Metamath Proof Explorer


Theorem proththdlem

Description: Lemma for proththd . (Contributed by AV, 4-Jul-2020)

Ref Expression
Hypotheses proththd.n φ N
proththd.k φ K
proththd.p φ P = K 2 N + 1
Assertion proththdlem φ P 1 < P P 1 2

Proof

Step Hyp Ref Expression
1 proththd.n φ N
2 proththd.k φ K
3 proththd.p φ P = K 2 N + 1
4 2nn 2
5 4 a1i φ 2
6 1 nnnn0d φ N 0
7 5 6 nnexpcld φ 2 N
8 2 7 nnmulcld φ K 2 N
9 8 peano2nnd φ K 2 N + 1
10 1m1e0 1 1 = 0
11 8 nngt0d φ 0 < K 2 N
12 10 11 eqbrtrid φ 1 1 < K 2 N
13 1red φ 1
14 8 nnred φ K 2 N
15 13 13 14 ltsubaddd φ 1 1 < K 2 N 1 < K 2 N + 1
16 12 15 mpbid φ 1 < K 2 N + 1
17 8 nncnd φ K 2 N
18 pncan1 K 2 N K 2 N + 1 - 1 = K 2 N
19 17 18 syl φ K 2 N + 1 - 1 = K 2 N
20 19 oveq1d φ K 2 N + 1 - 1 2 = K 2 N 2
21 2z 2
22 21 a1i φ 2
23 2 nnzd φ K
24 7 nnzd φ 2 N
25 22 23 24 3jca φ 2 K 2 N
26 iddvdsexp 2 N 2 2 N
27 22 1 26 syl2anc φ 2 2 N
28 dvdsmultr2 2 K 2 N 2 2 N 2 K 2 N
29 25 27 28 sylc φ 2 K 2 N
30 nndivdvds K 2 N 2 2 K 2 N K 2 N 2
31 8 5 30 syl2anc φ 2 K 2 N K 2 N 2
32 29 31 mpbid φ K 2 N 2
33 20 32 eqeltrd φ K 2 N + 1 - 1 2
34 9 16 33 3jca φ K 2 N + 1 1 < K 2 N + 1 K 2 N + 1 - 1 2
35 eleq1 P = K 2 N + 1 P K 2 N + 1
36 breq2 P = K 2 N + 1 1 < P 1 < K 2 N + 1
37 oveq1 P = K 2 N + 1 P 1 = K 2 N + 1 - 1
38 37 oveq1d P = K 2 N + 1 P 1 2 = K 2 N + 1 - 1 2
39 38 eleq1d P = K 2 N + 1 P 1 2 K 2 N + 1 - 1 2
40 35 36 39 3anbi123d P = K 2 N + 1 P 1 < P P 1 2 K 2 N + 1 1 < K 2 N + 1 K 2 N + 1 - 1 2
41 34 40 syl5ibrcom φ P = K 2 N + 1 P 1 < P P 1 2
42 3 41 mpd φ P 1 < P P 1 2