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 ˙