Metamath Proof Explorer


Theorem ismon1p

Description: Being a monic polynomial. (Contributed by Stefan O'Rear, 28-Mar-2015)

Ref Expression
Hypotheses uc1pval.p ⊢ P = Poly 1 ⁡ R
uc1pval.b ⊢ B = Base P
uc1pval.z ⊢ 0 ˙ = 0 P
uc1pval.d ⊢ D = deg 1 ⁡ R
mon1pval.m ⊢ M = Monic 1p ⁡ R
mon1pval.o ⊢ 1 ˙ = 1 R
Assertion ismon1p ⊢ F ∈ M ↔ F ∈ B ∧ F ≠ 0 ˙ ∧ coe 1 ⁡ F ⁡ D ⁡ F = 1 ˙

Proof

Step Hyp Ref Expression
1 uc1pval.p ⊢ P = Poly 1 ⁡ R
2 uc1pval.b ⊢ B = Base P
3 uc1pval.z ⊢ 0 ˙ = 0 P
4 uc1pval.d ⊢ D = deg 1 ⁡ R
5 mon1pval.m ⊢ M = Monic 1p ⁡ R
6 mon1pval.o ⊢ 1 ˙ = 1 R
7 neeq1 ⊢ f = F → f ≠ 0 ˙ ↔ F ≠ 0 ˙
8 fveq2 ⊢ f = F → coe 1 ⁡ f = coe 1 ⁡ F
9 fveq2 ⊢ f = F → D ⁡ f = D ⁡ F
10 8 9 fveq12d ⊢ f = F → coe 1 ⁡ f ⁡ D ⁡ f = coe 1 ⁡ F ⁡ D ⁡ F
11 10 eqeq1d ⊢ f = F → coe 1 ⁡ f ⁡ D ⁡ f = 1 ˙ ↔ coe 1 ⁡ F ⁡ D ⁡ F = 1 ˙
12 7 11 anbi12d ⊢ f = F → f ≠ 0 ˙ ∧ coe 1 ⁡ f ⁡ D ⁡ f = 1 ˙ ↔ F ≠ 0 ˙ ∧ coe 1 ⁡ F ⁡ D ⁡ F = 1 ˙
13 1 2 3 4 5 6 mon1pval ⊢ M = f ∈ B | f ≠ 0 ˙ ∧ coe 1 ⁡ f ⁡ D ⁡ f = 1 ˙
14 12 13 elrab2 ⊢ F ∈ M ↔ F ∈ B ∧ F ≠ 0 ˙ ∧ coe 1 ⁡ F ⁡ D ⁡ F = 1 ˙
15 3anass ⊢ F ∈ B ∧ F ≠ 0 ˙ ∧ coe 1 ⁡ F ⁡ D ⁡ F = 1 ˙ ↔ F ∈ B ∧ F ≠ 0 ˙ ∧ coe 1 ⁡ F ⁡ D ⁡ F = 1 ˙
16 14 15 bitr4i ⊢ F ∈ M ↔ F ∈ B ∧ F ≠ 0 ˙ ∧ coe 1 ⁡ F ⁡ D ⁡ F = 1 ˙