Certification Problem

Input (TPDB SRS_Standard/ICFP_2010/96334)

The rewrite relation of the following TRS is considered.

0(1(1(2(3(x1))))) → 0(1(3(3(4(x1))))) (1)
1(4(2(1(4(x1))))) → 1(1(1(0(x1)))) (2)
1(4(5(1(4(x1))))) → 5(5(5(2(x1)))) (3)
2(0(5(2(5(x1))))) → 2(2(3(2(1(x1))))) (4)
0(1(0(5(0(4(x1)))))) → 0(1(5(5(1(x1))))) (5)
1(0(5(5(1(2(4(x1))))))) → 2(5(4(5(1(1(4(x1))))))) (6)
4(0(2(5(4(3(5(x1))))))) → 1(1(3(5(1(5(x1)))))) (7)
0(4(0(0(3(1(4(1(2(x1))))))))) → 2(5(3(2(2(1(2(2(x1)))))))) (8)
5(4(5(4(2(5(0(1(5(4(x1)))))))))) → 0(0(1(2(1(2(3(2(2(x1))))))))) (9)
1(2(3(2(1(4(5(2(5(4(5(x1))))))))))) → 2(1(2(4(2(2(3(0(2(1(x1)))))))))) (10)
4(2(0(4(2(2(5(0(5(3(5(x1))))))))))) → 4(5(2(5(1(2(4(5(4(3(3(5(x1)))))))))))) (11)
2(4(0(0(0(3(3(3(3(0(0(4(x1)))))))))))) → 2(0(0(2(0(4(4(3(1(0(2(x1))))))))))) (12)
3(0(3(1(2(0(1(2(2(4(0(4(x1)))))))))))) → 0(4(1(4(3(2(0(2(5(4(3(3(x1)))))))))))) (13)
4(0(2(5(2(1(1(2(0(1(2(1(x1)))))))))))) → 0(2(0(0(0(1(0(1(5(2(1(x1))))))))))) (14)
4(3(4(5(2(4(4(5(4(3(4(3(x1)))))))))))) → 2(5(3(0(2(5(2(4(4(1(3(x1))))))))))) (15)
5(5(4(1(1(1(3(2(2(5(1(2(3(5(4(x1))))))))))))))) → 5(5(0(3(1(2(0(2(0(2(0(4(1(2(x1)))))))))))))) (16)
4(2(5(1(1(3(1(3(2(5(5(4(3(5(5(4(x1)))))))))))))))) → 4(2(2(5(1(0(2(5(1(0(3(0(3(5(3(x1))))))))))))))) (17)
1(5(5(5(3(2(0(1(3(4(0(2(5(0(0(3(4(3(x1)))))))))))))))))) → 5(4(3(1(4(2(4(5(4(5(0(3(1(0(2(2(1(3(x1)))))))))))))))))) (18)
3(0(3(4(3(3(5(1(0(4(5(0(0(0(2(2(1(4(x1)))))))))))))))))) → 0(1(5(5(0(1(5(0(5(0(0(4(0(4(3(4(x1)))))))))))))))) (19)
5(4(4(5(3(5(2(2(3(4(0(2(0(3(2(5(4(3(x1)))))))))))))))))) → 5(4(0(5(1(1(1(1(3(2(0(0(3(5(0(1(3(x1))))))))))))))))) (20)
1(5(4(3(4(4(3(3(4(5(4(1(5(5(0(1(4(0(2(x1))))))))))))))))))) → 0(4(3(3(2(2(0(5(1(0(4(4(3(2(3(5(4(2(x1)))))))))))))))))) (21)
3(2(0(3(1(4(3(1(0(5(5(1(3(4(2(1(5(4(3(x1))))))))))))))))))) → 3(0(3(1(1(3(0(4(1(1(2(1(5(2(0(0(4(2(3(x1))))))))))))))))))) (22)
1(1(1(3(3(0(0(1(2(2(5(1(4(4(4(5(4(3(4(1(x1)))))))))))))))))))) → 3(4(0(2(1(0(4(1(0(3(1(4(0(3(4(2(0(5(x1)))))))))))))))))) (23)
4(1(1(5(1(4(3(3(0(3(0(5(4(1(0(4(2(1(4(1(x1)))))))))))))))))))) → 4(3(4(5(3(1(1(4(1(3(0(2(0(2(5(1(1(3(1(x1))))))))))))))))))) (24)
5(4(2(4(0(4(1(4(4(2(0(1(3(4(1(3(5(4(4(4(x1)))))))))))))))))))) → 5(4(4(2(2(1(4(4(5(1(4(5(4(5(2(2(0(3(x1)))))))))))))))))) (25)

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 150 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 900 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 +
37885/67
[51(x1)] = x1 +
37818/67
[52(x1)] = x1 +
37818/67
[53(x1)] = x1 +
1
[54(x1)] = x1 +
30723/67
[55(x1)] = x1 +
80949/335
[40(x1)] = x1 +
125433/335
[41(x1)] = x1 +
38287/67
[42(x1)] = x1 +
3400/67
[43(x1)] = x1 +
37818/67
[44(x1)] = x1 +
37818/67
[45(x1)] = x1 +
38219/335
[30(x1)] = x1 +
38153/67
[31(x1)] = x1 +
37818/67
[32(x1)] = x1 +
0
[33(x1)] = x1 +
1
[34(x1)] = x1 +
5
[35(x1)] = x1 +
133881/335
[20(x1)] = x1 +
35607/67
[21(x1)] = x1 +
37818/67
[22(x1)] = x1 +
10660/67
[23(x1)] = x1 +
10593/67
[24(x1)] = x1 +
0
[25(x1)] = x1 +
43331/335
[10(x1)] = x1 +
0
[11(x1)] = x1 +
37818/67
[12(x1)] = x1 +
0
[13(x1)] = x1 +
37818/67
[14(x1)] = x1 +
27225/67
[15(x1)] = x1 +
114378/335
[00(x1)] = x1 +
37885/67
[01(x1)] = x1 +
57491/335
[02(x1)] = x1 +
27423/335
[03(x1)] = x1 +
0
[04(x1)] = x1 +
37818/67
[05(x1)] = x1 +
161304/335
all of the following rules can be deleted.

There are 900 ruless (increase limit for explicit display).

1.1.1.1 R is empty

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