Certification Problem

Input (TPDB SRS_Standard/ICFP_2010/96274)

The rewrite relation of the following TRS is considered.

0(1(2(3(x1)))) → 4(4(2(3(x1)))) (1)
5(3(3(5(4(x1))))) → 5(1(0(2(x1)))) (2)
2(3(1(5(0(5(x1)))))) → 2(1(3(5(0(5(x1)))))) (3)
5(3(3(5(5(4(x1)))))) → 4(2(4(3(2(x1))))) (4)
5(1(4(5(1(1(5(x1))))))) → 1(4(0(2(3(2(5(x1))))))) (5)
3(3(4(3(1(3(0(5(x1)))))))) → 3(5(2(4(5(0(5(2(x1)))))))) (6)
3(1(2(2(2(1(3(1(3(x1))))))))) → 1(4(3(1(5(0(2(2(x1)))))))) (7)
3(4(2(0(5(2(3(5(3(x1))))))))) → 3(5(4(4(2(2(0(5(1(x1))))))))) (8)
5(5(1(3(3(5(4(0(0(x1))))))))) → 3(1(0(1(4(2(4(3(x1)))))))) (9)
3(0(2(5(1(5(0(1(5(0(x1)))))))))) → 1(2(2(0(0(4(3(4(4(x1))))))))) (10)
3(5(5(4(4(4(2(0(0(3(x1)))))))))) → 1(1(2(3(2(3(4(1(x1)))))))) (11)
3(0(4(3(3(5(0(4(4(0(4(2(x1)))))))))))) → 3(4(5(5(3(2(0(5(1(4(2(x1))))))))))) (12)
5(2(0(4(5(0(2(1(1(1(2(0(x1)))))))))))) → 3(0(0(2(2(4(5(1(3(1(0(x1))))))))))) (13)
5(5(4(3(3(4(5(4(5(0(0(4(5(x1))))))))))))) → 5(0(1(0(3(1(4(1(2(3(1(x1))))))))))) (14)
5(2(1(3(0(2(2(4(5(2(2(0(0(1(x1)))))))))))))) → 3(4(5(1(4(3(3(5(0(3(0(1(x1)))))))))))) (15)
3(1(5(2(5(5(3(3(4(4(5(2(3(2(4(x1))))))))))))))) → 3(0(5(4(4(4(2(0(0(1(4(3(2(4(x1)))))))))))))) (16)
4(5(5(4(3(4(4(2(4(2(4(3(3(3(3(x1))))))))))))))) → 4(5(0(0(4(4(5(4(4(3(4(0(0(0(x1)))))))))))))) (17)
0(1(2(4(3(1(1(4(1(5(0(2(5(3(2(4(3(x1))))))))))))))))) → 4(2(2(1(3(1(3(0(4(5(1(2(2(5(5(4(1(x1))))))))))))))))) (18)
2(4(3(0(4(2(0(0(2(5(1(0(2(0(0(4(4(x1))))))))))))))))) → 5(4(1(2(1(2(1(0(2(0(4(3(1(0(0(2(x1)))))))))))))))) (19)
3(3(3(1(0(2(1(1(5(2(4(0(0(4(5(2(2(0(2(x1))))))))))))))))))) → 3(2(2(3(1(5(5(5(3(0(3(1(4(3(2(3(1(x1))))))))))))))))) (20)
5(3(2(2(5(2(1(3(0(2(4(3(2(5(3(3(0(5(4(x1))))))))))))))))))) → 1(3(0(3(3(4(5(5(0(5(5(4(0(2(1(1(0(0(2(x1))))))))))))))))))) (21)
5(4(5(5(5(2(0(1(2(1(0(1(2(1(5(3(1(3(1(x1))))))))))))))))))) → 0(0(3(5(3(0(2(0(1(4(0(5(4(3(0(2(4(1(x1)))))))))))))))))) (22)
4(0(4(0(5(1(0(3(2(5(3(1(3(0(2(5(3(5(0(0(x1)))))))))))))))))))) → 1(5(3(5(2(0(5(4(4(5(0(1(4(4(3(1(3(2(5(1(x1)))))))))))))))))))) (23)
5(4(2(1(3(2(5(4(2(2(0(0(5(5(1(0(5(1(3(0(x1)))))))))))))))))))) → 4(4(2(4(0(1(3(2(5(1(3(4(4(0(0(1(1(1(2(0(x1)))))))))))))))))))) (24)
3(0(4(5(4(1(4(3(5(5(3(5(4(0(1(4(3(5(0(3(2(x1))))))))))))))))))))) → 1(2(4(1(1(2(5(4(2(4(0(4(2(5(1(4(2(1(3(1(2(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 +
650
[4(x1)] = x1 +
1332/13
[3(x1)] = x1 +
7465/13
[2(x1)] = x1 +
4113/13
[1(x1)] = x1 +
2952/13
[0(x1)] = x1 +
7100/13
all of the following rules can be deleted.
0(1(2(3(x1)))) → 4(4(2(3(x1)))) (1)
5(3(3(5(4(x1))))) → 5(1(0(2(x1)))) (2)
5(3(3(5(5(4(x1)))))) → 4(2(4(3(2(x1))))) (4)
5(1(4(5(1(1(5(x1))))))) → 1(4(0(2(3(2(5(x1))))))) (5)
3(3(4(3(1(3(0(5(x1)))))))) → 3(5(2(4(5(0(5(2(x1)))))))) (6)
3(1(2(2(2(1(3(1(3(x1))))))))) → 1(4(3(1(5(0(2(2(x1)))))))) (7)
3(4(2(0(5(2(3(5(3(x1))))))))) → 3(5(4(4(2(2(0(5(1(x1))))))))) (8)
5(5(1(3(3(5(4(0(0(x1))))))))) → 3(1(0(1(4(2(4(3(x1)))))))) (9)
3(0(2(5(1(5(0(1(5(0(x1)))))))))) → 1(2(2(0(0(4(3(4(4(x1))))))))) (10)
3(5(5(4(4(4(2(0(0(3(x1)))))))))) → 1(1(2(3(2(3(4(1(x1)))))))) (11)
3(0(4(3(3(5(0(4(4(0(4(2(x1)))))))))))) → 3(4(5(5(3(2(0(5(1(4(2(x1))))))))))) (12)
5(2(0(4(5(0(2(1(1(1(2(0(x1)))))))))))) → 3(0(0(2(2(4(5(1(3(1(0(x1))))))))))) (13)
5(5(4(3(3(4(5(4(5(0(0(4(5(x1))))))))))))) → 5(0(1(0(3(1(4(1(2(3(1(x1))))))))))) (14)
5(2(1(3(0(2(2(4(5(2(2(0(0(1(x1)))))))))))))) → 3(4(5(1(4(3(3(5(0(3(0(1(x1)))))))))))) (15)
3(1(5(2(5(5(3(3(4(4(5(2(3(2(4(x1))))))))))))))) → 3(0(5(4(4(4(2(0(0(1(4(3(2(4(x1)))))))))))))) (16)
4(5(5(4(3(4(4(2(4(2(4(3(3(3(3(x1))))))))))))))) → 4(5(0(0(4(4(5(4(4(3(4(0(0(0(x1)))))))))))))) (17)
0(1(2(4(3(1(1(4(1(5(0(2(5(3(2(4(3(x1))))))))))))))))) → 4(2(2(1(3(1(3(0(4(5(1(2(2(5(5(4(1(x1))))))))))))))))) (18)
2(4(3(0(4(2(0(0(2(5(1(0(2(0(0(4(4(x1))))))))))))))))) → 5(4(1(2(1(2(1(0(2(0(4(3(1(0(0(2(x1)))))))))))))))) (19)
3(3(3(1(0(2(1(1(5(2(4(0(0(4(5(2(2(0(2(x1))))))))))))))))))) → 3(2(2(3(1(5(5(5(3(0(3(1(4(3(2(3(1(x1))))))))))))))))) (20)
5(3(2(2(5(2(1(3(0(2(4(3(2(5(3(3(0(5(4(x1))))))))))))))))))) → 1(3(0(3(3(4(5(5(0(5(5(4(0(2(1(1(0(0(2(x1))))))))))))))))))) (21)
5(4(5(5(5(2(0(1(2(1(0(1(2(1(5(3(1(3(1(x1))))))))))))))))))) → 0(0(3(5(3(0(2(0(1(4(0(5(4(3(0(2(4(1(x1)))))))))))))))))) (22)
4(0(4(0(5(1(0(3(2(5(3(1(3(0(2(5(3(5(0(0(x1)))))))))))))))))))) → 1(5(3(5(2(0(5(4(4(5(0(1(4(4(3(1(3(2(5(1(x1)))))))))))))))))))) (23)
5(4(2(1(3(2(5(4(2(2(0(0(5(5(1(0(5(1(3(0(x1)))))))))))))))))))) → 4(4(2(4(0(1(3(2(5(1(3(4(4(0(0(1(1(1(2(0(x1)))))))))))))))))))) (24)
3(0(4(5(4(1(4(3(5(5(3(5(4(0(1(4(3(5(0(3(2(x1))))))))))))))))))))) → 1(2(4(1(1(2(5(4(2(4(0(4(2(5(1(4(2(1(3(1(2(x1))))))))))))))))))))) (25)

1.1 Dependency Pair Transformation

The following set of initial dependency pairs has been identified.
2#(3(1(5(0(5(x1)))))) → 2#(1(3(5(0(5(x1)))))) (26)

1.1.1 Dependency Graph Processor

The dependency pairs are split into 0 components.