Theorem signshnz

Description: H is not the empty word. (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	signshnz	⊢ ( ( 𝐹 ∈ Word ℝ ∧ 𝐶 ∈ ℝ+ ) → 𝐻 ≠ ∅ )

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	signshlen	⊢ ( ( 𝐹 ∈ Word ℝ ∧ 𝐶 ∈ ℝ+ ) → ( ♯ ‘ 𝐻 ) = ( ( ♯ ‘ 𝐹 ) + 1 ) )
7		lencl	⊢ ( 𝐹 ∈ Word ℝ → ( ♯ ‘ 𝐹 ) ∈ ℕ0 )
8	7	adantr	⊢ ( ( 𝐹 ∈ Word ℝ ∧ 𝐶 ∈ ℝ+ ) → ( ♯ ‘ 𝐹 ) ∈ ℕ0 )
9		nn0p1nn	⊢ ( ( ♯ ‘ 𝐹 ) ∈ ℕ0 → ( ( ♯ ‘ 𝐹 ) + 1 ) ∈ ℕ )
10	8 9	syl	⊢ ( ( 𝐹 ∈ Word ℝ ∧ 𝐶 ∈ ℝ+ ) → ( ( ♯ ‘ 𝐹 ) + 1 ) ∈ ℕ )
11	6 10	eqeltrd	⊢ ( ( 𝐹 ∈ Word ℝ ∧ 𝐶 ∈ ℝ+ ) → ( ♯ ‘ 𝐻 ) ∈ ℕ )
12	11	nnne0d	⊢ ( ( 𝐹 ∈ Word ℝ ∧ 𝐶 ∈ ℝ+ ) → ( ♯ ‘ 𝐻 ) ≠ 0 )
13	1 2 3 4 5	signshwrd	⊢ ( ( 𝐹 ∈ Word ℝ ∧ 𝐶 ∈ ℝ+ ) → 𝐻 ∈ Word ℝ )
14		hasheq0	⊢ ( 𝐻 ∈ Word ℝ → ( ( ♯ ‘ 𝐻 ) = 0 ↔ 𝐻 = ∅ ) )
15	13 14	syl	⊢ ( ( 𝐹 ∈ Word ℝ ∧ 𝐶 ∈ ℝ+ ) → ( ( ♯ ‘ 𝐻 ) = 0 ↔ 𝐻 = ∅ ) )
16	15	necon3bid	⊢ ( ( 𝐹 ∈ Word ℝ ∧ 𝐶 ∈ ℝ+ ) → ( ( ♯ ‘ 𝐻 ) ≠ 0 ↔ 𝐻 ≠ ∅ ) )
17	12 16	mpbid	⊢ ( ( 𝐹 ∈ Word ℝ ∧ 𝐶 ∈ ℝ+ ) → 𝐻 ≠ ∅ )