Metamath Proof Explorer


Theorem signslema

Description: Computational part of signwlemn . (Contributed by Thierry Arnoux, 29-Sep-2018)

Ref Expression
Hypotheses signslema.1 φ E 0
signslema.2 φ F 0
signslema.3 φ G 0
signslema.4 φ H 0
signslema.5 φ E < G ¬ 2 G E
signslema.6 φ H - G - F E 0 2
Assertion signslema φ F < H ¬ 2 H F

Proof

Step Hyp Ref Expression
1 signslema.1 φ E 0
2 signslema.2 φ F 0
3 signslema.3 φ G 0
4 signslema.4 φ H 0
5 signslema.5 φ E < G ¬ 2 G E
6 signslema.6 φ H - G - F E 0 2
7 5 simpld φ E < G
8 7 adantr φ H - F - G E = 0 E < G
9 4 nn0cnd φ H
10 2 nn0cnd φ F
11 9 10 subcld φ H F
12 3 nn0cnd φ G
13 1 nn0cnd φ E
14 12 13 subcld φ G E
15 11 14 subeq0ad φ H - F - G E = 0 H F = G E
16 15 biimpa φ H - F - G E = 0 H F = G E
17 16 breq2d φ H - F - G E = 0 0 < H F 0 < G E
18 2 nn0red φ F
19 4 nn0red φ H
20 18 19 posdifd φ F < H 0 < H F
21 20 adantr φ H - F - G E = 0 F < H 0 < H F
22 1 nn0red φ E
23 3 nn0red φ G
24 22 23 posdifd φ E < G 0 < G E
25 24 adantr φ H - F - G E = 0 E < G 0 < G E
26 17 21 25 3bitr4rd φ H - F - G E = 0 E < G F < H
27 8 26 mpbid φ H - F - G E = 0 F < H
28 0red φ H - F - G E = 2 0
29 23 22 resubcld φ G E
30 29 adantr φ H - F - G E = 2 G E
31 19 18 resubcld φ H F
32 31 adantr φ H - F - G E = 2 H F
33 7 adantr φ H - F - G E = 2 E < G
34 24 adantr φ H - F - G E = 2 E < G 0 < G E
35 33 34 mpbid φ H - F - G E = 2 0 < G E
36 2pos 0 < 2
37 breq2 H - F - G E = 2 0 < H - F - G E 0 < 2
38 36 37 mpbiri H - F - G E = 2 0 < H - F - G E
39 29 31 posdifd φ G E < H F 0 < H - F - G E
40 39 biimpar φ 0 < H - F - G E G E < H F
41 38 40 sylan2 φ H - F - G E = 2 G E < H F
42 28 30 32 35 41 lttrd φ H - F - G E = 2 0 < H F
43 20 adantr φ H - F - G E = 2 F < H 0 < H F
44 42 43 mpbird φ H - F - G E = 2 F < H
45 9 12 10 13 sub4d φ H - G - F E = H - F - G E
46 45 6 eqeltrrd φ H - F - G E 0 2
47 ovex H - F - G E V
48 47 elpr H - F - G E 0 2 H - F - G E = 0 H - F - G E = 2
49 46 48 sylib φ H - F - G E = 0 H - F - G E = 2
50 27 44 49 mpjaodan φ F < H
51 5 simprd φ ¬ 2 G E
52 51 adantr φ H - F - G E = 0 ¬ 2 G E
53 16 breq2d φ H - F - G E = 0 2 H F 2 G E
54 52 53 mtbird φ H - F - G E = 0 ¬ 2 H F
55 2z 2
56 3 nn0zd φ G
57 1 nn0zd φ E
58 56 57 zsubcld φ G E
59 dvdsaddr 2 G E 2 G E 2 G - E + 2
60 55 58 59 sylancr φ 2 G E 2 G - E + 2
61 51 60 mtbid φ ¬ 2 G - E + 2
62 61 adantr φ H - F - G E = 2 ¬ 2 G - E + 2
63 2cnd φ 2
64 11 14 63 subaddd φ H - F - G E = 2 G - E + 2 = H F
65 64 biimpa φ H - F - G E = 2 G - E + 2 = H F
66 65 breq2d φ H - F - G E = 2 2 G - E + 2 2 H F
67 62 66 mtbid φ H - F - G E = 2 ¬ 2 H F
68 54 67 49 mpjaodan φ ¬ 2 H F
69 50 68 jca φ F < H ¬ 2 H F