Metamath Proof Explorer


Theorem proththdlem

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

Ref Expression
Hypotheses proththd.n φN
proththd.k φK
proththd.p φP=K2N+1
Assertion proththdlem φP1<PP12

Proof

Step Hyp Ref Expression
1 proththd.n φN
2 proththd.k φK
3 proththd.p φP=K2N+1
4 2nn 2
5 4 a1i φ2
6 1 nnnn0d φN0
7 5 6 nnexpcld φ2N
8 2 7 nnmulcld φK2N
9 8 peano2nnd φK2N+1
10 1m1e0 11=0
11 8 nngt0d φ0<K2N
12 10 11 eqbrtrid φ11<K2N
13 1red φ1
14 8 nnred φK2N
15 13 13 14 ltsubaddd φ11<K2N1<K2N+1
16 12 15 mpbid φ1<K2N+1
17 8 nncnd φK2N
18 pncan1 K2NK2N+1-1=K2N
19 17 18 syl φK2N+1-1=K2N
20 19 oveq1d φK2N+1-12=K2N2
21 2z 2
22 21 a1i φ2
23 2 nnzd φK
24 7 nnzd φ2N
25 22 23 24 3jca φ2K2N
26 iddvdsexp 2N22N
27 22 1 26 syl2anc φ22N
28 dvdsmultr2 2K2N22N2K2N
29 25 27 28 sylc φ2K2N
30 nndivdvds K2N22K2NK2N2
31 8 5 30 syl2anc φ2K2NK2N2
32 29 31 mpbid φK2N2
33 20 32 eqeltrd φK2N+1-12
34 9 16 33 3jca φK2N+11<K2N+1K2N+1-12
35 eleq1 P=K2N+1PK2N+1
36 breq2 P=K2N+11<P1<K2N+1
37 oveq1 P=K2N+1P1=K2N+1-1
38 37 oveq1d P=K2N+1P12=K2N+1-12
39 38 eleq1d P=K2N+1P12K2N+1-12
40 35 36 39 3anbi123d P=K2N+1P1<PP12K2N+11<K2N+1K2N+1-12
41 34 40 syl5ibrcom φP=K2N+1P1<PP12
42 3 41 mpd φP1<PP12