Metamath Proof Explorer


Theorem cdj3i

Description: Two ways to express " A and B are completely disjoint subspaces." (1) <=> (3) in Lemma 5 of Holland p. 1520. (Contributed by NM, 1-Jun-2005) (New usage is discouraged.)

Ref Expression
Hypotheses cdj3.1 A S
cdj3.2 B S
cdj3.3 S = x A + B ι z A | w B x = z + w
cdj3.4 T = x A + B ι w B | z A x = z + w
cdj3.5 φ v 0 < v u A + B norm S u v norm u
cdj3.6 ψ v 0 < v u A + B norm T u v norm u
Assertion cdj3i v 0 < v x A y B norm x + norm y v norm x + y A B = 0 φ ψ

Proof

Step Hyp Ref Expression
1 cdj3.1 A S
2 cdj3.2 B S
3 cdj3.3 S = x A + B ι z A | w B x = z + w
4 cdj3.4 T = x A + B ι w B | z A x = z + w
5 cdj3.5 φ v 0 < v u A + B norm S u v norm u
6 cdj3.6 ψ v 0 < v u A + B norm T u v norm u
7 1 2 cdj3lem1 v 0 < v x A y B norm x + norm y v norm x + y A B = 0
8 1 2 3 cdj3lem2b v 0 < v x A y B norm x + norm y v norm x + y v 0 < v u A + B norm S u v norm u
9 8 5 sylibr v 0 < v x A y B norm x + norm y v norm x + y φ
10 1 2 4 cdj3lem3b v 0 < v x A y B norm x + norm y v norm x + y v 0 < v u A + B norm T u v norm u
11 10 6 sylibr v 0 < v x A y B norm x + norm y v norm x + y ψ
12 7 9 11 3jca v 0 < v x A y B norm x + norm y v norm x + y A B = 0 φ ψ
13 breq2 v = f 0 < v 0 < f
14 oveq1 v = f v norm u = f norm u
15 14 breq2d v = f norm S u v norm u norm S u f norm u
16 15 ralbidv v = f u A + B norm S u v norm u u A + B norm S u f norm u
17 13 16 anbi12d v = f 0 < v u A + B norm S u v norm u 0 < f u A + B norm S u f norm u
18 17 cbvrexvw v 0 < v u A + B norm S u v norm u f 0 < f u A + B norm S u f norm u
19 5 18 bitri φ f 0 < f u A + B norm S u f norm u
20 breq2 v = g 0 < v 0 < g
21 oveq1 v = g v norm u = g norm u
22 21 breq2d v = g norm T u v norm u norm T u g norm u
23 22 ralbidv v = g u A + B norm T u v norm u u A + B norm T u g norm u
24 20 23 anbi12d v = g 0 < v u A + B norm T u v norm u 0 < g u A + B norm T u g norm u
25 24 cbvrexvw v 0 < v u A + B norm T u v norm u g 0 < g u A + B norm T u g norm u
26 6 25 bitri ψ g 0 < g u A + B norm T u g norm u
27 19 26 anbi12i φ ψ f 0 < f u A + B norm S u f norm u g 0 < g u A + B norm T u g norm u
28 reeanv f g 0 < f u A + B norm S u f norm u 0 < g u A + B norm T u g norm u f 0 < f u A + B norm S u f norm u g 0 < g u A + B norm T u g norm u
29 27 28 bitr4i φ ψ f g 0 < f u A + B norm S u f norm u 0 < g u A + B norm T u g norm u
30 an4 0 < f u A + B norm S u f norm u 0 < g u A + B norm T u g norm u 0 < f 0 < g u A + B norm S u f norm u u A + B norm T u g norm u
31 addgt0 f g 0 < f 0 < g 0 < f + g
32 31 ex f g 0 < f 0 < g 0 < f + g
33 32 adantl A B = 0 f g 0 < f 0 < g 0 < f + g
34 1 2 shsvai t A h B t + h A + B
35 2fveq3 u = t + h norm S u = norm S t + h
36 fveq2 u = t + h norm u = norm t + h
37 36 oveq2d u = t + h f norm u = f norm t + h
38 35 37 breq12d u = t + h norm S u f norm u norm S t + h f norm t + h
39 38 rspcv t + h A + B u A + B norm S u f norm u norm S t + h f norm t + h
40 2fveq3 u = t + h norm T u = norm T t + h
41 36 oveq2d u = t + h g norm u = g norm t + h
42 40 41 breq12d u = t + h norm T u g norm u norm T t + h g norm t + h
43 42 rspcv t + h A + B u A + B norm T u g norm u norm T t + h g norm t + h
44 39 43 anim12d t + h A + B u A + B norm S u f norm u u A + B norm T u g norm u norm S t + h f norm t + h norm T t + h g norm t + h
45 34 44 syl t A h B u A + B norm S u f norm u u A + B norm T u g norm u norm S t + h f norm t + h norm T t + h g norm t + h
46 45 adantl A B = 0 f g t A h B u A + B norm S u f norm u u A + B norm T u g norm u norm S t + h f norm t + h norm T t + h g norm t + h
47 1 sheli t A t
48 normcl t norm t
49 47 48 syl t A norm t
50 2 sheli h B h
51 normcl h norm h
52 50 51 syl h B norm h
53 49 52 anim12i t A h B norm t norm h
54 53 adantl f g t A h B norm t norm h
55 hvaddcl t h t + h
56 47 50 55 syl2an t A h B t + h
57 normcl t + h norm t + h
58 56 57 syl t A h B norm t + h
59 remulcl f norm t + h f norm t + h
60 58 59 sylan2 f t A h B f norm t + h
61 60 adantlr f g t A h B f norm t + h
62 remulcl g norm t + h g norm t + h
63 58 62 sylan2 g t A h B g norm t + h
64 63 adantll f g t A h B g norm t + h
65 le2add norm t norm h f norm t + h g norm t + h norm t f norm t + h norm h g norm t + h norm t + norm h f norm t + h + g norm t + h
66 54 61 64 65 syl12anc f g t A h B norm t f norm t + h norm h g norm t + h norm t + norm h f norm t + h + g norm t + h
67 66 adantll A B = 0 f g t A h B norm t f norm t + h norm h g norm t + h norm t + norm h f norm t + h + g norm t + h
68 1 2 3 cdj3lem2 t A h B A B = 0 S t + h = t
69 68 fveq2d t A h B A B = 0 norm S t + h = norm t
70 69 breq1d t A h B A B = 0 norm S t + h f norm t + h norm t f norm t + h
71 1 2 4 cdj3lem3 t A h B A B = 0 T t + h = h
72 71 fveq2d t A h B A B = 0 norm T t + h = norm h
73 72 breq1d t A h B A B = 0 norm T t + h g norm t + h norm h g norm t + h
74 70 73 anbi12d t A h B A B = 0 norm S t + h f norm t + h norm T t + h g norm t + h norm t f norm t + h norm h g norm t + h
75 74 3expa t A h B A B = 0 norm S t + h f norm t + h norm T t + h g norm t + h norm t f norm t + h norm h g norm t + h
76 75 ancoms A B = 0 t A h B norm S t + h f norm t + h norm T t + h g norm t + h norm t f norm t + h norm h g norm t + h
77 76 adantlr A B = 0 f g t A h B norm S t + h f norm t + h norm T t + h g norm t + h norm t f norm t + h norm h g norm t + h
78 recn f f
79 recn g g
80 58 recnd t A h B norm t + h
81 adddir f g norm t + h f + g norm t + h = f norm t + h + g norm t + h
82 78 79 80 81 syl3an f g t A h B f + g norm t + h = f norm t + h + g norm t + h
83 82 3expa f g t A h B f + g norm t + h = f norm t + h + g norm t + h
84 83 breq2d f g t A h B norm t + norm h f + g norm t + h norm t + norm h f norm t + h + g norm t + h
85 84 adantll A B = 0 f g t A h B norm t + norm h f + g norm t + h norm t + norm h f norm t + h + g norm t + h
86 67 77 85 3imtr4d A B = 0 f g t A h B norm S t + h f norm t + h norm T t + h g norm t + h norm t + norm h f + g norm t + h
87 46 86 syld A B = 0 f g t A h B u A + B norm S u f norm u u A + B norm T u g norm u norm t + norm h f + g norm t + h
88 87 ralrimdvva A B = 0 f g u A + B norm S u f norm u u A + B norm T u g norm u t A h B norm t + norm h f + g norm t + h
89 readdcl f g f + g
90 breq2 v = f + g 0 < v 0 < f + g
91 fveq2 x = t norm x = norm t
92 91 oveq1d x = t norm x + norm y = norm t + norm y
93 fvoveq1 x = t norm x + y = norm t + y
94 93 oveq2d x = t v norm x + y = v norm t + y
95 92 94 breq12d x = t norm x + norm y v norm x + y norm t + norm y v norm t + y
96 fveq2 y = h norm y = norm h
97 96 oveq2d y = h norm t + norm y = norm t + norm h
98 oveq2 y = h t + y = t + h
99 98 fveq2d y = h norm t + y = norm t + h
100 99 oveq2d y = h v norm t + y = v norm t + h
101 97 100 breq12d y = h norm t + norm y v norm t + y norm t + norm h v norm t + h
102 95 101 cbvral2vw x A y B norm x + norm y v norm x + y t A h B norm t + norm h v norm t + h
103 oveq1 v = f + g v norm t + h = f + g norm t + h
104 103 breq2d v = f + g norm t + norm h v norm t + h norm t + norm h f + g norm t + h
105 104 2ralbidv v = f + g t A h B norm t + norm h v norm t + h t A h B norm t + norm h f + g norm t + h
106 102 105 syl5bb v = f + g x A y B norm x + norm y v norm x + y t A h B norm t + norm h f + g norm t + h
107 90 106 anbi12d v = f + g 0 < v x A y B norm x + norm y v norm x + y 0 < f + g t A h B norm t + norm h f + g norm t + h
108 107 rspcev f + g 0 < f + g t A h B norm t + norm h f + g norm t + h v 0 < v x A y B norm x + norm y v norm x + y
109 108 ex f + g 0 < f + g t A h B norm t + norm h f + g norm t + h v 0 < v x A y B norm x + norm y v norm x + y
110 89 109 syl f g 0 < f + g t A h B norm t + norm h f + g norm t + h v 0 < v x A y B norm x + norm y v norm x + y
111 110 adantl A B = 0 f g 0 < f + g t A h B norm t + norm h f + g norm t + h v 0 < v x A y B norm x + norm y v norm x + y
112 33 88 111 syl2and A B = 0 f g 0 < f 0 < g u A + B norm S u f norm u u A + B norm T u g norm u v 0 < v x A y B norm x + norm y v norm x + y
113 30 112 syl5bi A B = 0 f g 0 < f u A + B norm S u f norm u 0 < g u A + B norm T u g norm u v 0 < v x A y B norm x + norm y v norm x + y
114 113 rexlimdvva A B = 0 f g 0 < f u A + B norm S u f norm u 0 < g u A + B norm T u g norm u v 0 < v x A y B norm x + norm y v norm x + y
115 29 114 syl5bi A B = 0 φ ψ v 0 < v x A y B norm x + norm y v norm x + y
116 115 3impib A B = 0 φ ψ v 0 < v x A y B norm x + norm y v norm x + y
117 12 116 impbii v 0 < v x A y B norm x + norm y v norm x + y A B = 0 φ ψ