Metamath Proof Explorer


Theorem 12gcd5e1

Description: The gcd of 12 and 5 is 1. (Contributed by metakunt, 25-Apr-2024)

Ref Expression
Assertion 12gcd5e1 12gcd5=1

Proof

Step Hyp Ref Expression
1 2lt5 2<5
2 1 olci 5<22<5
3 5re 5
4 2re 2
5 lttri2 52525<22<5
6 3 4 5 mp2an 525<22<5
7 2 6 mpbir 52
8 5prm 5
9 2prm 2
10 prmrp 525gcd2=152
11 8 9 10 mp2an 5gcd2=152
12 7 11 mpbir 5gcd2=1
13 5nn 5
14 2nn 2
15 14 nnzi 2
16 13 14 15 gcdaddmzz2nncomi 5gcd2=5gcd25+2
17 13 14 mulcomnni 52=25
18 5t2e10 52=10
19 17 18 eqtr3i 25=10
20 19 oveq1i 25+2=10+2
21 1nn0 10
22 0nn0 00
23 14 nnnn0i 20
24 eqid 10=10
25 23 dec0h 2=02
26 1p0e1 1+0=1
27 2cn 2
28 27 addlidi 0+2=2
29 21 22 22 23 24 25 26 28 decadd 10+2=12
30 20 29 eqtri 25+2=12
31 30 oveq2i 5gcd25+2=5gcd12
32 16 31 eqtri 5gcd2=5gcd12
33 12 32 eqtr3i 1=5gcd12
34 21 14 decnncl 12
35 13 34 gcdcomnni 5gcd12=12gcd5
36 33 35 eqtr2i 12gcd5=1