Certification Problem

Input (TPDB SRS_Standard/ICFP_2010/57453)

The rewrite relation of the following TRS is considered.

0(0(1(2(0(0(x1)))))) → 0(0(3(4(0(x1))))) (1)
1(5(5(3(1(4(x1)))))) → 2(1(4(1(0(x1))))) (2)
2(0(2(2(2(0(x1)))))) → 2(1(0(5(0(x1))))) (3)
2(5(2(2(3(0(x1)))))) → 2(5(5(1(0(x1))))) (4)
1(2(0(1(5(3(1(x1))))))) → 0(2(4(3(1(5(x1)))))) (5)
2(3(2(2(0(2(3(x1))))))) → 2(5(4(1(0(0(x1)))))) (6)
2(4(2(5(1(3(1(x1))))))) → 2(1(1(3(4(3(x1)))))) (7)
0(0(3(2(4(5(2(2(x1)))))))) → 5(1(2(0(2(5(2(2(x1)))))))) (8)
1(2(2(4(0(1(2(4(x1)))))))) → 4(2(0(4(3(3(4(x1))))))) (9)
0(0(2(0(3(5(4(2(0(x1))))))))) → 5(5(1(5(5(0(4(0(x1)))))))) (10)
4(1(3(4(1(2(3(4(2(x1))))))))) → 1(1(0(4(3(5(2(2(x1)))))))) (11)
0(2(2(0(3(5(2(4(1(2(x1)))))))))) → 0(4(5(3(3(0(5(5(3(x1))))))))) (12)
2(4(3(5(5(2(4(2(5(1(x1)))))))))) → 4(1(4(1(4(4(0(0(1(x1))))))))) (13)
1(3(2(5(1(1(1(1(0(1(3(x1))))))))))) → 0(3(5(5(5(0(0(1(2(4(2(0(x1)))))))))))) (14)
1(5(5(5(0(5(1(5(2(0(3(x1))))))))))) → 0(0(4(4(5(4(3(4(3(3(x1)))))))))) (15)
2(2(0(1(1(1(2(3(3(2(0(x1))))))))))) → 3(0(1(3(1(5(1(5(1(0(x1)))))))))) (16)
2(0(4(4(4(2(3(0(5(2(4(2(x1)))))))))))) → 2(5(5(5(3(4(1(3(4(4(2(x1))))))))))) (17)
2(2(2(1(1(0(5(3(1(4(4(3(x1)))))))))))) → 5(4(1(2(0(3(3(1(3(5(5(3(x1)))))))))))) (18)
2(2(4(2(1(1(2(0(0(5(2(1(x1)))))))))))) → 4(2(1(0(1(0(4(2(3(3(x1)))))))))) (19)
4(0(3(0(5(2(5(4(3(1(5(0(x1)))))))))))) → 1(3(1(1(2(1(5(2(0(4(3(0(x1)))))))))))) (20)
4(5(3(4(2(0(4(3(3(5(3(4(x1)))))))))))) → 2(1(4(5(3(3(1(0(3(2(2(4(x1)))))))))))) (21)
1(4(1(3(2(1(5(2(2(1(3(4(5(x1))))))))))))) → 0(3(1(5(3(0(1(2(3(3(1(0(x1)))))))))))) (22)
2(2(1(4(1(1(1(0(1(3(5(3(4(3(x1)))))))))))))) → 2(5(3(4(1(5(3(0(0(5(2(2(4(2(3(x1))))))))))))))) (23)
4(3(1(4(0(1(0(4(2(2(3(4(0(0(x1)))))))))))))) → 3(4(0(0(0(1(0(1(1(0(1(3(0(x1))))))))))))) (24)
0(1(2(0(5(3(4(5(2(4(3(4(1(3(5(x1))))))))))))))) → 0(2(1(1(3(3(3(3(0(5(2(3(2(0(0(x1))))))))))))))) (25)
1(5(5(4(3(3(5(1(2(0(0(2(4(3(0(x1))))))))))))))) → 1(1(1(0(3(4(5(1(3(1(1(0(x1)))))))))))) (26)
4(0(2(5(3(5(4(3(5(4(5(4(5(1(3(x1))))))))))))))) → 4(0(3(1(4(3(2(4(5(1(1(5(1(0(x1)))))))))))))) (27)
4(4(3(4(1(5(1(2(1(0(0(3(5(2(3(x1))))))))))))))) → 1(3(2(1(2(3(2(4(5(5(5(0(3(1(3(x1))))))))))))))) (28)
0(2(0(4(2(4(4(4(3(3(4(2(2(2(4(3(x1)))))))))))))))) → 0(4(2(1(5(4(3(3(1(3(3(0(1(0(3(x1))))))))))))))) (29)
2(5(0(0(1(3(4(1(0(5(3(2(2(3(0(3(x1)))))))))))))))) → 3(5(3(3(0(3(0(2(0(1(4(5(2(0(0(3(x1)))))))))))))))) (30)
2(4(5(5(3(2(0(0(1(4(2(1(3(2(2(0(5(x1))))))))))))))))) → 0(5(2(0(5(2(4(1(3(2(3(1(2(3(0(0(x1)))))))))))))))) (31)
0(5(5(3(4(0(1(4(1(4(4(2(3(2(5(4(1(3(x1)))))))))))))))))) → 0(3(0(0(5(0(3(0(1(1(0(3(3(3(4(5(5(x1))))))))))))))))) (32)
2(0(0(4(3(5(3(4(1(4(3(0(3(0(1(4(5(4(x1)))))))))))))))))) → 5(0(1(1(3(3(0(1(2(0(5(4(2(5(2(0(4(5(x1)))))))))))))))))) (33)
2(2(3(5(0(3(3(0(0(2(4(1(2(3(4(2(2(4(x1)))))))))))))))))) → 4(1(5(0(2(5(4(0(2(1(3(3(0(0(4(5(4(x1))))))))))))))))) (34)
4(0(4(1(5(4(5(5(2(4(5(4(4(0(5(3(5(0(x1)))))))))))))))))) → 3(5(5(4(4(2(2(0(3(0(4(5(0(1(5(4(4(0(x1)))))))))))))))))) (35)
4(2(1(4(0(0(2(1(1(4(3(4(1(5(4(0(1(4(x1)))))))))))))))))) → 2(5(4(2(5(2(1(4(0(3(1(3(1(4(0(0(1(4(x1)))))))))))))))))) (36)
4(5(1(4(5(3(3(5(1(2(4(3(4(3(5(1(3(3(1(x1))))))))))))))))))) → 0(1(2(0(0(1(5(1(5(1(4(0(0(4(3(1(1(1(x1)))))))))))))))))) (37)
4(1(5(2(2(4(2(3(3(1(1(4(2(0(5(1(0(0(3(5(x1)))))))))))))))))))) → 4(1(0(4(5(3(2(4(4(5(4(5(2(4(1(2(2(1(3(x1))))))))))))))))))) (38)
5(4(0(5(5(2(3(5(1(4(1(2(1(3(4(4(1(2(4(3(x1)))))))))))))))))))) → 5(2(5(4(3(3(4(2(5(3(5(3(4(1(0(2(4(4(3(x1))))))))))))))))))) (39)
3(2(1(5(5(4(5(3(4(2(2(4(5(5(4(4(0(4(2(5(5(x1))))))))))))))))))))) → 3(5(4(3(1(3(4(1(1(4(3(0(0(0(4(3(3(3(2(5(x1)))))))))))))))))))) (40)

Property / Task

Prove or disprove termination.

Answer / Result

Yes.

Proof (by matchbox @ termCOMP 2023)

1 Closure Under Flat Contexts

Using the flat contexts

{5(☐), 4(☐), 3(☐), 2(☐), 1(☐), 0(☐)}

We obtain the transformed TRS

There are 240 ruless (increase limit for explicit display).

1.1 Semantic Labeling

The following interpretations form a model of the rules.

As carrier we take the set {0,...,5}. Symbols are labeled by the interpretation of their arguments using the interpretations (modulo 6):

[5(x1)] = 6x1 + 0
[4(x1)] = 6x1 + 1
[3(x1)] = 6x1 + 2
[2(x1)] = 6x1 + 3
[1(x1)] = 6x1 + 4
[0(x1)] = 6x1 + 5

We obtain the labeled TRS

There are 1440 ruless (increase limit for explicit display).

1.1.1 Rule Removal

Using the matrix interpretations of dimension 1 with strict dimension 1 over the rationals with delta = 1
[50(x1)] = x1 +
0
[51(x1)] = x1 +
23579/2
[52(x1)] = x1 +
110151/8
[53(x1)] = x1 +
6811/8
[54(x1)] = x1 +
110143/8
[55(x1)] = x1 +
7/2
[40(x1)] = x1 +
50065/4
[41(x1)] = x1 +
28747/4
[42(x1)] = x1 +
49419/4
[43(x1)] = x1 +
110171/8
[44(x1)] = x1 +
74613/8
[45(x1)] = x1 +
17119/2
[30(x1)] = x1 +
110143/8
[31(x1)] = x1 +
34561/4
[32(x1)] = x1 +
0
[33(x1)] = x1 +
110287/8
[34(x1)] = x1 +
23229/8
[35(x1)] = x1 +
7/2
[20(x1)] = x1 +
12274
[21(x1)] = x1 +
110143/8
[22(x1)] = x1 +
13257/4
[23(x1)] = x1 +
110439/8
[24(x1)] = x1 +
71391/8
[25(x1)] = x1 +
110151/8
[10(x1)] = x1 +
110171/8
[11(x1)] = x1 +
110287/8
[12(x1)] = x1 +
110151/8
[13(x1)] = x1 +
8083/8
[14(x1)] = x1 +
110151/8
[15(x1)] = x1 +
18
[00(x1)] = x1 +
110171/8
[01(x1)] = x1 +
0
[02(x1)] = x1 +
14535/2
[03(x1)] = x1 +
0
[04(x1)] = x1 +
4847/2
[05(x1)] = x1 +
14535/2
all of the following rules can be deleted.

There are 1440 ruless (increase limit for explicit display).

1.1.1.1 R is empty

There are no rules in the TRS. Hence, it is terminating.