Certification Problem

Input (TPDB SRS_Standard/ICFP_2010/85675)

The rewrite relation of the following TRS is considered.

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

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 +
1135
[4(x1)] = x1 +
2755/2
[3(x1)] = x1 +
2459/2
[2(x1)] = x1 +
1131
[1(x1)] = x1 +
1262
[0(x1)] = x1 +
2375/2
all of the following rules can be deleted.
0(1(0(2(x1)))) → 3(4(4(x1))) (1)
2(0(0(3(x1)))) → 3(4(1(x1))) (2)
5(3(5(0(1(x1))))) → 5(2(3(0(x1)))) (3)
0(0(1(2(2(3(x1)))))) → 4(0(2(2(3(x1))))) (4)
0(5(3(1(4(3(x1)))))) → 1(4(0(5(2(x1))))) (5)
2(0(4(3(5(3(3(1(x1)))))))) → 2(1(2(2(2(0(3(1(x1)))))))) (6)
2(4(0(2(3(0(0(2(x1)))))))) → 2(4(0(3(4(4(2(x1))))))) (7)
2(1(4(0(4(1(5(0(2(x1))))))))) → 3(3(2(2(3(0(3(3(4(x1))))))))) (8)
5(0(4(0(0(1(3(5(0(x1))))))))) → 5(0(0(0(3(5(5(3(2(x1))))))))) (9)
5(3(5(4(4(2(2(2(1(x1))))))))) → 5(0(3(0(0(4(5(2(1(x1))))))))) (10)
4(4(3(2(0(1(1(4(0(2(x1)))))))))) → 2(3(4(0(0(1(2(1(3(4(x1)))))))))) (12)
1(5(5(2(4(2(4(0(3(3(0(x1))))))))))) → 1(3(5(4(5(5(5(3(2(4(2(x1))))))))))) (13)
2(5(5(0(1(1(5(1(4(2(3(3(x1)))))))))))) → 4(2(3(0(3(0(0(2(3(2(3(x1))))))))))) (14)
3(0(2(4(5(2(1(2(0(2(5(1(x1)))))))))))) → 3(1(3(0(5(0(2(2(4(0(4(x1))))))))))) (15)
3(3(3(4(3(0(0(4(4(5(0(5(x1)))))))))))) → 3(1(5(5(0(3(1(0(5(2(5(x1))))))))))) (16)
0(5(3(5(1(0(5(1(2(4(4(5(0(x1))))))))))))) → 0(3(1(0(4(5(5(2(5(5(2(4(0(x1))))))))))))) (17)
2(5(2(4(1(5(3(3(1(0(2(5(3(x1))))))))))))) → 2(3(1(3(3(5(1(2(0(3(2(3(0(x1))))))))))))) (18)
2(5(5(3(5(3(1(3(1(2(5(0(0(x1))))))))))))) → 2(5(5(5(0(4(5(5(1(1(3(2(2(x1))))))))))))) (19)
3(5(3(3(3(2(3(0(2(4(2(1(3(x1))))))))))))) → 5(1(3(1(4(0(0(1(5(0(4(5(x1)))))))))))) (20)
2(0(3(0(2(3(1(0(0(3(0(3(5(3(x1)))))))))))))) → 3(4(0(2(1(4(4(4(0(5(5(5(2(x1))))))))))))) (21)
3(5(5(0(2(2(0(0(3(5(1(3(3(5(3(5(x1)))))))))))))))) → 3(1(4(2(0(3(4(0(4(1(5(4(2(5(x1)))))))))))))) (23)
5(5(3(5(0(5(3(5(2(3(1(3(3(4(0(2(x1)))))))))))))))) → 5(2(4(2(4(4(0(5(5(1(3(4(0(4(0(x1))))))))))))))) (24)
2(2(1(3(4(0(1(1(4(2(2(2(0(4(4(2(4(4(x1)))))))))))))))))) → 4(3(4(4(5(5(0(1(0(1(3(2(5(5(1(1(4(4(x1)))))))))))))))))) (25)
0(3(0(2(2(2(5(5(2(4(4(3(5(1(0(0(0(3(3(x1))))))))))))))))))) → 3(3(1(3(4(5(5(2(0(4(2(3(1(1(5(1(2(3(x1)))))))))))))))))) (26)
3(3(4(3(3(4(3(0(2(5(3(1(4(5(2(5(2(4(3(5(x1)))))))))))))))))))) → 5(2(5(0(5(0(1(1(1(3(2(4(4(0(3(2(4(1(5(x1))))))))))))))))))) (27)
2(1(5(5(3(0(1(3(3(3(1(2(0(5(2(0(3(5(2(2(2(x1))))))))))))))))))))) → 3(0(3(3(4(0(1(4(5(1(3(3(3(4(1(2(1(3(2(1(x1)))))))))))))))))))) (28)
5(1(3(4(0(3(0(2(5(3(0(2(2(0(1(2(3(3(4(1(1(x1))))))))))))))))))))) → 5(0(1(1(4(2(0(1(1(2(1(4(0(4(3(1(2(0(0(2(x1)))))))))))))))))))) (29)
5(4(0(0(3(2(5(3(0(3(0(4(5(0(4(4(5(1(1(1(3(x1))))))))))))))))))))) → 5(2(5(1(0(3(5(3(2(1(0(0(5(5(4(5(4(4(2(2(2(x1))))))))))))))))))))) (30)

1.1 String Reversal

Since only unary symbols occur, one can reverse all terms and obtains the TRS
2(4(5(3(3(3(2(4(0(2(x1)))))))))) → 2(5(3(3(2(0(2(4(4(3(x1)))))))))) (31)
1(2(4(5(3(4(0(4(2(1(3(3(5(3(x1)))))))))))))) → 2(1(4(3(5(3(3(3(4(4(1(0(2(5(x1)))))))))))))) (32)

1.1.1 Dependency Pair Transformation

The following set of initial dependency pairs has been identified.
2#(4(5(3(3(3(2(4(0(2(x1)))))))))) → 2#(5(3(3(2(0(2(4(4(3(x1)))))))))) (33)
2#(4(5(3(3(3(2(4(0(2(x1)))))))))) → 2#(4(4(3(x1)))) (34)
2#(4(5(3(3(3(2(4(0(2(x1)))))))))) → 2#(0(2(4(4(3(x1)))))) (35)
1#(2(4(5(3(4(0(4(2(1(3(3(5(3(x1)))))))))))))) → 2#(5(x1)) (36)
1#(2(4(5(3(4(0(4(2(1(3(3(5(3(x1)))))))))))))) → 2#(1(4(3(5(3(3(3(4(4(1(0(2(5(x1)))))))))))))) (37)
1#(2(4(5(3(4(0(4(2(1(3(3(5(3(x1)))))))))))))) → 1#(4(3(5(3(3(3(4(4(1(0(2(5(x1))))))))))))) (38)
1#(2(4(5(3(4(0(4(2(1(3(3(5(3(x1)))))))))))))) → 1#(0(2(5(x1)))) (39)

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
[5(x1)] = x1 +
1
[4(x1)] = x1 +
1
[3(x1)] = x1 +
1
[2(x1)] = x1 +
1
[1(x1)] = x1 +
0
[0(x1)] = x1 +
1
[2#(x1)] = x1 +
0
[1#(x1)] = x1 +
1
together with the usable rules
2(4(5(3(3(3(2(4(0(2(x1)))))))))) → 2(5(3(3(2(0(2(4(4(3(x1)))))))))) (31)
1(2(4(5(3(4(0(4(2(1(3(3(5(3(x1)))))))))))))) → 2(1(4(3(5(3(3(3(4(4(1(0(2(5(x1)))))))))))))) (32)
(w.r.t. the implicit argument filter of the reduction pair), the pairs
2#(4(5(3(3(3(2(4(0(2(x1)))))))))) → 2#(4(4(3(x1)))) (34)
2#(4(5(3(3(3(2(4(0(2(x1)))))))))) → 2#(0(2(4(4(3(x1)))))) (35)
1#(2(4(5(3(4(0(4(2(1(3(3(5(3(x1)))))))))))))) → 2#(5(x1)) (36)
1#(2(4(5(3(4(0(4(2(1(3(3(5(3(x1)))))))))))))) → 2#(1(4(3(5(3(3(3(4(4(1(0(2(5(x1)))))))))))))) (37)
1#(2(4(5(3(4(0(4(2(1(3(3(5(3(x1)))))))))))))) → 1#(4(3(5(3(3(3(4(4(1(0(2(5(x1))))))))))))) (38)
1#(2(4(5(3(4(0(4(2(1(3(3(5(3(x1)))))))))))))) → 1#(0(2(5(x1)))) (39)
and no rules could be deleted.

1.1.1.1.1 Dependency Graph Processor

The dependency pairs are split into 0 components.