Certification Problem

Input (TPDB SRS_Standard/ICFP_2010/88156)

The rewrite relation of the following TRS is considered.

0(0(1(1(2(0(3(0(1(2(0(1(1(x1))))))))))))) → 0(0(3(0(0(2(0(2(2(0(0(1(0(0(0(1(0(x1))))))))))))))))) (1)
0(1(0(3(1(0(2(2(1(1(0(2(1(x1))))))))))))) → 0(0(2(2(3(0(0(0(0(1(0(1(0(3(0(1(2(x1))))))))))))))))) (2)
0(1(1(0(0(1(3(1(2(0(3(1(2(x1))))))))))))) → 0(2(0(2(0(1(3(0(2(0(0(0(0(1(2(2(0(x1))))))))))))))))) (3)
0(1(3(2(1(0(3(0(0(1(1(1(1(x1))))))))))))) → 0(3(1(3(1(0(0(3(2(0(3(0(3(0(0(1(0(x1))))))))))))))))) (4)
0(2(0(2(0(2(3(2(3(1(1(3(1(x1))))))))))))) → 0(0(1(3(0(0(3(0(3(2(3(0(0(2(2(1(0(x1))))))))))))))))) (5)
0(2(2(3(2(2(1(2(0(3(2(0(3(x1))))))))))))) → 0(3(2(1(0(2(3(0(0(1(0(2(1(0(0(3(0(x1))))))))))))))))) (6)
0(2(3(0(2(2(3(2(2(1(1(2(3(x1))))))))))))) → 1(0(0(2(3(2(0(2(0(1(3(0(2(0(1(1(2(x1))))))))))))))))) (7)
0(2(3(1(1(0(2(0(0(2(1(3(2(x1))))))))))))) → 0(2(2(0(0(3(2(2(0(1(2(2(0(0(2(2(0(x1))))))))))))))))) (8)
0(2(3(2(2(3(1(0(2(0(3(1(3(x1))))))))))))) → 0(0(3(0(2(1(1(0(0(2(2(0(2(0(2(2(3(x1))))))))))))))))) (9)
1(0(0(3(2(0(1(0(1(2(2(1(1(x1))))))))))))) → 0(0(3(3(1(0(0(0(2(0(2(0(1(0(0(1(2(x1))))))))))))))))) (10)
1(0(3(0(2(1(1(0(1(1(1(2(2(x1))))))))))))) → 0(3(2(0(0(2(0(0(3(0(0(0(3(3(3(3(3(x1))))))))))))))))) (11)
1(1(2(0(2(2(0(0(1(3(2(3(2(x1))))))))))))) → 2(0(0(0(0(1(2(3(0(1(0(0(3(2(0(0(1(x1))))))))))))))))) (12)
1(2(2(1(2(2(0(0(1(2(2(0(1(x1))))))))))))) → 0(0(3(0(0(1(2(2(0(3(2(0(2(0(1(0(3(x1))))))))))))))))) (13)
1(2(3(1(0(2(1(0(0(1(1(1(0(x1))))))))))))) → 0(3(0(1(0(1(2(0(3(0(0(0(1(0(1(3(0(x1))))))))))))))))) (14)
1(3(0(0(3(2(2(2(2(1(0(2(3(x1))))))))))))) → 3(0(0(2(2(0(3(2(0(3(0(2(3(1(2(0(0(x1))))))))))))))))) (15)
1(3(1(0(1(3(1(2(0(1(3(1(0(x1))))))))))))) → 1(2(0(3(1(3(0(0(3(3(1(0(3(0(0(0(0(x1))))))))))))))))) (16)
1(3(1(1(3(0(0(1(0(0(2(3(0(x1))))))))))))) → 2(1(0(2(0(3(2(0(0(0(0(2(0(1(1(3(0(x1))))))))))))))))) (17)
1(3(1(3(1(0(2(0(1(3(0(0(1(x1))))))))))))) → 2(2(0(0(0(1(0(0(2(0(0(1(3(3(3(0(1(x1))))))))))))))))) (18)
1(3(2(1(0(1(0(3(0(1(3(0(0(x1))))))))))))) → 0(3(1(0(0(0(3(0(0(2(3(2(1(0(1(0(0(x1))))))))))))))))) (19)
1(3(3(0(2(3(0(3(2(0(0(1(1(x1))))))))))))) → 3(2(0(3(0(3(0(0(2(0(0(0(1(0(2(3(1(x1))))))))))))))))) (20)
1(3(3(2(2(2(3(2(2(0(2(3(0(x1))))))))))))) → 3(0(3(2(2(0(2(2(1(0(2(2(3(1(2(0(0(x1))))))))))))))))) (21)
2(0(2(1(2(2(3(2(2(2(2(1(0(x1))))))))))))) → 1(0(0(0(0(1(0(0(2(0(3(1(0(0(2(3(0(x1))))))))))))))))) (22)
2(0(2(2(1(2(2(3(2(0(1(1(2(x1))))))))))))) → 0(2(3(1(3(1(0(0(0(0(0(0(1(2(2(1(0(x1))))))))))))))))) (23)
2(0(3(3(1(2(2(0(0(2(1(0(1(x1))))))))))))) → 3(2(0(1(0(0(2(0(3(1(0(0(0(2(1(0(2(x1))))))))))))))))) (24)
2(1(1(0(3(2(1(2(0(0(3(1(3(x1))))))))))))) → 2(2(0(1(0(0(0(0(2(2(0(0(2(2(1(3(3(x1))))))))))))))))) (25)
2(2(0(1(0(1(0(3(3(2(1(2(3(x1))))))))))))) → 0(3(0(1(2(0(1(2(0(2(2(1(2(1(0(0(0(x1))))))))))))))))) (26)
2(3(0(0(0(2(3(3(2(0(3(0(3(x1))))))))))))) → 2(2(0(0(0(1(2(0(0(0(3(0(2(0(3(2(0(x1))))))))))))))))) (27)
2(3(0(2(2(0(2(0(3(2(3(2(3(x1))))))))))))) → 1(0(0(3(2(0(3(3(0(0(3(0(0(2(0(0(2(x1))))))))))))))))) (28)
3(0(0(1(3(1(2(0(2(0(3(3(3(x1))))))))))))) → 0(0(1(0(1(2(0(0(3(0(2(2(2(0(0(1(3(x1))))))))))))))))) (29)
3(2(2(2(0(1(3(0(2(2(3(3(0(x1))))))))))))) → 1(1(0(1(0(0(3(0(1(2(1(1(0(0(1(0(0(x1))))))))))))))))) (30)

Property / Task

Prove or disprove termination.

Answer / Result

Yes.

Proof (by matchbox @ termCOMP 2023)

1 Closure Under Flat Contexts

Using the flat contexts

{3(☐), 2(☐), 1(☐), 0(☐)}

We obtain the transformed TRS

There are 120 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,...,3}. Symbols are labeled by the interpretation of their arguments using the interpretations (modulo 4):

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

We obtain the labeled TRS

There are 480 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
[30(x1)] = x1 +
13
[31(x1)] = x1 +
21
[32(x1)] = x1 +
17
[33(x1)] = x1 +
0
[20(x1)] = x1 +
21
[21(x1)] = x1 +
20
[22(x1)] = x1 +
20
[23(x1)] = x1 +
0
[10(x1)] = x1 +
21
[11(x1)] = x1 +
16
[12(x1)] = x1 +
21
[13(x1)] = x1 +
0
[00(x1)] = x1 +
0
[01(x1)] = x1 +
1
[02(x1)] = x1 +
0
[03(x1)] = x1 +
0
all of the following rules can be deleted.

There are 480 ruless (increase limit for explicit display).

1.1.1.1 R is empty

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