Metamath Proof Explorer


Theorem quart1cl

Description: Closure lemmas for quart . (Contributed by Mario Carneiro, 7-May-2015)

Ref Expression
Hypotheses quart1.a φ A
quart1.b φ B
quart1.c φ C
quart1.d φ D
quart1.p φ P = B 3 8 A 2
quart1.q φ Q = C - A B 2 + A 3 8
quart1.r φ R = D C A 4 + A 2 B 16 - 3 256 A 4
Assertion quart1cl φ P Q R

Proof

Step Hyp Ref Expression
1 quart1.a φ A
2 quart1.b φ B
3 quart1.c φ C
4 quart1.d φ D
5 quart1.p φ P = B 3 8 A 2
6 quart1.q φ Q = C - A B 2 + A 3 8
7 quart1.r φ R = D C A 4 + A 2 B 16 - 3 256 A 4
8 3cn 3
9 8cn 8
10 8nn 8
11 10 nnne0i 8 0
12 8 9 11 divcli 3 8
13 1 sqcld φ A 2
14 mulcl 3 8 A 2 3 8 A 2
15 12 13 14 sylancr φ 3 8 A 2
16 2 15 subcld φ B 3 8 A 2
17 5 16 eqeltrd φ P
18 1 2 mulcld φ A B
19 18 halfcld φ A B 2
20 3 19 subcld φ C A B 2
21 3nn0 3 0
22 expcl A 3 0 A 3
23 1 21 22 sylancl φ A 3
24 9 a1i φ 8
25 11 a1i φ 8 0
26 23 24 25 divcld φ A 3 8
27 20 26 addcld φ C - A B 2 + A 3 8
28 6 27 eqeltrd φ Q
29 3 1 mulcld φ C A
30 4cn 4
31 30 a1i φ 4
32 4ne0 4 0
33 32 a1i φ 4 0
34 29 31 33 divcld φ C A 4
35 4 34 subcld φ D C A 4
36 13 2 mulcld φ A 2 B
37 1nn0 1 0
38 6nn 6
39 37 38 decnncl 16
40 39 nncni 16
41 40 a1i φ 16
42 39 nnne0i 16 0
43 42 a1i φ 16 0
44 36 41 43 divcld φ A 2 B 16
45 25nn0 25 0
46 45 38 decnncl 256
47 46 nncni 256
48 46 nnne0i 256 0
49 8 47 48 divcli 3 256
50 4nn0 4 0
51 expcl A 4 0 A 4
52 1 50 51 sylancl φ A 4
53 mulcl 3 256 A 4 3 256 A 4
54 49 52 53 sylancr φ 3 256 A 4
55 44 54 subcld φ A 2 B 16 3 256 A 4
56 35 55 addcld φ D C A 4 + A 2 B 16 - 3 256 A 4
57 7 56 eqeltrd φ R
58 17 28 57 3jca φ P Q R