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 e. NN0
2 2nn
 |-  2 e. NN
3 1 2 decnncl
 |-  ; 1 2 e. NN
4 5nn
 |-  5 e. NN
5 1nn
 |-  1 e. NN
6 6nn
 |-  6 e. NN
7 6 decnncl2
 |-  ; 6 0 e. NN
8 12gcd5e1
 |-  ( ; 1 2 gcd 5 ) = 1
9 6nn0
 |-  6 e. NN0
10 0nn0
 |-  0 e. NN0
11 9 10 deccl
 |-  ; 6 0 e. NN0
12 11 nn0cni
 |-  ; 6 0 e. CC
13 12 mulid2i
 |-  ( 1 x. ; 6 0 ) = ; 6 0
14 5nn0
 |-  5 e. NN0
15 2nn0
 |-  2 e. NN0
16 eqid
 |-  ; 1 2 = ; 1 2
17 5cn
 |-  5 e. CC
18 17 mulid2i
 |-  ( 1 x. 5 ) = 5
19 18 oveq1i
 |-  ( ( 1 x. 5 ) + 1 ) = ( 5 + 1 )
20 5p1e6
 |-  ( 5 + 1 ) = 6
21 19 20 eqtri
 |-  ( ( 1 x. 5 ) + 1 ) = 6
22 2cn
 |-  2 e. CC
23 5t2e10
 |-  ( 5 x. 2 ) = ; 1 0
24 17 22 23 mulcomli
 |-  ( 2 x. 5 ) = ; 1 0
25 14 1 15 16 10 1 21 24 decmul1c
 |-  ( ; 1 2 x. 5 ) = ; 6 0
26 3 4 5 7 8 13 25 lcmeprodgcdi
 |-  ( ; 1 2 lcm 5 ) = ; 6 0