Metamath Proof Explorer


Theorem fproddvdsd

Description: A finite product of integers is divisible by any of its factors. (Contributed by AV, 14-Aug-2020) (Proof shortened by AV, 2-Aug-2021)

Ref Expression
Hypotheses fproddvdsd.f ( 𝜑𝐴 ∈ Fin )
fproddvdsd.s ( 𝜑𝐴 ⊆ ℤ )
Assertion fproddvdsd ( 𝜑 → ∀ 𝑥𝐴 𝑥 ∥ ∏ 𝑘𝐴 𝑘 )

Proof

Step Hyp Ref Expression
1 fproddvdsd.f ( 𝜑𝐴 ∈ Fin )
2 fproddvdsd.s ( 𝜑𝐴 ⊆ ℤ )
3 f1oi ( I ↾ ℤ ) : ℤ –1-1-onto→ ℤ
4 f1of ( ( I ↾ ℤ ) : ℤ –1-1-onto→ ℤ → ( I ↾ ℤ ) : ℤ ⟶ ℤ )
5 3 4 mp1i ( 𝜑 → ( I ↾ ℤ ) : ℤ ⟶ ℤ )
6 1 2 5 fprodfvdvdsd ( 𝜑 → ∀ 𝑥𝐴 ( ( I ↾ ℤ ) ‘ 𝑥 ) ∥ ∏ 𝑘𝐴 ( ( I ↾ ℤ ) ‘ 𝑘 ) )
7 2 sselda ( ( 𝜑𝑥𝐴 ) → 𝑥 ∈ ℤ )
8 fvresi ( 𝑥 ∈ ℤ → ( ( I ↾ ℤ ) ‘ 𝑥 ) = 𝑥 )
9 7 8 syl ( ( 𝜑𝑥𝐴 ) → ( ( I ↾ ℤ ) ‘ 𝑥 ) = 𝑥 )
10 9 eqcomd ( ( 𝜑𝑥𝐴 ) → 𝑥 = ( ( I ↾ ℤ ) ‘ 𝑥 ) )
11 2 sseld ( 𝜑 → ( 𝑘𝐴𝑘 ∈ ℤ ) )
12 11 adantr ( ( 𝜑𝑥𝐴 ) → ( 𝑘𝐴𝑘 ∈ ℤ ) )
13 12 imp ( ( ( 𝜑𝑥𝐴 ) ∧ 𝑘𝐴 ) → 𝑘 ∈ ℤ )
14 fvresi ( 𝑘 ∈ ℤ → ( ( I ↾ ℤ ) ‘ 𝑘 ) = 𝑘 )
15 13 14 syl ( ( ( 𝜑𝑥𝐴 ) ∧ 𝑘𝐴 ) → ( ( I ↾ ℤ ) ‘ 𝑘 ) = 𝑘 )
16 15 eqcomd ( ( ( 𝜑𝑥𝐴 ) ∧ 𝑘𝐴 ) → 𝑘 = ( ( I ↾ ℤ ) ‘ 𝑘 ) )
17 16 prodeq2dv ( ( 𝜑𝑥𝐴 ) → ∏ 𝑘𝐴 𝑘 = ∏ 𝑘𝐴 ( ( I ↾ ℤ ) ‘ 𝑘 ) )
18 10 17 breq12d ( ( 𝜑𝑥𝐴 ) → ( 𝑥 ∥ ∏ 𝑘𝐴 𝑘 ↔ ( ( I ↾ ℤ ) ‘ 𝑥 ) ∥ ∏ 𝑘𝐴 ( ( I ↾ ℤ ) ‘ 𝑘 ) ) )
19 18 ralbidva ( 𝜑 → ( ∀ 𝑥𝐴 𝑥 ∥ ∏ 𝑘𝐴 𝑘 ↔ ∀ 𝑥𝐴 ( ( I ↾ ℤ ) ‘ 𝑥 ) ∥ ∏ 𝑘𝐴 ( ( I ↾ ℤ ) ‘ 𝑘 ) ) )
20 6 19 mpbird ( 𝜑 → ∀ 𝑥𝐴 𝑥 ∥ ∏ 𝑘𝐴 𝑘 )