Certification Problem

Input (TPDB SRS_Standard/ICFP_2010/5011)

The rewrite relation of the following TRS is considered.

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

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 138 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 828 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 +
183
[53(x1)] = x1 +
0
[54(x1)] = x1 +
0
[55(x1)] = x1 +
0
[40(x1)] = x1 +
0
[41(x1)] = x1 +
0
[42(x1)] = x1 +
58
[43(x1)] = x1 +
9
[44(x1)] = x1 +
0
[45(x1)] = x1 +
0
[30(x1)] = x1 +
549
[31(x1)] = x1 +
192
[32(x1)] = x1 +
366
[33(x1)] = x1 +
192
[34(x1)] = x1 +
184
[35(x1)] = x1 +
183
[20(x1)] = x1 +
0
[21(x1)] = x1 +
0
[22(x1)] = x1 +
0
[23(x1)] = x1 +
1
[24(x1)] = x1 +
1
[25(x1)] = x1 +
0
[10(x1)] = x1 +
549
[11(x1)] = x1 +
0
[12(x1)] = x1 +
550
[13(x1)] = x1 +
1
[14(x1)] = x1 +
0
[15(x1)] = x1 +
1
[00(x1)] = x1 +
1358
[01(x1)] = x1 +
0
[02(x1)] = x1 +
37
[03(x1)] = x1 +
558
[04(x1)] = x1 +
0
[05(x1)] = x1 +
549
all of the following rules can be deleted.

There are 828 ruless (increase limit for explicit display).

1.1.1.1 R is empty

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