Metamath Proof Explorer


Theorem gpg3nbgrvtxlem

Description: Lemma for gpg3nbgrvtx0ALT and gpg3nbgrvtx1 . For this theorem, it is essential that 2 < N and K < ( N / 2 ) ! (Contributed by AV, 3-Sep-2025) (Proof shortened by AV, 9-Sep-2025)

Ref Expression
Assertion gpg3nbgrvtxlem N 3 K 1 ..^ N 2 A 0 ..^ N A + K mod N A K mod N

Proof

Step Hyp Ref Expression
1 eluzge3nn N 3 N
2 1 3ad2ant1 N 3 K 1 ..^ N 2 A 0 ..^ N N
3 elfzoelz A 0 ..^ N A
4 3 3ad2ant3 N 3 K 1 ..^ N 2 A 0 ..^ N A
5 elfzoelz K 1 ..^ N 2 K
6 5 3ad2ant2 N 3 K 1 ..^ N 2 A 0 ..^ N K
7 5 zcnd K 1 ..^ N 2 K
8 2times K 2 K = K + K
9 8 eqcomd K K + K = 2 K
10 7 9 syl K 1 ..^ N 2 K + K = 2 K
11 10 adantl N 3 K 1 ..^ N 2 K + K = 2 K
12 1red K 1 ..^ N 2 1
13 5 zred K 1 ..^ N 2 K
14 2z 2
15 14 a1i K 1 ..^ N 2 2
16 15 5 zmulcld K 1 ..^ N 2 2 K
17 16 zred K 1 ..^ N 2 2 K
18 elfzole1 K 1 ..^ N 2 1 K
19 elfzo1 K 1 ..^ N 2 K N 2 K < N 2
20 19 simp1bi K 1 ..^ N 2 K
21 20 nnnn0d K 1 ..^ N 2 K 0
22 nn0le2x K 0 K 2 K
23 21 22 syl K 1 ..^ N 2 K 2 K
24 12 13 17 18 23 letrd K 1 ..^ N 2 1 2 K
25 24 adantl N 3 K 1 ..^ N 2 1 2 K
26 2tceilhalfelfzo1 N 3 K 1 ..^ N 2 2 K < N
27 25 26 jca N 3 K 1 ..^ N 2 1 2 K 2 K < N
28 breq2 K + K = 2 K 1 K + K 1 2 K
29 breq1 K + K = 2 K K + K < N 2 K < N
30 28 29 anbi12d K + K = 2 K 1 K + K K + K < N 1 2 K 2 K < N
31 27 30 syl5ibrcom N 3 K 1 ..^ N 2 K + K = 2 K 1 K + K K + K < N
32 11 31 mpd N 3 K 1 ..^ N 2 1 K + K K + K < N
33 32 3adant3 N 3 K 1 ..^ N 2 A 0 ..^ N 1 K + K K + K < N
34 submodneaddmod N A K K 1 K + K K + K < N A + K mod N A K mod N
35 2 4 6 6 33 34 syl131anc N 3 K 1 ..^ N 2 A 0 ..^ N A + K mod N A K mod N