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 = coeff F
ftalem.2 N = deg F
ftalem.3 φ F Poly S
ftalem.4 φ N
ftalem6.5 φ F 0 0
Assertion ftalem6 φ x F x < F 0

Proof

Step Hyp Ref Expression
1 ftalem.1 A = coeff F
2 ftalem.2 N = deg F
3 ftalem.3 φ F Poly S
4 ftalem.4 φ N
5 ftalem6.5 φ F 0 0
6 fveq2 k = n A k = A n
7 6 neeq1d k = n A k 0 A n 0
8 7 cbvrabv k | A k 0 = n | A n 0
9 8 infeq1i sup k | A k 0 < = sup n | A n 0 <
10 eqid F 0 A sup k | A k 0 < 1 sup k | A k 0 < = F 0 A sup k | A k 0 < 1 sup k | A k 0 <
11 fveq2 r = s A r = A s
12 oveq2 r = s F 0 A sup k | A k 0 < 1 sup k | A k 0 < r = F 0 A sup k | A k 0 < 1 sup k | A k 0 < s
13 11 12 oveq12d r = s A r F 0 A sup k | A k 0 < 1 sup k | A k 0 < r = A s F 0 A sup k | A k 0 < 1 sup k | A k 0 < s
14 13 fveq2d r = s A r F 0 A sup k | A k 0 < 1 sup k | A k 0 < r = A s F 0 A sup k | A k 0 < 1 sup k | A k 0 < s
15 14 cbvsumv r = sup k | A k 0 < + 1 N A r F 0 A sup k | A k 0 < 1 sup k | A k 0 < r = s = sup k | A k 0 < + 1 N A s F 0 A sup k | A k 0 < 1 sup k | A k 0 < s
16 15 oveq1i r = sup k | A k 0 < + 1 N A r F 0 A sup k | A k 0 < 1 sup k | A k 0 < r + 1 = s = sup k | A k 0 < + 1 N A s F 0 A sup k | A k 0 < 1 sup k | A k 0 < s + 1
17 16 oveq2i F 0 r = sup k | A k 0 < + 1 N A r F 0 A sup k | A k 0 < 1 sup k | A k 0 < r + 1 = F 0 s = sup k | A k 0 < + 1 N A s F 0 A sup k | A k 0 < 1 sup k | A k 0 < s + 1
18 eqid if 1 F 0 r = sup k | A k 0 < + 1 N A r F 0 A sup k | A k 0 < 1 sup k | A k 0 < r + 1 1 F 0 r = sup k | A k 0 < + 1 N A r F 0 A sup k | A k 0 < 1 sup k | A k 0 < r + 1 = if 1 F 0 r = sup k | A k 0 < + 1 N A r F 0 A sup k | A k 0 < 1 sup k | A k 0 < r + 1 1 F 0 r = sup k | A k 0 < + 1 N A r F 0 A sup k | A k 0 < 1 sup k | A k 0 < r + 1
19 1 2 3 4 5 9 10 17 18 ftalem5 φ x F x < F 0