Certification Problem

Input (TPDB SRS_Standard/ICFP_2010/96256)

The rewrite relation of the following TRS is considered.

0(1(2(3(1(x1))))) → 0(0(1(1(x1)))) (1)
4(5(2(3(5(x1))))) → 4(4(0(5(x1)))) (2)
5(3(2(4(3(2(x1)))))) → 5(3(0(5(4(x1))))) (3)
2(2(5(5(2(2(1(x1))))))) → 3(2(3(3(1(0(x1)))))) (4)
4(5(2(1(1(3(4(x1))))))) → 1(1(0(5(0(x1))))) (5)
0(2(2(3(3(2(5(0(4(x1))))))))) → 0(2(2(1(3(4(1(4(x1)))))))) (6)
2(3(0(4(0(0(5(2(4(x1))))))))) → 3(2(2(2(3(0(4(0(4(x1))))))))) (7)
2(4(3(3(3(5(0(1(3(x1))))))))) → 0(1(0(0(5(0(5(x1))))))) (8)
2(3(5(3(3(4(1(5(1(2(x1)))))))))) → 2(0(3(5(2(5(1(3(4(2(x1)))))))))) (9)
3(0(1(2(4(0(0(3(3(4(x1)))))))))) → 3(1(5(0(4(2(0(2(5(4(x1)))))))))) (10)
1(1(4(0(4(2(1(2(4(5(3(x1))))))))))) → 1(1(0(1(5(2(2(0(2(4(x1)))))))))) (11)
3(1(2(4(0(0(3(3(1(4(5(x1))))))))))) → 2(3(1(1(4(0(4(2(2(2(5(x1))))))))))) (12)
4(3(0(4(1(0(4(1(5(0(3(x1))))))))))) → 0(4(2(3(0(0(1(5(5(1(x1)))))))))) (13)
0(5(1(0(1(1(2(2(1(1(2(2(2(2(x1)))))))))))))) → 0(3(0(5(3(1(3(0(3(4(2(3(3(1(4(x1))))))))))))))) (14)
0(5(2(1(1(4(2(5(2(0(1(3(3(3(x1)))))))))))))) → 0(4(0(4(4(4(5(0(2(4(0(3(2(2(x1)))))))))))))) (15)
3(0(2(3(2(0(3(3(2(1(2(2(0(1(3(x1))))))))))))))) → 2(1(5(3(2(1(2(2(0(4(0(1(3(1(x1)))))))))))))) (16)
0(5(2(3(4(0(5(3(2(4(2(2(4(0(3(4(x1)))))))))))))))) → 0(0(0(5(1(3(2(2(3(0(1(3(2(1(x1)))))))))))))) (17)
4(4(2(5(3(3(1(1(3(1(1(0(5(3(1(4(x1)))))))))))))))) → 4(2(4(4(4(2(1(5(5(4(3(3(0(5(1(5(4(x1))))))))))))))))) (18)
5(3(5(1(1(5(4(2(2(4(1(5(0(5(4(3(x1)))))))))))))))) → 0(0(4(2(2(1(3(2(1(0(5(1(1(1(x1)))))))))))))) (19)
2(2(0(1(4(3(5(4(2(2(5(1(0(5(0(5(4(5(2(x1))))))))))))))))))) → 3(3(4(2(0(4(5(2(5(2(0(0(4(4(3(5(1(4(5(2(x1)))))))))))))))))))) (20)
2(5(3(4(1(3(2(5(3(1(1(2(2(1(0(2(4(3(3(x1))))))))))))))))))) → 0(3(5(2(3(1(1(0(3(4(3(4(0(0(5(2(2(3(x1)))))))))))))))))) (21)
5(3(1(5(0(1(4(4(2(4(3(2(5(0(4(1(0(4(4(x1))))))))))))))))))) → 5(2(2(4(1(1(5(2(3(4(2(5(2(0(2(0(2(5(4(x1))))))))))))))))))) (22)
2(3(0(3(2(3(1(1(4(2(5(3(4(2(5(2(5(0(2(3(1(x1))))))))))))))))))))) → 3(2(4(3(2(1(3(3(1(0(1(1(3(3(0(3(0(5(1(x1))))))))))))))))))) (23)
3(1(2(1(3(1(2(2(4(1(1(3(2(4(4(3(2(0(3(4(5(x1))))))))))))))))))))) → 2(5(5(1(1(2(1(0(1(3(1(0(5(2(4(1(1(5(x1)))))))))))))))))) (24)
5(4(3(1(3(4(5(3(0(1(3(4(0(2(2(4(3(3(0(5(3(x1))))))))))))))))))))) → 5(5(1(1(5(2(4(0(0(1(0(1(2(0(5(1(3(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 +
0
[51(x1)] = x1 +
620
[52(x1)] = x1 +
620
[53(x1)] = x1 +
620
[54(x1)] = x1 +
0
[55(x1)] = x1 +
628
[40(x1)] = x1 +
620
[41(x1)] = x1 +
620
[42(x1)] = x1 +
620
[43(x1)] = x1 +
240
[44(x1)] = x1 +
620
[45(x1)] = x1 +
0
[30(x1)] = x1 +
620
[31(x1)] = x1 +
620
[32(x1)] = x1 +
621
[33(x1)] = x1 +
241
[34(x1)] = x1 +
0
[35(x1)] = x1 +
1
[20(x1)] = x1 +
620
[21(x1)] = x1 +
620
[22(x1)] = x1 +
1
[23(x1)] = x1 +
520
[24(x1)] = x1 +
0
[25(x1)] = x1 +
1
[10(x1)] = x1 +
0
[11(x1)] = x1 +
620
[12(x1)] = x1 +
101
[13(x1)] = x1 +
620
[14(x1)] = x1 +
380
[15(x1)] = x1 +
8
[00(x1)] = x1 +
428
[01(x1)] = x1 +
0
[02(x1)] = x1 +
501
[03(x1)] = x1 +
161
[04(x1)] = x1 +
368
[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.