Metamath Proof Explorer


Theorem knoppndvlem9

Description: Lemma for knoppndv . (Contributed by Asger C. Ipsen, 15-Jun-2021) (Revised by Asger C. Ipsen, 5-Jul-2021)

Ref Expression
Hypotheses knoppndvlem9.t T = x x + 1 2 x
knoppndvlem9.f F = y n 0 C n T 2 N n y
knoppndvlem9.a A = 2 N J 2 M
knoppndvlem9.c φ C 1 1
knoppndvlem9.j φ J 0
knoppndvlem9.m φ M
knoppndvlem9.n φ N
knoppndvlem9.1 φ ¬ 2 M
Assertion knoppndvlem9 φ F A J = C J 2

Proof

Step Hyp Ref Expression
1 knoppndvlem9.t T = x x + 1 2 x
2 knoppndvlem9.f F = y n 0 C n T 2 N n y
3 knoppndvlem9.a A = 2 N J 2 M
4 knoppndvlem9.c φ C 1 1
5 knoppndvlem9.j φ J 0
6 knoppndvlem9.m φ M
7 knoppndvlem9.n φ N
8 knoppndvlem9.1 φ ¬ 2 M
9 1 2 3 5 6 7 knoppndvlem7 φ F A J = C J T M 2
10 odd2np1 M ¬ 2 M m 2 m + 1 = M
11 6 10 syl φ ¬ 2 M m 2 m + 1 = M
12 8 11 mpbid φ m 2 m + 1 = M
13 eqcom 2 m + 1 = M M = 2 m + 1
14 13 biimpi 2 m + 1 = M M = 2 m + 1
15 14 oveq1d 2 m + 1 = M M 2 = 2 m + 1 2
16 15 adantl m 2 m + 1 = M M 2 = 2 m + 1 2
17 16 adantl φ m 2 m + 1 = M M 2 = 2 m + 1 2
18 2cnd φ m 2
19 zcn m m
20 19 adantl φ m m
21 18 20 mulcld φ m 2 m
22 1cnd φ m 1
23 2ne0 2 0
24 23 a1i φ m 2 0
25 21 22 18 24 divdird φ m 2 m + 1 2 = 2 m 2 + 1 2
26 20 18 24 divcan3d φ m 2 m 2 = m
27 26 oveq1d φ m 2 m 2 + 1 2 = m + 1 2
28 25 27 eqtrd φ m 2 m + 1 2 = m + 1 2
29 28 adantrr φ m 2 m + 1 = M 2 m + 1 2 = m + 1 2
30 17 29 eqtrd φ m 2 m + 1 = M M 2 = m + 1 2
31 30 fveq2d φ m 2 m + 1 = M T M 2 = T m + 1 2
32 id m m
33 1 32 dnizphlfeqhlf m T m + 1 2 = 1 2
34 33 adantl φ m T m + 1 2 = 1 2
35 34 adantrr φ m 2 m + 1 = M T m + 1 2 = 1 2
36 31 35 eqtrd φ m 2 m + 1 = M T M 2 = 1 2
37 12 36 rexlimddv φ T M 2 = 1 2
38 37 oveq2d φ C J T M 2 = C J 1 2
39 4 knoppndvlem3 φ C C < 1
40 39 simpld φ C
41 40 recnd φ C
42 41 5 expcld φ C J
43 1cnd φ 1
44 2cnd φ 2
45 23 a1i φ 2 0
46 42 43 44 45 div12d φ C J 1 2 = 1 C J 2
47 42 44 45 divcld φ C J 2
48 47 mulid2d φ 1 C J 2 = C J 2
49 46 48 eqtrd φ C J 1 2 = C J 2
50 9 38 49 3eqtrd φ F A J = C J 2