Metamath Proof Explorer


Theorem nmpropd2

Description: Strong property deduction for a norm. (Contributed by Mario Carneiro, 4-Oct-2015)

Ref Expression
Hypotheses nmpropd2.1 φ B = Base K
nmpropd2.2 φ B = Base L
nmpropd2.3 φ K Grp
nmpropd2.4 φ x B y B x + K y = x + L y
nmpropd2.5 φ dist K B × B = dist L B × B
Assertion nmpropd2 φ norm K = norm L

Proof

Step Hyp Ref Expression
1 nmpropd2.1 φ B = Base K
2 nmpropd2.2 φ B = Base L
3 nmpropd2.3 φ K Grp
4 nmpropd2.4 φ x B y B x + K y = x + L y
5 nmpropd2.5 φ dist K B × B = dist L B × B
6 1 2 eqtr3d φ Base K = Base L
7 1 sqxpeqd φ B × B = Base K × Base K
8 7 reseq2d φ dist K B × B = dist K Base K × Base K
9 5 8 eqtr3d φ dist L B × B = dist K Base K × Base K
10 2 sqxpeqd φ B × B = Base L × Base L
11 10 reseq2d φ dist L B × B = dist L Base L × Base L
12 9 11 eqtr3d φ dist K Base K × Base K = dist L Base L × Base L
13 eqidd φ a = a
14 1 2 4 grpidpropd φ 0 K = 0 L
15 12 13 14 oveq123d φ a dist K Base K × Base K 0 K = a dist L Base L × Base L 0 L
16 6 15 mpteq12dv φ a Base K a dist K Base K × Base K 0 K = a Base L a dist L Base L × Base L 0 L
17 eqid norm K = norm K
18 eqid Base K = Base K
19 eqid 0 K = 0 K
20 eqid dist K = dist K
21 eqid dist K Base K × Base K = dist K Base K × Base K
22 17 18 19 20 21 nmfval2 K Grp norm K = a Base K a dist K Base K × Base K 0 K
23 3 22 syl φ norm K = a Base K a dist K Base K × Base K 0 K
24 1 2 4 grppropd φ K Grp L Grp
25 3 24 mpbid φ L Grp
26 eqid norm L = norm L
27 eqid Base L = Base L
28 eqid 0 L = 0 L
29 eqid dist L = dist L
30 eqid dist L Base L × Base L = dist L Base L × Base L
31 26 27 28 29 30 nmfval2 L Grp norm L = a Base L a dist L Base L × Base L 0 L
32 25 31 syl φ norm L = a Base L a dist L Base L × Base L 0 L
33 16 23 32 3eqtr4d φ norm K = norm L