Certification Problem

Input (TPDB SRS_Standard/ICFP_2010/3336)

The rewrite relation of the following TRS is considered.

2(5(x1)) → 1(3(3(0(1(0(x1)))))) (1)
2(5(x1)) → 2(2(0(5(0(1(x1)))))) (2)
3(5(x1)) → 1(3(2(0(0(1(x1)))))) (3)
3(5(x1)) → 3(2(0(5(3(0(x1)))))) (4)
4(5(x1)) → 2(2(1(3(2(1(x1)))))) (5)
4(5(x1)) → 3(2(0(5(0(0(x1)))))) (6)
1(2(5(x1))) → 1(0(5(0(5(4(x1)))))) (7)
1(2(5(x1))) → 1(2(2(1(0(1(x1)))))) (8)
1(2(5(x1))) → 2(0(1(3(1(0(x1)))))) (9)
1(4(5(x1))) → 1(2(4(0(2(1(x1)))))) (10)
2(5(1(x1))) → 2(2(2(1(2(3(x1)))))) (11)
2(5(2(x1))) → 4(0(2(2(3(3(x1)))))) (12)
2(5(3(x1))) → 2(0(4(1(3(3(x1)))))) (13)
2(5(4(x1))) → 2(0(5(1(0(1(x1)))))) (14)
3(2(5(x1))) → 3(2(0(1(0(5(x1)))))) (15)
3(4(2(x1))) → 3(4(0(2(2(2(x1)))))) (16)
3(5(1(x1))) → 0(4(2(0(0(5(x1)))))) (17)
3(5(1(x1))) → 0(4(2(2(3(4(x1)))))) (18)
3(5(1(x1))) → 2(1(4(1(0(1(x1)))))) (19)
3(5(2(x1))) → 0(4(3(2(2(2(x1)))))) (20)
3(5(2(x1))) → 2(0(2(2(3(0(x1)))))) (21)
3(5(2(x1))) → 2(3(3(2(1(2(x1)))))) (22)
3(5(3(x1))) → 0(2(4(3(3(0(x1)))))) (23)
3(5(3(x1))) → 0(5(4(3(3(0(x1)))))) (24)
3(5(3(x1))) → 2(3(4(0(4(2(x1)))))) (25)
3(5(4(x1))) → 0(2(0(5(0(0(x1)))))) (26)
3(5(4(x1))) → 0(5(0(0(1(2(x1)))))) (27)
3(5(5(x1))) → 0(5(4(1(0(5(x1)))))) (28)
4(5(1(x1))) → 2(1(0(5(3(3(x1)))))) (29)
4(5(2(x1))) → 0(5(1(0(0(4(x1)))))) (30)
4(5(4(x1))) → 2(2(1(0(4(2(x1)))))) (31)
4(5(4(x1))) → 3(2(0(3(2(0(x1)))))) (32)
5(5(3(x1))) → 5(1(0(1(2(2(x1)))))) (33)
5(5(4(x1))) → 5(1(0(4(2(2(x1)))))) (34)

Property / Task

Prove or disprove termination.

Answer / Result

Yes.

Proof (by NaTT @ termCOMP 2023)

1 Dependency Pair Transformation

The following set of initial dependency pairs has been identified.

There are 142 ruless (increase limit for explicit display).

1.1 Dependency Graph Processor

The dependency pairs are split into 1 component.