Certification Problem

Input (TPDB SRS_Standard/ICFP_2010/85267)

The rewrite relation of the following TRS is considered.

0(1(1(2(x1)))) → 0(0(1(x1))) (1)
3(0(2(2(x1)))) → 3(4(3(2(x1)))) (2)
5(2(1(5(x1)))) → 4(2(5(x1))) (3)
5(2(3(0(x1)))) → 2(0(0(x1))) (4)
4(3(5(3(2(5(x1)))))) → 2(3(1(2(5(x1))))) (5)
3(1(0(0(0(1(1(x1))))))) → 5(3(2(0(2(0(1(x1))))))) (6)
3(2(5(3(1(0(3(x1))))))) → 3(5(3(4(0(1(x1)))))) (7)
5(2(2(2(4(0(3(x1))))))) → 4(5(4(0(1(5(x1)))))) (8)
0(1(2(5(1(5(3(0(x1)))))))) → 0(0(5(2(5(4(4(x1))))))) (9)
3(0(2(2(2(1(0(3(x1)))))))) → 3(5(2(3(0(0(4(x1))))))) (10)
3(3(1(5(0(3(2(4(x1)))))))) → 4(3(3(1(0(4(4(x1))))))) (11)
1(2(4(5(4(0(5(2(1(x1))))))))) → 1(5(0(5(4(4(0(4(1(x1))))))))) (12)
2(3(1(3(1(3(1(5(3(x1))))))))) → 5(4(1(0(4(1(3(x1))))))) (13)
3(5(5(4(4(1(5(1(4(x1))))))))) → 3(2(5(0(4(0(0(5(4(x1))))))))) (14)
1(0(1(5(3(3(4(5(5(1(x1)))))))))) → 1(1(5(4(3(5(1(1(5(1(x1)))))))))) (15)
2(3(0(3(0(5(3(2(1(3(x1)))))))))) → 2(5(0(2(2(5(0(1(x1)))))))) (16)
5(5(1(1(4(5(0(5(1(3(x1)))))))))) → 5(1(1(2(5(3(3(4(3(3(x1)))))))))) (17)
5(0(5(3(0(5(0(4(0(5(5(x1))))))))))) → 3(4(4(3(4(0(3(5(3(5(x1)))))))))) (18)
5(3(2(5(5(5(4(3(3(4(5(x1))))))))))) → 3(0(2(3(2(5(3(0(2(4(x1)))))))))) (19)
5(4(2(1(4(4(1(3(2(1(2(x1))))))))))) → 5(4(1(2(3(1(3(0(0(0(1(x1))))))))))) (20)
0(0(2(1(0(1(2(1(0(3(2(2(x1)))))))))))) → 0(1(2(5(1(4(5(2(1(4(1(x1))))))))))) (21)
5(3(2(0(5(4(5(2(2(3(4(5(x1)))))))))))) → 2(4(3(5(0(2(0(2(0(4(4(x1))))))))))) (22)
0(1(3(3(0(4(1(0(2(2(4(3(2(x1))))))))))))) → 1(3(4(2(4(1(3(3(4(2(1(x1))))))))))) (23)
2(4(4(5(1(5(1(0(2(5(4(1(2(x1))))))))))))) → 5(1(5(2(1(2(3(2(0(3(1(2(3(3(x1)))))))))))))) (24)
0(0(0(4(0(0(3(2(0(5(4(4(2(1(2(x1))))))))))))))) → 0(4(4(2(1(0(2(1(5(0(2(3(4(2(x1)))))))))))))) (25)
3(0(2(5(1(1(4(1(3(4(2(5(3(1(0(x1))))))))))))))) → 3(5(5(4(0(5(2(1(5(0(4(0(4(4(3(0(x1)))))))))))))))) (26)
5(3(4(3(2(1(4(2(2(1(3(3(1(4(4(x1))))))))))))))) → 1(1(5(3(4(5(4(3(3(3(3(2(0(2(4(x1))))))))))))))) (27)
4(3(3(4(2(0(0(1(3(3(0(1(3(1(2(1(x1)))))))))))))))) → 4(3(3(1(1(4(4(4(5(2(0(5(0(5(4(1(x1)))))))))))))))) (28)
0(3(3(4(2(4(3(1(4(1(0(4(1(4(0(4(1(1(x1)))))))))))))))))) → 0(2(2(1(1(2(3(1(3(2(2(2(5(5(5(0(3(4(2(x1))))))))))))))))))) (29)
3(1(3(2(4(2(3(3(1(5(4(0(3(1(0(4(1(0(x1)))))))))))))))))) → 5(1(3(5(0(2(3(2(0(4(0(1(2(2(2(2(0(x1))))))))))))))))) (30)
5(5(5(3(2(1(0(4(1(3(4(4(3(1(1(4(3(2(x1)))))))))))))))))) → 5(5(4(3(2(2(3(0(3(4(4(5(5(4(4(3(0(3(x1)))))))))))))))))) (31)
1(1(1(0(3(0(1(4(3(1(1(2(5(2(1(2(1(1(0(x1))))))))))))))))))) → 1(5(5(2(3(1(3(3(2(0(1(5(5(4(1(2(4(x1))))))))))))))))) (32)
1(1(5(3(2(0(4(1(3(4(5(3(0(2(5(2(3(3(0(x1))))))))))))))))))) → 0(1(3(4(3(5(0(2(2(0(2(4(5(0(3(3(2(0(x1)))))))))))))))))) (33)
5(2(2(2(1(1(4(0(4(3(5(3(4(4(4(4(0(5(0(x1))))))))))))))))))) → 3(4(5(0(2(4(3(5(1(1(5(5(4(1(4(5(5(0(0(x1))))))))))))))))))) (34)
5(5(4(3(2(1(3(3(1(4(1(1(3(3(3(4(0(1(2(x1))))))))))))))))))) → 4(0(1(3(1(2(4(5(3(5(0(0(3(2(4(5(5(3(x1)))))))))))))))))) (35)
2(0(4(4(2(4(3(3(2(0(0(3(3(0(1(0(2(2(2(5(x1)))))))))))))))))))) → 2(3(2(3(5(1(1(0(1(3(0(4(3(2(4(2(5(2(5(x1))))))))))))))))))) (36)
2(2(2(3(0(3(1(4(4(1(1(0(5(1(0(0(5(4(1(2(x1)))))))))))))))))))) → 2(3(4(3(5(5(2(1(1(2(5(3(5(0(4(1(2(2(1(4(x1)))))))))))))))))))) (37)
4(4(0(2(4(0(0(2(3(0(1(4(1(0(0(0(3(0(2(3(x1)))))))))))))))))))) → 5(2(4(4(0(4(1(5(0(5(3(3(1(4(1(0(2(4(2(x1))))))))))))))))))) (38)
0(3(4(1(5(3(4(5(1(5(0(2(3(1(1(0(1(0(5(4(2(x1))))))))))))))))))))) → 0(5(5(0(2(4(1(0(1(4(1(5(3(0(1(5(2(5(2(x1))))))))))))))))))) (39)
0(4(5(3(5(2(4(4(1(4(2(0(5(5(1(0(1(5(0(4(3(x1))))))))))))))))))))) → 4(4(2(2(3(2(0(5(5(0(2(4(0(0(0(1(3(5(2(3(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 +
72068507/914
[51(x1)] = x1 +
224626225/1828
[52(x1)] = x1 +
193744863/1828
[53(x1)] = x1 +
13654290/457
[54(x1)] = x1 +
225393985/1828
[55(x1)] = x1 +
420
[40(x1)] = x1 +
45333005/914
[41(x1)] = x1 +
7968703/1828
[42(x1)] = x1 +
209343025/1828
[43(x1)] = x1 +
421
[44(x1)] = x1 +
225393985/1828
[45(x1)] = x1 +
225393985/1828
[30(x1)] = x1 +
23637835/1828
[31(x1)] = x1 +
103477767/914
[32(x1)] = x1 +
41063650/457
[33(x1)] = x1 +
44015196/457
[34(x1)] = x1 +
196896697/1828
[35(x1)] = x1 +
224626225/1828
[20(x1)] = x1 +
134001625/1828
[21(x1)] = x1 +
20470040/457
[22(x1)] = x1 +
1
[23(x1)] = x1 +
225221239/1828
[24(x1)] = x1 +
112229597/914
[25(x1)] = x1 +
60377109/1828
[10(x1)] = x1 +
20657919/914
[11(x1)] = x1 +
225393985/1828
[12(x1)] = x1 +
41256047/457
[13(x1)] = x1 +
783/2
[14(x1)] = x1 +
225202045/1828
[15(x1)] = x1 +
225393985/1828
[00(x1)] = x1 +
224626225/1828
[01(x1)] = x1 +
63/2
[02(x1)] = x1 +
224631709/1828
[03(x1)] = x1 +
143863685/1828
[04(x1)] = x1 +
37396361/457
[05(x1)] = x1 +
60889863/1828
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.