Metamath Proof Explorer


Theorem ex-lcm

Description: Example for df-lcm . (Contributed by AV, 5-Sep-2021)

Ref Expression
Assertion ex-lcm ⊢ 6 lcm 9 = 18

Proof

Step Hyp Ref Expression
1 6nn ⊢ 6 ∈ ℕ
2 9nn ⊢ 9 ∈ ℕ
3 1 2 nnmulcli ⊢ 6 ⋅ 9 ∈ ℕ
4 3 nncni ⊢ 6 ⋅ 9 ∈ ℂ
5 1 nnzi ⊢ 6 ∈ ℤ
6 2 nnzi ⊢ 9 ∈ ℤ
7 5 6 pm3.2i ⊢ 6 ∈ ℤ ∧ 9 ∈ ℤ
8 lcmcl ⊢ 6 ∈ ℤ ∧ 9 ∈ ℤ → 6 lcm 9 ∈ ℕ 0
9 8 nn0cnd ⊢ 6 ∈ ℤ ∧ 9 ∈ ℤ → 6 lcm 9 ∈ ℂ
10 7 9 ax-mp ⊢ 6 lcm 9 ∈ ℂ
11 neggcd ⊢ 6 ∈ ℤ ∧ 9 ∈ ℤ → -6 gcd 9 = 6 gcd 9
12 7 11 ax-mp ⊢ -6 gcd 9 = 6 gcd 9
13 12 eqcomi ⊢ 6 gcd 9 = -6 gcd 9
14 ex-gcd ⊢ -6 gcd 9 = 3
15 13 14 eqtri ⊢ 6 gcd 9 = 3
16 3cn ⊢ 3 ∈ ℂ
17 15 16 eqeltri ⊢ 6 gcd 9 ∈ ℂ
18 3ne0 ⊢ 3 ≠ 0
19 15 18 eqnetri ⊢ 6 gcd 9 ≠ 0
20 17 19 pm3.2i ⊢ 6 gcd 9 ∈ ℂ ∧ 6 gcd 9 ≠ 0
21 1 2 pm3.2i ⊢ 6 ∈ ℕ ∧ 9 ∈ ℕ
22 lcmgcdnn ⊢ 6 ∈ ℕ ∧ 9 ∈ ℕ → 6 lcm 9 ⁢ 6 gcd 9 = 6 ⋅ 9
23 21 22 mp1i ⊢ 6 ⋅ 9 ∈ ℂ ∧ 6 lcm 9 ∈ ℂ ∧ 6 gcd 9 ∈ ℂ ∧ 6 gcd 9 ≠ 0 → 6 lcm 9 ⁢ 6 gcd 9 = 6 ⋅ 9
24 23 eqcomd ⊢ 6 ⋅ 9 ∈ ℂ ∧ 6 lcm 9 ∈ ℂ ∧ 6 gcd 9 ∈ ℂ ∧ 6 gcd 9 ≠ 0 → 6 ⋅ 9 = 6 lcm 9 ⁢ 6 gcd 9
25 divmul3 ⊢ 6 ⋅ 9 ∈ ℂ ∧ 6 lcm 9 ∈ ℂ ∧ 6 gcd 9 ∈ ℂ ∧ 6 gcd 9 ≠ 0 → 6 ⋅ 9 6 gcd 9 = 6 lcm 9 ↔ 6 ⋅ 9 = 6 lcm 9 ⁢ 6 gcd 9
26 24 25 mpbird ⊢ 6 ⋅ 9 ∈ ℂ ∧ 6 lcm 9 ∈ ℂ ∧ 6 gcd 9 ∈ ℂ ∧ 6 gcd 9 ≠ 0 → 6 ⋅ 9 6 gcd 9 = 6 lcm 9
27 26 eqcomd ⊢ 6 ⋅ 9 ∈ ℂ ∧ 6 lcm 9 ∈ ℂ ∧ 6 gcd 9 ∈ ℂ ∧ 6 gcd 9 ≠ 0 → 6 lcm 9 = 6 ⋅ 9 6 gcd 9
28 4 10 20 27 mp3an ⊢ 6 lcm 9 = 6 ⋅ 9 6 gcd 9
29 15 oveq2i ⊢ 6 ⋅ 9 6 gcd 9 = 6 ⋅ 9 3
30 6cn ⊢ 6 ∈ ℂ
31 9cn ⊢ 9 ∈ ℂ
32 30 31 16 18 divassi ⊢ 6 ⋅ 9 3 = 6 ⁢ 9 3
33 3t3e9 ⊢ 3 ⋅ 3 = 9
34 33 eqcomi ⊢ 9 = 3 ⋅ 3
35 34 oveq1i ⊢ 9 3 = 3 ⋅ 3 3
36 16 16 18 divcan3i ⊢ 3 ⋅ 3 3 = 3
37 35 36 eqtri ⊢ 9 3 = 3
38 37 oveq2i ⊢ 6 ⁢ 9 3 = 6 ⋅ 3
39 6t3e18 ⊢ 6 ⋅ 3 = 18
40 32 38 39 3eqtri ⊢ 6 ⋅ 9 3 = 18
41 28 29 40 3eqtri ⊢ 6 lcm 9 = 18