Metamath Proof Explorer


Theorem 12gcd5e1

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

Ref Expression
Assertion 12gcd5e1 ⊢ 12 gcd 5 = 1

Proof

Step Hyp Ref Expression
1 2lt5 ⊢ 2 < 5
2 1 olci ⊢ 5 < 2 ∨ 2 < 5
3 5re ⊢ 5 ∈ ℝ
4 2re ⊢ 2 ∈ ℝ
5 lttri2 ⊢ 5 ∈ ℝ ∧ 2 ∈ ℝ → 5 ≠ 2 ↔ 5 < 2 ∨ 2 < 5
6 3 4 5 mp2an ⊢ 5 ≠ 2 ↔ 5 < 2 ∨ 2 < 5
7 2 6 mpbir ⊢ 5 ≠ 2
8 5prm ⊢ 5 ∈ ℙ
9 2prm ⊢ 2 ∈ ℙ
10 prmrp ⊢ 5 ∈ ℙ ∧ 2 ∈ ℙ → 5 gcd 2 = 1 ↔ 5 ≠ 2
11 8 9 10 mp2an ⊢ 5 gcd 2 = 1 ↔ 5 ≠ 2
12 7 11 mpbir ⊢ 5 gcd 2 = 1
13 5nn ⊢ 5 ∈ ℕ
14 2nn ⊢ 2 ∈ ℕ
15 14 nnzi ⊢ 2 ∈ ℤ
16 13 14 15 gcdaddmzz2nncomi ⊢ 5 gcd 2 = 5 gcd 2 ⋅ 5 + 2
17 13 14 mulcomnni ⊢ 5 ⋅ 2 = 2 ⋅ 5
18 5t2e10 ⊢ 5 ⋅ 2 = 10
19 17 18 eqtr3i ⊢ 2 ⋅ 5 = 10
20 19 oveq1i ⊢ 2 ⋅ 5 + 2 = 10 + 2
21 1nn0 ⊢ 1 ∈ ℕ 0
22 0nn0 ⊢ 0 ∈ ℕ 0
23 14 nnnn0i ⊢ 2 ∈ ℕ 0
24 eqid ⊢ 10 = 10
25 23 dec0h ⊢ 2 = 02
26 1p0e1 ⊢ 1 + 0 = 1
27 2cn ⊢ 2 ∈ ℂ
28 27 addlidi ⊢ 0 + 2 = 2
29 21 22 22 23 24 25 26 28 decadd ⊢ 10 + 2 = 12
30 20 29 eqtri ⊢ 2 ⋅ 5 + 2 = 12
31 30 oveq2i ⊢ 5 gcd 2 ⋅ 5 + 2 = 5 gcd 12
32 16 31 eqtri ⊢ 5 gcd 2 = 5 gcd 12
33 12 32 eqtr3i ⊢ 1 = 5 gcd 12
34 21 14 decnncl ⊢ 12 ∈ ℕ
35 13 34 gcdcomnni ⊢ 5 gcd 12 = 12 gcd 5
36 33 35 eqtr2i ⊢ 12 gcd 5 = 1