Metamath Proof Explorer


Theorem pntrlog2bndlem6a

Description: Lemma for pntrlog2bndlem6 . (Contributed by Mario Carneiro, 7-Jun-2016)

Ref Expression
Hypotheses pntsval.1 S = a i = 1 a Λ i log i + ψ a i
pntrlog2bnd.r R = a + ψ a a
pntrlog2bnd.t T = a if a + a log a 0
pntrlog2bndlem5.1 φ B +
pntrlog2bndlem5.2 φ y + R y y B
pntrlog2bndlem6.1 φ A
pntrlog2bndlem6.2 φ 1 A
Assertion pntrlog2bndlem6a φ x 1 +∞ 1 x = 1 x A x A + 1 x

Proof

Step Hyp Ref Expression
1 pntsval.1 S = a i = 1 a Λ i log i + ψ a i
2 pntrlog2bnd.r R = a + ψ a a
3 pntrlog2bnd.t T = a if a + a log a 0
4 pntrlog2bndlem5.1 φ B +
5 pntrlog2bndlem5.2 φ y + R y y B
6 pntrlog2bndlem6.1 φ A
7 pntrlog2bndlem6.2 φ 1 A
8 elioore x 1 +∞ x
9 8 adantl φ x 1 +∞ x
10 1rp 1 +
11 10 a1i φ x 1 +∞ 1 +
12 11 rpred φ x 1 +∞ 1
13 eliooord x 1 +∞ 1 < x x < +∞
14 13 adantl φ x 1 +∞ 1 < x x < +∞
15 14 simpld φ x 1 +∞ 1 < x
16 12 9 15 ltled φ x 1 +∞ 1 x
17 9 11 16 rpgecld φ x 1 +∞ x +
18 10 a1i φ 1 +
19 6 18 7 rpgecld φ A +
20 19 adantr φ x 1 +∞ A +
21 17 20 rpdivcld φ x 1 +∞ x A +
22 21 rprege0d φ x 1 +∞ x A 0 x A
23 flge0nn0 x A 0 x A x A 0
24 nn0p1nn x A 0 x A + 1
25 22 23 24 3syl φ x 1 +∞ x A + 1
26 nnuz = 1
27 25 26 eleqtrdi φ x 1 +∞ x A + 1 1
28 21 rpred φ x 1 +∞ x A
29 17 rpge0d φ x 1 +∞ 0 x
30 7 adantr φ x 1 +∞ 1 A
31 11 20 9 29 30 lediv2ad φ x 1 +∞ x A x 1
32 9 recnd φ x 1 +∞ x
33 32 div1d φ x 1 +∞ x 1 = x
34 31 33 breqtrd φ x 1 +∞ x A x
35 flword2 x A x x A x x x A
36 28 9 34 35 syl3anc φ x 1 +∞ x x A
37 fzsplit2 x A + 1 1 x x A 1 x = 1 x A x A + 1 x
38 27 36 37 syl2anc φ x 1 +∞ 1 x = 1 x A x A + 1 x