Certification Problem

Input (TPDB SRS_Standard/ICFP_2010/85294)

The rewrite relation of the following TRS is considered.

0(1(2(3(x1)))) → 3(4(5(x1))) (1)
1(2(0(5(5(x1))))) → 1(5(4(0(x1)))) (2)
2(0(3(4(2(x1))))) → 3(1(2(4(x1)))) (3)
5(5(5(5(0(x1))))) → 5(4(4(3(x1)))) (4)
0(1(2(4(3(5(x1)))))) → 2(4(5(3(0(x1))))) (5)
0(3(0(4(2(0(x1)))))) → 2(3(4(1(2(x1))))) (6)
4(5(0(4(0(0(0(x1))))))) → 4(3(4(2(5(0(x1)))))) (7)
0(1(0(2(1(4(4(2(x1)))))))) → 0(2(0(3(3(4(5(4(x1)))))))) (8)
2(0(0(1(5(0(5(2(x1)))))))) → 2(0(5(2(1(1(x1)))))) (9)
3(4(1(4(3(0(3(2(x1)))))))) → 5(3(5(4(3(1(1(x1))))))) (10)
5(3(0(1(5(5(0(5(x1)))))))) → 5(0(4(2(1(3(4(x1))))))) (11)
5(5(5(3(1(3(4(4(x1)))))))) → 1(5(2(4(4(5(4(x1))))))) (12)
0(1(5(4(2(5(3(3(4(x1))))))))) → 2(0(5(4(3(0(1(0(4(x1))))))))) (13)
0(3(3(1(2(2(0(4(5(x1))))))))) → 2(3(4(4(4(0(3(1(x1)))))))) (14)
4(3(4(0(5(3(0(0(5(4(x1)))))))))) → 3(1(1(2(1(2(0(4(2(x1))))))))) (15)
4(2(5(4(5(2(1(1(0(2(1(x1))))))))))) → 4(2(3(4(1(0(5(4(5(2(1(x1))))))))))) (16)
5(5(4(2(5(3(1(0(5(3(1(x1))))))))))) → 2(2(2(2(2(0(3(5(1(2(1(x1))))))))))) (17)
5(5(5(0(3(5(2(4(1(2(4(x1))))))))))) → 1(0(3(3(5(0(3(1(4(4(x1)))))))))) (18)
0(3(3(2(0(1(1(2(1(3(0(5(x1)))))))))))) → 2(3(0(2(5(4(1(0(1(2(4(3(x1)))))))))))) (19)
3(3(2(5(2(5(4(5(0(5(4(2(1(x1))))))))))))) → 2(1(2(3(3(4(4(2(0(3(4(1(x1)))))))))))) (20)
4(3(4(2(2(5(4(5(0(0(0(2(5(x1))))))))))))) → 4(4(5(1(4(5(0(1(0(5(3(x1))))))))))) (21)
3(4(0(4(1(2(3(5(2(1(4(4(0(3(x1)))))))))))))) → 3(2(2(5(0(4(5(5(0(5(5(4(1(2(x1)))))))))))))) (22)
4(0(5(4(4(4(5(0(2(5(2(3(4(5(x1)))))))))))))) → 4(1(0(4(3(4(2(3(1(3(2(3(3(x1))))))))))))) (23)
5(3(5(4(2(4(4(5(0(5(4(1(2(2(x1)))))))))))))) → 5(0(2(5(5(1(5(3(4(1(1(0(2(x1))))))))))))) (24)
2(2(4(3(4(4(2(1(5(2(1(4(0(2(0(x1))))))))))))))) → 3(5(3(3(1(2(1(3(0(4(5(3(4(0(x1)))))))))))))) (25)
0(2(0(2(5(2(2(1(4(5(0(3(1(5(0(2(x1)))))))))))))))) → 0(4(3(5(5(5(1(3(0(4(5(0(2(2(4(x1))))))))))))))) (26)
5(0(0(2(0(2(2(0(1(1(3(4(4(3(3(5(x1)))))))))))))))) → 5(4(0(3(4(4(1(0(0(3(5(1(4(5(5(5(x1)))))))))))))))) (27)
5(0(2(3(2(4(4(3(4(1(0(0(0(4(4(3(x1)))))))))))))))) → 5(2(4(4(4(0(1(1(4(2(5(5(1(3(x1)))))))))))))) (28)
0(2(0(2(0(1(1(3(5(2(0(4(4(1(5(5(5(x1))))))))))))))))) → 4(1(3(4(1(1(3(1(0(4(2(2(5(3(3(x1))))))))))))))) (29)
0(4(3(0(2(1(3(3(3(0(5(1(2(3(2(5(4(0(x1)))))))))))))))))) → 1(5(5(5(2(2(1(2(0(4(0(1(4(1(5(4(4(x1))))))))))))))))) (30)
2(2(3(2(3(2(2(4(1(2(4(5(1(4(4(0(4(0(x1)))))))))))))))))) → 5(2(2(0(4(4(5(4(2(3(2(3(1(0(5(3(3(4(x1)))))))))))))))))) (31)
2(4(1(1(1(2(5(4(4(2(0(0(1(4(5(5(5(0(x1)))))))))))))))))) → 3(3(2(4(5(2(5(4(5(4(5(3(3(3(0(5(0(2(3(x1))))))))))))))))))) (32)
3(2(0(2(1(0(3(1(3(4(3(5(2(1(0(0(3(0(x1)))))))))))))))))) → 5(2(4(0(1(5(2(3(3(5(4(4(5(2(3(3(3(0(x1)))))))))))))))))) (33)
2(5(1(2(0(4(1(2(4(3(2(0(4(3(1(3(5(2(5(x1))))))))))))))))))) → 2(4(5(1(2(3(0(4(1(3(2(3(4(2(0(3(3(3(x1)))))))))))))))))) (34)
4(4(4(1(1(1(2(1(1(0(4(3(2(3(1(3(0(3(5(x1))))))))))))))))))) → 4(1(0(2(3(3(0(1(1(4(5(3(1(1(5(3(2(2(3(5(x1)))))))))))))))))))) (35)
3(5(3(3(0(0(4(4(4(4(0(1(1(5(2(5(0(5(0(2(x1)))))))))))))))))))) → 2(0(5(2(4(3(3(4(5(4(3(1(4(0(0(2(5(1(1(x1))))))))))))))))))) (36)
4(1(5(3(2(5(5(5(2(5(5(3(5(5(5(3(5(0(2(0(x1)))))))))))))))))))) → 4(1(4(0(5(2(1(3(4(3(1(1(0(4(3(3(2(0(x1)))))))))))))))))) (37)
0(0(4(0(0(0(0(1(0(4(4(5(3(2(2(4(0(5(3(0(5(x1))))))))))))))))))))) → 0(2(2(2(1(2(4(4(5(0(5(1(0(0(5(5(2(5(3(2(x1)))))))))))))))))))) (38)
2(2(0(3(4(0(5(2(5(2(2(5(2(5(5(1(5(2(4(1(4(x1))))))))))))))))))))) → 0(0(5(2(4(1(4(3(4(3(2(4(4(3(4(4(1(1(4(x1))))))))))))))))))) (39)
2(5(5(5(0(2(3(2(4(0(0(1(5(2(3(2(3(0(1(0(5(x1))))))))))))))))))))) → 2(0(3(4(5(3(3(2(2(1(5(2(2(2(4(2(5(4(0(1(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 +
1939379/181
[51(x1)] = x1 +
576963/181
[52(x1)] = x1 +
1587061/181
[53(x1)] = x1 +
1264678/181
[54(x1)] = x1 +
577144/181
[55(x1)] = x1 +
1675555/181
[40(x1)] = x1 +
2
[41(x1)] = x1 +
1885147/181
[42(x1)] = x1 +
148556/181
[43(x1)] = x1 +
2527910/181
[44(x1)] = x1 +
482353/181
[45(x1)] = x1 +
2535150/181
[30(x1)] = x1 +
2535150/181
[31(x1)] = x1 +
431874/181
[32(x1)] = x1 +
1449212/181
[33(x1)] = x1 +
2528272/181
[34(x1)] = x1 +
40
[35(x1)] = x1 +
2527910/181
[20(x1)] = x1 +
2340326/181
[21(x1)] = x1 +
2535150/181
[22(x1)] = x1 +
0
[23(x1)] = x1 +
0
[24(x1)] = x1 +
2110516/181
[25(x1)] = x1 +
2531711/181
[10(x1)] = x1 +
2527910/181
[11(x1)] = x1 +
1229025/181
[12(x1)] = x1 +
1482845/181
[13(x1)] = x1 +
883443/181
[14(x1)] = x1 +
0
[15(x1)] = x1 +
883262/181
[00(x1)] = x1 +
2244401/181
[01(x1)] = x1 +
2527910/181
[02(x1)] = x1 +
2535150/181
[03(x1)] = x1 +
2532435/181
[04(x1)] = x1 +
2535150/181
[05(x1)] = x1 +
2546915/181
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.