Metamath Proof Explorer


Theorem assamulgscmlem1

Description: Lemma 1 for assamulgscm (induction base). (Contributed by AV, 26-Aug-2019)

Ref Expression
Hypotheses assamulgscm.v ⊢ V = Base W
assamulgscm.f ⊢ F = Scalar ⁡ W
assamulgscm.b ⊢ B = Base F
assamulgscm.s ⊢ · ˙ = ⋅ W
assamulgscm.g ⊢ G = mulGrp F
assamulgscm.p ⊢ × ˙ = ⋅ G
assamulgscm.h ⊢ H = mulGrp W
assamulgscm.e ⊢ E = ⋅ H
Assertion assamulgscmlem1 ⊢ A ∈ B ∧ X ∈ V ∧ W ∈ AssAlg → 0 E A · ˙ X = 0 × ˙ A · ˙ 0 E X

Proof

Step Hyp Ref Expression
1 assamulgscm.v ⊢ V = Base W
2 assamulgscm.f ⊢ F = Scalar ⁡ W
3 assamulgscm.b ⊢ B = Base F
4 assamulgscm.s ⊢ · ˙ = ⋅ W
5 assamulgscm.g ⊢ G = mulGrp F
6 assamulgscm.p ⊢ × ˙ = ⋅ G
7 assamulgscm.h ⊢ H = mulGrp W
8 assamulgscm.e ⊢ E = ⋅ H
9 assalmod ⊢ W ∈ AssAlg → W ∈ LMod
10 assaring ⊢ W ∈ AssAlg → W ∈ Ring
11 eqid ⊢ 1 W = 1 W
12 1 11 ringidcl ⊢ W ∈ Ring → 1 W ∈ V
13 10 12 syl ⊢ W ∈ AssAlg → 1 W ∈ V
14 eqid ⊢ 1 F = 1 F
15 1 2 4 14 lmodvs1 ⊢ W ∈ LMod ∧ 1 W ∈ V → 1 F · ˙ 1 W = 1 W
16 15 eqcomd ⊢ W ∈ LMod ∧ 1 W ∈ V → 1 W = 1 F · ˙ 1 W
17 9 13 16 syl2anc ⊢ W ∈ AssAlg → 1 W = 1 F · ˙ 1 W
18 17 adantl ⊢ A ∈ B ∧ X ∈ V ∧ W ∈ AssAlg → 1 W = 1 F · ˙ 1 W
19 9 adantl ⊢ A ∈ B ∧ X ∈ V ∧ W ∈ AssAlg → W ∈ LMod
20 simpll ⊢ A ∈ B ∧ X ∈ V ∧ W ∈ AssAlg → A ∈ B
21 simplr ⊢ A ∈ B ∧ X ∈ V ∧ W ∈ AssAlg → X ∈ V
22 1 2 4 3 lmodvscl ⊢ W ∈ LMod ∧ A ∈ B ∧ X ∈ V → A · ˙ X ∈ V
23 19 20 21 22 syl3anc ⊢ A ∈ B ∧ X ∈ V ∧ W ∈ AssAlg → A · ˙ X ∈ V
24 7 1 mgpbas ⊢ V = Base H
25 7 11 ringidval ⊢ 1 W = 0 H
26 24 25 8 mulg0 ⊢ A · ˙ X ∈ V → 0 E A · ˙ X = 1 W
27 23 26 syl ⊢ A ∈ B ∧ X ∈ V ∧ W ∈ AssAlg → 0 E A · ˙ X = 1 W
28 5 3 mgpbas ⊢ B = Base G
29 5 14 ringidval ⊢ 1 F = 0 G
30 28 29 6 mulg0 ⊢ A ∈ B → 0 × ˙ A = 1 F
31 20 30 syl ⊢ A ∈ B ∧ X ∈ V ∧ W ∈ AssAlg → 0 × ˙ A = 1 F
32 24 25 8 mulg0 ⊢ X ∈ V → 0 E X = 1 W
33 21 32 syl ⊢ A ∈ B ∧ X ∈ V ∧ W ∈ AssAlg → 0 E X = 1 W
34 31 33 oveq12d ⊢ A ∈ B ∧ X ∈ V ∧ W ∈ AssAlg → 0 × ˙ A · ˙ 0 E X = 1 F · ˙ 1 W
35 18 27 34 3eqtr4d ⊢ A ∈ B ∧ X ∈ V ∧ W ∈ AssAlg → 0 E A · ˙ X = 0 × ˙ A · ˙ 0 E X