Metamath Proof Explorer


Theorem deg1submon1p

Description: The difference of two monic polynomials of the same degree is a polynomial of lesser degree. (Contributed by Stefan O'Rear, 28-Mar-2015)

Ref Expression
Hypotheses deg1submon1p.d D = deg 1 R
deg1submon1p.o O = Monic 1p R
deg1submon1p.p P = Poly 1 R
deg1submon1p.m - ˙ = - P
deg1submon1p.r φ R Ring
deg1submon1p.f1 φ F O
deg1submon1p.f2 φ D F = X
deg1submon1p.g1 φ G O
deg1submon1p.g2 φ D G = X
Assertion deg1submon1p φ D F - ˙ G < X

Proof

Step Hyp Ref Expression
1 deg1submon1p.d D = deg 1 R
2 deg1submon1p.o O = Monic 1p R
3 deg1submon1p.p P = Poly 1 R
4 deg1submon1p.m - ˙ = - P
5 deg1submon1p.r φ R Ring
6 deg1submon1p.f1 φ F O
7 deg1submon1p.f2 φ D F = X
8 deg1submon1p.g1 φ G O
9 deg1submon1p.g2 φ D G = X
10 eqid Base P = Base P
11 3 10 2 mon1pcl F O F Base P
12 6 11 syl φ F Base P
13 eqid 0 P = 0 P
14 3 13 2 mon1pn0 F O F 0 P
15 6 14 syl φ F 0 P
16 1 3 13 10 deg1nn0cl R Ring F Base P F 0 P D F 0
17 5 12 15 16 syl3anc φ D F 0
18 7 17 eqeltrrd φ X 0
19 18 nn0red φ X
20 19 leidd φ X X
21 7 20 eqbrtrd φ D F X
22 3 10 2 mon1pcl G O G Base P
23 8 22 syl φ G Base P
24 9 20 eqbrtrd φ D G X
25 eqid coe 1 F = coe 1 F
26 eqid coe 1 G = coe 1 G
27 7 fveq2d φ coe 1 F D F = coe 1 F X
28 eqid 1 R = 1 R
29 1 28 2 mon1pldg F O coe 1 F D F = 1 R
30 6 29 syl φ coe 1 F D F = 1 R
31 27 30 eqtr3d φ coe 1 F X = 1 R
32 1 28 2 mon1pldg G O coe 1 G D G = 1 R
33 8 32 syl φ coe 1 G D G = 1 R
34 9 fveq2d φ coe 1 G D G = coe 1 G X
35 31 33 34 3eqtr2d φ coe 1 F X = coe 1 G X
36 1 3 10 4 18 5 12 21 23 24 25 26 35 deg1sublt φ D F - ˙ G < X