Metamath Proof Explorer


Theorem 0mod

Description: Special case: 0 modulo a positive real number is 0. (Contributed by Mario Carneiro, 22-Feb-2014)

Ref Expression
Assertion 0mod ( 𝑁 ∈ ℝ+ → ( 0 mod 𝑁 ) = 0 )

Proof

Step Hyp Ref Expression
1 0re 0 ∈ ℝ
2 1 jctl ( 𝑁 ∈ ℝ+ → ( 0 ∈ ℝ ∧ 𝑁 ∈ ℝ+ ) )
3 rpgt0 ( 𝑁 ∈ ℝ+ → 0 < 𝑁 )
4 0le0 0 ≤ 0
5 3 4 jctil ( 𝑁 ∈ ℝ+ → ( 0 ≤ 0 ∧ 0 < 𝑁 ) )
6 modid ( ( ( 0 ∈ ℝ ∧ 𝑁 ∈ ℝ+ ) ∧ ( 0 ≤ 0 ∧ 0 < 𝑁 ) ) → ( 0 mod 𝑁 ) = 0 )
7 2 5 6 syl2anc ( 𝑁 ∈ ℝ+ → ( 0 mod 𝑁 ) = 0 )