Metamath Proof Explorer


Theorem dvds0

Description: Any integer divides 0. Theorem 1.1(g) in ApostolNT p. 14. (Contributed by Paul Chapman, 21-Mar-2011)

Ref Expression
Assertion dvds0 ( 𝑁 ∈ ℤ → 𝑁 ∥ 0 )

Proof

Step Hyp Ref Expression
1 zcn ( 𝑁 ∈ ℤ → 𝑁 ∈ ℂ )
2 1 mul02d ( 𝑁 ∈ ℤ → ( 0 · 𝑁 ) = 0 )
3 0z 0 ∈ ℤ
4 dvds0lem ( ( ( 0 ∈ ℤ ∧ 𝑁 ∈ ℤ ∧ 0 ∈ ℤ ) ∧ ( 0 · 𝑁 ) = 0 ) → 𝑁 ∥ 0 )
5 4 ex ( ( 0 ∈ ℤ ∧ 𝑁 ∈ ℤ ∧ 0 ∈ ℤ ) → ( ( 0 · 𝑁 ) = 0 → 𝑁 ∥ 0 ) )
6 3 3 5 mp3an13 ( 𝑁 ∈ ℤ → ( ( 0 · 𝑁 ) = 0 → 𝑁 ∥ 0 ) )
7 2 6 mpd ( 𝑁 ∈ ℤ → 𝑁 ∥ 0 )