Theorem signsvf1

Description: In a single-letter word, which represents a constant polynomial, there is no change of sign. (Contributed by Thierry Arnoux, 8-Oct-2018)

	Ref	Expression
Hypotheses	signsv.p	⊢ ⨣ = ( 𝑎 ∈ { - 1 , 0 , 1 } , 𝑏 ∈ { - 1 , 0 , 1 } ↦ if ( 𝑏 = 0 , 𝑎 , 𝑏 ) )
	signsv.w	⊢ 𝑊 = { ⟨ ( Base ‘ ndx ) , { - 1 , 0 , 1 } ⟩ , ⟨ ( +g ‘ ndx ) , ⨣ ⟩ }
	signsv.t	⊢ 𝑇 = ( 𝑓 ∈ Word ℝ ↦ ( 𝑛 ∈ ( 0 ..^ ( ♯ ‘ 𝑓 ) ) ↦ ( 𝑊 Σg ( 𝑖 ∈ ( 0 ... 𝑛 ) ↦ ( sgn ‘ ( 𝑓 ‘ 𝑖 ) ) ) ) ) )
	signsv.v	⊢ 𝑉 = ( 𝑓 ∈ Word ℝ ↦ Σ 𝑗 ∈ ( 1 ..^ ( ♯ ‘ 𝑓 ) ) if ( ( ( 𝑇 ‘ 𝑓 ) ‘ 𝑗 ) ≠ ( ( 𝑇 ‘ 𝑓 ) ‘ ( 𝑗 − 1 ) ) , 1 , 0 ) )
Assertion	signsvf1	⊢ ( 𝐾 ∈ ℝ → ( 𝑉 ‘ ⟨“ 𝐾 ”⟩ ) = 0 )

Proof

Step	Hyp	Ref	Expression
1		signsv.p	⊢ ⨣ = ( 𝑎 ∈ { - 1 , 0 , 1 } , 𝑏 ∈ { - 1 , 0 , 1 } ↦ if ( 𝑏 = 0 , 𝑎 , 𝑏 ) )
2		signsv.w	⊢ 𝑊 = { ⟨ ( Base ‘ ndx ) , { - 1 , 0 , 1 } ⟩ , ⟨ ( +g ‘ ndx ) , ⨣ ⟩ }
3		signsv.t	⊢ 𝑇 = ( 𝑓 ∈ Word ℝ ↦ ( 𝑛 ∈ ( 0 ..^ ( ♯ ‘ 𝑓 ) ) ↦ ( 𝑊 Σg ( 𝑖 ∈ ( 0 ... 𝑛 ) ↦ ( sgn ‘ ( 𝑓 ‘ 𝑖 ) ) ) ) ) )
4		signsv.v	⊢ 𝑉 = ( 𝑓 ∈ Word ℝ ↦ Σ 𝑗 ∈ ( 1 ..^ ( ♯ ‘ 𝑓 ) ) if ( ( ( 𝑇 ‘ 𝑓 ) ‘ 𝑗 ) ≠ ( ( 𝑇 ‘ 𝑓 ) ‘ ( 𝑗 − 1 ) ) , 1 , 0 ) )
5		s1cl	⊢ ( 𝐾 ∈ ℝ → ⟨“ 𝐾 ”⟩ ∈ Word ℝ )
6	1 2 3 4	signsvvfval	⊢ ( ⟨“ 𝐾 ”⟩ ∈ Word ℝ → ( 𝑉 ‘ ⟨“ 𝐾 ”⟩ ) = Σ 𝑗 ∈ ( 1 ..^ ( ♯ ‘ ⟨“ 𝐾 ”⟩ ) ) if ( ( ( 𝑇 ‘ ⟨“ 𝐾 ”⟩ ) ‘ 𝑗 ) ≠ ( ( 𝑇 ‘ ⟨“ 𝐾 ”⟩ ) ‘ ( 𝑗 − 1 ) ) , 1 , 0 ) )
7	5 6	syl	⊢ ( 𝐾 ∈ ℝ → ( 𝑉 ‘ ⟨“ 𝐾 ”⟩ ) = Σ 𝑗 ∈ ( 1 ..^ ( ♯ ‘ ⟨“ 𝐾 ”⟩ ) ) if ( ( ( 𝑇 ‘ ⟨“ 𝐾 ”⟩ ) ‘ 𝑗 ) ≠ ( ( 𝑇 ‘ ⟨“ 𝐾 ”⟩ ) ‘ ( 𝑗 − 1 ) ) , 1 , 0 ) )
8		s1len	⊢ ( ♯ ‘ ⟨“ 𝐾 ”⟩ ) = 1
9	8	oveq2i	⊢ ( 1 ..^ ( ♯ ‘ ⟨“ 𝐾 ”⟩ ) ) = ( 1 ..^ 1 )
10		fzo0	⊢ ( 1 ..^ 1 ) = ∅
11	9 10	eqtri	⊢ ( 1 ..^ ( ♯ ‘ ⟨“ 𝐾 ”⟩ ) ) = ∅
12	11	sumeq1i	⊢ Σ 𝑗 ∈ ( 1 ..^ ( ♯ ‘ ⟨“ 𝐾 ”⟩ ) ) if ( ( ( 𝑇 ‘ ⟨“ 𝐾 ”⟩ ) ‘ 𝑗 ) ≠ ( ( 𝑇 ‘ ⟨“ 𝐾 ”⟩ ) ‘ ( 𝑗 − 1 ) ) , 1 , 0 ) = Σ 𝑗 ∈ ∅ if ( ( ( 𝑇 ‘ ⟨“ 𝐾 ”⟩ ) ‘ 𝑗 ) ≠ ( ( 𝑇 ‘ ⟨“ 𝐾 ”⟩ ) ‘ ( 𝑗 − 1 ) ) , 1 , 0 )
13		sum0	⊢ Σ 𝑗 ∈ ∅ if ( ( ( 𝑇 ‘ ⟨“ 𝐾 ”⟩ ) ‘ 𝑗 ) ≠ ( ( 𝑇 ‘ ⟨“ 𝐾 ”⟩ ) ‘ ( 𝑗 − 1 ) ) , 1 , 0 ) = 0
14	12 13	eqtri	⊢ Σ 𝑗 ∈ ( 1 ..^ ( ♯ ‘ ⟨“ 𝐾 ”⟩ ) ) if ( ( ( 𝑇 ‘ ⟨“ 𝐾 ”⟩ ) ‘ 𝑗 ) ≠ ( ( 𝑇 ‘ ⟨“ 𝐾 ”⟩ ) ‘ ( 𝑗 − 1 ) ) , 1 , 0 ) = 0
15	7 14	eqtrdi	⊢ ( 𝐾 ∈ ℝ → ( 𝑉 ‘ ⟨“ 𝐾 ”⟩ ) = 0 )