Metamath Proof Explorer


Theorem oddge22np1

Description: An integer greater than one is odd iff it is one plus twice a positive integer. (Contributed by AV, 16-Aug-2021) (Proof shortened by AV, 9-Jul-2022)

Ref Expression
Assertion oddge22np1 N2¬2Nn2n+1=N

Proof

Step Hyp Ref Expression
1 eleq1 2n+1=N2n+12N2
2 nn0z n0n
3 2 adantl 2n+12n0n
4 eluz2 2n+1222n+122n+1
5 2re 2
6 5 a1i n02
7 1red n01
8 2nn0 20
9 8 a1i n020
10 id n0n0
11 9 10 nn0mulcld n02n0
12 11 nn0red n02n
13 6 7 12 lesubaddd n0212n22n+1
14 2m1e1 21=1
15 14 breq1i 212n12n
16 nn0re n0n
17 2rp 2+
18 17 a1i n02+
19 7 16 18 ledivmuld n012n12n
20 halfgt0 0<12
21 0red n00
22 halfre 12
23 22 a1i n012
24 ltletr 012n0<1212n0<n
25 21 23 16 24 syl3anc n00<1212n0<n
26 20 25 mpani n012n0<n
27 19 26 sylbird n012n0<n
28 15 27 biimtrid n0212n0<n
29 13 28 sylbird n022n+10<n
30 29 com12 22n+1n00<n
31 30 3ad2ant3 22n+122n+1n00<n
32 4 31 sylbi 2n+12n00<n
33 32 imp 2n+12n00<n
34 elnnz nn0<n
35 3 33 34 sylanbrc 2n+12n0n
36 35 ex 2n+12n0n
37 1 36 syl6bir 2n+1=NN2n0n
38 37 com13 n0N22n+1=Nn
39 38 impcom N2n02n+1=Nn
40 39 pm4.71rd N2n02n+1=Nn2n+1=N
41 40 bicomd N2n0n2n+1=N2n+1=N
42 41 rexbidva N2n0n2n+1=Nn02n+1=N
43 nnssnn0 0
44 rexss 0n2n+1=Nn0n2n+1=N
45 43 44 mp1i N2n2n+1=Nn0n2n+1=N
46 eluzge2nn0 N2N0
47 oddnn02np1 N0¬2Nn02n+1=N
48 46 47 syl N2¬2Nn02n+1=N
49 42 45 48 3bitr4rd N2¬2Nn2n+1=N