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Metamath Proof Explorer

Theorem 6gcd4e2

Description: The greatest common divisor of six and four is two. To calculate this gcd, a simple form of Euclid's algorithm is used: ( 6 gcd 4 ) = ( ( 4 + 2 ) gcd 4 ) = ( 2 gcd 4 ) and ( 2 gcd 4 ) = ( 2 gcd ( 2 + 2 ) ) = ( 2 gcd 2 ) = 2 . (Contributed by AV, 27-Aug-2020)

Ref Expression
Assertion 6gcd4e2 ( 6 gcd 4 ) = 2

Proof

Step Hyp Ref Expression
1 6nn ⊢ 6 ∈ ℕ
2 1 nnzi ⊢ 6 ∈ ℤ
3 4z ⊢ 4 ∈ ℤ
4 gcdcom ⊢ ( ( 6 ∈ ℤ ∧ 4 ∈ ℤ ) → ( 6 gcd 4 ) = ( 4 gcd 6 ) )
5 2 3 4 mp2an ⊢ ( 6 gcd 4 ) = ( 4 gcd 6 )
6 4cn ⊢ 4 ∈ ℂ
7 2cn ⊢ 2 ∈ ℂ
8 4p2e6 ⊢ ( 4 + 2 ) = 6
9 6 7 8 addcomli ⊢ ( 2 + 4 ) = 6
10 9 oveq2i ⊢ ( 4 gcd ( 2 + 4 ) ) = ( 4 gcd 6 )
11 2z ⊢ 2 ∈ ℤ
12 gcdadd ⊢ ( ( 2 ∈ ℤ ∧ 2 ∈ ℤ ) → ( 2 gcd 2 ) = ( 2 gcd ( 2 + 2 ) ) )
13 11 11 12 mp2an ⊢ ( 2 gcd 2 ) = ( 2 gcd ( 2 + 2 ) )
14 2p2e4 ⊢ ( 2 + 2 ) = 4
15 14 oveq2i ⊢ ( 2 gcd ( 2 + 2 ) ) = ( 2 gcd 4 )
16 gcdcom ⊢ ( ( 2 ∈ ℤ ∧ 4 ∈ ℤ ) → ( 2 gcd 4 ) = ( 4 gcd 2 ) )
17 11 3 16 mp2an ⊢ ( 2 gcd 4 ) = ( 4 gcd 2 )
18 15 17 eqtri ⊢ ( 2 gcd ( 2 + 2 ) ) = ( 4 gcd 2 )
19 13 18 eqtri ⊢ ( 2 gcd 2 ) = ( 4 gcd 2 )
20 gcdid ⊢ ( 2 ∈ ℤ → ( 2 gcd 2 ) = ( abs ‘ 2 ) )
21 11 20 ax-mp ⊢ ( 2 gcd 2 ) = ( abs ‘ 2 )
22 2re ⊢ 2 ∈ ℝ
23 0le2 ⊢ 0 ≤ 2
24 absid ⊢ ( ( 2 ∈ ℝ ∧ 0 ≤ 2 ) → ( abs ‘ 2 ) = 2 )
25 22 23 24 mp2an ⊢ ( abs ‘ 2 ) = 2
26 21 25 eqtri ⊢ ( 2 gcd 2 ) = 2
27 gcdadd ⊢ ( ( 4 ∈ ℤ ∧ 2 ∈ ℤ ) → ( 4 gcd 2 ) = ( 4 gcd ( 2 + 4 ) ) )
28 3 11 27 mp2an ⊢ ( 4 gcd 2 ) = ( 4 gcd ( 2 + 4 ) )
29 19 26 28 3eqtr3ri ⊢ ( 4 gcd ( 2 + 4 ) ) = 2
30 5 10 29 3eqtr2i ⊢ ( 6 gcd 4 ) = 2