Certification Problem

Input (TPDB SRS_Standard/ICFP_2010/96224)

The rewrite relation of the following TRS is considered.

0(1(2(2(3(0(4(4(x1)))))))) → 0(4(5(0(3(4(0(4(x1)))))))) (1)
0(2(2(2(0(5(2(5(4(x1))))))))) → 0(0(0(3(4(3(0(1(0(x1))))))))) (2)
2(1(1(3(4(3(1(1(5(x1))))))))) → 1(0(2(3(1(0(5(1(5(x1))))))))) (3)
5(4(0(4(3(3(1(2(5(3(0(x1))))))))))) → 5(5(2(5(3(1(5(0(3(5(2(x1))))))))))) (4)
2(4(5(0(1(1(3(3(5(3(0(0(x1)))))))))))) → 4(4(2(2(1(0(4(0(1(3(2(0(x1)))))))))))) (5)
4(4(3(0(3(1(5(3(5(1(3(1(5(3(x1)))))))))))))) → 0(2(4(5(0(0(0(5(1(0(5(4(4(x1))))))))))))) (6)
2(2(0(0(2(1(0(5(3(2(2(1(4(0(5(x1))))))))))))))) → 0(0(3(5(3(0(4(3(1(3(0(2(5(5(x1)))))))))))))) (7)
3(0(0(4(2(5(5(1(3(0(2(3(3(5(1(4(5(x1))))))))))))))))) → 3(4(0(1(0(5(5(3(1(4(0(3(5(3(2(2(5(x1))))))))))))))))) (8)
5(1(1(4(1(5(3(0(4(3(2(5(4(1(3(3(5(x1))))))))))))))))) → 5(0(1(0(4(0(2(4(5(1(5(4(1(5(3(3(5(x1))))))))))))))))) (9)
0(1(4(4(3(2(0(4(1(4(3(4(4(1(5(3(4(4(x1)))))))))))))))))) → 0(2(4(2(3(1(0(1(1(1(3(0(2(4(4(1(1(2(x1)))))))))))))))))) (10)
1(1(4(1(0(1(0(3(3(4(4(1(5(4(0(4(4(5(5(3(x1)))))))))))))))))))) → 1(3(0(3(2(2(4(4(2(0(3(3(4(0(3(0(4(3(4(0(x1)))))))))))))))))))) (11)
2(0(3(3(3(4(1(1(0(4(4(0(3(3(3(0(0(1(5(3(x1)))))))))))))))))))) → 5(3(3(1(4(0(4(5(4(4(4(2(4(3(1(1(1(5(4(x1))))))))))))))))))) (12)
2(3(3(2(1(5(0(5(0(1(3(3(2(5(1(5(0(3(0(5(x1)))))))))))))))))))) → 3(4(0(2(5(5(2(4(2(4(3(1(1(4(4(5(5(3(5(x1))))))))))))))))))) (13)
3(3(2(2(3(3(4(0(0(0(2(5(0(5(3(0(0(1(1(4(x1)))))))))))))))))))) → 3(1(1(0(3(3(5(4(2(2(1(0(1(1(0(5(0(3(4(0(x1)))))))))))))))))))) (14)
4(2(4(1(0(5(0(4(1(0(3(0(2(5(4(3(5(3(5(3(x1)))))))))))))))))))) → 4(2(5(4(4(2(4(5(0(0(3(1(5(0(2(0(2(2(1(x1))))))))))))))))))) (15)

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 +
617459/406
[4(x1)] = x1 +
1284417/812
[3(x1)] = x1 +
377452/203
[2(x1)] = x1 +
1160459/812
[1(x1)] = x1 +
82615/58
[0(x1)] = x1 +
801999/812
all of the following rules can be deleted.
0(1(2(2(3(0(4(4(x1)))))))) → 0(4(5(0(3(4(0(4(x1)))))))) (1)
0(2(2(2(0(5(2(5(4(x1))))))))) → 0(0(0(3(4(3(0(1(0(x1))))))))) (2)
2(1(1(3(4(3(1(1(5(x1))))))))) → 1(0(2(3(1(0(5(1(5(x1))))))))) (3)
5(4(0(4(3(3(1(2(5(3(0(x1))))))))))) → 5(5(2(5(3(1(5(0(3(5(2(x1))))))))))) (4)
2(4(5(0(1(1(3(3(5(3(0(0(x1)))))))))))) → 4(4(2(2(1(0(4(0(1(3(2(0(x1)))))))))))) (5)
4(4(3(0(3(1(5(3(5(1(3(1(5(3(x1)))))))))))))) → 0(2(4(5(0(0(0(5(1(0(5(4(4(x1))))))))))))) (6)
2(2(0(0(2(1(0(5(3(2(2(1(4(0(5(x1))))))))))))))) → 0(0(3(5(3(0(4(3(1(3(0(2(5(5(x1)))))))))))))) (7)
5(1(1(4(1(5(3(0(4(3(2(5(4(1(3(3(5(x1))))))))))))))))) → 5(0(1(0(4(0(2(4(5(1(5(4(1(5(3(3(5(x1))))))))))))))))) (9)
0(1(4(4(3(2(0(4(1(4(3(4(4(1(5(3(4(4(x1)))))))))))))))))) → 0(2(4(2(3(1(0(1(1(1(3(0(2(4(4(1(1(2(x1)))))))))))))))))) (10)
1(1(4(1(0(1(0(3(3(4(4(1(5(4(0(4(4(5(5(3(x1)))))))))))))))))))) → 1(3(0(3(2(2(4(4(2(0(3(3(4(0(3(0(4(3(4(0(x1)))))))))))))))))))) (11)
2(0(3(3(3(4(1(1(0(4(4(0(3(3(3(0(0(1(5(3(x1)))))))))))))))))))) → 5(3(3(1(4(0(4(5(4(4(4(2(4(3(1(1(1(5(4(x1))))))))))))))))))) (12)
2(3(3(2(1(5(0(5(0(1(3(3(2(5(1(5(0(3(0(5(x1)))))))))))))))))))) → 3(4(0(2(5(5(2(4(2(4(3(1(1(4(4(5(5(3(5(x1))))))))))))))))))) (13)
3(3(2(2(3(3(4(0(0(0(2(5(0(5(3(0(0(1(1(4(x1)))))))))))))))))))) → 3(1(1(0(3(3(5(4(2(2(1(0(1(1(0(5(0(3(4(0(x1)))))))))))))))))))) (14)
4(2(4(1(0(5(0(4(1(0(3(0(2(5(4(3(5(3(5(3(x1)))))))))))))))))))) → 4(2(5(4(4(2(4(5(0(0(3(1(5(0(2(0(2(2(1(x1))))))))))))))))))) (15)

1.1 Dependency Pair Transformation

The following set of initial dependency pairs has been identified.
3#(0(0(4(2(5(5(1(3(0(2(3(3(5(1(4(5(x1))))))))))))))))) → 3#(5(3(2(2(5(x1)))))) (16)
3#(0(0(4(2(5(5(1(3(0(2(3(3(5(1(4(5(x1))))))))))))))))) → 3#(4(0(1(0(5(5(3(1(4(0(3(5(3(2(2(5(x1))))))))))))))))) (17)
3#(0(0(4(2(5(5(1(3(0(2(3(3(5(1(4(5(x1))))))))))))))))) → 3#(2(2(5(x1)))) (18)
3#(0(0(4(2(5(5(1(3(0(2(3(3(5(1(4(5(x1))))))))))))))))) → 3#(1(4(0(3(5(3(2(2(5(x1)))))))))) (19)

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 +
1
[3(x1)] = x1 +
1
[2(x1)] = x1 +
0
[1(x1)] = x1 +
1
[0(x1)] = x1 +
1
[3#(x1)] = x1 +
0
together with the usable rule
3(0(0(4(2(5(5(1(3(0(2(3(3(5(1(4(5(x1))))))))))))))))) → 3(4(0(1(0(5(5(3(1(4(0(3(5(3(2(2(5(x1))))))))))))))))) (8)
(w.r.t. the implicit argument filter of the reduction pair), the pairs
3#(0(0(4(2(5(5(1(3(0(2(3(3(5(1(4(5(x1))))))))))))))))) → 3#(5(3(2(2(5(x1)))))) (16)
3#(0(0(4(2(5(5(1(3(0(2(3(3(5(1(4(5(x1))))))))))))))))) → 3#(2(2(5(x1)))) (18)
3#(0(0(4(2(5(5(1(3(0(2(3(3(5(1(4(5(x1))))))))))))))))) → 3#(1(4(0(3(5(3(2(2(5(x1)))))))))) (19)
and no rules could be deleted.

1.1.1.1 Dependency Graph Processor

The dependency pairs are split into 0 components.