Metamath Proof Explorer


Theorem lcmfdvds

Description: The least common multiple of a set of integers divides any integer which is divisible by all elements of the set. (Contributed by AV, 26-Aug-2020)

Ref Expression
Assertion lcmfdvds ( ( 𝐾 ∈ ℤ ∧ 𝑍 ⊆ ℤ ∧ 𝑍 ∈ Fin ) → ( ∀ 𝑚𝑍 𝑚𝐾 → ( lcm𝑍 ) ∥ 𝐾 ) )

Proof

Step Hyp Ref Expression
1 breq2 ( 𝑘 = 𝐾 → ( 𝑚𝑘𝑚𝐾 ) )
2 1 ralbidv ( 𝑘 = 𝐾 → ( ∀ 𝑚𝑍 𝑚𝑘 ↔ ∀ 𝑚𝑍 𝑚𝐾 ) )
3 breq2 ( 𝑘 = 𝐾 → ( ( lcm𝑍 ) ∥ 𝑘 ↔ ( lcm𝑍 ) ∥ 𝐾 ) )
4 2 3 imbi12d ( 𝑘 = 𝐾 → ( ( ∀ 𝑚𝑍 𝑚𝑘 → ( lcm𝑍 ) ∥ 𝑘 ) ↔ ( ∀ 𝑚𝑍 𝑚𝐾 → ( lcm𝑍 ) ∥ 𝐾 ) ) )
5 4 rspccv ( ∀ 𝑘 ∈ ℤ ( ∀ 𝑚𝑍 𝑚𝑘 → ( lcm𝑍 ) ∥ 𝑘 ) → ( 𝐾 ∈ ℤ → ( ∀ 𝑚𝑍 𝑚𝐾 → ( lcm𝑍 ) ∥ 𝐾 ) ) )
6 5 adantr ( ( ∀ 𝑘 ∈ ℤ ( ∀ 𝑚𝑍 𝑚𝑘 → ( lcm𝑍 ) ∥ 𝑘 ) ∧ ∀ 𝑛 ∈ ℤ ( lcm ‘ ( 𝑍 ∪ { 𝑛 } ) ) = ( ( lcm𝑍 ) lcm 𝑛 ) ) → ( 𝐾 ∈ ℤ → ( ∀ 𝑚𝑍 𝑚𝐾 → ( lcm𝑍 ) ∥ 𝐾 ) ) )
7 lcmfunsnlem ( ( 𝑍 ⊆ ℤ ∧ 𝑍 ∈ Fin ) → ( ∀ 𝑘 ∈ ℤ ( ∀ 𝑚𝑍 𝑚𝑘 → ( lcm𝑍 ) ∥ 𝑘 ) ∧ ∀ 𝑛 ∈ ℤ ( lcm ‘ ( 𝑍 ∪ { 𝑛 } ) ) = ( ( lcm𝑍 ) lcm 𝑛 ) ) )
8 6 7 syl11 ( 𝐾 ∈ ℤ → ( ( 𝑍 ⊆ ℤ ∧ 𝑍 ∈ Fin ) → ( ∀ 𝑚𝑍 𝑚𝐾 → ( lcm𝑍 ) ∥ 𝐾 ) ) )
9 8 3impib ( ( 𝐾 ∈ ℤ ∧ 𝑍 ⊆ ℤ ∧ 𝑍 ∈ Fin ) → ( ∀ 𝑚𝑍 𝑚𝐾 → ( lcm𝑍 ) ∥ 𝐾 ) )