Metamath Proof Explorer


Theorem dvhvaddcomN

Description: Commutativity of vector sum. (Contributed by NM, 26-Oct-2013) (Revised by Mario Carneiro, 23-Jun-2014) (New usage is discouraged.)

Ref Expression
Hypotheses dvhvaddcl.h H = LHyp K
dvhvaddcl.t T = LTrn K W
dvhvaddcl.e E = TEndo K W
dvhvaddcl.u U = DVecH K W
dvhvaddcl.d D = Scalar U
dvhvaddcl.p ˙ = + D
dvhvaddcl.a + ˙ = + U
Assertion dvhvaddcomN K HL W H F T × E G T × E F + ˙ G = G + ˙ F

Proof

Step Hyp Ref Expression
1 dvhvaddcl.h H = LHyp K
2 dvhvaddcl.t T = LTrn K W
3 dvhvaddcl.e E = TEndo K W
4 dvhvaddcl.u U = DVecH K W
5 dvhvaddcl.d D = Scalar U
6 dvhvaddcl.p ˙ = + D
7 dvhvaddcl.a + ˙ = + U
8 simpl K HL W H F T × E G T × E K HL W H
9 xp1st F T × E 1 st F T
10 9 ad2antrl K HL W H F T × E G T × E 1 st F T
11 xp1st G T × E 1 st G T
12 11 ad2antll K HL W H F T × E G T × E 1 st G T
13 1 2 ltrncom K HL W H 1 st F T 1 st G T 1 st F 1 st G = 1 st G 1 st F
14 8 10 12 13 syl3anc K HL W H F T × E G T × E 1 st F 1 st G = 1 st G 1 st F
15 xp2nd F T × E 2 nd F E
16 xp2nd G T × E 2 nd G E
17 15 16 anim12i F T × E G T × E 2 nd F E 2 nd G E
18 eqid a E , b E c T a c b c = a E , b E c T a c b c
19 1 2 3 18 tendoplcom K HL W H 2 nd F E 2 nd G E 2 nd F a E , b E c T a c b c 2 nd G = 2 nd G a E , b E c T a c b c 2 nd F
20 19 3expb K HL W H 2 nd F E 2 nd G E 2 nd F a E , b E c T a c b c 2 nd G = 2 nd G a E , b E c T a c b c 2 nd F
21 17 20 sylan2 K HL W H F T × E G T × E 2 nd F a E , b E c T a c b c 2 nd G = 2 nd G a E , b E c T a c b c 2 nd F
22 1 2 3 4 5 18 6 dvhfplusr K HL W H ˙ = a E , b E c T a c b c
23 22 adantr K HL W H F T × E G T × E ˙ = a E , b E c T a c b c
24 23 oveqd K HL W H F T × E G T × E 2 nd F ˙ 2 nd G = 2 nd F a E , b E c T a c b c 2 nd G
25 23 oveqd K HL W H F T × E G T × E 2 nd G ˙ 2 nd F = 2 nd G a E , b E c T a c b c 2 nd F
26 21 24 25 3eqtr4d K HL W H F T × E G T × E 2 nd F ˙ 2 nd G = 2 nd G ˙ 2 nd F
27 14 26 opeq12d K HL W H F T × E G T × E 1 st F 1 st G 2 nd F ˙ 2 nd G = 1 st G 1 st F 2 nd G ˙ 2 nd F
28 1 2 3 4 5 7 6 dvhvadd K HL W H F T × E G T × E F + ˙ G = 1 st F 1 st G 2 nd F ˙ 2 nd G
29 1 2 3 4 5 7 6 dvhvadd K HL W H G T × E F T × E G + ˙ F = 1 st G 1 st F 2 nd G ˙ 2 nd F
30 29 ancom2s K HL W H F T × E G T × E G + ˙ F = 1 st G 1 st F 2 nd G ˙ 2 nd F
31 27 28 30 3eqtr4d K HL W H F T × E G T × E F + ˙ G = G + ˙ F