Certification Problem

Input (TPDB SRS_Standard/ICFP_2010/85749)

The rewrite relation of the following TRS is considered.

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

Property / Task

Prove or disprove termination.

Answer / Result

Yes.

Proof (by AProVE @ termCOMP 2023)

1 Rule Removal

Using the linear polynomial interpretation over the naturals
[0(x1)] = 1 · x1 + 23
[1(x1)] = 1 · x1 + 21
[2(x1)] = 1 · x1 + 26
[3(x1)] = 1 · x1 + 27
[4(x1)] = 1 · x1 + 25
[5(x1)] = 1 · x1 + 25
all of the following rules can be deleted.
0(1(1(1(x1)))) → 2(1(3(x1))) (1)
1(2(3(1(x1)))) → 1(1(1(1(x1)))) (2)
1(2(0(3(4(x1))))) → 3(0(2(2(x1)))) (3)
3(2(0(5(3(5(x1)))))) → 3(1(2(4(2(5(x1)))))) (4)
1(3(4(3(1(5(3(x1))))))) → 5(2(0(0(4(0(3(x1))))))) (5)
4(2(2(4(2(4(4(x1))))))) → 4(2(2(3(0(4(x1)))))) (6)
1(3(3(3(4(4(1(1(x1)))))))) → 1(2(4(4(3(4(0(1(x1)))))))) (7)
3(0(0(2(5(0(0(1(x1)))))))) → 0(2(4(1(4(2(0(1(x1)))))))) (8)
4(0(1(0(0(1(2(2(1(x1))))))))) → 2(4(0(0(4(5(5(5(x1)))))))) (9)
2(1(3(3(0(1(3(2(5(1(x1)))))))))) → 2(2(0(1(2(1(2(1(1(3(x1)))))))))) (10)
3(4(3(5(1(3(1(2(2(5(x1)))))))))) → 1(0(4(0(1(5(3(5(2(1(x1)))))))))) (12)
5(0(1(1(0(2(2(0(1(1(x1)))))))))) → 4(5(5(5(2(3(2(0(3(x1))))))))) (13)
4(0(1(1(0(0(1(5(3(5(0(1(x1)))))))))))) → 2(3(4(4(5(4(5(5(0(4(2(x1))))))))))) (14)
5(5(4(2(0(5(2(4(3(2(5(5(x1)))))))))))) → 3(4(0(0(1(1(3(5(1(0(1(4(x1)))))))))))) (15)
3(4(0(1(1(5(3(5(0(4(4(3(4(3(x1)))))))))))))) → 3(5(5(4(0(1(1(1(1(4(4(0(0(3(x1)))))))))))))) (16)
4(0(0(2(4(5(4(2(1(4(3(4(4(3(x1)))))))))))))) → 4(4(0(1(5(2(2(1(1(0(1(3(1(3(x1)))))))))))))) (17)
1(5(2(4(4(0(2(4(5(5(1(2(3(3(1(x1))))))))))))))) → 3(1(2(4(5(4(0(3(2(1(5(5(1(3(1(x1))))))))))))))) (18)
2(0(2(0(5(5(5(1(4(4(4(3(3(3(0(x1))))))))))))))) → 5(3(3(3(2(5(0(5(4(4(5(3(3(0(x1)))))))))))))) (19)
5(0(4(2(1(1(2(4(1(0(3(1(0(3(5(x1))))))))))))))) → 4(5(3(2(3(1(1(0(1(3(5(3(0(1(x1)))))))))))))) (20)
5(0(4(5(4(0(4(0(4(5(2(0(1(2(1(x1))))))))))))))) → 1(2(2(2(0(4(0(5(1(3(4(4(1(1(4(x1))))))))))))))) (21)
5(3(1(2(3(2(2(2(4(1(0(4(3(3(4(0(x1)))))))))))))))) → 5(5(4(0(2(5(0(0(1(3(0(3(4(3(4(0(x1)))))))))))))))) (22)
1(3(4(1(3(3(0(2(3(3(3(2(5(4(4(4(4(x1))))))))))))))))) → 1(1(3(2(5(0(4(2(4(3(4(0(3(3(3(0(4(x1))))))))))))))))) (23)
3(4(3(3(4(3(2(4(1(1(2(1(4(5(3(0(0(x1))))))))))))))))) → 1(1(3(0(5(3(2(0(4(4(1(3(5(0(0(2(0(x1))))))))))))))))) (24)
5(3(3(4(5(3(5(1(2(5(5(0(5(5(1(5(3(x1))))))))))))))))) → 5(5(3(5(2(1(3(3(5(1(3(5(0(5(4(3(x1)))))))))))))))) (25)
2(0(1(2(0(3(5(1(4(5(2(3(5(2(4(1(0(3(x1)))))))))))))))))) → 1(4(5(4(0(5(5(1(3(0(1(4(4(0(3(3(4(3(x1)))))))))))))))))) (26)
5(1(1(0(2(2(1(1(5(5(2(2(4(0(4(3(2(2(x1)))))))))))))))))) → 3(1(1(1(2(2(1(5(4(3(0(2(3(5(4(2(2(x1))))))))))))))))) (27)
2(2(3(0(5(2(3(4(4(0(3(3(0(2(0(2(4(1(1(x1))))))))))))))))))) → 2(1(5(2(1(1(5(4(5(4(3(3(5(4(5(5(3(4(x1)))))))))))))))))) (28)
1(3(5(3(3(3(4(0(1(2(3(3(0(4(3(0(3(4(4(5(x1)))))))))))))))))))) → 4(2(4(5(4(5(1(4(4(5(5(0(2(1(1(1(1(4(2(5(1(x1))))))))))))))))))))) (29)
5(0(4(2(0(0(0(3(3(1(5(2(2(5(1(3(4(3(0(5(1(x1))))))))))))))))))))) → 0(3(1(5(4(3(1(1(3(0(3(2(0(0(2(2(2(1(4(4(1(x1))))))))))))))))))))) (30)

1.1 Dependency Pair Transformation

The following set of initial dependency pairs has been identified.
3#(3(3(5(3(0(2(1(4(3(x1)))))))))) → 3#(5(2(3(3(1(3(0(3(x1))))))))) (31)
3#(3(3(5(3(0(2(1(4(3(x1)))))))))) → 3#(3(1(3(0(3(x1)))))) (32)
3#(3(3(5(3(0(2(1(4(3(x1)))))))))) → 3#(1(3(0(3(x1))))) (33)
3#(3(3(5(3(0(2(1(4(3(x1)))))))))) → 3#(0(3(x1))) (34)

1.1.1 Dependency Graph Processor

The dependency pairs are split into 0 components.