Metamath Proof Explorer


Theorem aks6d1c2p1

Description: In the AKS-theorem the subset defined by E takes values in the positive integers. (Contributed by metakunt, 7-Jan-2025)

Ref Expression
Hypotheses aks6d1c2p1.1 φN
aks6d1c2p1.2 φP
aks6d1c2p1.3 φPN
aks6d1c2p1.4 E=k0,l0PkNPl
Assertion aks6d1c2p1 φE:0×0

Proof

Step Hyp Ref Expression
1 aks6d1c2p1.1 φN
2 aks6d1c2p1.2 φP
3 aks6d1c2p1.3 φPN
4 aks6d1c2p1.4 E=k0,l0PkNPl
5 prmnn PP
6 2 5 syl φP
7 6 adantr φa0×0P
8 simpr φa0×0a0×0
9 xp1st a0×01sta0
10 8 9 syl φa0×01sta0
11 7 10 nnexpcld φa0×0P1sta
12 1 6 jca φNP
13 nndivdvds NPPNNP
14 12 13 syl φPNNP
15 3 14 mpbid φNP
16 15 adantr φa0×0NP
17 xp2nd a0×02nda0
18 8 17 syl φa0×02nda0
19 16 18 nnexpcld φa0×0NP2nda
20 11 19 nnmulcld φa0×0P1staNP2nda
21 vex kV
22 vex lV
23 21 22 op1std a=kl1sta=k
24 23 oveq2d a=klP1sta=Pk
25 21 22 op2ndd a=kl2nda=l
26 25 oveq2d a=klNP2nda=NPl
27 24 26 oveq12d a=klP1staNP2nda=PkNPl
28 27 mpompt a0×0P1staNP2nda=k0,l0PkNPl
29 28 eqcomi k0,l0PkNPl=a0×0P1staNP2nda
30 4 29 eqtri E=a0×0P1staNP2nda
31 20 30 fmptd φE:0×0