Certification Problem

Input (TPDB SRS_Standard/ICFP_2010/96485)

The rewrite relation of the following TRS is considered.

0(1(0(2(x1)))) → 2(2(2(x1))) (1)
0(1(1(1(x1)))) → 3(1(0(x1))) (2)
4(5(1(4(x1)))) → 1(5(4(4(x1)))) (3)
1(2(0(4(1(x1))))) → 2(5(2(2(x1)))) (4)
1(5(1(0(3(x1))))) → 5(5(0(3(x1)))) (5)
1(3(5(1(0(4(x1)))))) → 4(0(0(0(1(4(x1)))))) (6)
3(3(1(4(5(4(5(5(x1)))))))) → 3(5(3(2(0(5(2(x1))))))) (7)
4(2(5(3(1(0(0(5(2(x1))))))))) → 4(4(1(5(3(5(3(2(x1)))))))) (8)
2(2(1(1(0(2(0(1(2(3(x1)))))))))) → 5(4(4(5(2(5(5(2(3(x1))))))))) (9)
2(2(2(3(3(4(3(3(5(4(5(x1))))))))))) → 5(5(4(2(5(3(5(3(2(0(5(x1))))))))))) (10)
4(2(1(1(5(1(0(0(0(2(5(x1))))))))))) → 5(5(0(3(0(1(4(5(2(5(x1)))))))))) (11)
5(4(5(4(2(0(2(2(3(0(0(2(x1)))))))))))) → 5(3(4(5(0(3(2(1(1(2(2(x1))))))))))) (12)
0(2(1(1(3(1(5(0(5(2(3(1(4(x1))))))))))))) → 3(0(1(5(1(4(4(5(0(5(1(1(4(x1))))))))))))) (13)
1(1(2(4(2(2(2(1(1(3(4(1(2(x1))))))))))))) → 1(5(0(3(4(1(0(5(2(2(2(3(x1)))))))))))) (14)
1(4(0(0(2(4(0(3(3(2(0(3(1(x1))))))))))))) → 1(5(0(5(0(2(0(5(1(0(1(2(5(5(x1)))))))))))))) (15)
4(1(4(1(3(0(0(2(1(5(4(1(0(0(x1)))))))))))))) → 2(3(0(2(5(4(4(3(0(4(1(2(0(x1))))))))))))) (16)
4(2(4(4(0(0(0(4(0(3(5(0(3(3(x1)))))))))))))) → 5(1(0(1(0(1(3(5(3(5(5(0(0(3(x1)))))))))))))) (17)
1(2(5(3(3(0(2(2(5(3(2(3(3(3(2(1(x1)))))))))))))))) → 1(1(0(0(4(1(2(1(0(5(0(0(3(5(0(4(1(x1))))))))))))))))) (18)
3(1(2(5(0(3(4(3(1(5(4(1(5(2(0(5(x1)))))))))))))))) → 3(1(0(4(2(0(5(2(4(4(2(2(1(1(1(5(x1)))))))))))))))) (19)
1(3(4(5(1(0(0(3(1(2(4(2(3(5(2(0(4(1(x1)))))))))))))))))) → 1(3(0(0(4(2(2(5(3(1(0(1(2(1(5(0(1(4(1(x1))))))))))))))))))) (20)
1(0(3(3(2(5(0(0(3(0(3(2(4(1(4(0(2(4(2(x1))))))))))))))))))) → 4(3(4(2(1(4(2(4(3(3(2(2(2(1(1(1(5(3(2(x1))))))))))))))))))) (21)
5(3(2(1(0(1(3(1(3(3(0(0(3(2(5(3(0(3(0(x1))))))))))))))))))) → 5(1(0(3(2(4(0(0(3(2(1(5(1(3(0(5(1(3(1(x1))))))))))))))))))) (22)
0(2(2(3(3(5(0(0(5(0(3(1(3(0(1(2(1(5(5(1(x1)))))))))))))))))))) → 0(0(3(2(5(2(5(2(3(2(1(2(5(4(3(4(5(0(4(x1))))))))))))))))))) (23)
1(0(4(2(3(3(5(4(3(5(0(2(0(4(5(0(2(0(2(4(x1)))))))))))))))))))) → 4(5(4(5(4(1(1(2(5(0(4(3(1(5(4(3(1(5(4(0(4(x1))))))))))))))))))))) (24)
1(4(2(3(1(3(4(2(4(1(5(1(4(0(4(5(2(0(0(3(4(x1))))))))))))))))))))) → 4(3(4(1(3(5(1(1(4(3(1(5(1(3(1(2(4(2(1(3(x1)))))))))))))))))))) (25)

Property / Task

Prove or disprove termination.

Answer / Result

Yes.

Proof (by matchbox @ termCOMP 2023)

1 Rule Removal

Using the matrix interpretations of dimension 1 with strict dimension 1 over the rationals with delta = 1
[5(x1)] = x1 +
131
[4(x1)] = x1 +
1847/6
[3(x1)] = x1 +
1429/3
[2(x1)] = x1 +
279
[1(x1)] = x1 +
1519/6
[0(x1)] = x1 +
1655/6
all of the following rules can be deleted.
0(1(0(2(x1)))) → 2(2(2(x1))) (1)
0(1(1(1(x1)))) → 3(1(0(x1))) (2)
1(2(0(4(1(x1))))) → 2(5(2(2(x1)))) (4)
1(5(1(0(3(x1))))) → 5(5(0(3(x1)))) (5)
1(3(5(1(0(4(x1)))))) → 4(0(0(0(1(4(x1)))))) (6)
3(3(1(4(5(4(5(5(x1)))))))) → 3(5(3(2(0(5(2(x1))))))) (7)
4(2(5(3(1(0(0(5(2(x1))))))))) → 4(4(1(5(3(5(3(2(x1)))))))) (8)
2(2(1(1(0(2(0(1(2(3(x1)))))))))) → 5(4(4(5(2(5(5(2(3(x1))))))))) (9)
2(2(2(3(3(4(3(3(5(4(5(x1))))))))))) → 5(5(4(2(5(3(5(3(2(0(5(x1))))))))))) (10)
4(2(1(1(5(1(0(0(0(2(5(x1))))))))))) → 5(5(0(3(0(1(4(5(2(5(x1)))))))))) (11)
5(4(5(4(2(0(2(2(3(0(0(2(x1)))))))))))) → 5(3(4(5(0(3(2(1(1(2(2(x1))))))))))) (12)
0(2(1(1(3(1(5(0(5(2(3(1(4(x1))))))))))))) → 3(0(1(5(1(4(4(5(0(5(1(1(4(x1))))))))))))) (13)
1(1(2(4(2(2(2(1(1(3(4(1(2(x1))))))))))))) → 1(5(0(3(4(1(0(5(2(2(2(3(x1)))))))))))) (14)
1(4(0(0(2(4(0(3(3(2(0(3(1(x1))))))))))))) → 1(5(0(5(0(2(0(5(1(0(1(2(5(5(x1)))))))))))))) (15)
4(1(4(1(3(0(0(2(1(5(4(1(0(0(x1)))))))))))))) → 2(3(0(2(5(4(4(3(0(4(1(2(0(x1))))))))))))) (16)
4(2(4(4(0(0(0(4(0(3(5(0(3(3(x1)))))))))))))) → 5(1(0(1(0(1(3(5(3(5(5(0(0(3(x1)))))))))))))) (17)
1(2(5(3(3(0(2(2(5(3(2(3(3(3(2(1(x1)))))))))))))))) → 1(1(0(0(4(1(2(1(0(5(0(0(3(5(0(4(1(x1))))))))))))))))) (18)
3(1(2(5(0(3(4(3(1(5(4(1(5(2(0(5(x1)))))))))))))))) → 3(1(0(4(2(0(5(2(4(4(2(2(1(1(1(5(x1)))))))))))))))) (19)
1(3(4(5(1(0(0(3(1(2(4(2(3(5(2(0(4(1(x1)))))))))))))))))) → 1(3(0(0(4(2(2(5(3(1(0(1(2(1(5(0(1(4(1(x1))))))))))))))))))) (20)
1(0(3(3(2(5(0(0(3(0(3(2(4(1(4(0(2(4(2(x1))))))))))))))))))) → 4(3(4(2(1(4(2(4(3(3(2(2(2(1(1(1(5(3(2(x1))))))))))))))))))) (21)
5(3(2(1(0(1(3(1(3(3(0(0(3(2(5(3(0(3(0(x1))))))))))))))))))) → 5(1(0(3(2(4(0(0(3(2(1(5(1(3(0(5(1(3(1(x1))))))))))))))))))) (22)
0(2(2(3(3(5(0(0(5(0(3(1(3(0(1(2(1(5(5(1(x1)))))))))))))))))))) → 0(0(3(2(5(2(5(2(3(2(1(2(5(4(3(4(5(0(4(x1))))))))))))))))))) (23)
1(0(4(2(3(3(5(4(3(5(0(2(0(4(5(0(2(0(2(4(x1)))))))))))))))))))) → 4(5(4(5(4(1(1(2(5(0(4(3(1(5(4(3(1(5(4(0(4(x1))))))))))))))))))))) (24)
1(4(2(3(1(3(4(2(4(1(5(1(4(0(4(5(2(0(0(3(4(x1))))))))))))))))))))) → 4(3(4(1(3(5(1(1(4(3(1(5(1(3(1(2(4(2(1(3(x1)))))))))))))))))))) (25)

1.1 Dependency Pair Transformation

The following set of initial dependency pairs has been identified.
4#(5(1(4(x1)))) → 4#(4(x1)) (26)

1.1.1 Monotonic Reduction Pair Processor with Usable Rules

Using the matrix interpretations of dimension 1 with strict dimension 1 over the rationals with delta = 1
[5(x1)] = x1 +
1
[4(x1)] = x1 +
0
[1(x1)] = x1 +
1
[4#(x1)] = x1 +
0
together with the usable rule
4(5(1(4(x1)))) → 1(5(4(4(x1)))) (3)
(w.r.t. the implicit argument filter of the reduction pair), the pair
4#(5(1(4(x1)))) → 4#(4(x1)) (26)
and no rules could be deleted.

1.1.1.1 Dependency Graph Processor

The dependency pairs are split into 0 components.