Metamath Proof Explorer


Theorem cdj3lem3b

Description: Lemma for cdj3i . The second-component function T is bounded if the subspaces are completely disjoint. (Contributed by NM, 31-May-2005) (New usage is discouraged.)

Ref Expression
Hypotheses cdj3lem2.1 A S
cdj3lem2.2 B S
cdj3lem3.3 T = x A + B ι w B | z A x = z + w
Assertion 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

Proof

Step Hyp Ref Expression
1 cdj3lem2.1 A S
2 cdj3lem2.2 B S
3 cdj3lem3.3 T = x A + B ι w B | z A x = z + w
4 2 1 shscomi B + A = A + B
5 2 sheli w B w
6 1 sheli z A z
7 ax-hvcom w z w + z = z + w
8 5 6 7 syl2an w B z A w + z = z + w
9 8 eqeq2d w B z A x = w + z x = z + w
10 9 rexbidva w B z A x = w + z z A x = z + w
11 10 riotabiia ι w B | z A x = w + z = ι w B | z A x = z + w
12 4 11 mpteq12i x B + A ι w B | z A x = w + z = x A + B ι w B | z A x = z + w
13 3 12 eqtr4i T = x B + A ι w B | z A x = w + z
14 2 1 13 cdj3lem2b v 0 < v x B y A norm x + norm y v norm x + y v 0 < v u B + A norm T u v norm u
15 fveq2 x = t norm x = norm t
16 15 oveq1d x = t norm x + norm y = norm t + norm y
17 fvoveq1 x = t norm x + y = norm t + y
18 17 oveq2d x = t v norm x + y = v norm t + y
19 16 18 breq12d x = t norm x + norm y v norm x + y norm t + norm y v norm t + y
20 fveq2 y = h norm y = norm h
21 20 oveq2d y = h norm t + norm y = norm t + norm h
22 oveq2 y = h t + y = t + h
23 22 fveq2d y = h norm t + y = norm t + h
24 23 oveq2d y = h v norm t + y = v norm t + h
25 21 24 breq12d y = h norm t + norm y v norm t + y norm t + norm h v norm t + h
26 19 25 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
27 ralcom t A h B norm t + norm h v norm t + h h B t A norm t + norm h v norm t + h
28 2 sheli x B x
29 normcl x norm x
30 28 29 syl x B norm x
31 30 recnd x B norm x
32 1 sheli y A y
33 normcl y norm y
34 32 33 syl y A norm y
35 34 recnd y A norm y
36 addcom norm x norm y norm x + norm y = norm y + norm x
37 31 35 36 syl2an x B y A norm x + norm y = norm y + norm x
38 ax-hvcom x y x + y = y + x
39 28 32 38 syl2an x B y A x + y = y + x
40 39 fveq2d x B y A norm x + y = norm y + x
41 40 oveq2d x B y A v norm x + y = v norm y + x
42 37 41 breq12d x B y A norm x + norm y v norm x + y norm y + norm x v norm y + x
43 42 ralbidva x B y A norm x + norm y v norm x + y y A norm y + norm x v norm y + x
44 43 ralbiia x B y A norm x + norm y v norm x + y x B y A norm y + norm x v norm y + x
45 fveq2 x = h norm x = norm h
46 45 oveq2d x = h norm y + norm x = norm y + norm h
47 oveq2 x = h y + x = y + h
48 47 fveq2d x = h norm y + x = norm y + h
49 48 oveq2d x = h v norm y + x = v norm y + h
50 46 49 breq12d x = h norm y + norm x v norm y + x norm y + norm h v norm y + h
51 fveq2 y = t norm y = norm t
52 51 oveq1d y = t norm y + norm h = norm t + norm h
53 fvoveq1 y = t norm y + h = norm t + h
54 53 oveq2d y = t v norm y + h = v norm t + h
55 52 54 breq12d y = t norm y + norm h v norm y + h norm t + norm h v norm t + h
56 50 55 cbvral2vw x B y A norm y + norm x v norm y + x h B t A norm t + norm h v norm t + h
57 44 56 bitr2i h B t A norm t + norm h v norm t + h x B y A norm x + norm y v norm x + y
58 26 27 57 3bitri x A y B norm x + norm y v norm x + y x B y A norm x + norm y v norm x + y
59 58 anbi2i 0 < v x A y B norm x + norm y v norm x + y 0 < v x B y A norm x + norm y v norm x + y
60 59 rexbii v 0 < v x A y B norm x + norm y v norm x + y v 0 < v x B y A norm x + norm y v norm x + y
61 1 2 shscomi A + B = B + A
62 61 raleqi u A + B norm T u v norm u u B + A norm T u v norm u
63 62 anbi2i 0 < v u A + B norm T u v norm u 0 < v u B + A norm T u v norm u
64 63 rexbii v 0 < v u A + B norm T u v norm u v 0 < v u B + A norm T u v norm u
65 14 60 64 3imtr4i 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