Certification Problem

Input (TPDB SRS_Standard/ICFP_2010/96523)

The rewrite relation of the following TRS is considered.

0(1(1(2(x1)))) → 0(0(1(x1))) (1)
1(2(3(3(4(5(x1)))))) → 4(0(0(1(5(x1))))) (2)
4(4(3(4(3(1(3(x1))))))) → 0(1(3(1(4(4(x1)))))) (3)
5(2(5(2(5(1(1(x1))))))) → 5(4(5(5(1(x1))))) (4)
1(0(5(5(5(3(4(1(x1)))))))) → 4(3(0(3(4(3(0(5(0(x1))))))))) (5)
3(4(3(0(4(0(3(2(x1)))))))) → 5(1(5(4(2(x1))))) (6)
0(5(3(2(3(1(5(4(4(x1))))))))) → 0(0(4(4(1(0(3(4(x1)))))))) (7)
1(0(4(3(5(2(1(0(1(x1))))))))) → 1(4(2(0(1(1(3(4(1(x1))))))))) (8)
5(3(2(1(1(2(1(5(3(x1))))))))) → 5(1(4(3(3(5(0(4(x1)))))))) (9)
1(2(4(4(4(5(5(0(2(2(x1)))))))))) → 4(4(1(1(0(5(5(5(0(x1))))))))) (10)
2(1(3(1(5(5(5(1(1(1(x1)))))))))) → 5(4(2(3(5(4(2(2(1(x1))))))))) (11)
2(0(3(1(1(0(2(5(3(3(2(3(x1)))))))))))) → 5(4(0(4(4(4(1(4(0(3(x1)))))))))) (12)
3(4(1(1(4(0(1(2(2(4(5(3(2(x1))))))))))))) → 5(3(1(3(4(5(4(0(3(0(1(x1))))))))))) (13)
1(2(0(3(1(0(4(3(3(0(3(0(5(3(x1)))))))))))))) → 4(2(0(5(3(3(4(4(3(1(5(0(4(x1))))))))))))) (14)
1(5(4(4(3(2(4(0(1(5(2(0(5(2(x1)))))))))))))) → 1(5(5(2(2(2(4(0(0(5(3(3(2(2(2(0(x1)))))))))))))))) (15)
3(5(0(1(4(0(0(1(3(5(4(1(0(2(x1)))))))))))))) → 1(1(4(1(1(5(0(3(0(0(4(5(0(x1))))))))))))) (16)
3(2(3(4(3(5(5(3(4(0(5(4(3(5(2(x1))))))))))))))) → 3(5(5(0(1(4(3(4(1(1(5(1(0(x1))))))))))))) (17)
0(0(2(2(5(0(3(0(4(0(4(0(2(3(1(5(x1)))))))))))))))) → 0(4(2(5(2(5(3(0(4(3(2(0(2(4(5(x1))))))))))))))) (18)
2(1(3(0(5(1(2(2(5(5(1(0(2(1(3(2(x1)))))))))))))))) → 5(5(2(3(2(4(5(0(2(0(3(3(4(1(1(x1))))))))))))))) (19)
2(4(4(1(5(2(3(3(2(0(4(5(3(5(0(2(x1)))))))))))))))) → 0(5(1(4(5(2(4(1(1(5(4(3(3(x1))))))))))))) (20)
1(5(0(0(1(4(5(3(5(4(1(0(1(2(1(2(0(x1))))))))))))))))) → 1(4(0(1(3(5(3(3(4(1(0(4(3(3(0(5(1(0(x1)))))))))))))))))) (21)
4(0(4(4(2(1(4(2(0(3(1(2(5(5(5(5(3(1(2(x1))))))))))))))))))) → 3(5(4(4(2(0(4(4(4(0(1(0(2(4(0(4(1(5(x1)))))))))))))))))) (22)
1(1(0(1(1(4(5(0(4(1(3(0(4(4(0(5(5(0(2(0(x1)))))))))))))))))))) → 4(0(5(3(5(3(2(2(5(3(5(1(2(4(2(4(5(0(5(5(x1)))))))))))))))))))) (23)
3(1(2(1(3(1(1(4(3(5(2(0(3(3(2(3(1(3(0(1(x1)))))))))))))))))))) → 0(4(3(5(1(4(2(4(0(5(3(2(4(1(5(5(1(x1))))))))))))))))) (24)
5(0(1(3(4(2(1(3(1(4(2(0(0(2(3(2(5(2(2(4(x1)))))))))))))))))))) → 5(4(1(5(5(0(1(3(1(2(1(1(2(4(3(2(5(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 +
110/3
[51(x1)] = x1 +
0
[52(x1)] = x1 +
0
[53(x1)] = x1 +
113/3
[54(x1)] = x1 +
0
[55(x1)] = x1 +
0
[40(x1)] = x1 +
0
[41(x1)] = x1 +
103/3
[42(x1)] = x1 +
113/3
[43(x1)] = x1 +
0
[44(x1)] = x1 +
103/3
[45(x1)] = x1 +
70/3
[30(x1)] = x1 +
0
[31(x1)] = x1 +
113/3
[32(x1)] = x1 +
0
[33(x1)] = x1 +
113/3
[34(x1)] = x1 +
110/3
[35(x1)] = x1 +
40/3
[20(x1)] = x1 +
110/3
[21(x1)] = x1 +
1
[22(x1)] = x1 +
128/3
[23(x1)] = x1 +
70/3
[24(x1)] = x1 +
128/3
[25(x1)] = x1 +
110/3
[10(x1)] = x1 +
1
[11(x1)] = x1 +
0
[12(x1)] = x1 +
110/3
[13(x1)] = x1 +
128/3
[14(x1)] = x1 +
11
[15(x1)] = x1 +
110/3
[00(x1)] = x1 +
0
[01(x1)] = x1 +
0
[02(x1)] = x1 +
10/3
[03(x1)] = x1 +
110/3
[04(x1)] = x1 +
110/3
[05(x1)] = x1 +
0
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.