Metamath Proof Explorer


Theorem 2lgslem1c

Description: Lemma 3 for 2lgslem1 . (Contributed by AV, 19-Jun-2021)

Ref Expression
Assertion 2lgslem1c P¬2PP4P12

Proof

Step Hyp Ref Expression
1 prmnn PP
2 nnnn0 PP0
3 oddnn02np1 P0¬2Pn02n+1=P
4 1 2 3 3syl P¬2Pn02n+1=P
5 iftrue 2nif2nn2n12=n2
6 5 adantr 2nn0if2nn2n12=n2
7 2nn 2
8 nn0ledivnn n02n2n
9 7 8 mpan2 n0n2n
10 9 adantl 2nn0n2n
11 6 10 eqbrtrd 2nn0if2nn2n12n
12 iffalse ¬2nif2nn2n12=n12
13 12 adantr ¬2nn0if2nn2n12=n12
14 nn0re n0n
15 peano2rem nn1
16 15 rehalfcld nn12
17 14 16 syl n0n12
18 14 rehalfcld n0n2
19 14 lem1d n0n1n
20 14 15 syl n0n1
21 2re 2
22 2pos 0<2
23 21 22 pm3.2i 20<2
24 23 a1i n020<2
25 lediv1 n1n20<2n1nn12n2
26 20 14 24 25 syl3anc n0n1nn12n2
27 19 26 mpbid n0n12n2
28 17 18 14 27 9 letrd n0n12n
29 28 adantl ¬2nn0n12n
30 13 29 eqbrtrd ¬2nn0if2nn2n12n
31 11 30 pm2.61ian n0if2nn2n12n
32 31 ad2antlr Pn02n+1=Pif2nn2n12n
33 nn0z n0n
34 33 adantl Pn0n
35 eqcom 2n+1=PP=2n+1
36 35 biimpi 2n+1=PP=2n+1
37 flodddiv4 nP=2n+1P4=if2nn2n12
38 34 36 37 syl2an Pn02n+1=PP4=if2nn2n12
39 oveq1 P=2n+1P1=2n+1-1
40 39 eqcoms 2n+1=PP1=2n+1-1
41 40 adantl Pn02n+1=PP1=2n+1-1
42 2nn0 20
43 42 a1i n020
44 id n0n0
45 43 44 nn0mulcld n02n0
46 45 nn0cnd n02n
47 pncan1 2n2n+1-1=2n
48 46 47 syl n02n+1-1=2n
49 48 ad2antlr Pn02n+1=P2n+1-1=2n
50 41 49 eqtrd Pn02n+1=PP1=2n
51 50 oveq1d Pn02n+1=PP12=2n2
52 nn0cn n0n
53 2cnd n02
54 2ne0 20
55 54 a1i n020
56 52 53 55 divcan3d n02n2=n
57 56 ad2antlr Pn02n+1=P2n2=n
58 51 57 eqtrd Pn02n+1=PP12=n
59 32 38 58 3brtr4d Pn02n+1=PP4P12
60 59 rexlimdva2 Pn02n+1=PP4P12
61 4 60 sylbid P¬2PP4P12
62 61 imp P¬2PP4P12