Metamath Proof Explorer


Theorem psdascl

Description: The derivative of a constant polynomial is zero. (Contributed by SN, 25-Apr-2025)

Ref Expression
Hypotheses psdascl.s ⊢ S = I mPwSer R
psdascl.z ⊢ 0 ˙ = 0 S
psdascl.a ⊢ A = algSc ⁡ S
psdascl.b ⊢ B = Base R
psdascl.i ⊢ φ → I ∈ V
psdascl.r ⊢ φ → R ∈ CRing
psdascl.x ⊢ φ → X ∈ I
psdascl.c ⊢ φ → C ∈ B
Assertion psdascl ⊢ φ → I mPSDer R ⁡ X ⁡ A ⁡ C = 0 ˙

Proof

Step Hyp Ref Expression
1 psdascl.s ⊢ S = I mPwSer R
2 psdascl.z ⊢ 0 ˙ = 0 S
3 psdascl.a ⊢ A = algSc ⁡ S
4 psdascl.b ⊢ B = Base R
5 psdascl.i ⊢ φ → I ∈ V
6 psdascl.r ⊢ φ → R ∈ CRing
7 psdascl.x ⊢ φ → X ∈ I
8 psdascl.c ⊢ φ → C ∈ B
9 1 5 6 psrsca ⊢ φ → R = Scalar ⁡ S
10 9 fveq2d ⊢ φ → Base R = Base Scalar ⁡ S
11 4 10 eqtrid ⊢ φ → B = Base Scalar ⁡ S
12 8 11 eleqtrd ⊢ φ → C ∈ Base Scalar ⁡ S
13 eqid ⊢ Scalar ⁡ S = Scalar ⁡ S
14 eqid ⊢ Base Scalar ⁡ S = Base Scalar ⁡ S
15 eqid ⊢ ⋅ S = ⋅ S
16 eqid ⊢ 1 S = 1 S
17 3 13 14 15 16 asclval ⊢ C ∈ Base Scalar ⁡ S → A ⁡ C = C ⋅ S 1 S
18 12 17 syl ⊢ φ → A ⁡ C = C ⋅ S 1 S
19 18 fveq2d ⊢ φ → I mPSDer R ⁡ X ⁡ A ⁡ C = I mPSDer R ⁡ X ⁡ C ⋅ S 1 S
20 eqid ⊢ Base S = Base S
21 6 crngringd ⊢ φ → R ∈ Ring
22 1 5 21 psrring ⊢ φ → S ∈ Ring
23 20 16 ringidcl ⊢ S ∈ Ring → 1 S ∈ Base S
24 22 23 syl ⊢ φ → 1 S ∈ Base S
25 1 20 15 4 6 7 24 8 psdvsca ⊢ φ → I mPSDer R ⁡ X ⁡ C ⋅ S 1 S = C ⋅ S I mPSDer R ⁡ X ⁡ 1 S
26 1 16 2 5 6 7 psd1 ⊢ φ → I mPSDer R ⁡ X ⁡ 1 S = 0 ˙
27 26 oveq2d ⊢ φ → C ⋅ S I mPSDer R ⁡ X ⁡ 1 S = C ⋅ S 0 ˙
28 1 5 21 psrlmod ⊢ φ → S ∈ LMod
29 13 15 14 2 lmodvs0 ⊢ S ∈ LMod ∧ C ∈ Base Scalar ⁡ S → C ⋅ S 0 ˙ = 0 ˙
30 28 12 29 syl2anc ⊢ φ → C ⋅ S 0 ˙ = 0 ˙
31 27 30 eqtrd ⊢ φ → C ⋅ S I mPSDer R ⁡ X ⁡ 1 S = 0 ˙
32 19 25 31 3eqtrd ⊢ φ → I mPSDer R ⁡ X ⁡ A ⁡ C = 0 ˙