Certification Problem

Input (TPDB SRS_Standard/ICFP_2010/86816)

The rewrite relation of the following TRS is considered.

0(1(2(x1))) → 3(2(2(x1))) (1)
4(2(3(2(1(x1))))) → 0(5(3(4(x1)))) (2)
2(2(5(1(1(2(x1)))))) → 4(1(4(5(3(x1))))) (3)
3(2(5(0(0(2(x1)))))) → 4(3(1(5(1(2(x1)))))) (4)
2(3(3(1(3(1(4(2(3(x1))))))))) → 5(2(5(5(0(2(3(x1))))))) (5)
2(4(3(0(5(2(4(4(1(2(4(2(x1)))))))))))) → 4(1(5(0(4(0(0(5(5(0(2(x1))))))))))) (6)
2(2(5(4(0(2(0(0(4(1(3(3(1(x1))))))))))))) → 2(2(4(4(0(5(4(4(5(4(5(3(x1)))))))))))) (7)
4(0(3(5(5(5(4(5(0(4(1(3(0(x1))))))))))))) → 4(0(3(3(5(2(4(4(4(0(0(3(4(4(x1)))))))))))))) (8)
4(3(2(2(5(5(2(0(0(4(3(2(4(x1))))))))))))) → 1(2(4(4(1(1(3(5(1(2(4(4(x1)))))))))))) (9)
5(2(0(0(0(3(3(5(1(4(0(0(5(x1))))))))))))) → 3(1(4(1(3(1(5(1(0(3(1(2(2(x1))))))))))))) (10)
4(0(2(5(4(0(1(2(1(0(2(5(1(1(x1)))))))))))))) → 4(0(1(4(2(3(1(3(3(2(4(4(4(x1))))))))))))) (11)
3(1(0(1(3(4(0(4(1(1(3(0(4(1(1(x1))))))))))))))) → 3(1(2(0(4(0(5(1(4(1(3(5(0(5(4(x1))))))))))))))) (12)
0(1(0(5(5(1(0(0(3(0(1(0(1(4(3(1(x1)))))))))))))))) → 3(4(5(4(3(5(1(5(1(4(3(0(1(3(5(0(x1)))))))))))))))) (13)
2(4(5(3(0(3(0(2(2(0(5(5(5(1(1(0(x1)))))))))))))))) → 3(3(1(1(4(4(1(4(4(2(2(4(1(5(3(x1))))))))))))))) (14)
5(2(1(2(1(4(3(2(4(3(3(1(5(1(3(3(x1)))))))))))))))) → 5(5(3(3(3(3(3(5(2(4(3(4(4(0(3(x1))))))))))))))) (15)
2(0(1(2(5(5(0(3(5(2(2(3(0(3(1(0(1(x1))))))))))))))))) → 5(3(2(4(0(1(2(0(2(1(5(3(4(1(0(2(2(x1))))))))))))))))) (16)
0(2(0(3(3(4(0(5(2(5(5(5(1(5(2(0(1(0(x1)))))))))))))))))) → 3(2(2(2(1(0(5(3(5(0(1(2(0(2(1(4(4(0(x1)))))))))))))))))) (17)
4(3(0(0(3(0(4(3(5(3(2(1(4(5(5(3(3(1(x1)))))))))))))))))) → 4(3(3(5(5(0(3(5(3(5(0(0(5(4(2(3(0(x1))))))))))))))))) (18)
4(4(0(4(2(0(5(4(1(1(5(0(3(2(4(2(0(3(x1)))))))))))))))))) → 4(3(0(2(4(1(0(4(3(1(1(1(4(0(2(5(3(3(x1)))))))))))))))))) (19)
5(5(3(5(2(4(0(4(2(1(3(0(5(3(1(2(1(2(x1)))))))))))))))))) → 5(0(1(2(3(5(1(2(3(0(5(0(3(0(3(3(4(2(x1)))))))))))))))))) (20)
1(4(5(2(0(2(2(2(4(1(4(3(1(3(0(2(0(5(2(x1))))))))))))))))))) → 1(4(2(3(5(0(2(0(2(0(0(4(0(4(4(1(2(1(x1)))))))))))))))))) (21)
2(0(0(4(0(2(3(0(2(5(5(3(4(5(4(0(4(0(2(x1))))))))))))))))))) → 1(4(2(2(2(2(0(2(4(1(0(1(4(1(0(4(2(1(x1)))))))))))))))))) (22)
2(1(5(4(4(1(4(0(3(5(3(0(1(5(5(5(3(2(1(x1))))))))))))))))))) → 2(2(1(4(2(2(1(5(3(2(4(4(5(4(2(1(0(1(2(0(x1)))))))))))))))))))) (23)
3(3(2(0(4(4(0(4(2(3(1(5(3(2(0(0(2(1(1(x1))))))))))))))))))) → 5(5(3(1(1(1(0(5(5(0(4(1(1(1(5(0(x1)))))))))))))))) (24)
5(3(4(3(3(3(2(4(2(0(3(1(1(2(4(2(5(3(2(x1))))))))))))))))))) → 0(3(2(2(4(3(5(2(4(0(4(5(3(5(5(4(0(2(4(x1))))))))))))))))))) (25)
5(3(5(3(2(5(0(2(0(3(0(2(1(1(5(5(4(0(0(x1))))))))))))))))))) → 5(5(2(3(5(2(1(1(4(0(4(2(1(4(3(5(1(1(0(x1))))))))))))))))))) (26)
3(5(2(5(4(4(0(1(5(5(5(0(5(1(0(0(0(4(5(1(x1)))))))))))))))))))) → 5(3(1(0(5(1(4(4(2(1(4(3(0(5(4(0(0(5(0(3(x1)))))))))))))))))))) (27)
4(4(2(0(5(1(4(4(0(1(0(4(5(0(3(4(0(5(4(1(x1)))))))))))))))))))) → 4(4(4(5(5(2(4(1(1(4(4(5(5(2(4(4(1(4(4(x1))))))))))))))))))) (28)
4(3(4(5(2(5(2(2(1(4(0(0(4(3(0(4(5(4(1(4(3(x1))))))))))))))))))))) → 4(0(1(0(0(1(5(2(5(4(5(5(4(5(4(1(1(5(5(x1))))))))))))))))))) (29)
4(5(2(3(0(2(3(4(3(1(3(3(4(0(2(4(3(0(4(1(1(x1))))))))))))))))))))) → 4(4(5(4(2(4(2(4(1(3(4(4(2(0(1(4(1(2(2(1(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 +
380359/629
[51(x1)] = x1 +
0
[52(x1)] = x1 +
0
[53(x1)] = x1 +
381617/629
[54(x1)] = x1 +
380359/629
[55(x1)] = x1 +
0
[40(x1)] = x1 +
98743/629
[41(x1)] = x1 +
0
[42(x1)] = x1 +
112679/629
[43(x1)] = x1 +
119889/629
[44(x1)] = x1 +
19
[45(x1)] = x1 +
338216/629
[30(x1)] = x1 +
0
[31(x1)] = x1 +
0
[32(x1)] = x1 +
380359/629
[33(x1)] = x1 +
380359/629
[34(x1)] = x1 +
365403/629
[35(x1)] = x1 +
380359/629
[20(x1)] = x1 +
380359/629
[21(x1)] = x1 +
183029/629
[22(x1)] = x1 +
165851/629
[23(x1)] = x1 +
32428/629
[24(x1)] = x1 +
104144/629
[25(x1)] = x1 +
385391/629
[10(x1)] = x1 +
0
[11(x1)] = x1 +
2
[12(x1)] = x1 +
380359/629
[13(x1)] = x1 +
323569/629
[14(x1)] = x1 +
0
[15(x1)] = x1 +
90476/629
[00(x1)] = x1 +
139628/629
[01(x1)] = x1 +
380359/629
[02(x1)] = x1 +
380359/629
[03(x1)] = x1 +
231016/629
[04(x1)] = x1 +
380988/629
[05(x1)] = x1 +
380359/629
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.