Metamath Proof Explorer


Theorem lcmineqlem15

Description: F times the least common multiple of 1 to n is a natural number. (Contributed by metakunt, 10-May-2024)

Ref Expression
Hypotheses lcmineqlem15.1 ⊢ F = ∫ 0 1 x M − 1 ⁢ 1 − x N − M dx
lcmineqlem15.2 ⊢ φ → N ∈ ℕ
lcmineqlem15.3 ⊢ φ → M ∈ ℕ
lcmineqlem15.4 ⊢ φ → M ≤ N
Assertion lcmineqlem15 ⊢ φ → lcm _ ⁡ 1 … N ⁢ F ∈ ℕ

Proof

Step Hyp Ref Expression
1 lcmineqlem15.1 ⊢ F = ∫ 0 1 x M − 1 ⁢ 1 − x N − M dx
2 lcmineqlem15.2 ⊢ φ → N ∈ ℕ
3 lcmineqlem15.3 ⊢ φ → M ∈ ℕ
4 lcmineqlem15.4 ⊢ φ → M ≤ N
5 1 2 3 4 lcmineqlem6 ⊢ φ → lcm _ ⁡ 1 … N ⁢ F ∈ ℤ
6 fz1ssnn ⊢ 1 … N ⊆ ℕ
7 fzfi ⊢ 1 … N ∈ Fin
8 lcmfnncl ⊢ 1 … N ⊆ ℕ ∧ 1 … N ∈ Fin → lcm _ ⁡ 1 … N ∈ ℕ
9 6 7 8 mp2an ⊢ lcm _ ⁡ 1 … N ∈ ℕ
10 9 a1i ⊢ φ → lcm _ ⁡ 1 … N ∈ ℕ
11 10 nnred ⊢ φ → lcm _ ⁡ 1 … N ∈ ℝ
12 1 3 2 4 lcmineqlem13 ⊢ φ → F = 1 M ⁢ ( N M)
13 1red ⊢ φ → 1 ∈ ℝ
14 3 nnnn0d ⊢ φ → M ∈ ℕ 0
15 2 14 4 bccl2d ⊢ φ → ( N M) ∈ ℕ
16 3 15 nnmulcld ⊢ φ → M ⁢ ( N M) ∈ ℕ
17 16 nnred ⊢ φ → M ⁢ ( N M) ∈ ℝ
18 16 nnne0d ⊢ φ → M ⁢ ( N M) ≠ 0
19 13 17 18 redivcld ⊢ φ → 1 M ⁢ ( N M) ∈ ℝ
20 12 19 eqeltrd ⊢ φ → F ∈ ℝ
21 10 nngt0d ⊢ φ → 0 < lcm _ ⁡ 1 … N
22 nnrecgt0 ⊢ M ⁢ ( N M) ∈ ℕ → 0 < 1 M ⁢ ( N M)
23 16 22 syl ⊢ φ → 0 < 1 M ⁢ ( N M)
24 23 12 breqtrrd ⊢ φ → 0 < F
25 11 20 21 24 mulgt0d ⊢ φ → 0 < lcm _ ⁡ 1 … N ⁢ F
26 elnnz ⊢ lcm _ ⁡ 1 … N ⁢ F ∈ ℕ ↔ lcm _ ⁡ 1 … N ⁢ F ∈ ℤ ∧ 0 < lcm _ ⁡ 1 … N ⁢ F
27 5 25 26 sylanbrc ⊢ φ → lcm _ ⁡ 1 … N ⁢ F ∈ ℕ