Metamath Proof Explorer


Theorem lcm3un

Description: Least common multiple of natural numbers up to 3 equals 6. (Contributed by metakunt, 25-Apr-2024)

Ref Expression
Assertion lcm3un ( lcm ‘ ( 1 ... 3 ) ) = 6

Proof

Step Hyp Ref Expression
1 3nn 3 ∈ ℕ
2 id ( 3 ∈ ℕ → 3 ∈ ℕ )
3 2 lcmfunnnd ( 3 ∈ ℕ → ( lcm ‘ ( 1 ... 3 ) ) = ( ( lcm ‘ ( 1 ... ( 3 − 1 ) ) ) lcm 3 ) )
4 1 3 ax-mp ( lcm ‘ ( 1 ... 3 ) ) = ( ( lcm ‘ ( 1 ... ( 3 − 1 ) ) ) lcm 3 )
5 3m1e2 ( 3 − 1 ) = 2
6 5 oveq2i ( 1 ... ( 3 − 1 ) ) = ( 1 ... 2 )
7 6 fveq2i ( lcm ‘ ( 1 ... ( 3 − 1 ) ) ) = ( lcm ‘ ( 1 ... 2 ) )
8 lcm2un ( lcm ‘ ( 1 ... 2 ) ) = 2
9 7 8 eqtri ( lcm ‘ ( 1 ... ( 3 − 1 ) ) ) = 2
10 9 oveq1i ( ( lcm ‘ ( 1 ... ( 3 − 1 ) ) ) lcm 3 ) = ( 2 lcm 3 )
11 2z 2 ∈ ℤ
12 3z 3 ∈ ℤ
13 11 12 pm3.2i ( 2 ∈ ℤ ∧ 3 ∈ ℤ )
14 lcmcom ( ( 2 ∈ ℤ ∧ 3 ∈ ℤ ) → ( 2 lcm 3 ) = ( 3 lcm 2 ) )
15 13 14 ax-mp ( 2 lcm 3 ) = ( 3 lcm 2 )
16 3lcm2e6 ( 3 lcm 2 ) = 6
17 15 16 eqtri ( 2 lcm 3 ) = 6
18 10 17 eqtri ( ( lcm ‘ ( 1 ... ( 3 − 1 ) ) ) lcm 3 ) = 6
19 4 18 eqtri ( lcm ‘ ( 1 ... 3 ) ) = 6