Metamath Proof Explorer


Theorem muldvds2

Description: If a product divides an integer, so does one of its factors. (Contributed by Paul Chapman, 21-Mar-2011)

Ref Expression
Assertion muldvds2 ⊢ K ∈ ℤ ∧ M ∈ ℤ ∧ N ∈ ℤ → K ⋅ M ∥ N → M ∥ N

Proof

Step Hyp Ref Expression
1 zmulcl ⊢ K ∈ ℤ ∧ M ∈ ℤ → K ⋅ M ∈ ℤ
2 1 anim1i ⊢ K ∈ ℤ ∧ M ∈ ℤ ∧ N ∈ ℤ → K ⋅ M ∈ ℤ ∧ N ∈ ℤ
3 2 3impa ⊢ K ∈ ℤ ∧ M ∈ ℤ ∧ N ∈ ℤ → K ⋅ M ∈ ℤ ∧ N ∈ ℤ
4 3simpc ⊢ K ∈ ℤ ∧ M ∈ ℤ ∧ N ∈ ℤ → M ∈ ℤ ∧ N ∈ ℤ
5 zmulcl ⊢ x ∈ ℤ ∧ K ∈ ℤ → x ⁢ K ∈ ℤ
6 5 ancoms ⊢ K ∈ ℤ ∧ x ∈ ℤ → x ⁢ K ∈ ℤ
7 6 3ad2antl1 ⊢ K ∈ ℤ ∧ M ∈ ℤ ∧ N ∈ ℤ ∧ x ∈ ℤ → x ⁢ K ∈ ℤ
8 zcn ⊢ x ∈ ℤ → x ∈ ℂ
9 zcn ⊢ K ∈ ℤ → K ∈ ℂ
10 zcn ⊢ M ∈ ℤ → M ∈ ℂ
11 mulass ⊢ x ∈ ℂ ∧ K ∈ ℂ ∧ M ∈ ℂ → x ⁢ K ⋅ M = x ⁢ K ⋅ M
12 8 9 10 11 syl3an ⊢ x ∈ ℤ ∧ K ∈ ℤ ∧ M ∈ ℤ → x ⁢ K ⋅ M = x ⁢ K ⋅ M
13 12 3coml ⊢ K ∈ ℤ ∧ M ∈ ℤ ∧ x ∈ ℤ → x ⁢ K ⋅ M = x ⁢ K ⋅ M
14 13 3expa ⊢ K ∈ ℤ ∧ M ∈ ℤ ∧ x ∈ ℤ → x ⁢ K ⋅ M = x ⁢ K ⋅ M
15 14 3adantl3 ⊢ K ∈ ℤ ∧ M ∈ ℤ ∧ N ∈ ℤ ∧ x ∈ ℤ → x ⁢ K ⋅ M = x ⁢ K ⋅ M
16 15 eqeq1d ⊢ K ∈ ℤ ∧ M ∈ ℤ ∧ N ∈ ℤ ∧ x ∈ ℤ → x ⁢ K ⋅ M = N ↔ x ⁢ K ⋅ M = N
17 16 biimprd ⊢ K ∈ ℤ ∧ M ∈ ℤ ∧ N ∈ ℤ ∧ x ∈ ℤ → x ⁢ K ⋅ M = N → x ⁢ K ⋅ M = N
18 3 4 7 17 dvds1lem ⊢ K ∈ ℤ ∧ M ∈ ℤ ∧ N ∈ ℤ → K ⋅ M ∥ N → M ∥ N