Metamath Proof Explorer


Theorem ftalem6

Description: Lemma for fta : Discharge the auxiliary variables in ftalem5 . (Contributed by Mario Carneiro, 20-Sep-2014) (Proof shortened by AV, 28-Sep-2020)

Ref Expression
Hypotheses ftalem.1 A=coeffF
ftalem.2 N=degF
ftalem.3 φFPolyS
ftalem.4 φN
ftalem6.5 φF00
Assertion ftalem6 φxFx<F0

Proof

Step Hyp Ref Expression
1 ftalem.1 A=coeffF
2 ftalem.2 N=degF
3 ftalem.3 φFPolyS
4 ftalem.4 φN
5 ftalem6.5 φF00
6 fveq2 k=nAk=An
7 6 neeq1d k=nAk0An0
8 7 cbvrabv k|Ak0=n|An0
9 8 infeq1i supk|Ak0<=supn|An0<
10 eqid F0Asupk|Ak0<1supk|Ak0<=F0Asupk|Ak0<1supk|Ak0<
11 fveq2 r=sAr=As
12 oveq2 r=sF0Asupk|Ak0<1supk|Ak0<r=F0Asupk|Ak0<1supk|Ak0<s
13 11 12 oveq12d r=sArF0Asupk|Ak0<1supk|Ak0<r=AsF0Asupk|Ak0<1supk|Ak0<s
14 13 fveq2d r=sArF0Asupk|Ak0<1supk|Ak0<r=AsF0Asupk|Ak0<1supk|Ak0<s
15 14 cbvsumv r=supk|Ak0<+1NArF0Asupk|Ak0<1supk|Ak0<r=s=supk|Ak0<+1NAsF0Asupk|Ak0<1supk|Ak0<s
16 15 oveq1i r=supk|Ak0<+1NArF0Asupk|Ak0<1supk|Ak0<r+1=s=supk|Ak0<+1NAsF0Asupk|Ak0<1supk|Ak0<s+1
17 16 oveq2i F0r=supk|Ak0<+1NArF0Asupk|Ak0<1supk|Ak0<r+1=F0s=supk|Ak0<+1NAsF0Asupk|Ak0<1supk|Ak0<s+1
18 eqid if1F0r=supk|Ak0<+1NArF0Asupk|Ak0<1supk|Ak0<r+11F0r=supk|Ak0<+1NArF0Asupk|Ak0<1supk|Ak0<r+1=if1F0r=supk|Ak0<+1NArF0Asupk|Ak0<1supk|Ak0<r+11F0r=supk|Ak0<+1NArF0Asupk|Ak0<1supk|Ak0<r+1
19 1 2 3 4 5 9 10 17 18 ftalem5 φxFx<F0