Metamath Proof Explorer


Theorem normlem0

Description: Lemma used to derive properties of norm. Part of Theorem 3.3(ii) of Beran p. 97. (Contributed by NM, 7-Oct-1999) (New usage is discouraged.)

Ref Expression
Hypotheses normlem1.1 S
normlem1.2 F
normlem1.3 G
Assertion normlem0 F - S G ih F - S G = F ih F + S F ih G + S G ih F + S S G ih G

Proof

Step Hyp Ref Expression
1 normlem1.1 S
2 normlem1.2 F
3 normlem1.3 G
4 1 3 hvmulcli S G
5 2 4 hvsubvali F - S G = F + -1 S G
6 1 mulm1i -1 S = S
7 6 oveq1i -1 S G = S G
8 neg1cn 1
9 8 1 3 hvmulassi -1 S G = -1 S G
10 7 9 eqtr3i S G = -1 S G
11 10 oveq2i F + S G = F + -1 S G
12 5 11 eqtr4i F - S G = F + S G
13 12 12 oveq12i F - S G ih F - S G = F + S G ih F + S G
14 1 negcli S
15 14 3 hvmulcli S G
16 2 15 hvaddcli F + S G
17 ax-his2 F S G F + S G F + S G ih F + S G = F ih F + S G + S G ih F + S G
18 2 15 16 17 mp3an F + S G ih F + S G = F ih F + S G + S G ih F + S G
19 his7 F F S G F ih F + S G = F ih F + F ih S G
20 2 2 15 19 mp3an F ih F + S G = F ih F + F ih S G
21 his5 S F G F ih S G = S F ih G
22 14 2 3 21 mp3an F ih S G = S F ih G
23 1 cjnegi S = S
24 23 oveq1i S F ih G = S F ih G
25 22 24 eqtri F ih S G = S F ih G
26 25 oveq2i F ih F + F ih S G = F ih F + S F ih G
27 20 26 eqtri F ih F + S G = F ih F + S F ih G
28 ax-his3 S G F + S G S G ih F + S G = S G ih F + S G
29 14 3 16 28 mp3an S G ih F + S G = S G ih F + S G
30 his7 G F S G G ih F + S G = G ih F + G ih S G
31 3 2 15 30 mp3an G ih F + S G = G ih F + G ih S G
32 his5 S G G G ih S G = S G ih G
33 14 3 3 32 mp3an G ih S G = S G ih G
34 33 oveq2i G ih F + G ih S G = G ih F + S G ih G
35 31 34 eqtri G ih F + S G = G ih F + S G ih G
36 35 oveq2i S G ih F + S G = S G ih F + S G ih G
37 3 2 hicli G ih F
38 14 cjcli S
39 3 3 hicli G ih G
40 38 39 mulcli S G ih G
41 14 37 40 adddii S G ih F + S G ih G = S G ih F + S S G ih G
42 14 38 39 mulassi S S G ih G = S S G ih G
43 23 oveq2i S S = S S
44 1 cjcli S
45 1 44 mul2negi S S = S S
46 43 45 eqtri S S = S S
47 46 oveq1i S S G ih G = S S G ih G
48 42 47 eqtr3i S S G ih G = S S G ih G
49 48 oveq2i S G ih F + S S G ih G = S G ih F + S S G ih G
50 41 49 eqtri S G ih F + S G ih G = S G ih F + S S G ih G
51 29 36 50 3eqtri S G ih F + S G = S G ih F + S S G ih G
52 27 51 oveq12i F ih F + S G + S G ih F + S G = F ih F + S F ih G + S G ih F + S S G ih G
53 13 18 52 3eqtri F - S G ih F - S G = F ih F + S F ih G + S G ih F + S S G ih G