Certification Problem

Input (TPDB SRS_Standard/ICFP_2010/65081)

The rewrite relation of the following TRS is considered.

0(1(2(2(x1)))) → 0(1(0(1(x1)))) (1)
0(0(0(0(0(x1))))) → 2(0(2(0(x1)))) (2)
3(2(3(4(0(4(1(0(x1)))))))) → 2(3(2(2(5(3(1(x1))))))) (3)
1(4(1(0(4(1(2(2(5(3(x1)))))))))) → 1(4(4(2(5(3(1(5(2(x1))))))))) (4)
1(4(2(3(1(4(0(2(1(1(x1)))))))))) → 1(1(4(1(0(4(2(3(2(1(x1)))))))))) (5)
2(5(1(2(4(5(1(3(1(5(x1)))))))))) → 2(4(3(0(3(4(2(4(5(x1))))))))) (6)
2(3(2(4(3(2(3(4(4(0(0(0(x1)))))))))))) → 2(4(2(0(2(2(2(2(4(3(0(x1))))))))))) (7)
5(0(2(4(2(4(1(4(4(5(1(4(x1)))))))))))) → 1(4(5(3(3(2(3(2(3(3(4(x1))))))))))) (8)
1(0(1(3(5(5(1(2(5(2(3(5(4(x1))))))))))))) → 1(0(3(4(1(2(3(4(5(3(3(5(4(x1))))))))))))) (9)
1(4(4(2(5(3(1(5(1(2(1(5(0(x1))))))))))))) → 3(5(2(5(4(1(5(2(4(1(3(2(x1)))))))))))) (10)
4(1(2(5(1(1(0(0(5(4(1(3(1(x1))))))))))))) → 4(1(1(2(2(3(5(1(4(2(3(1(x1)))))))))))) (11)
5(5(3(2(0(3(4(0(0(3(1(4(3(x1))))))))))))) → 5(5(3(5(0(3(1(5(2(3(1(3(4(x1))))))))))))) (12)
0(5(3(2(4(0(2(1(2(3(3(4(3(3(x1)))))))))))))) → 0(5(4(4(5(3(4(0(1(0(1(3(1(2(3(x1))))))))))))))) (13)
4(3(0(5(5(2(5(2(3(5(3(0(2(2(4(x1))))))))))))))) → 4(3(3(4(3(1(4(1(5(0(0(5(1(5(3(4(x1)))))))))))))))) (14)
3(1(5(4(1(2(0(0(1(0(0(0(2(0(4(5(x1)))))))))))))))) → 3(0(2(2(3(3(1(3(2(2(1(2(2(5(5(x1))))))))))))))) (15)
4(1(1(0(5(4(2(0(4(0(5(1(2(0(3(1(x1)))))))))))))))) → 2(4(3(1(4(1(4(0(1(1(0(5(4(0(5(0(x1)))))))))))))))) (16)
4(0(2(2(4(4(1(1(1(1(0(4(1(5(1(2(0(1(x1)))))))))))))))))) → 4(5(5(2(5(0(2(1(5(2(4(1(1(1(5(3(2(x1))))))))))))))))) (17)
4(4(4(4(3(1(1(3(3(4(2(2(4(0(3(5(4(2(5(2(3(x1))))))))))))))))))))) → 2(5(5(4(0(1(5(3(3(5(0(1(5(1(5(4(2(4(2(2(x1)))))))))))))))))))) (18)

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 +
25264/93
[4(x1)] = x1 +
47224/93
[3(x1)] = x1 +
130546/155
[2(x1)] = x1 +
570232/465
[1(x1)] = x1 +
28969/31
[0(x1)] = x1 +
4463/3
all of the following rules can be deleted.
0(1(2(2(x1)))) → 0(1(0(1(x1)))) (1)
0(0(0(0(0(x1))))) → 2(0(2(0(x1)))) (2)
3(2(3(4(0(4(1(0(x1)))))))) → 2(3(2(2(5(3(1(x1))))))) (3)
1(4(1(0(4(1(2(2(5(3(x1)))))))))) → 1(4(4(2(5(3(1(5(2(x1))))))))) (4)
2(5(1(2(4(5(1(3(1(5(x1)))))))))) → 2(4(3(0(3(4(2(4(5(x1))))))))) (6)
2(3(2(4(3(2(3(4(4(0(0(0(x1)))))))))))) → 2(4(2(0(2(2(2(2(4(3(0(x1))))))))))) (7)
5(0(2(4(2(4(1(4(4(5(1(4(x1)))))))))))) → 1(4(5(3(3(2(3(2(3(3(4(x1))))))))))) (8)
1(0(1(3(5(5(1(2(5(2(3(5(4(x1))))))))))))) → 1(0(3(4(1(2(3(4(5(3(3(5(4(x1))))))))))))) (9)
1(4(4(2(5(3(1(5(1(2(1(5(0(x1))))))))))))) → 3(5(2(5(4(1(5(2(4(1(3(2(x1)))))))))))) (10)
4(1(2(5(1(1(0(0(5(4(1(3(1(x1))))))))))))) → 4(1(1(2(2(3(5(1(4(2(3(1(x1)))))))))))) (11)
5(5(3(2(0(3(4(0(0(3(1(4(3(x1))))))))))))) → 5(5(3(5(0(3(1(5(2(3(1(3(4(x1))))))))))))) (12)
0(5(3(2(4(0(2(1(2(3(3(4(3(3(x1)))))))))))))) → 0(5(4(4(5(3(4(0(1(0(1(3(1(2(3(x1))))))))))))))) (13)
4(3(0(5(5(2(5(2(3(5(3(0(2(2(4(x1))))))))))))))) → 4(3(3(4(3(1(4(1(5(0(0(5(1(5(3(4(x1)))))))))))))))) (14)
3(1(5(4(1(2(0(0(1(0(0(0(2(0(4(5(x1)))))))))))))))) → 3(0(2(2(3(3(1(3(2(2(1(2(2(5(5(x1))))))))))))))) (15)
4(1(1(0(5(4(2(0(4(0(5(1(2(0(3(1(x1)))))))))))))))) → 2(4(3(1(4(1(4(0(1(1(0(5(4(0(5(0(x1)))))))))))))))) (16)
4(0(2(2(4(4(1(1(1(1(0(4(1(5(1(2(0(1(x1)))))))))))))))))) → 4(5(5(2(5(0(2(1(5(2(4(1(1(1(5(3(2(x1))))))))))))))))) (17)
4(4(4(4(3(1(1(3(3(4(2(2(4(0(3(5(4(2(5(2(3(x1))))))))))))))))))))) → 2(5(5(4(0(1(5(3(3(5(0(1(5(1(5(4(2(4(2(2(x1)))))))))))))))))))) (18)

1.1 String Reversal

Since only unary symbols occur, one can reverse all terms and obtains the TRS
1(1(2(0(4(1(3(2(4(1(x1)))))))))) → 1(2(3(2(4(0(1(4(1(1(x1)))))))))) (19)

1.1.1 Dependency Pair Transformation

The following set of initial dependency pairs has been identified.
1#(1(2(0(4(1(3(2(4(1(x1)))))))))) → 1#(4(1(1(x1)))) (20)
1#(1(2(0(4(1(3(2(4(1(x1)))))))))) → 1#(2(3(2(4(0(1(4(1(1(x1)))))))))) (21)
1#(1(2(0(4(1(3(2(4(1(x1)))))))))) → 1#(1(x1)) (22)

1.1.1.1 Monotonic Reduction Pair Processor with Usable Rules

Using the matrix interpretations of dimension 1 with strict dimension 1 over the rationals with delta = 1
[4(x1)] = x1 +
1
[3(x1)] = x1 +
1
[2(x1)] = x1 +
1
[1(x1)] = x1 +
1
[0(x1)] = x1 +
1
[1#(x1)] = x1 +
0
together with the usable rule
1(1(2(0(4(1(3(2(4(1(x1)))))))))) → 1(2(3(2(4(0(1(4(1(1(x1)))))))))) (19)
(w.r.t. the implicit argument filter of the reduction pair), the pairs
1#(1(2(0(4(1(3(2(4(1(x1)))))))))) → 1#(4(1(1(x1)))) (20)
1#(1(2(0(4(1(3(2(4(1(x1)))))))))) → 1#(1(x1)) (22)
and no rules could be deleted.

1.1.1.1.1 Dependency Graph Processor

The dependency pairs are split into 0 components.