Metamath Proof Explorer


Theorem jensenlem1

Description: Lemma for jensen . (Contributed by Mario Carneiro, 4-Jun-2016)

Ref Expression
Hypotheses jensen.1 φ D
jensen.2 φ F : D
jensen.3 φ a D b D a b D
jensen.4 φ A Fin
jensen.5 φ T : A 0 +∞
jensen.6 φ X : A D
jensen.7 φ 0 < fld T
jensen.8 φ x D y D t 0 1 F t x + 1 t y t F x + 1 t F y
jensenlem.1 φ ¬ z B
jensenlem.2 φ B z A
jensenlem.s S = fld T B
jensenlem.l L = fld T B z
Assertion jensenlem1 φ L = S + T z

Proof

Step Hyp Ref Expression
1 jensen.1 φ D
2 jensen.2 φ F : D
3 jensen.3 φ a D b D a b D
4 jensen.4 φ A Fin
5 jensen.5 φ T : A 0 +∞
6 jensen.6 φ X : A D
7 jensen.7 φ 0 < fld T
8 jensen.8 φ x D y D t 0 1 F t x + 1 t y t F x + 1 t F y
9 jensenlem.1 φ ¬ z B
10 jensenlem.2 φ B z A
11 jensenlem.s S = fld T B
12 jensenlem.l L = fld T B z
13 cnfldbas = Base fld
14 cnfldadd + = + fld
15 cnring fld Ring
16 ringcmn fld Ring fld CMnd
17 15 16 mp1i φ fld CMnd
18 10 unssad φ B A
19 4 18 ssfid φ B Fin
20 rge0ssre 0 +∞
21 ax-resscn
22 20 21 sstri 0 +∞
23 18 sselda φ x B x A
24 5 ffvelrnda φ x A T x 0 +∞
25 23 24 syldan φ x B T x 0 +∞
26 22 25 sselid φ x B T x
27 10 unssbd φ z A
28 vex z V
29 28 snss z A z A
30 27 29 sylibr φ z A
31 5 30 ffvelrnd φ T z 0 +∞
32 22 31 sselid φ T z
33 fveq2 x = z T x = T z
34 13 14 17 19 26 30 9 32 33 gsumunsn φ fld x B z T x = fld x B T x + T z
35 5 10 feqresmpt φ T B z = x B z T x
36 35 oveq2d φ fld T B z = fld x B z T x
37 5 18 feqresmpt φ T B = x B T x
38 37 oveq2d φ fld T B = fld x B T x
39 38 oveq1d φ fld T B + T z = fld x B T x + T z
40 34 36 39 3eqtr4d φ fld T B z = fld T B + T z
41 11 oveq1i S + T z = fld T B + T z
42 40 12 41 3eqtr4g φ L = S + T z