Metamath Proof Explorer


Theorem ply1plusgpropd

Description: Property deduction for univariate polynomial addition. (Contributed by Stefan O'Rear, 27-Mar-2015)

Ref Expression
Hypotheses ply1baspropd.b1 ⊢ φ → B = Base R
ply1baspropd.b2 ⊢ φ → B = Base S
ply1baspropd.p ⊢ φ ∧ x ∈ B ∧ y ∈ B → x + R y = x + S y
Assertion ply1plusgpropd ⊢ φ → + Poly 1 ⁡ R = + Poly 1 ⁡ S

Proof

Step Hyp Ref Expression
1 ply1baspropd.b1 ⊢ φ → B = Base R
2 ply1baspropd.b2 ⊢ φ → B = Base S
3 ply1baspropd.p ⊢ φ ∧ x ∈ B ∧ y ∈ B → x + R y = x + S y
4 1 2 3 psrplusgpropd ⊢ φ → + 1 𝑜 mPwSer R = + 1 𝑜 mPwSer S
5 eqid ⊢ 1 𝑜 mPoly R = 1 𝑜 mPoly R
6 eqid ⊢ 1 𝑜 mPwSer R = 1 𝑜 mPwSer R
7 eqid ⊢ + 1 𝑜 mPoly R = + 1 𝑜 mPoly R
8 5 6 7 mplplusg ⊢ + 1 𝑜 mPoly R = + 1 𝑜 mPwSer R
9 eqid ⊢ 1 𝑜 mPoly S = 1 𝑜 mPoly S
10 eqid ⊢ 1 𝑜 mPwSer S = 1 𝑜 mPwSer S
11 eqid ⊢ + 1 𝑜 mPoly S = + 1 𝑜 mPoly S
12 9 10 11 mplplusg ⊢ + 1 𝑜 mPoly S = + 1 𝑜 mPwSer S
13 4 8 12 3eqtr4g ⊢ φ → + 1 𝑜 mPoly R = + 1 𝑜 mPoly S
14 eqid ⊢ Poly 1 ⁡ R = Poly 1 ⁡ R
15 eqid ⊢ + Poly 1 ⁡ R = + Poly 1 ⁡ R
16 14 5 15 ply1plusg ⊢ + Poly 1 ⁡ R = + 1 𝑜 mPoly R
17 eqid ⊢ Poly 1 ⁡ S = Poly 1 ⁡ S
18 eqid ⊢ + Poly 1 ⁡ S = + Poly 1 ⁡ S
19 17 9 18 ply1plusg ⊢ + Poly 1 ⁡ S = + 1 𝑜 mPoly S
20 13 16 19 3eqtr4g ⊢ φ → + Poly 1 ⁡ R = + Poly 1 ⁡ S