Metamath Proof Explorer


Theorem 12lcm5e60

Description: The lcm of 12 and 5 is 60. (Contributed by metakunt, 25-Apr-2024)

Ref Expression
Assertion 12lcm5e60 ( 1 2 lcm 5 ) = 6 0

Proof

Step Hyp Ref Expression
1 1nn0 1 ∈ ℕ0
2 2nn 2 ∈ ℕ
3 1 2 decnncl 1 2 ∈ ℕ
4 5nn 5 ∈ ℕ
5 1nn 1 ∈ ℕ
6 6nn 6 ∈ ℕ
7 6 decnncl2 6 0 ∈ ℕ
8 12gcd5e1 ( 1 2 gcd 5 ) = 1
9 6nn0 6 ∈ ℕ0
10 0nn0 0 ∈ ℕ0
11 9 10 deccl 6 0 ∈ ℕ0
12 11 nn0cni 6 0 ∈ ℂ
13 12 mulid2i ( 1 · 6 0 ) = 6 0
14 5nn0 5 ∈ ℕ0
15 2nn0 2 ∈ ℕ0
16 eqid 1 2 = 1 2
17 5cn 5 ∈ ℂ
18 17 mulid2i ( 1 · 5 ) = 5
19 18 oveq1i ( ( 1 · 5 ) + 1 ) = ( 5 + 1 )
20 5p1e6 ( 5 + 1 ) = 6
21 19 20 eqtri ( ( 1 · 5 ) + 1 ) = 6
22 2cn 2 ∈ ℂ
23 5t2e10 ( 5 · 2 ) = 1 0
24 17 22 23 mulcomli ( 2 · 5 ) = 1 0
25 14 1 15 16 10 1 21 24 decmul1c ( 1 2 · 5 ) = 6 0
26 3 4 5 7 8 13 25 lcmeprodgcdi ( 1 2 lcm 5 ) = 6 0