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
2 9nn 9
3 1 2 nnmulcli 6 9
4 3 nncni 6 9
5 1 nnzi 6
6 2 nnzi 9
7 5 6 pm3.2i 6 9
8 lcmcl 6 9 6 lcm 9 0
9 8 nn0cnd 6 9 6 lcm 9
10 7 9 ax-mp 6 lcm 9
11 neggcd 6 9 -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
17 15 16 eqeltri 6 gcd 9
18 3ne0 3 0
19 15 18 eqnetri 6 gcd 9 0
20 17 19 pm3.2i 6 gcd 9 6 gcd 9 0
21 1 2 pm3.2i 6 9
22 lcmgcdnn 6 9 6 lcm 9 6 gcd 9 = 6 9
23 21 22 mp1i 6 9 6 lcm 9 6 gcd 9 6 gcd 9 0 6 lcm 9 6 gcd 9 = 6 9
24 23 eqcomd 6 9 6 lcm 9 6 gcd 9 6 gcd 9 0 6 9 = 6 lcm 9 6 gcd 9
25 divmul3 6 9 6 lcm 9 6 gcd 9 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 6 lcm 9 6 gcd 9 6 gcd 9 0 6 9 6 gcd 9 = 6 lcm 9
27 26 eqcomd 6 9 6 lcm 9 6 gcd 9 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
31 9cn 9
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