Metamath Proof Explorer


Theorem lcmfunsn

Description: The _lcm function for a union of a set of integer and a singleton. (Contributed by AV, 26-Aug-2020)

Ref Expression
Assertion lcmfunsn ( ( 𝑌 ⊆ ℤ ∧ 𝑌 ∈ Fin ∧ 𝑁 ∈ ℤ ) → ( lcm ‘ ( 𝑌 ∪ { 𝑁 } ) ) = ( ( lcm𝑌 ) lcm 𝑁 ) )

Proof

Step Hyp Ref Expression
1 lcmfunsnlem ( ( 𝑌 ⊆ ℤ ∧ 𝑌 ∈ Fin ) → ( ∀ 𝑘 ∈ ℤ ( ∀ 𝑚𝑌 𝑚𝑘 → ( lcm𝑌 ) ∥ 𝑘 ) ∧ ∀ 𝑛 ∈ ℤ ( lcm ‘ ( 𝑌 ∪ { 𝑛 } ) ) = ( ( lcm𝑌 ) lcm 𝑛 ) ) )
2 sneq ( 𝑛 = 𝑁 → { 𝑛 } = { 𝑁 } )
3 2 uneq2d ( 𝑛 = 𝑁 → ( 𝑌 ∪ { 𝑛 } ) = ( 𝑌 ∪ { 𝑁 } ) )
4 3 fveq2d ( 𝑛 = 𝑁 → ( lcm ‘ ( 𝑌 ∪ { 𝑛 } ) ) = ( lcm ‘ ( 𝑌 ∪ { 𝑁 } ) ) )
5 oveq2 ( 𝑛 = 𝑁 → ( ( lcm𝑌 ) lcm 𝑛 ) = ( ( lcm𝑌 ) lcm 𝑁 ) )
6 4 5 eqeq12d ( 𝑛 = 𝑁 → ( ( lcm ‘ ( 𝑌 ∪ { 𝑛 } ) ) = ( ( lcm𝑌 ) lcm 𝑛 ) ↔ ( lcm ‘ ( 𝑌 ∪ { 𝑁 } ) ) = ( ( lcm𝑌 ) lcm 𝑁 ) ) )
7 6 rspccv ( ∀ 𝑛 ∈ ℤ ( lcm ‘ ( 𝑌 ∪ { 𝑛 } ) ) = ( ( lcm𝑌 ) lcm 𝑛 ) → ( 𝑁 ∈ ℤ → ( lcm ‘ ( 𝑌 ∪ { 𝑁 } ) ) = ( ( lcm𝑌 ) lcm 𝑁 ) ) )
8 1 7 simpl2im ( ( 𝑌 ⊆ ℤ ∧ 𝑌 ∈ Fin ) → ( 𝑁 ∈ ℤ → ( lcm ‘ ( 𝑌 ∪ { 𝑁 } ) ) = ( ( lcm𝑌 ) lcm 𝑁 ) ) )
9 8 3impia ( ( 𝑌 ⊆ ℤ ∧ 𝑌 ∈ Fin ∧ 𝑁 ∈ ℤ ) → ( lcm ‘ ( 𝑌 ∪ { 𝑁 } ) ) = ( ( lcm𝑌 ) lcm 𝑁 ) )