Theorem signshlen

Description: Length of H , corresponding to the word F multiplied by ( x - C ) . (Contributed by Thierry Arnoux, 14-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 ) )
	signs.h	⊢ 𝐻 = ( ( ⟨“ 0 ”⟩ ++ 𝐹 ) ∘f − ( ( 𝐹 ++ ⟨“ 0 ”⟩ ) ∘f/c · 𝐶 ) )
Assertion	signshlen	⊢ ( ( 𝐹 ∈ Word ℝ ∧ 𝐶 ∈ ℝ+ ) → ( ♯ ‘ 𝐻 ) = ( ( ♯ ‘ 𝐹 ) + 1 ) )

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		signs.h	⊢ 𝐻 = ( ( ⟨“ 0 ”⟩ ++ 𝐹 ) ∘f − ( ( 𝐹 ++ ⟨“ 0 ”⟩ ) ∘f/c · 𝐶 ) )
6	1 2 3 4 5	signshf	⊢ ( ( 𝐹 ∈ Word ℝ ∧ 𝐶 ∈ ℝ+ ) → 𝐻 : ( 0 ..^ ( ( ♯ ‘ 𝐹 ) + 1 ) ) ⟶ ℝ )
7		ffn	⊢ ( 𝐻 : ( 0 ..^ ( ( ♯ ‘ 𝐹 ) + 1 ) ) ⟶ ℝ → 𝐻 Fn ( 0 ..^ ( ( ♯ ‘ 𝐹 ) + 1 ) ) )
8		hashfn	⊢ ( 𝐻 Fn ( 0 ..^ ( ( ♯ ‘ 𝐹 ) + 1 ) ) → ( ♯ ‘ 𝐻 ) = ( ♯ ‘ ( 0 ..^ ( ( ♯ ‘ 𝐹 ) + 1 ) ) ) )
9	6 7 8	3syl	⊢ ( ( 𝐹 ∈ Word ℝ ∧ 𝐶 ∈ ℝ+ ) → ( ♯ ‘ 𝐻 ) = ( ♯ ‘ ( 0 ..^ ( ( ♯ ‘ 𝐹 ) + 1 ) ) ) )
10		lencl	⊢ ( 𝐹 ∈ Word ℝ → ( ♯ ‘ 𝐹 ) ∈ ℕ0 )
11	10	adantr	⊢ ( ( 𝐹 ∈ Word ℝ ∧ 𝐶 ∈ ℝ+ ) → ( ♯ ‘ 𝐹 ) ∈ ℕ0 )
12		1nn0	⊢ 1 ∈ ℕ0
13	12	a1i	⊢ ( ( 𝐹 ∈ Word ℝ ∧ 𝐶 ∈ ℝ+ ) → 1 ∈ ℕ0 )
14	11 13	nn0addcld	⊢ ( ( 𝐹 ∈ Word ℝ ∧ 𝐶 ∈ ℝ+ ) → ( ( ♯ ‘ 𝐹 ) + 1 ) ∈ ℕ0 )
15		hashfzo0	⊢ ( ( ( ♯ ‘ 𝐹 ) + 1 ) ∈ ℕ0 → ( ♯ ‘ ( 0 ..^ ( ( ♯ ‘ 𝐹 ) + 1 ) ) ) = ( ( ♯ ‘ 𝐹 ) + 1 ) )
16	14 15	syl	⊢ ( ( 𝐹 ∈ Word ℝ ∧ 𝐶 ∈ ℝ+ ) → ( ♯ ‘ ( 0 ..^ ( ( ♯ ‘ 𝐹 ) + 1 ) ) ) = ( ( ♯ ‘ 𝐹 ) + 1 ) )
17	9 16	eqtrd	⊢ ( ( 𝐹 ∈ Word ℝ ∧ 𝐶 ∈ ℝ+ ) → ( ♯ ‘ 𝐻 ) = ( ( ♯ ‘ 𝐹 ) + 1 ) )