Theorem signsvf0

Description: There is no change of sign in the empty word. (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	signsvf0	⊢ ( 𝑉 ‘ ∅ ) = 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		wrd0	⊢ ∅ ∈ Word ℝ
6	1 2 3 4	signsvvfval	⊢ ( ∅ ∈ Word ℝ → ( 𝑉 ‘ ∅ ) = Σ 𝑗 ∈ ( 1 ..^ ( ♯ ‘ ∅ ) ) if ( ( ( 𝑇 ‘ ∅ ) ‘ 𝑗 ) ≠ ( ( 𝑇 ‘ ∅ ) ‘ ( 𝑗 − 1 ) ) , 1 , 0 ) )
7	5 6	ax-mp	⊢ ( 𝑉 ‘ ∅ ) = Σ 𝑗 ∈ ( 1 ..^ ( ♯ ‘ ∅ ) ) if ( ( ( 𝑇 ‘ ∅ ) ‘ 𝑗 ) ≠ ( ( 𝑇 ‘ ∅ ) ‘ ( 𝑗 − 1 ) ) , 1 , 0 )
8		hash0	⊢ ( ♯ ‘ ∅ ) = 0
9	8	oveq2i	⊢ ( 1 ..^ ( ♯ ‘ ∅ ) ) = ( 1 ..^ 0 )
10		0le1	⊢ 0 ≤ 1
11		1z	⊢ 1 ∈ ℤ
12		0z	⊢ 0 ∈ ℤ
13		fzon	⊢ ( ( 1 ∈ ℤ ∧ 0 ∈ ℤ ) → ( 0 ≤ 1 ↔ ( 1 ..^ 0 ) = ∅ ) )
14	11 12 13	mp2an	⊢ ( 0 ≤ 1 ↔ ( 1 ..^ 0 ) = ∅ )
15	10 14	mpbi	⊢ ( 1 ..^ 0 ) = ∅
16	9 15	eqtri	⊢ ( 1 ..^ ( ♯ ‘ ∅ ) ) = ∅
17	16	sumeq1i	⊢ Σ 𝑗 ∈ ( 1 ..^ ( ♯ ‘ ∅ ) ) if ( ( ( 𝑇 ‘ ∅ ) ‘ 𝑗 ) ≠ ( ( 𝑇 ‘ ∅ ) ‘ ( 𝑗 − 1 ) ) , 1 , 0 ) = Σ 𝑗 ∈ ∅ if ( ( ( 𝑇 ‘ ∅ ) ‘ 𝑗 ) ≠ ( ( 𝑇 ‘ ∅ ) ‘ ( 𝑗 − 1 ) ) , 1 , 0 )
18		sum0	⊢ Σ 𝑗 ∈ ∅ if ( ( ( 𝑇 ‘ ∅ ) ‘ 𝑗 ) ≠ ( ( 𝑇 ‘ ∅ ) ‘ ( 𝑗 − 1 ) ) , 1 , 0 ) = 0
19	7 17 18	3eqtri	⊢ ( 𝑉 ‘ ∅ ) = 0