Metamath Proof Explorer


Theorem deg1tm

Description: Exact degree of a polynomial term. (Contributed by Stefan O'Rear, 27-Mar-2015)

Ref Expression
Hypotheses deg1tm.d 𝐷 = ( deg1𝑅 )
deg1tm.k 𝐾 = ( Base ‘ 𝑅 )
deg1tm.p 𝑃 = ( Poly1𝑅 )
deg1tm.x 𝑋 = ( var1𝑅 )
deg1tm.m · = ( ·𝑠𝑃 )
deg1tm.n 𝑁 = ( mulGrp ‘ 𝑃 )
deg1tm.e = ( .g𝑁 )
deg1tm.z 0 = ( 0g𝑅 )
Assertion deg1tm ( ( 𝑅 ∈ Ring ∧ ( 𝐶𝐾𝐶0 ) ∧ 𝐹 ∈ ℕ0 ) → ( 𝐷 ‘ ( 𝐶 · ( 𝐹 𝑋 ) ) ) = 𝐹 )

Proof

Step Hyp Ref Expression
1 deg1tm.d 𝐷 = ( deg1𝑅 )
2 deg1tm.k 𝐾 = ( Base ‘ 𝑅 )
3 deg1tm.p 𝑃 = ( Poly1𝑅 )
4 deg1tm.x 𝑋 = ( var1𝑅 )
5 deg1tm.m · = ( ·𝑠𝑃 )
6 deg1tm.n 𝑁 = ( mulGrp ‘ 𝑃 )
7 deg1tm.e = ( .g𝑁 )
8 deg1tm.z 0 = ( 0g𝑅 )
9 eqid ( Base ‘ 𝑃 ) = ( Base ‘ 𝑃 )
10 2 3 4 5 6 7 9 ply1tmcl ( ( 𝑅 ∈ Ring ∧ 𝐶𝐾𝐹 ∈ ℕ0 ) → ( 𝐶 · ( 𝐹 𝑋 ) ) ∈ ( Base ‘ 𝑃 ) )
11 10 3adant2r ( ( 𝑅 ∈ Ring ∧ ( 𝐶𝐾𝐶0 ) ∧ 𝐹 ∈ ℕ0 ) → ( 𝐶 · ( 𝐹 𝑋 ) ) ∈ ( Base ‘ 𝑃 ) )
12 1 3 9 deg1xrcl ( ( 𝐶 · ( 𝐹 𝑋 ) ) ∈ ( Base ‘ 𝑃 ) → ( 𝐷 ‘ ( 𝐶 · ( 𝐹 𝑋 ) ) ) ∈ ℝ* )
13 11 12 syl ( ( 𝑅 ∈ Ring ∧ ( 𝐶𝐾𝐶0 ) ∧ 𝐹 ∈ ℕ0 ) → ( 𝐷 ‘ ( 𝐶 · ( 𝐹 𝑋 ) ) ) ∈ ℝ* )
14 simp3 ( ( 𝑅 ∈ Ring ∧ ( 𝐶𝐾𝐶0 ) ∧ 𝐹 ∈ ℕ0 ) → 𝐹 ∈ ℕ0 )
15 14 nn0red ( ( 𝑅 ∈ Ring ∧ ( 𝐶𝐾𝐶0 ) ∧ 𝐹 ∈ ℕ0 ) → 𝐹 ∈ ℝ )
16 15 rexrd ( ( 𝑅 ∈ Ring ∧ ( 𝐶𝐾𝐶0 ) ∧ 𝐹 ∈ ℕ0 ) → 𝐹 ∈ ℝ* )
17 1 2 3 4 5 6 7 deg1tmle ( ( 𝑅 ∈ Ring ∧ 𝐶𝐾𝐹 ∈ ℕ0 ) → ( 𝐷 ‘ ( 𝐶 · ( 𝐹 𝑋 ) ) ) ≤ 𝐹 )
18 17 3adant2r ( ( 𝑅 ∈ Ring ∧ ( 𝐶𝐾𝐶0 ) ∧ 𝐹 ∈ ℕ0 ) → ( 𝐷 ‘ ( 𝐶 · ( 𝐹 𝑋 ) ) ) ≤ 𝐹 )
19 8 2 3 4 5 6 7 coe1tmfv1 ( ( 𝑅 ∈ Ring ∧ 𝐶𝐾𝐹 ∈ ℕ0 ) → ( ( coe1 ‘ ( 𝐶 · ( 𝐹 𝑋 ) ) ) ‘ 𝐹 ) = 𝐶 )
20 19 3adant2r ( ( 𝑅 ∈ Ring ∧ ( 𝐶𝐾𝐶0 ) ∧ 𝐹 ∈ ℕ0 ) → ( ( coe1 ‘ ( 𝐶 · ( 𝐹 𝑋 ) ) ) ‘ 𝐹 ) = 𝐶 )
21 simp2r ( ( 𝑅 ∈ Ring ∧ ( 𝐶𝐾𝐶0 ) ∧ 𝐹 ∈ ℕ0 ) → 𝐶0 )
22 20 21 eqnetrd ( ( 𝑅 ∈ Ring ∧ ( 𝐶𝐾𝐶0 ) ∧ 𝐹 ∈ ℕ0 ) → ( ( coe1 ‘ ( 𝐶 · ( 𝐹 𝑋 ) ) ) ‘ 𝐹 ) ≠ 0 )
23 eqid ( coe1 ‘ ( 𝐶 · ( 𝐹 𝑋 ) ) ) = ( coe1 ‘ ( 𝐶 · ( 𝐹 𝑋 ) ) )
24 1 3 9 8 23 deg1ge ( ( ( 𝐶 · ( 𝐹 𝑋 ) ) ∈ ( Base ‘ 𝑃 ) ∧ 𝐹 ∈ ℕ0 ∧ ( ( coe1 ‘ ( 𝐶 · ( 𝐹 𝑋 ) ) ) ‘ 𝐹 ) ≠ 0 ) → 𝐹 ≤ ( 𝐷 ‘ ( 𝐶 · ( 𝐹 𝑋 ) ) ) )
25 11 14 22 24 syl3anc ( ( 𝑅 ∈ Ring ∧ ( 𝐶𝐾𝐶0 ) ∧ 𝐹 ∈ ℕ0 ) → 𝐹 ≤ ( 𝐷 ‘ ( 𝐶 · ( 𝐹 𝑋 ) ) ) )
26 13 16 18 25 xrletrid ( ( 𝑅 ∈ Ring ∧ ( 𝐶𝐾𝐶0 ) ∧ 𝐹 ∈ ℕ0 ) → ( 𝐷 ‘ ( 𝐶 · ( 𝐹 𝑋 ) ) ) = 𝐹 )