Metamath Proof Explorer


Theorem fsummsnunz

Description: A finite sum all of whose summands are integers is itself an integer (case where the summation set is the union of a finite set and a singleton). (Contributed by Alexander van der Vekens, 1-Sep-2018) (Revised by AV, 17-Dec-2021)

Ref Expression
Assertion fsummsnunz ( ( 𝐴 ∈ Fin ∧ ∀ 𝑘 ∈ ( 𝐴 ∪ { 𝑍 } ) 𝐵 ∈ ℤ ) → Σ 𝑘 ∈ ( 𝐴 ∪ { 𝑍 } ) 𝐵 ∈ ℤ )

Proof

Step Hyp Ref Expression
1 nfcv 𝑥 𝐵
2 nfcsb1v 𝑘 𝑥 / 𝑘 𝐵
3 csbeq1a ( 𝑘 = 𝑥𝐵 = 𝑥 / 𝑘 𝐵 )
4 1 2 3 cbvsumi Σ 𝑘 ∈ ( 𝐴 ∪ { 𝑍 } ) 𝐵 = Σ 𝑥 ∈ ( 𝐴 ∪ { 𝑍 } ) 𝑥 / 𝑘 𝐵
5 snfi { 𝑍 } ∈ Fin
6 5 a1i ( ( 𝐴 ∈ Fin ∧ ∀ 𝑘 ∈ ( 𝐴 ∪ { 𝑍 } ) 𝐵 ∈ ℤ ) → { 𝑍 } ∈ Fin )
7 unfi ( ( 𝐴 ∈ Fin ∧ { 𝑍 } ∈ Fin ) → ( 𝐴 ∪ { 𝑍 } ) ∈ Fin )
8 6 7 syldan ( ( 𝐴 ∈ Fin ∧ ∀ 𝑘 ∈ ( 𝐴 ∪ { 𝑍 } ) 𝐵 ∈ ℤ ) → ( 𝐴 ∪ { 𝑍 } ) ∈ Fin )
9 rspcsbela ( ( 𝑥 ∈ ( 𝐴 ∪ { 𝑍 } ) ∧ ∀ 𝑘 ∈ ( 𝐴 ∪ { 𝑍 } ) 𝐵 ∈ ℤ ) → 𝑥 / 𝑘 𝐵 ∈ ℤ )
10 9 expcom ( ∀ 𝑘 ∈ ( 𝐴 ∪ { 𝑍 } ) 𝐵 ∈ ℤ → ( 𝑥 ∈ ( 𝐴 ∪ { 𝑍 } ) → 𝑥 / 𝑘 𝐵 ∈ ℤ ) )
11 10 adantl ( ( 𝐴 ∈ Fin ∧ ∀ 𝑘 ∈ ( 𝐴 ∪ { 𝑍 } ) 𝐵 ∈ ℤ ) → ( 𝑥 ∈ ( 𝐴 ∪ { 𝑍 } ) → 𝑥 / 𝑘 𝐵 ∈ ℤ ) )
12 11 imp ( ( ( 𝐴 ∈ Fin ∧ ∀ 𝑘 ∈ ( 𝐴 ∪ { 𝑍 } ) 𝐵 ∈ ℤ ) ∧ 𝑥 ∈ ( 𝐴 ∪ { 𝑍 } ) ) → 𝑥 / 𝑘 𝐵 ∈ ℤ )
13 8 12 fsumzcl ( ( 𝐴 ∈ Fin ∧ ∀ 𝑘 ∈ ( 𝐴 ∪ { 𝑍 } ) 𝐵 ∈ ℤ ) → Σ 𝑥 ∈ ( 𝐴 ∪ { 𝑍 } ) 𝑥 / 𝑘 𝐵 ∈ ℤ )
14 4 13 eqeltrid ( ( 𝐴 ∈ Fin ∧ ∀ 𝑘 ∈ ( 𝐴 ∪ { 𝑍 } ) 𝐵 ∈ ℤ ) → Σ 𝑘 ∈ ( 𝐴 ∪ { 𝑍 } ) 𝐵 ∈ ℤ )