Metamath Proof Explorer


Theorem jensenlem1

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

Ref Expression
Hypotheses jensen.1 φD
jensen.2 φF:D
jensen.3 φaDbDabD
jensen.4 φAFin
jensen.5 φT:A0+∞
jensen.6 φX:AD
jensen.7 φ0<fldT
jensen.8 φxDyDt01Ftx+1tytFx+1tFy
jensenlem.1 φ¬zB
jensenlem.2 φBzA
jensenlem.s S=fldTB
jensenlem.l L=fldTBz
Assertion jensenlem1 φL=S+Tz

Proof

Step Hyp Ref Expression
1 jensen.1 φD
2 jensen.2 φF:D
3 jensen.3 φaDbDabD
4 jensen.4 φAFin
5 jensen.5 φT:A0+∞
6 jensen.6 φX:AD
7 jensen.7 φ0<fldT
8 jensen.8 φxDyDt01Ftx+1tytFx+1tFy
9 jensenlem.1 φ¬zB
10 jensenlem.2 φBzA
11 jensenlem.s S=fldTB
12 jensenlem.l L=fldTBz
13 cnfldbas =Basefld
14 cnfldadd +=+fld
15 cnring fldRing
16 ringcmn fldRingfldCMnd
17 15 16 mp1i φfldCMnd
18 10 unssad φBA
19 4 18 ssfid φBFin
20 rge0ssre 0+∞
21 ax-resscn
22 20 21 sstri 0+∞
23 18 sselda φxBxA
24 5 ffvelcdmda φxATx0+∞
25 23 24 syldan φxBTx0+∞
26 22 25 sselid φxBTx
27 10 unssbd φzA
28 vex zV
29 28 snss zAzA
30 27 29 sylibr φzA
31 5 30 ffvelcdmd φTz0+∞
32 22 31 sselid φTz
33 fveq2 x=zTx=Tz
34 13 14 17 19 26 30 9 32 33 gsumunsn φfldxBzTx=fldxBTx+Tz
35 5 10 feqresmpt φTBz=xBzTx
36 35 oveq2d φfldTBz=fldxBzTx
37 5 18 feqresmpt φTB=xBTx
38 37 oveq2d φfldTB=fldxBTx
39 38 oveq1d φfldTB+Tz=fldxBTx+Tz
40 34 36 39 3eqtr4d φfldTBz=fldTB+Tz
41 11 oveq1i S+Tz=fldTB+Tz
42 40 12 41 3eqtr4g φL=S+Tz