Metamath Proof Explorer


Theorem srgbinomlem1

Description: Lemma 1 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 srgbinomlem1 ⊢ φ ∧ D ∈ ℕ 0 ∧ E ∈ ℕ 0 → 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 7 adantr ⊢ φ ∧ D ∈ ℕ 0 ∧ E ∈ ℕ 0 → R ∈ SRing
13 5 1 mgpbas ⊢ S = Base G
14 5 srgmgp ⊢ R ∈ SRing → G ∈ Mnd
15 7 14 syl ⊢ φ → G ∈ Mnd
16 15 adantr ⊢ φ ∧ D ∈ ℕ 0 ∧ E ∈ ℕ 0 → G ∈ Mnd
17 simprl ⊢ φ ∧ D ∈ ℕ 0 ∧ E ∈ ℕ 0 → D ∈ ℕ 0
18 8 adantr ⊢ φ ∧ D ∈ ℕ 0 ∧ E ∈ ℕ 0 → A ∈ S
19 13 6 16 17 18 mulgnn0cld ⊢ φ ∧ D ∈ ℕ 0 ∧ E ∈ ℕ 0 → D × ˙ A ∈ S
20 simprr ⊢ φ ∧ D ∈ ℕ 0 ∧ E ∈ ℕ 0 → E ∈ ℕ 0
21 9 adantr ⊢ φ ∧ D ∈ ℕ 0 ∧ E ∈ ℕ 0 → B ∈ S
22 13 6 16 20 21 mulgnn0cld ⊢ φ ∧ D ∈ ℕ 0 ∧ E ∈ ℕ 0 → E × ˙ B ∈ S
23 1 2 srgcl ⊢ R ∈ SRing ∧ D × ˙ A ∈ S ∧ E × ˙ B ∈ S → D × ˙ A × ˙ E × ˙ B ∈ S
24 12 19 22 23 syl3anc ⊢ φ ∧ D ∈ ℕ 0 ∧ E ∈ ℕ 0 → D × ˙ A × ˙ E × ˙ B ∈ S