊢ B ∈ ℕ ∧ K ∈ ℤ → B ≠ 0
8 znegcl ⊢ K ∈ ℤ → − K ∈ ℤ
9 8 adantl ⊢ B ∈ ℕ ∧ K ∈ ℤ → − K ∈ ℤ
10 5 7 9 expclzd ⊢ B ∈ ℕ ∧ K ∈ ℤ → B − K ∈ ℂ
11 10 mul01d ⊢ B ∈ ℕ ∧ K ∈ ℤ → B − K ⋅ 0 = 0
12 11 fveq2d ⊢ B ∈ ℕ ∧ K ∈ ℤ → B − K ⋅ 0 = 0
13 0zd ⊢ B ∈ ℕ ∧ K ∈ ℤ → 0 ∈ ℤ
14 flid ⊢ 0 ∈ ℤ → 0 = 0
15 13 14 syl ⊢ B ∈ ℕ ∧ K ∈ ℤ → 0 = 0
16 12 15 eqtrd ⊢ B ∈ ℕ ∧ K ∈ ℤ → B − K ⋅ 0 = 0
17 16 oveq1d ⊢ B ∈ ℕ ∧ K ∈ ℤ → B − K ⋅ 0 mod B = 0 mod B
18 nnrp ⊢ B ∈ ℕ → B ∈ ℝ +
19 0mod ⊢ B ∈ ℝ + → 0 mod B = 0
20 18 19 syl ⊢ B ∈ ℕ → 0 mod B = 0
21 20 adantr ⊢ B ∈ ℕ ∧ K ∈ ℤ → 0 mod B = 0
22 17 21 eqtrd ⊢ B ∈ ℕ ∧ K ∈ ℤ → B − K ⋅ 0 mod B = 0
23 3 22 eqtrd ⊢ B ∈ ℕ ∧ K ∈ ℤ → K digit ⁡ B 0 = 0