Metamath Proof Explorer


Theorem deg1ldgn

Description: An index at which a polynomial is zero, cannot be its degree. (Contributed by Stefan O'Rear, 23-Mar-2015)

Ref Expression
Hypotheses deg1z.d D = deg 1 R
deg1z.p P = Poly 1 R
deg1z.z 0 ˙ = 0 P
deg1nn0cl.b B = Base P
deg1ldg.y Y = 0 R
deg1ldg.a A = coe 1 F
deg1ldgn.r φ R Ring
deg1ldgn.f φ F B
deg1ldgn.x φ X 0
deg1ldgn.e φ A X = Y
Assertion deg1ldgn φ D F X

Proof

Step Hyp Ref Expression
1 deg1z.d D = deg 1 R
2 deg1z.p P = Poly 1 R
3 deg1z.z 0 ˙ = 0 P
4 deg1nn0cl.b B = Base P
5 deg1ldg.y Y = 0 R
6 deg1ldg.a A = coe 1 F
7 deg1ldgn.r φ R Ring
8 deg1ldgn.f φ F B
9 deg1ldgn.x φ X 0
10 deg1ldgn.e φ A X = Y
11 fveq2 D F = X A D F = A X
12 11 adantl φ D F = X A D F = A X
13 7 adantr φ D F = X R Ring
14 8 adantr φ D F = X F B
15 eleq1a X 0 D F = X D F 0
16 9 15 syl φ D F = X D F 0
17 16 imp φ D F = X D F 0
18 1 2 3 4 deg1nn0clb R Ring F B F 0 ˙ D F 0
19 7 8 18 syl2anc φ F 0 ˙ D F 0
20 19 adantr φ D F = X F 0 ˙ D F 0
21 17 20 mpbird φ D F = X F 0 ˙
22 1 2 3 4 5 6 deg1ldg R Ring F B F 0 ˙ A D F Y
23 13 14 21 22 syl3anc φ D F = X A D F Y
24 12 23 eqnetrrd φ D F = X A X Y
25 24 ex φ D F = X A X Y
26 25 necon2d φ A X = Y D F X
27 10 26 mpd φ D F X