Metamath Proof Explorer


Theorem coe1z

Description: The coefficient vector of 0. (Contributed by Stefan O'Rear, 23-Mar-2015)

Ref Expression
Hypotheses coe1z.p ⊢ P = Poly 1 ⁡ R
coe1z.z ⊢ 0 ˙ = 0 P
coe1z.y ⊢ Y = 0 R
Assertion coe1z ⊢ R ∈ Ring → coe 1 ⁡ 0 ˙ = ℕ 0 × Y

Proof

Step Hyp Ref Expression
1 coe1z.p ⊢ P = Poly 1 ⁡ R
2 coe1z.z ⊢ 0 ˙ = 0 P
3 coe1z.y ⊢ Y = 0 R
4 fconst6g ⊢ a ∈ ℕ 0 → 1 𝑜 × a : 1 𝑜 ⟶ ℕ 0
5 4 adantl ⊢ R ∈ Ring ∧ a ∈ ℕ 0 → 1 𝑜 × a : 1 𝑜 ⟶ ℕ 0
6 nn0ex ⊢ ℕ 0 ∈ V
7 1oex ⊢ 1 𝑜 ∈ V
8 6 7 elmap ⊢ 1 𝑜 × a ∈ ℕ 0 1 𝑜 ↔ 1 𝑜 × a : 1 𝑜 ⟶ ℕ 0
9 5 8 sylibr ⊢ R ∈ Ring ∧ a ∈ ℕ 0 → 1 𝑜 × a ∈ ℕ 0 1 𝑜
10 eqidd ⊢ R ∈ Ring → a ∈ ℕ 0 ⟼ 1 𝑜 × a = a ∈ ℕ 0 ⟼ 1 𝑜 × a
11 eqid ⊢ 1 𝑜 mPoly R = 1 𝑜 mPoly R
12 psr1baslem ⊢ ℕ 0 1 𝑜 = c ∈ ℕ 0 1 𝑜 | c -1 ℕ ∈ Fin
13 11 1 2 ply1mpl0 ⊢ 0 ˙ = 0 1 𝑜 mPoly R
14 1on ⊢ 1 𝑜 ∈ On
15 14 a1i ⊢ R ∈ Ring → 1 𝑜 ∈ On
16 ringgrp ⊢ R ∈ Ring → R ∈ Grp
17 11 12 3 13 15 16 mpl0 ⊢ R ∈ Ring → 0 ˙ = ℕ 0 1 𝑜 × Y
18 fconstmpt ⊢ ℕ 0 1 𝑜 × Y = b ∈ ℕ 0 1 𝑜 ⟼ Y
19 17 18 eqtrdi ⊢ R ∈ Ring → 0 ˙ = b ∈ ℕ 0 1 𝑜 ⟼ Y
20 eqidd ⊢ b = 1 𝑜 × a → Y = Y
21 9 10 19 20 fmptco ⊢ R ∈ Ring → 0 ˙ ∘ a ∈ ℕ 0 ⟼ 1 𝑜 × a = a ∈ ℕ 0 ⟼ Y
22 1 ply1ring ⊢ R ∈ Ring → P ∈ Ring
23 eqid ⊢ Base P = Base P
24 23 2 ring0cl ⊢ P ∈ Ring → 0 ˙ ∈ Base P
25 eqid ⊢ coe 1 ⁡ 0 ˙ = coe 1 ⁡ 0 ˙
26 eqid ⊢ a ∈ ℕ 0 ⟼ 1 𝑜 × a = a ∈ ℕ 0 ⟼ 1 𝑜 × a
27 25 23 1 26 coe1fval2 ⊢ 0 ˙ ∈ Base P → coe 1 ⁡ 0 ˙ = 0 ˙ ∘ a ∈ ℕ 0 ⟼ 1 𝑜 × a
28 22 24 27 3syl ⊢ R ∈ Ring → coe 1 ⁡ 0 ˙ = 0 ˙ ∘ a ∈ ℕ 0 ⟼ 1 𝑜 × a
29 fconstmpt ⊢ ℕ 0 × Y = a ∈ ℕ 0 ⟼ Y
30 29 a1i ⊢ R ∈ Ring → ℕ 0 × Y = a ∈ ℕ 0 ⟼ Y
31 21 28 30 3eqtr4d ⊢ R ∈ Ring → coe 1 ⁡ 0 ˙ = ℕ 0 × Y