Certification Problem

Input (TPDB SRS_Standard/ICFP_2010/68535)

The rewrite relation of the following TRS is considered.

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

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 + 1
[1(x1)] = 1 · x1 + 1
[2(x1)] = 1 · x1 + 1
[3(x1)] = 1 · x1 + 1
[4(x1)] = 1 · x1 + 1
[5(x1)] = 1 · x1 + 1
all of the following rules can be deleted.
4(3(0(2(5(3(1(2(3(x1))))))))) 4(1(1(0(0(0(5(3(x1)))))))) (9)
3(0(4(4(4(5(3(1(2(5(0(4(x1)))))))))))) 5(3(3(0(3(5(1(0(0(1(0(x1))))))))))) (12)
5(4(2(3(0(4(3(2(2(2(4(1(x1)))))))))))) 5(1(0(2(4(4(4(5(1(4(1(x1))))))))))) (14)
3(0(0(2(1(1(3(1(5(0(5(1(0(5(x1)))))))))))))) 5(1(0(1(2(2(1(3(3(1(0(2(5(x1))))))))))))) (15)
0(4(2(4(1(1(5(3(0(2(2(0(5(1(5(x1))))))))))))))) 1(1(0(0(3(5(5(1(1(2(5(3(5(5(x1)))))))))))))) (16)
0(0(0(2(0(3(0(5(1(0(5(0(5(1(0(4(3(x1))))))))))))))))) 0(0(1(4(5(0(0(2(0(0(1(1(3(2(5(3(x1)))))))))))))))) (19)
2(4(2(1(5(5(0(1(0(4(2(2(2(3(4(1(0(4(x1)))))))))))))))))) 2(1(3(2(1(2(2(5(1(1(3(5(4(4(5(5(5(x1))))))))))))))))) (23)

1.1 Dependency Pair Transformation

The following set of initial dependency pairs has been identified.

There are 191 ruless (increase limit for explicit display).

1.1.1 Reduction Pair Processor

Using the linear polynomial interpretation over the naturals
[0#(x1)] = 1 · x1
[0(x1)] = 1 + 1 · x1
[1(x1)] = 1 + 1 · x1
[2#(x1)] = 1 · x1
[2(x1)] = 1 + 1 · x1
[3(x1)] = 1 + 1 · x1
[1#(x1)] = 1 · x1
[3#(x1)] = 1 · x1
[4(x1)] = 1 + 1 · x1
[4#(x1)] = 1 · x1
[5(x1)] = 1 + 1 · x1
[5#(x1)] = 1 · x1
the pairs

There are 173 ruless (increase limit for explicit display).

could be deleted.

1.1.1.1 Dependency Graph Processor

The dependency pairs are split into 4 components.