Certification Problem

Input (TPDB SRS_Standard/ICFP_2010/84886)

The rewrite relation of the following TRS is considered.

0(1(2(0(x1)))) → 3(4(0(x1))) (1)
0(5(1(5(x1)))) → 2(3(1(x1))) (2)
2(3(3(5(1(x1))))) → 3(5(5(3(x1)))) (3)
4(5(1(2(0(x1))))) → 4(5(5(3(x1)))) (4)
0(5(3(1(1(1(x1)))))) → 0(1(0(1(5(1(x1)))))) (5)
0(0(1(2(4(1(1(x1))))))) → 0(0(3(5(5(1(x1)))))) (6)
0(2(4(3(1(1(5(x1))))))) → 0(2(0(0(1(3(1(x1))))))) (7)
0(3(5(0(4(0(0(x1))))))) → 1(3(4(0(3(4(x1)))))) (8)
1(5(4(5(5(4(5(x1))))))) → 4(2(2(5(1(0(x1)))))) (9)
4(4(3(5(1(0(5(x1))))))) → 2(5(1(1(1(4(x1)))))) (10)
3(2(5(2(2(4(0(5(1(1(x1)))))))))) → 3(1(3(5(5(5(0(2(5(x1))))))))) (11)
1(2(2(1(4(3(2(2(1(3(0(x1))))))))))) → 4(2(3(3(3(5(3(1(0(1(5(x1))))))))))) (12)
4(5(1(3(0(0(0(5(3(1(5(x1))))))))))) → 2(4(0(5(2(3(0(0(5(4(x1)))))))))) (13)
2(0(1(2(5(2(4(3(5(2(0(5(x1)))))))))))) → 2(4(4(4(2(1(0(4(3(3(2(5(x1)))))))))))) (14)
5(5(3(0(2(2(4(2(5(0(1(5(x1)))))))))))) → 3(2(3(1(5(0(4(4(2(3(5(5(x1)))))))))))) (15)
0(1(5(1(0(2(5(5(1(4(1(0(1(x1))))))))))))) → 0(0(5(1(0(5(3(1(0(1(5(3(x1)))))))))))) (16)
4(5(1(5(1(5(3(4(3(3(0(4(1(x1))))))))))))) → 4(5(2(3(2(5(3(2(1(0(2(1(x1)))))))))))) (17)
5(1(3(5(0(2(5(5(1(0(3(2(2(x1))))))))))))) → 1(2(3(4(4(3(3(5(0(1(1(5(2(x1))))))))))))) (18)
0(2(4(1(5(3(3(3(1(1(1(1(0(2(x1)))))))))))))) → 0(2(2(2(4(5(2(0(2(0(0(3(2(x1))))))))))))) (19)
0(4(4(5(2(0(3(0(2(4(2(5(1(1(0(x1))))))))))))))) → 2(1(0(0(3(3(2(1(1(3(0(5(0(5(0(x1))))))))))))))) (20)
0(2(5(3(4(2(0(5(0(0(0(4(3(4(2(3(x1)))))))))))))))) → 0(0(1(0(5(2(0(1(4(1(4(2(2(1(1(3(x1)))))))))))))))) (21)
1(3(3(4(0(2(5(1(2(2(2(3(1(0(3(4(x1)))))))))))))))) → 0(5(2(4(3(1(3(2(5(2(5(5(2(2(4(x1))))))))))))))) (22)
5(0(5(1(0(0(1(4(3(5(1(4(4(2(1(0(x1)))))))))))))))) → 2(0(4(3(2(1(0(4(4(2(4(3(3(5(0(5(x1)))))))))))))))) (23)
5(0(5(3(0(0(3(1(5(1(4(0(4(3(4(5(x1)))))))))))))))) → 2(0(5(1(5(4(1(2(3(4(3(5(2(4(5(x1))))))))))))))) (24)
5(2(4(0(4(4(3(2(3(5(4(5(4(3(4(5(x1)))))))))))))))) → 4(3(5(5(2(2(0(1(1(1(2(5(3(1(5(x1))))))))))))))) (25)
1(4(0(0(5(4(1(4(2(2(3(2(0(0(5(1(1(x1))))))))))))))))) → 1(2(1(0(5(3(3(0(1(5(1(0(4(5(0(3(1(x1))))))))))))))))) (26)
2(1(2(0(1(1(2(5(3(3(1(5(5(0(4(3(0(x1))))))))))))))))) → 2(1(0(2(1(1(3(2(1(1(2(3(1(5(3(4(x1)))))))))))))))) (27)
4(5(5(0(5(5(2(5(1(5(3(5(0(4(4(5(3(x1))))))))))))))))) → 4(5(0(1(3(5(0(0(0(3(2(3(5(4(2(3(3(x1))))))))))))))))) (28)
5(5(5(2(1(5(3(0(2(5(4(2(5(5(0(0(5(x1))))))))))))))))) → 1(3(4(4(4(4(5(1(0(3(2(5(0(1(2(0(3(1(x1)))))))))))))))))) (29)
1(4(1(5(2(4(0(2(3(0(0(1(5(5(1(1(0(3(x1)))))))))))))))))) → 0(3(3(2(1(5(5(0(5(2(0(2(2(3(0(0(3(x1))))))))))))))))) (30)
2(2(1(0(1(2(0(2(4(2(3(2(0(3(0(0(0(2(x1)))))))))))))))))) → 0(0(3(4(2(5(3(0(1(4(5(2(0(5(5(1(2(x1))))))))))))))))) (31)
2(3(0(3(0(3(3(3(4(2(4(4(5(4(4(5(0(0(x1)))))))))))))))))) → 3(3(1(4(3(1(5(2(5(5(1(4(5(2(0(4(x1)))))))))))))))) (32)
4(1(4(3(4(0(0(1(1(0(4(1(0(4(0(5(1(1(x1)))))))))))))))))) → 2(0(0(0(5(0(3(2(4(4(4(2(1(4(4(4(1(x1))))))))))))))))) (33)
0(5(0(2(1(4(0(1(1(0(0(2(4(2(5(2(2(2(0(x1))))))))))))))))))) → 0(4(1(2(2(0(3(0(3(3(4(5(0(4(2(2(1(5(4(x1))))))))))))))))))) (34)
5(0(4(0(4(5(2(5(5(0(4(0(2(5(4(1(4(3(4(5(x1)))))))))))))))))))) → 5(0(4(0(5(4(5(3(2(4(1(5(5(1(3(3(2(3(1(1(x1)))))))))))))))))))) (35)
0(3(3(4(4(0(3(1(5(5(0(1(4(2(4(3(0(2(1(1(5(x1))))))))))))))))))))) → 0(3(2(0(5(3(3(5(0(1(0(4(0(1(2(0(1(1(4(2(5(x1))))))))))))))))))))) (36)
4(1(1(0(0(3(1(5(1(1(5(4(2(0(3(4(4(5(1(3(3(x1))))))))))))))))))))) → 3(0(1(0(0(4(1(2(0(5(5(3(4(4(5(3(5(2(5(0(3(x1))))))))))))))))))))) (37)
5(0(1(0(0(2(2(0(3(1(1(2(3(2(4(1(0(0(5(1(3(x1))))))))))))))))))))) → 5(4(0(5(5(5(3(1(5(1(0(5(5(1(4(4(3(2(4(3(x1)))))))))))))))))))) (38)
5(0(5(4(4(1(4(4(1(1(5(1(0(0(1(5(4(0(1(3(0(x1))))))))))))))))))))) → 1(5(3(4(5(2(5(4(2(1(4(0(3(5(0(2(3(1(x1)))))))))))))))))) (39)
5(1(4(0(4(3(5(3(0(3(4(2(0(2(1(5(2(0(1(1(0(x1))))))))))))))))))))) → 5(4(5(3(1(4(1(3(0(2(3(4(2(4(5(2(4(3(0(4(4(x1))))))))))))))))))))) (40)

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 240 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 1440 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 +
2288/5
[51(x1)] = x1 +
2288/5
[52(x1)] = x1 +
1
[53(x1)] = x1 +
5
[54(x1)] = x1 +
2308/5
[55(x1)] = x1 +
2293/5
[40(x1)] = x1 +
2308/5
[41(x1)] = x1 +
342
[42(x1)] = x1 +
5
[43(x1)] = x1 +
0
[44(x1)] = x1 +
1326/5
[45(x1)] = x1 +
2288/5
[30(x1)] = x1 +
103/5
[31(x1)] = x1 +
0
[32(x1)] = x1 +
0
[33(x1)] = x1 +
4
[34(x1)] = x1 +
5
[35(x1)] = x1 +
2288/5
[20(x1)] = x1 +
104/5
[21(x1)] = x1 +
649/5
[22(x1)] = x1 +
4
[23(x1)] = x1 +
2288/5
[24(x1)] = x1 +
4
[25(x1)] = x1 +
1612/5
[10(x1)] = x1 +
2288/5
[11(x1)] = x1 +
4
[12(x1)] = x1 +
4
[13(x1)] = x1 +
264
[14(x1)] = x1 +
2293/5
[15(x1)] = x1 +
2308/5
[00(x1)] = x1 +
2313/5
[01(x1)] = x1 +
1222/5
[02(x1)] = x1 +
18
[03(x1)] = x1 +
2378/5
[04(x1)] = x1 +
0
[05(x1)] = x1 +
2308/5
all of the following rules can be deleted.

There are 1440 ruless (increase limit for explicit display).

1.1.1.1 R is empty

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