Metamath Proof Explorer


Theorem etransclem9

Description: If K divides N but K does not divide M then M + N cannot be zero. (Contributed by Glauco Siliprandi, 5-Apr-2020)

Ref Expression
Hypotheses etransclem9.k ⊢ φ → K ∈ ℤ
etransclem9.kn0 ⊢ φ → K ≠ 0
etransclem9.m ⊢ φ → M ∈ ℤ
etransclem9.n ⊢ φ → N ∈ ℤ
etransclem9.km ⊢ φ → ¬ K ∥ M
etransclem9.kn ⊢ φ → K ∥ N
Assertion etransclem9 ⊢ φ → M + N ≠ 0

Proof

Step Hyp Ref Expression
1 etransclem9.k ⊢ φ → K ∈ ℤ
2 etransclem9.kn0 ⊢ φ → K ≠ 0
3 etransclem9.m ⊢ φ → M ∈ ℤ
4 etransclem9.n ⊢ φ → N ∈ ℤ
5 etransclem9.km ⊢ φ → ¬ K ∥ M
6 etransclem9.kn ⊢ φ → K ∥ N
7 dvdsval2 ⊢ K ∈ ℤ ∧ K ≠ 0 ∧ M ∈ ℤ → K ∥ M ↔ M K ∈ ℤ
8 1 2 3 7 syl3anc ⊢ φ → K ∥ M ↔ M K ∈ ℤ
9 5 8 mtbid ⊢ φ → ¬ M K ∈ ℤ
10 df-neg ⊢ − N = 0 − N
11 10 a1i ⊢ φ ∧ M + N = 0 → − N = 0 − N
12 oveq1 ⊢ M + N = 0 → M + N - N = 0 − N
13 12 eqcomd ⊢ M + N = 0 → 0 − N = M + N - N
14 13 adantl ⊢ φ ∧ M + N = 0 → 0 − N = M + N - N
15 3 zcnd ⊢ φ → M ∈ ℂ
16 4 zcnd ⊢ φ → N ∈ ℂ
17 15 16 pncand ⊢ φ → M + N - N = M
18 17 adantr ⊢ φ ∧ M + N = 0 → M + N - N = M
19 11 14 18 3eqtrrd ⊢ φ ∧ M + N = 0 → M = − N
20 19 oveq1d ⊢ φ ∧ M + N = 0 → M K = − N K
21 dvdsnegb ⊢ K ∈ ℤ ∧ N ∈ ℤ → K ∥ N ↔ K ∥ -N
22 1 4 21 syl2anc ⊢ φ → K ∥ N ↔ K ∥ -N
23 6 22 mpbid ⊢ φ → K ∥ -N
24 4 znegcld ⊢ φ → − N ∈ ℤ
25 dvdsval2 ⊢ K ∈ ℤ ∧ K ≠ 0 ∧ − N ∈ ℤ → K ∥ -N ↔ − N K ∈ ℤ
26 1 2 24 25 syl3anc ⊢ φ → K ∥ -N ↔ − N K ∈ ℤ
27 23 26 mpbid ⊢ φ → − N K ∈ ℤ
28 27 adantr ⊢ φ ∧ M + N = 0 → − N K ∈ ℤ
29 20 28 eqeltrd ⊢ φ ∧ M + N = 0 → M K ∈ ℤ
30 9 29 mtand ⊢ φ → ¬ M + N = 0
31 30 neqned ⊢ φ → M + N ≠ 0