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 0
Assertion srgbinomlem1 φ D 0 E 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 0
12 7 adantr φ D 0 E 0 R SRing
13 5 srgmgp R SRing G Mnd
14 7 13 syl φ G Mnd
15 14 adantr φ D 0 E 0 G Mnd
16 simprl φ D 0 E 0 D 0
17 8 adantr φ D 0 E 0 A S
18 5 1 mgpbas S = Base G
19 18 6 mulgnn0cl G Mnd D 0 A S D × ˙ A S
20 15 16 17 19 syl3anc φ D 0 E 0 D × ˙ A S
21 simprr φ D 0 E 0 E 0
22 9 adantr φ D 0 E 0 B S
23 18 6 mulgnn0cl G Mnd E 0 B S E × ˙ B S
24 15 21 22 23 syl3anc φ D 0 E 0 E × ˙ B S
25 1 2 srgcl R SRing D × ˙ A S E × ˙ B S D × ˙ A × ˙ E × ˙ B S
26 12 20 24 25 syl3anc φ D 0 E 0 D × ˙ A × ˙ E × ˙ B S