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 φ N
aks6d1c2p1.2 φ P
aks6d1c2p1.3 φ P N
aks6d1c2p1.4 E = k 0 , l 0 P k N P l
Assertion aks6d1c2p1 φ E : 0 × 0

Proof

Step Hyp Ref Expression
1 aks6d1c2p1.1 φ N
2 aks6d1c2p1.2 φ P
3 aks6d1c2p1.3 φ P N
4 aks6d1c2p1.4 E = k 0 , l 0 P k N P l
5 prmnn P P
6 2 5 syl φ P
7 6 adantr φ a 0 × 0 P
8 simpr φ a 0 × 0 a 0 × 0
9 xp1st a 0 × 0 1 st a 0
10 8 9 syl φ a 0 × 0 1 st a 0
11 7 10 nnexpcld φ a 0 × 0 P 1 st a
12 1 6 jca φ N P
13 nndivdvds N P P N N P
14 12 13 syl φ P N N P
15 3 14 mpbid φ N P
16 15 adantr φ a 0 × 0 N P
17 xp2nd a 0 × 0 2 nd a 0
18 8 17 syl φ a 0 × 0 2 nd a 0
19 16 18 nnexpcld φ a 0 × 0 N P 2 nd a
20 11 19 nnmulcld φ a 0 × 0 P 1 st a N P 2 nd a
21 vex k V
22 vex l V
23 21 22 op1std a = k l 1 st a = k
24 23 oveq2d a = k l P 1 st a = P k
25 21 22 op2ndd a = k l 2 nd a = l
26 25 oveq2d a = k l N P 2 nd a = N P l
27 24 26 oveq12d a = k l P 1 st a N P 2 nd a = P k N P l
28 27 mpompt a 0 × 0 P 1 st a N P 2 nd a = k 0 , l 0 P k N P l
29 28 eqcomi k 0 , l 0 P k N P l = a 0 × 0 P 1 st a N P 2 nd a
30 4 29 eqtri E = a 0 × 0 P 1 st a N P 2 nd a
31 20 30 fmptd φ E : 0 × 0