Metamath Proof Explorer


Theorem redwlklem

Description: Lemma for redwlk . (Contributed by Alexander van der Vekens, 1-Nov-2017) (Revised by AV, 29-Jan-2021)

Ref Expression
Assertion redwlklem ⊢ F ∈ Word S ∧ 1 ≤ F ∧ P : 0 … F ⟶ V → P ↾ 0 ..^ F : 0 … F ↾ 0 ..^ F − 1 ⟶ V

Proof

Step Hyp Ref Expression
1 simpr ⊢ F ∈ Word S ∧ 1 ≤ F ∧ P : 0 … F ⟶ V → P : 0 … F ⟶ V
2 fzossfz ⊢ 0 ..^ F ⊆ 0 … F
3 fssres ⊢ P : 0 … F ⟶ V ∧ 0 ..^ F ⊆ 0 … F → P ↾ 0 ..^ F : 0 ..^ F ⟶ V
4 1 2 3 sylancl ⊢ F ∈ Word S ∧ 1 ≤ F ∧ P : 0 … F ⟶ V → P ↾ 0 ..^ F : 0 ..^ F ⟶ V
5 4 ex ⊢ F ∈ Word S ∧ 1 ≤ F → P : 0 … F ⟶ V → P ↾ 0 ..^ F : 0 ..^ F ⟶ V
6 lencl ⊢ F ∈ Word S → F ∈ ℕ 0
7 6 nn0zd ⊢ F ∈ Word S → F ∈ ℤ
8 fzoval ⊢ F ∈ ℤ → 0 ..^ F = 0 … F − 1
9 7 8 syl ⊢ F ∈ Word S → 0 ..^ F = 0 … F − 1
10 9 adantr ⊢ F ∈ Word S ∧ 1 ≤ F → 0 ..^ F = 0 … F − 1
11 wrdred1hash ⊢ F ∈ Word S ∧ 1 ≤ F → F ↾ 0 ..^ F − 1 = F − 1
12 oveq2 ⊢ F ↾ 0 ..^ F − 1 = F − 1 → 0 … F ↾ 0 ..^ F − 1 = 0 … F − 1
13 12 eqeq2d ⊢ F ↾ 0 ..^ F − 1 = F − 1 → 0 ..^ F = 0 … F ↾ 0 ..^ F − 1 ↔ 0 ..^ F = 0 … F − 1
14 11 13 syl ⊢ F ∈ Word S ∧ 1 ≤ F → 0 ..^ F = 0 … F ↾ 0 ..^ F − 1 ↔ 0 ..^ F = 0 … F − 1
15 10 14 mpbird ⊢ F ∈ Word S ∧ 1 ≤ F → 0 ..^ F = 0 … F ↾ 0 ..^ F − 1
16 15 feq2d ⊢ F ∈ Word S ∧ 1 ≤ F → P ↾ 0 ..^ F : 0 ..^ F ⟶ V ↔ P ↾ 0 ..^ F : 0 … F ↾ 0 ..^ F − 1 ⟶ V
17 5 16 sylibd ⊢ F ∈ Word S ∧ 1 ≤ F → P : 0 … F ⟶ V → P ↾ 0 ..^ F : 0 … F ↾ 0 ..^ F − 1 ⟶ V
18 17 3impia ⊢ F ∈ Word S ∧ 1 ≤ F ∧ P : 0 … F ⟶ V → P ↾ 0 ..^ F : 0 … F ↾ 0 ..^ F − 1 ⟶ V