Metamath Proof Explorer


Theorem clmvsubval

Description: Value of vector subtraction in terms of addition in a subcomplex module. Analogue of lmodvsubval2 . (Contributed by NM, 31-Mar-2014) (Revised by AV, 7-Oct-2021)

Ref Expression
Hypotheses clmvsubval.v V = Base W
clmvsubval.p + ˙ = + W
clmvsubval.m - ˙ = - W
clmvsubval.f F = Scalar W
clmvsubval.s · ˙ = W
Assertion clmvsubval W CMod A V B V A - ˙ B = A + ˙ -1 · ˙ B

Proof

Step Hyp Ref Expression
1 clmvsubval.v V = Base W
2 clmvsubval.p + ˙ = + W
3 clmvsubval.m - ˙ = - W
4 clmvsubval.f F = Scalar W
5 clmvsubval.s · ˙ = W
6 clmlmod W CMod W LMod
7 eqid inv g F = inv g F
8 eqid 1 F = 1 F
9 1 2 3 4 5 7 8 lmodvsubval2 W LMod A V B V A - ˙ B = A + ˙ inv g F 1 F · ˙ B
10 6 9 syl3an1 W CMod A V B V A - ˙ B = A + ˙ inv g F 1 F · ˙ B
11 4 clm1 W CMod 1 = 1 F
12 11 eqcomd W CMod 1 F = 1
13 12 fveq2d W CMod inv g F 1 F = inv g F 1
14 4 clmring W CMod F Ring
15 eqid Base F = Base F
16 15 8 ringidcl F Ring 1 F Base F
17 14 16 syl W CMod 1 F Base F
18 11 17 eqeltrd W CMod 1 Base F
19 4 15 clmneg W CMod 1 Base F 1 = inv g F 1
20 18 19 mpdan W CMod 1 = inv g F 1
21 13 20 eqtr4d W CMod inv g F 1 F = 1
22 21 3ad2ant1 W CMod A V B V inv g F 1 F = 1
23 22 oveq1d W CMod A V B V inv g F 1 F · ˙ B = -1 · ˙ B
24 23 oveq2d W CMod A V B V A + ˙ inv g F 1 F · ˙ B = A + ˙ -1 · ˙ B
25 10 24 eqtrd W CMod A V B V A - ˙ B = A + ˙ -1 · ˙ B