Metamath Proof Explorer


Theorem fprod1p

Description: Separate out the first term in a finite product. (Contributed by Scott Fenton, 24-Dec-2017)

Ref Expression
Hypotheses fprod1p.1 ( 𝜑𝑁 ∈ ( ℤ𝑀 ) )
fprod1p.2 ( ( 𝜑𝑘 ∈ ( 𝑀 ... 𝑁 ) ) → 𝐴 ∈ ℂ )
fprod1p.3 ( 𝑘 = 𝑀𝐴 = 𝐵 )
Assertion fprod1p ( 𝜑 → ∏ 𝑘 ∈ ( 𝑀 ... 𝑁 ) 𝐴 = ( 𝐵 · ∏ 𝑘 ∈ ( ( 𝑀 + 1 ) ... 𝑁 ) 𝐴 ) )

Proof

Step Hyp Ref Expression
1 fprod1p.1 ( 𝜑𝑁 ∈ ( ℤ𝑀 ) )
2 fprod1p.2 ( ( 𝜑𝑘 ∈ ( 𝑀 ... 𝑁 ) ) → 𝐴 ∈ ℂ )
3 fprod1p.3 ( 𝑘 = 𝑀𝐴 = 𝐵 )
4 eluzfz1 ( 𝑁 ∈ ( ℤ𝑀 ) → 𝑀 ∈ ( 𝑀 ... 𝑁 ) )
5 1 4 syl ( 𝜑𝑀 ∈ ( 𝑀 ... 𝑁 ) )
6 5 elfzelzd ( 𝜑𝑀 ∈ ℤ )
7 fzsn ( 𝑀 ∈ ℤ → ( 𝑀 ... 𝑀 ) = { 𝑀 } )
8 6 7 syl ( 𝜑 → ( 𝑀 ... 𝑀 ) = { 𝑀 } )
9 8 ineq1d ( 𝜑 → ( ( 𝑀 ... 𝑀 ) ∩ ( ( 𝑀 + 1 ) ... 𝑁 ) ) = ( { 𝑀 } ∩ ( ( 𝑀 + 1 ) ... 𝑁 ) ) )
10 6 zred ( 𝜑𝑀 ∈ ℝ )
11 10 ltp1d ( 𝜑𝑀 < ( 𝑀 + 1 ) )
12 fzdisj ( 𝑀 < ( 𝑀 + 1 ) → ( ( 𝑀 ... 𝑀 ) ∩ ( ( 𝑀 + 1 ) ... 𝑁 ) ) = ∅ )
13 11 12 syl ( 𝜑 → ( ( 𝑀 ... 𝑀 ) ∩ ( ( 𝑀 + 1 ) ... 𝑁 ) ) = ∅ )
14 9 13 eqtr3d ( 𝜑 → ( { 𝑀 } ∩ ( ( 𝑀 + 1 ) ... 𝑁 ) ) = ∅ )
15 fzsplit ( 𝑀 ∈ ( 𝑀 ... 𝑁 ) → ( 𝑀 ... 𝑁 ) = ( ( 𝑀 ... 𝑀 ) ∪ ( ( 𝑀 + 1 ) ... 𝑁 ) ) )
16 5 15 syl ( 𝜑 → ( 𝑀 ... 𝑁 ) = ( ( 𝑀 ... 𝑀 ) ∪ ( ( 𝑀 + 1 ) ... 𝑁 ) ) )
17 8 uneq1d ( 𝜑 → ( ( 𝑀 ... 𝑀 ) ∪ ( ( 𝑀 + 1 ) ... 𝑁 ) ) = ( { 𝑀 } ∪ ( ( 𝑀 + 1 ) ... 𝑁 ) ) )
18 16 17 eqtrd ( 𝜑 → ( 𝑀 ... 𝑁 ) = ( { 𝑀 } ∪ ( ( 𝑀 + 1 ) ... 𝑁 ) ) )
19 fzfid ( 𝜑 → ( 𝑀 ... 𝑁 ) ∈ Fin )
20 14 18 19 2 fprodsplit ( 𝜑 → ∏ 𝑘 ∈ ( 𝑀 ... 𝑁 ) 𝐴 = ( ∏ 𝑘 ∈ { 𝑀 } 𝐴 · ∏ 𝑘 ∈ ( ( 𝑀 + 1 ) ... 𝑁 ) 𝐴 ) )
21 3 eleq1d ( 𝑘 = 𝑀 → ( 𝐴 ∈ ℂ ↔ 𝐵 ∈ ℂ ) )
22 2 ralrimiva ( 𝜑 → ∀ 𝑘 ∈ ( 𝑀 ... 𝑁 ) 𝐴 ∈ ℂ )
23 21 22 5 rspcdva ( 𝜑𝐵 ∈ ℂ )
24 3 prodsn ( ( 𝑀 ∈ ( 𝑀 ... 𝑁 ) ∧ 𝐵 ∈ ℂ ) → ∏ 𝑘 ∈ { 𝑀 } 𝐴 = 𝐵 )
25 5 23 24 syl2anc ( 𝜑 → ∏ 𝑘 ∈ { 𝑀 } 𝐴 = 𝐵 )
26 25 oveq1d ( 𝜑 → ( ∏ 𝑘 ∈ { 𝑀 } 𝐴 · ∏ 𝑘 ∈ ( ( 𝑀 + 1 ) ... 𝑁 ) 𝐴 ) = ( 𝐵 · ∏ 𝑘 ∈ ( ( 𝑀 + 1 ) ... 𝑁 ) 𝐴 ) )
27 20 26 eqtrd ( 𝜑 → ∏ 𝑘 ∈ ( 𝑀 ... 𝑁 ) 𝐴 = ( 𝐵 · ∏ 𝑘 ∈ ( ( 𝑀 + 1 ) ... 𝑁 ) 𝐴 ) )