Metamath Proof Explorer


Theorem srgbinomlem2

Description: Lemma 2 for srgbinomlem . (Contributed by AV, 23-Aug-2019)

Ref Expression
Hypotheses srgbinom.s ⊢ S = Base R
srgbinom.m ⊢ × ˙ = ⋅ R
srgbinom.t ⊢ · ˙ = ⋅ R
srgbinom.a ⊢ + ˙ = + R
srgbinom.g ⊢ G = mulGrp R
srgbinom.e ⊢ × ˙ = ⋅ G
srgbinomlem.r ⊢ φ → R ∈ SRing
srgbinomlem.a ⊢ φ → A ∈ S
srgbinomlem.b ⊢ φ → B ∈ S
srgbinomlem.c ⊢ φ → A × ˙ B = B × ˙ A
srgbinomlem.n ⊢ φ → N ∈ ℕ 0
Assertion srgbinomlem2 ⊢ φ ∧ C ∈ ℕ 0 ∧ D ∈ ℕ 0 ∧ E ∈ ℕ 0 → C · ˙ D × ˙ A × ˙ E × ˙ B ∈ S

Proof

Step Hyp Ref Expression
1 srgbinom.s ⊢ S = Base R
2 srgbinom.m ⊢ × ˙ = ⋅ R
3 srgbinom.t ⊢ · ˙ = ⋅ R
4 srgbinom.a ⊢ + ˙ = + R
5 srgbinom.g ⊢ G = mulGrp R
6 srgbinom.e ⊢ × ˙ = ⋅ G
7 srgbinomlem.r ⊢ φ → R ∈ SRing
8 srgbinomlem.a ⊢ φ → A ∈ S
9 srgbinomlem.b ⊢ φ → B ∈ S
10 srgbinomlem.c ⊢ φ → A × ˙ B = B × ˙ A
11 srgbinomlem.n ⊢ φ → N ∈ ℕ 0
12 srgmnd ⊢ R ∈ SRing → R ∈ Mnd
13 7 12 syl ⊢ φ → R ∈ Mnd
14 13 adantr ⊢ φ ∧ C ∈ ℕ 0 ∧ D ∈ ℕ 0 ∧ E ∈ ℕ 0 → R ∈ Mnd
15 simpr1 ⊢ φ ∧ C ∈ ℕ 0 ∧ D ∈ ℕ 0 ∧ E ∈ ℕ 0 → C ∈ ℕ 0
16 1 2 3 4 5 6 7 8 9 10 11 srgbinomlem1 ⊢ φ ∧ D ∈ ℕ 0 ∧ E ∈ ℕ 0 → D × ˙ A × ˙ E × ˙ B ∈ S
17 16 3adantr1 ⊢ φ ∧ C ∈ ℕ 0 ∧ D ∈ ℕ 0 ∧ E ∈ ℕ 0 → D × ˙ A × ˙ E × ˙ B ∈ S
18 1 3 14 15 17 mulgnn0cld ⊢ φ ∧ C ∈ ℕ 0 ∧ D ∈ ℕ 0 ∧ E ∈ ℕ 0 → C · ˙ D × ˙ A × ˙ E × ˙ B ∈ S