Certification Problem

Input (TPDB SRS_Standard/ICFP_2010/85590)

The rewrite relation of the following TRS is considered.

0(0(0(1(x1)))) → 2(2(1(x1))) (1)
0(1(1(3(x1)))) → 4(2(3(x1))) (2)
2(3(0(2(1(x1))))) → 2(2(1(2(1(x1))))) (3)
3(0(0(1(4(x1))))) → 2(4(0(4(x1)))) (4)
4(4(5(0(1(x1))))) → 3(0(0(2(x1)))) (5)
5(0(5(1(1(x1))))) → 3(2(2(2(x1)))) (6)
1(0(5(4(1(0(x1)))))) → 0(5(3(0(0(x1))))) (7)
1(1(5(2(3(2(x1)))))) → 1(3(3(3(0(x1))))) (8)
4(0(5(0(1(0(x1)))))) → 4(0(0(0(3(x1))))) (9)
2(2(5(5(1(5(2(x1))))))) → 4(0(1(1(3(4(3(x1))))))) (10)
4(3(3(0(0(1(5(4(x1)))))))) → 4(3(2(1(1(5(3(4(x1)))))))) (11)
0(1(4(3(4(1(4(0(0(1(x1)))))))))) → 0(3(1(2(2(5(0(4(1(x1))))))))) (12)
3(2(2(5(2(0(1(5(4(0(4(1(x1)))))))))))) → 3(4(1(4(3(1(1(0(4(1(3(1(x1)))))))))))) (13)
4(1(5(2(2(2(3(5(3(0(1(4(x1)))))))))))) → 4(5(3(5(5(5(0(0(3(1(4(x1))))))))))) (14)
0(2(1(3(3(4(0(1(5(2(2(2(3(x1))))))))))))) → 0(3(4(2(4(2(0(5(4(2(2(3(x1)))))))))))) (15)
2(0(5(4(3(2(2(3(3(2(5(1(2(x1))))))))))))) → 2(5(2(3(0(4(3(3(3(5(0(0(x1)))))))))))) (16)
1(5(3(1(1(3(1(5(1(5(4(3(1(3(4(x1))))))))))))))) → 2(3(5(2(2(1(4(1(2(5(4(3(4(x1))))))))))))) (17)
5(0(5(0(3(3(0(2(5(0(2(1(4(4(0(x1))))))))))))))) → 3(5(4(0(3(1(4(2(4(3(0(2(1(4(0(x1))))))))))))))) (18)
5(1(5(4(3(0(4(0(2(0(2(4(3(1(5(x1))))))))))))))) → 0(1(2(0(0(1(1(0(2(0(0(5(4(3(4(3(x1)))))))))))))))) (19)
2(0(0(1(5(4(3(5(3(5(0(1(3(5(1(4(5(x1))))))))))))))))) → 4(0(4(3(4(1(3(3(5(3(3(1(4(1(4(5(x1)))))))))))))))) (20)
4(1(3(4(0(2(3(3(4(0(1(3(2(1(3(5(5(x1))))))))))))))))) → 4(0(0(4(1(5(0(2(3(0(5(2(1(5(5(5(x1)))))))))))))))) (21)
0(5(3(3(3(5(3(1(4(5(5(3(5(2(5(1(0(0(x1)))))))))))))))))) → 3(5(1(0(1(5(5(0(4(2(5(2(5(5(3(5(0(x1))))))))))))))))) (22)
5(4(2(0(5(1(3(2(0(2(1(4(2(2(1(3(1(0(x1)))))))))))))))))) → 0(5(0(5(5(2(2(5(1(5(1(0(5(5(3(0(1(x1))))))))))))))))) (23)
1(1(1(2(0(2(4(2(2(0(3(0(4(5(2(2(1(0(1(x1))))))))))))))))))) → 3(5(3(0(4(5(0(3(5(3(2(4(1(1(3(1(1(5(4(x1))))))))))))))))))) (24)
5(5(0(1(3(2(5(3(4(1(4(1(1(2(5(4(1(5(5(x1))))))))))))))))))) → 5(3(3(3(2(0(5(4(1(2(2(4(2(0(4(3(5(x1))))))))))))))))) (25)
0(4(3(2(2(5(1(5(0(0(0(1(1(1(2(4(2(1(0(5(x1)))))))))))))))))))) → 0(4(0(2(2(5(0(0(1(5(3(4(5(2(5(0(5(3(3(x1))))))))))))))))))) (26)
4(2(1(2(2(3(3(3(5(1(2(4(4(1(5(1(3(2(1(0(x1)))))))))))))))))))) → 4(0(1(3(0(4(4(1(2(3(0(4(0(4(1(4(5(4(2(x1))))))))))))))))))) (27)
5(4(4(1(2(1(5(1(5(2(1(2(5(5(2(1(5(1(1(5(x1)))))))))))))))))))) → 4(2(0(3(2(5(1(1(5(5(1(2(4(4(0(1(3(4(5(x1))))))))))))))))))) (28)
2(5(1(1(5(1(2(4(5(0(1(2(3(3(1(5(5(3(2(3(4(x1))))))))))))))))))))) → 3(3(1(4(3(2(2(2(0(0(3(3(1(1(5(1(2(4(0(2(x1)))))))))))))))))))) (29)
3(1(0(1(4(5(0(2(5(4(5(0(4(1(5(1(1(2(2(1(5(x1))))))))))))))))))))) → 4(1(2(3(3(4(1(5(0(1(5(4(1(2(3(4(0(0(0(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 +
1
[51(x1)] = x1 +
364
[52(x1)] = x1 +
0
[53(x1)] = x1 +
336
[54(x1)] = x1 +
370
[55(x1)] = x1 +
127
[40(x1)] = x1 +
370
[41(x1)] = x1 +
364
[42(x1)] = x1 +
0
[43(x1)] = x1 +
28
[44(x1)] = x1 +
0
[45(x1)] = x1 +
14
[30(x1)] = x1 +
0
[31(x1)] = x1 +
0
[32(x1)] = x1 +
1
[33(x1)] = x1 +
6
[34(x1)] = x1 +
364
[35(x1)] = x1 +
1
[20(x1)] = x1 +
365
[21(x1)] = x1 +
364
[22(x1)] = x1 +
364
[23(x1)] = x1 +
0
[24(x1)] = x1 +
364
[25(x1)] = x1 +
364
[10(x1)] = x1 +
365
[11(x1)] = x1 +
364
[12(x1)] = x1 +
364
[13(x1)] = x1 +
364
[14(x1)] = x1 +
113
[15(x1)] = x1 +
364
[00(x1)] = x1 +
29
[01(x1)] = x1 +
0
[02(x1)] = x1 +
0
[03(x1)] = x1 +
364
[04(x1)] = x1 +
267
[05(x1)] = x1 +
155
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.