Metamath Proof Explorer


Theorem peano5uzi

Description: Peano's inductive postulate for upper integers. (Contributed by NM, 6-Jul-2005) (Revised by Mario Carneiro, 3-May-2014)

Ref Expression
Hypothesis peano5uzi.1 N
Assertion peano5uzi NAxAx+1Ak|NkA

Proof

Step Hyp Ref Expression
1 peano5uzi.1 N
2 breq2 k=nNkNn
3 2 elrab nk|NknNn
4 zcn nn
5 4 ad2antrl NAxAx+1AnNnn
6 zcn NN
7 1 6 ax-mp N
8 ax-1cn 1
9 7 8 subcli N1
10 npcan nN1nN1+N-1=n
11 5 9 10 sylancl NAxAx+1AnNnnN1+N-1=n
12 subsub nN1nN1=n-N+1
13 7 8 12 mp3an23 nnN1=n-N+1
14 5 13 syl NAxAx+1AnNnnN1=n-N+1
15 znn0sub NnNnnN0
16 1 15 mpan nNnnN0
17 16 biimpa nNnnN0
18 17 adantl NAxAx+1AnNnnN0
19 nn0p1nn nN0n-N+1
20 18 19 syl NAxAx+1AnNnn-N+1
21 14 20 eqeltrd NAxAx+1AnNnnN1
22 simpl NAxAx+1AnNnNAxAx+1A
23 oveq1 k=1k+N-1=1+N-1
24 23 eleq1d k=1k+N-1A1+N-1A
25 24 imbi2d k=1NAxAx+1Ak+N-1ANAxAx+1A1+N-1A
26 oveq1 k=nk+N-1=n+N-1
27 26 eleq1d k=nk+N-1An+N-1A
28 27 imbi2d k=nNAxAx+1Ak+N-1ANAxAx+1An+N-1A
29 oveq1 k=n+1k+N-1=n+1+N1
30 29 eleq1d k=n+1k+N-1An+1+N1A
31 30 imbi2d k=n+1NAxAx+1Ak+N-1ANAxAx+1An+1+N1A
32 oveq1 k=nN1k+N-1=nN1+N-1
33 32 eleq1d k=nN1k+N-1AnN1+N-1A
34 33 imbi2d k=nN1NAxAx+1Ak+N-1ANAxAx+1AnN1+N-1A
35 8 7 pncan3i 1+N-1=N
36 simpl NAxAx+1ANA
37 35 36 eqeltrid NAxAx+1A1+N-1A
38 oveq1 x=n+N-1x+1=n+N1+1
39 38 eleq1d x=n+N-1x+1An+N1+1A
40 39 rspccv xAx+1An+N-1An+N1+1A
41 40 ad2antll nNAxAx+1An+N-1An+N1+1A
42 nncn nn
43 42 adantr nNAxAx+1An
44 add32 nN11n+N1+1=n+1+N1
45 9 8 44 mp3an23 nn+N1+1=n+1+N1
46 43 45 syl nNAxAx+1An+N1+1=n+1+N1
47 46 eleq1d nNAxAx+1An+N1+1An+1+N1A
48 41 47 sylibd nNAxAx+1An+N-1An+1+N1A
49 48 ex nNAxAx+1An+N-1An+1+N1A
50 49 a2d nNAxAx+1An+N-1ANAxAx+1An+1+N1A
51 25 28 31 34 37 50 nnind nN1NAxAx+1AnN1+N-1A
52 21 22 51 sylc NAxAx+1AnNnnN1+N-1A
53 11 52 eqeltrrd NAxAx+1AnNnnA
54 53 ex NAxAx+1AnNnnA
55 3 54 biimtrid NAxAx+1Ank|NknA
56 55 ssrdv NAxAx+1Ak|NkA