Metamath Proof Explorer


Theorem fsum1p

Description: Separate out the first term in a finite sum. (Contributed by NM, 3-Jan-2006) (Revised by Mario Carneiro, 23-Apr-2014)

Ref Expression
Hypotheses fsumm1.1 ( 𝜑𝑁 ∈ ( ℤ𝑀 ) )
fsumm1.2 ( ( 𝜑𝑘 ∈ ( 𝑀 ... 𝑁 ) ) → 𝐴 ∈ ℂ )
fsum1p.3 ( 𝑘 = 𝑀𝐴 = 𝐵 )
Assertion fsum1p ( 𝜑 → Σ 𝑘 ∈ ( 𝑀 ... 𝑁 ) 𝐴 = ( 𝐵 + Σ 𝑘 ∈ ( ( 𝑀 + 1 ) ... 𝑁 ) 𝐴 ) )

Proof

Step Hyp Ref Expression
1 fsumm1.1 ( 𝜑𝑁 ∈ ( ℤ𝑀 ) )
2 fsumm1.2 ( ( 𝜑𝑘 ∈ ( 𝑀 ... 𝑁 ) ) → 𝐴 ∈ ℂ )
3 fsum1p.3 ( 𝑘 = 𝑀𝐴 = 𝐵 )
4 eluzel2 ( 𝑁 ∈ ( ℤ𝑀 ) → 𝑀 ∈ ℤ )
5 1 4 syl ( 𝜑𝑀 ∈ ℤ )
6 fzsn ( 𝑀 ∈ ℤ → ( 𝑀 ... 𝑀 ) = { 𝑀 } )
7 5 6 syl ( 𝜑 → ( 𝑀 ... 𝑀 ) = { 𝑀 } )
8 7 ineq1d ( 𝜑 → ( ( 𝑀 ... 𝑀 ) ∩ ( ( 𝑀 + 1 ) ... 𝑁 ) ) = ( { 𝑀 } ∩ ( ( 𝑀 + 1 ) ... 𝑁 ) ) )
9 5 zred ( 𝜑𝑀 ∈ ℝ )
10 9 ltp1d ( 𝜑𝑀 < ( 𝑀 + 1 ) )
11 fzdisj ( 𝑀 < ( 𝑀 + 1 ) → ( ( 𝑀 ... 𝑀 ) ∩ ( ( 𝑀 + 1 ) ... 𝑁 ) ) = ∅ )
12 10 11 syl ( 𝜑 → ( ( 𝑀 ... 𝑀 ) ∩ ( ( 𝑀 + 1 ) ... 𝑁 ) ) = ∅ )
13 8 12 eqtr3d ( 𝜑 → ( { 𝑀 } ∩ ( ( 𝑀 + 1 ) ... 𝑁 ) ) = ∅ )
14 eluzfz1 ( 𝑁 ∈ ( ℤ𝑀 ) → 𝑀 ∈ ( 𝑀 ... 𝑁 ) )
15 1 14 syl ( 𝜑𝑀 ∈ ( 𝑀 ... 𝑁 ) )
16 fzsplit ( 𝑀 ∈ ( 𝑀 ... 𝑁 ) → ( 𝑀 ... 𝑁 ) = ( ( 𝑀 ... 𝑀 ) ∪ ( ( 𝑀 + 1 ) ... 𝑁 ) ) )
17 15 16 syl ( 𝜑 → ( 𝑀 ... 𝑁 ) = ( ( 𝑀 ... 𝑀 ) ∪ ( ( 𝑀 + 1 ) ... 𝑁 ) ) )
18 7 uneq1d ( 𝜑 → ( ( 𝑀 ... 𝑀 ) ∪ ( ( 𝑀 + 1 ) ... 𝑁 ) ) = ( { 𝑀 } ∪ ( ( 𝑀 + 1 ) ... 𝑁 ) ) )
19 17 18 eqtrd ( 𝜑 → ( 𝑀 ... 𝑁 ) = ( { 𝑀 } ∪ ( ( 𝑀 + 1 ) ... 𝑁 ) ) )
20 fzfid ( 𝜑 → ( 𝑀 ... 𝑁 ) ∈ Fin )
21 13 19 20 2 fsumsplit ( 𝜑 → Σ 𝑘 ∈ ( 𝑀 ... 𝑁 ) 𝐴 = ( Σ 𝑘 ∈ { 𝑀 } 𝐴 + Σ 𝑘 ∈ ( ( 𝑀 + 1 ) ... 𝑁 ) 𝐴 ) )
22 3 eleq1d ( 𝑘 = 𝑀 → ( 𝐴 ∈ ℂ ↔ 𝐵 ∈ ℂ ) )
23 2 ralrimiva ( 𝜑 → ∀ 𝑘 ∈ ( 𝑀 ... 𝑁 ) 𝐴 ∈ ℂ )
24 22 23 15 rspcdva ( 𝜑𝐵 ∈ ℂ )
25 3 sumsn ( ( 𝑀 ∈ ℤ ∧ 𝐵 ∈ ℂ ) → Σ 𝑘 ∈ { 𝑀 } 𝐴 = 𝐵 )
26 5 24 25 syl2anc ( 𝜑 → Σ 𝑘 ∈ { 𝑀 } 𝐴 = 𝐵 )
27 26 oveq1d ( 𝜑 → ( Σ 𝑘 ∈ { 𝑀 } 𝐴 + Σ 𝑘 ∈ ( ( 𝑀 + 1 ) ... 𝑁 ) 𝐴 ) = ( 𝐵 + Σ 𝑘 ∈ ( ( 𝑀 + 1 ) ... 𝑁 ) 𝐴 ) )
28 21 27 eqtrd ( 𝜑 → Σ 𝑘 ∈ ( 𝑀 ... 𝑁 ) 𝐴 = ( 𝐵 + Σ 𝑘 ∈ ( ( 𝑀 + 1 ) ... 𝑁 ) 𝐴 ) )