Certification Problem

Input (TPDB SRS_Standard/ICFP_2010/86636)

The rewrite relation of the following TRS is considered.

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

{5(☐), 4(☐), 3(☐), 2(☐), 1(☐), 0(☐)}

We obtain the transformed TRS

There are 180 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 1080 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 +
0
[52(x1)] = x1 +
244518/31
[53(x1)] = x1 +
144767/31
[54(x1)] = x1 +
318171/31
[55(x1)] = x1 +
0
[40(x1)] = x1 +
318481/31
[41(x1)] = x1 +
318171/31
[42(x1)] = x1 +
318481/31
[43(x1)] = x1 +
318171/31
[44(x1)] = x1 +
0
[45(x1)] = x1 +
1
[30(x1)] = x1 +
24852/31
[31(x1)] = x1 +
159607/31
[32(x1)] = x1 +
318171/31
[33(x1)] = x1 +
166334/31
[34(x1)] = x1 +
318171/31
[35(x1)] = x1 +
318946/31
[20(x1)] = x1 +
1
[21(x1)] = x1 +
318481/31
[22(x1)] = x1 +
238601/31
[23(x1)] = x1 +
83385/31
[24(x1)] = x1 +
319349/31
[25(x1)] = x1 +
318295/31
[10(x1)] = x1 +
0
[11(x1)] = x1 +
0
[12(x1)] = x1 +
293319/31
[13(x1)] = x1 +
138878/31
[14(x1)] = x1 +
293443/31
[15(x1)] = x1 +
184428/31
[00(x1)] = x1 +
172308/31
[01(x1)] = x1 +
318171/31
[02(x1)] = x1 +
244487/31
[03(x1)] = x1 +
232310/31
[04(x1)] = x1 +
318171/31
[05(x1)] = x1 +
215602/31
all of the following rules can be deleted.

There are 1080 ruless (increase limit for explicit display).

1.1.1.1 R is empty

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