Metamath Proof Explorer


Theorem pcfaclem

Description: Lemma for pcfac . (Contributed by Mario Carneiro, 20-May-2014)

Ref Expression
Assertion pcfaclem ⊢ N ∈ ℕ 0 ∧ M ∈ ℤ ≥ N ∧ P ∈ ℙ → N P M = 0

Proof

Step Hyp Ref Expression
1 nn0ge0 ⊢ N ∈ ℕ 0 → 0 ≤ N
2 1 3ad2ant1 ⊢ N ∈ ℕ 0 ∧ M ∈ ℤ ≥ N ∧ P ∈ ℙ → 0 ≤ N
3 nn0re ⊢ N ∈ ℕ 0 → N ∈ ℝ
4 3 3ad2ant1 ⊢ N ∈ ℕ 0 ∧ M ∈ ℤ ≥ N ∧ P ∈ ℙ → N ∈ ℝ
5 prmnn ⊢ P ∈ ℙ → P ∈ ℕ
6 5 3ad2ant3 ⊢ N ∈ ℕ 0 ∧ M ∈ ℤ ≥ N ∧ P ∈ ℙ → P ∈ ℕ
7 eluznn0 ⊢ N ∈ ℕ 0 ∧ M ∈ ℤ ≥ N → M ∈ ℕ 0
8 7 3adant3 ⊢ N ∈ ℕ 0 ∧ M ∈ ℤ ≥ N ∧ P ∈ ℙ → M ∈ ℕ 0
9 6 8 nnexpcld ⊢ N ∈ ℕ 0 ∧ M ∈ ℤ ≥ N ∧ P ∈ ℙ → P M ∈ ℕ
10 9 nnred ⊢ N ∈ ℕ 0 ∧ M ∈ ℤ ≥ N ∧ P ∈ ℙ → P M ∈ ℝ
11 9 nngt0d ⊢ N ∈ ℕ 0 ∧ M ∈ ℤ ≥ N ∧ P ∈ ℙ → 0 < P M
12 ge0div ⊢ N ∈ ℝ ∧ P M ∈ ℝ ∧ 0 < P M → 0 ≤ N ↔ 0 ≤ N P M
13 4 10 11 12 syl3anc ⊢ N ∈ ℕ 0 ∧ M ∈ ℤ ≥ N ∧ P ∈ ℙ → 0 ≤ N ↔ 0 ≤ N P M
14 2 13 mpbid ⊢ N ∈ ℕ 0 ∧ M ∈ ℤ ≥ N ∧ P ∈ ℙ → 0 ≤ N P M
15 8 nn0red ⊢ N ∈ ℕ 0 ∧ M ∈ ℤ ≥ N ∧ P ∈ ℙ → M ∈ ℝ
16 eluzle ⊢ M ∈ ℤ ≥ N → N ≤ M
17 16 3ad2ant2 ⊢ N ∈ ℕ 0 ∧ M ∈ ℤ ≥ N ∧ P ∈ ℙ → N ≤ M
18 prmuz2 ⊢ P ∈ ℙ → P ∈ ℤ ≥ 2
19 18 3ad2ant3 ⊢ N ∈ ℕ 0 ∧ M ∈ ℤ ≥ N ∧ P ∈ ℙ → P ∈ ℤ ≥ 2
20 bernneq3 ⊢ P ∈ ℤ ≥ 2 ∧ M ∈ ℕ 0 → M < P M
21 19 8 20 syl2anc ⊢ N ∈ ℕ 0 ∧ M ∈ ℤ ≥ N ∧ P ∈ ℙ → M < P M
22 4 15 10 17 21 lelttrd ⊢ N ∈ ℕ 0 ∧ M ∈ ℤ ≥ N ∧ P ∈ ℙ → N < P M
23 9 nncnd ⊢ N ∈ ℕ 0 ∧ M ∈ ℤ ≥ N ∧ P ∈ ℙ → P M ∈ ℂ
24 23 mulridd ⊢ N ∈ ℕ 0 ∧ M ∈ ℤ ≥ N ∧ P ∈ ℙ → P M ⋅ 1 = P M
25 22 24 breqtrrd ⊢ N ∈ ℕ 0 ∧ M ∈ ℤ ≥ N ∧ P ∈ ℙ → N < P M ⋅ 1
26 1red ⊢ N ∈ ℕ 0 ∧ M ∈ ℤ ≥ N ∧ P ∈ ℙ → 1 ∈ ℝ
27 ltdivmul ⊢ N ∈ ℝ ∧ 1 ∈ ℝ ∧ P M ∈ ℝ ∧ 0 < P M → N P M < 1 ↔ N < P M ⋅ 1
28 4 26 10 11 27 syl112anc ⊢ N ∈ ℕ 0 ∧ M ∈ ℤ ≥ N ∧ P ∈ ℙ → N P M < 1 ↔ N < P M ⋅ 1
29 25 28 mpbird ⊢ N ∈ ℕ 0 ∧ M ∈ ℤ ≥ N ∧ P ∈ ℙ → N P M < 1
30 0p1e1 ⊢ 0 + 1 = 1
31 29 30 breqtrrdi ⊢ N ∈ ℕ 0 ∧ M ∈ ℤ ≥ N ∧ P ∈ ℙ → N P M < 0 + 1
32 4 9 nndivred ⊢ N ∈ ℕ 0 ∧ M ∈ ℤ ≥ N ∧ P ∈ ℙ → N P M ∈ ℝ
33 0z ⊢ 0 ∈ ℤ
34 flbi ⊢ N P M ∈ ℝ ∧ 0 ∈ ℤ → N P M = 0 ↔ 0 ≤ N P M ∧ N P M < 0 + 1
35 32 33 34 sylancl ⊢ N ∈ ℕ 0 ∧ M ∈ ℤ ≥ N ∧ P ∈ ℙ → N P M = 0 ↔ 0 ≤ N P M ∧ N P M < 0 + 1
36 14 31 35 mpbir2and ⊢ N ∈ ℕ 0 ∧ M ∈ ℤ ≥ N ∧ P ∈ ℙ → N P M = 0