Certification Problem

Input (TPDB SRS_Standard/ICFP_2010/4181)

The rewrite relation of the following TRS is considered.

0(1(0(x1))) → 2(3(3(2(2(3(2(4(2(4(x1)))))))))) (1)
0(0(0(4(0(x1))))) → 3(4(4(4(1(2(2(0(3(4(x1)))))))))) (2)
0(4(4(1(1(x1))))) → 0(4(2(3(2(2(2(2(2(2(x1)))))))))) (3)
0(5(4(1(2(x1))))) → 3(2(1(3(3(4(2(3(2(1(x1)))))))))) (4)
1(0(0(3(0(x1))))) → 1(1(2(3(2(3(2(4(3(0(x1)))))))))) (5)
0(0(0(0(4(5(x1)))))) → 0(2(3(2(4(1(5(2(3(5(x1)))))))))) (6)
0(0(0(2(0(4(x1)))))) → 3(2(2(1(2(4(1(1(0(4(x1)))))))))) (7)
0(0(4(5(3(0(x1)))))) → 1(2(2(3(5(2(2(1(1(4(x1)))))))))) (8)
0(1(5(3(1(1(x1)))))) → 3(2(1(3(2(1(2(5(5(3(x1)))))))))) (9)
0(2(0(0(1(1(x1)))))) → 3(4(2(3(1(2(3(2(3(3(x1)))))))))) (10)
0(5(0(0(0(2(x1)))))) → 2(3(2(0(5(4(3(1(2(1(x1)))))))))) (11)
1(4(0(3(0(4(x1)))))) → 3(2(2(5(3(3(2(1(4(4(x1)))))))))) (12)
3(0(0(1(3(5(x1)))))) → 2(1(3(3(2(0(5(3(1(5(x1)))))))))) (13)
3(1(4(0(1(2(x1)))))) → 2(3(2(2(1(0(3(1(1(2(x1)))))))))) (14)
4(0(0(1(3(1(x1)))))) → 3(2(2(1(2(0(5(4(4(1(x1)))))))))) (15)
4(5(0(2(4(1(x1)))))) → 2(1(4(2(3(2(2(3(4(1(x1)))))))))) (16)
5(4(3(0(1(5(x1)))))) → 5(2(3(3(3(2(1(5(3(2(x1)))))))))) (17)
0(0(1(2(3(0(5(x1))))))) → 3(2(1(3(2(1(4(3(5(5(x1)))))))))) (18)
0(0(5(1(5(1(3(x1))))))) → 3(3(1(3(3(5(0(3(2(2(x1)))))))))) (19)
0(0(5(2(5(2(1(x1))))))) → 2(3(5(3(4(2(2(1(2(0(x1)))))))))) (20)
0(0(5(4(2(0(2(x1))))))) → 3(2(3(2(2(0(0(3(1(3(x1)))))))))) (21)
0(1(0(5(5(2(0(x1))))))) → 3(2(2(4(3(3(3(0(2(0(x1)))))))))) (22)
0(2(0(3(0(0(2(x1))))))) → 3(2(3(1(3(4(4(5(2(3(x1)))))))))) (23)
0(2(2(5(0(4(3(x1))))))) → 0(4(1(1(2(2(3(2(5(3(x1)))))))))) (24)
0(2(4(0(1(5(4(x1))))))) → 3(2(0(1(3(2(1(5(3(4(x1)))))))))) (25)
0(3(0(0(0(0(0(x1))))))) → 2(0(5(2(3(1(0(2(4(4(x1)))))))))) (26)
0(4(0(0(0(4(3(x1))))))) → 0(5(5(2(1(3(2(3(3(3(x1)))))))))) (27)
0(4(5(5(5(0(4(x1))))))) → 2(3(5(1(2(3(0(2(4(4(x1)))))))))) (28)
0(5(1(1(5(0(0(x1))))))) → 3(2(1(0(5(2(0(3(3(4(x1)))))))))) (29)
0(5(2(2(4(1(0(x1))))))) → 2(3(3(1(2(3(2(3(0(4(x1)))))))))) (30)
0(5(3(1(4(3(1(x1))))))) → 2(3(2(1(3(4(4(1(0(1(x1)))))))))) (31)
1(0(3(1(0(0(0(x1))))))) → 2(4(2(2(5(3(2(4(4(4(x1)))))))))) (32)
1(1(2(4(4(0(2(x1))))))) → 1(1(2(2(3(2(1(5(2(2(x1)))))))))) (33)
1(2(4(4(0(5(1(x1))))))) → 1(3(3(2(2(3(5(1(0(3(x1)))))))))) (34)
1(3(0(0(3(3(5(x1))))))) → 3(2(2(1(2(4(5(4(3(5(x1)))))))))) (35)
1(4(1(3(0(4(3(x1))))))) → 1(3(5(1(2(3(2(2(5(1(x1)))))))))) (36)
1(4(4(0(0(0(0(x1))))))) → 2(1(2(4(3(3(5(3(1(0(x1)))))))))) (37)
1(5(0(0(5(3(3(x1))))))) → 1(5(4(3(2(1(1(3(2(1(x1)))))))))) (38)
4(0(0(0(4(0(2(x1))))))) → 4(4(2(2(3(2(4(1(2(2(x1)))))))))) (39)
4(0(0(0(4(1(4(x1))))))) → 4(4(3(2(1(1(2(1(0(0(x1)))))))))) (40)
4(0(0(4(0(0(2(x1))))))) → 3(0(3(2(3(3(5(4(1(5(x1)))))))))) (41)
4(0(0(4(5(2(4(x1))))))) → 3(3(5(2(2(2(3(4(4(0(x1)))))))))) (42)
4(0(3(0(2(5(1(x1))))))) → 4(3(3(2(3(4(3(1(0(3(x1)))))))))) (43)
4(0(4(0(1(1(2(x1))))))) → 2(2(0(3(1(4(3(2(2(2(x1)))))))))) (44)
4(0(4(1(4(0(0(x1))))))) → 3(1(2(2(0(0(2(1(1(4(x1)))))))))) (45)
4(1(0(5(4(1(4(x1))))))) → 2(5(1(2(1(3(2(4(3(4(x1)))))))))) (46)
4(1(2(5(4(0(0(x1))))))) → 2(4(5(1(3(0(3(2(0(4(x1)))))))))) (47)
4(1(4(0(3(1(0(x1))))))) → 5(3(2(0(2(2(2(5(1(4(x1)))))))))) (48)
4(3(0(5(5(0(2(x1))))))) → 3(0(3(2(3(2(2(4(5(2(x1)))))))))) (49)
4(3(5(5(4(1(0(x1))))))) → 3(5(1(3(4(5(2(3(3(4(x1)))))))))) (50)
5(0(0(0(1(4(0(x1))))))) → 5(2(5(3(2(2(3(0(5(4(x1)))))))))) (51)
5(0(2(1(5(1(5(x1))))))) → 5(3(2(3(3(3(4(3(3(2(x1)))))))))) (52)
5(0(2(5(4(4(0(x1))))))) → 5(4(0(3(2(2(1(1(3(4(x1)))))))))) (53)
5(0(5(0(1(5(2(x1))))))) → 5(3(2(3(2(4(3(2(0(2(x1)))))))))) (54)
5(0(5(5(5(4(5(x1))))))) → 5(2(3(3(2(3(3(0(3(2(x1)))))))))) (55)
5(3(0(1(4(3(1(x1))))))) → 3(2(2(1(2(1(0(0(3(1(x1)))))))))) (56)
5(3(5(0(1(0(1(x1))))))) → 5(3(2(2(1(5(5(3(5(1(x1)))))))))) (57)
5(4(0(4(1(0(3(x1))))))) → 5(3(2(2(4(4(3(2(4(3(x1)))))))))) (58)

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 +
4
[4(x1)] = x1 +
3
[3(x1)] = x1 +
0
[2(x1)] = x1 +
0
[1(x1)] = x1 +
0
[0(x1)] = x1 +
7
all of the following rules can be deleted.
0(1(0(x1))) → 2(3(3(2(2(3(2(4(2(4(x1)))))))))) (1)
0(0(0(4(0(x1))))) → 3(4(4(4(1(2(2(0(3(4(x1)))))))))) (2)
0(4(4(1(1(x1))))) → 0(4(2(3(2(2(2(2(2(2(x1)))))))))) (3)
0(5(4(1(2(x1))))) → 3(2(1(3(3(4(2(3(2(1(x1)))))))))) (4)
1(0(0(3(0(x1))))) → 1(1(2(3(2(3(2(4(3(0(x1)))))))))) (5)
0(0(0(0(4(5(x1)))))) → 0(2(3(2(4(1(5(2(3(5(x1)))))))))) (6)
0(0(0(2(0(4(x1)))))) → 3(2(2(1(2(4(1(1(0(4(x1)))))))))) (7)
0(0(4(5(3(0(x1)))))) → 1(2(2(3(5(2(2(1(1(4(x1)))))))))) (8)
0(1(5(3(1(1(x1)))))) → 3(2(1(3(2(1(2(5(5(3(x1)))))))))) (9)
0(2(0(0(1(1(x1)))))) → 3(4(2(3(1(2(3(2(3(3(x1)))))))))) (10)
0(5(0(0(0(2(x1)))))) → 2(3(2(0(5(4(3(1(2(1(x1)))))))))) (11)
1(4(0(3(0(4(x1)))))) → 3(2(2(5(3(3(2(1(4(4(x1)))))))))) (12)
3(0(0(1(3(5(x1)))))) → 2(1(3(3(2(0(5(3(1(5(x1)))))))))) (13)
3(1(4(0(1(2(x1)))))) → 2(3(2(2(1(0(3(1(1(2(x1)))))))))) (14)
4(5(0(2(4(1(x1)))))) → 2(1(4(2(3(2(2(3(4(1(x1)))))))))) (16)
5(4(3(0(1(5(x1)))))) → 5(2(3(3(3(2(1(5(3(2(x1)))))))))) (17)
0(0(1(2(3(0(5(x1))))))) → 3(2(1(3(2(1(4(3(5(5(x1)))))))))) (18)
0(0(5(1(5(1(3(x1))))))) → 3(3(1(3(3(5(0(3(2(2(x1)))))))))) (19)
0(0(5(2(5(2(1(x1))))))) → 2(3(5(3(4(2(2(1(2(0(x1)))))))))) (20)
0(0(5(4(2(0(2(x1))))))) → 3(2(3(2(2(0(0(3(1(3(x1)))))))))) (21)
0(1(0(5(5(2(0(x1))))))) → 3(2(2(4(3(3(3(0(2(0(x1)))))))))) (22)
0(2(0(3(0(0(2(x1))))))) → 3(2(3(1(3(4(4(5(2(3(x1)))))))))) (23)
0(2(2(5(0(4(3(x1))))))) → 0(4(1(1(2(2(3(2(5(3(x1)))))))))) (24)
0(2(4(0(1(5(4(x1))))))) → 3(2(0(1(3(2(1(5(3(4(x1)))))))))) (25)
0(3(0(0(0(0(0(x1))))))) → 2(0(5(2(3(1(0(2(4(4(x1)))))))))) (26)
0(4(0(0(0(4(3(x1))))))) → 0(5(5(2(1(3(2(3(3(3(x1)))))))))) (27)
0(4(5(5(5(0(4(x1))))))) → 2(3(5(1(2(3(0(2(4(4(x1)))))))))) (28)
0(5(1(1(5(0(0(x1))))))) → 3(2(1(0(5(2(0(3(3(4(x1)))))))))) (29)
0(5(2(2(4(1(0(x1))))))) → 2(3(3(1(2(3(2(3(0(4(x1)))))))))) (30)
0(5(3(1(4(3(1(x1))))))) → 2(3(2(1(3(4(4(1(0(1(x1)))))))))) (31)
1(0(3(1(0(0(0(x1))))))) → 2(4(2(2(5(3(2(4(4(4(x1)))))))))) (32)
1(1(2(4(4(0(2(x1))))))) → 1(1(2(2(3(2(1(5(2(2(x1)))))))))) (33)
1(2(4(4(0(5(1(x1))))))) → 1(3(3(2(2(3(5(1(0(3(x1)))))))))) (34)
1(3(0(0(3(3(5(x1))))))) → 3(2(2(1(2(4(5(4(3(5(x1)))))))))) (35)
1(4(1(3(0(4(3(x1))))))) → 1(3(5(1(2(3(2(2(5(1(x1)))))))))) (36)
1(4(4(0(0(0(0(x1))))))) → 2(1(2(4(3(3(5(3(1(0(x1)))))))))) (37)
1(5(0(0(5(3(3(x1))))))) → 1(5(4(3(2(1(1(3(2(1(x1)))))))))) (38)
4(0(0(0(4(0(2(x1))))))) → 4(4(2(2(3(2(4(1(2(2(x1)))))))))) (39)
4(0(0(0(4(1(4(x1))))))) → 4(4(3(2(1(1(2(1(0(0(x1)))))))))) (40)
4(0(0(4(0(0(2(x1))))))) → 3(0(3(2(3(3(5(4(1(5(x1)))))))))) (41)
4(0(0(4(5(2(4(x1))))))) → 3(3(5(2(2(2(3(4(4(0(x1)))))))))) (42)
4(0(3(0(2(5(1(x1))))))) → 4(3(3(2(3(4(3(1(0(3(x1)))))))))) (43)
4(0(4(0(1(1(2(x1))))))) → 2(2(0(3(1(4(3(2(2(2(x1)))))))))) (44)
4(0(4(1(4(0(0(x1))))))) → 3(1(2(2(0(0(2(1(1(4(x1)))))))))) (45)
4(1(0(5(4(1(4(x1))))))) → 2(5(1(2(1(3(2(4(3(4(x1)))))))))) (46)
4(1(4(0(3(1(0(x1))))))) → 5(3(2(0(2(2(2(5(1(4(x1)))))))))) (48)
4(3(0(5(5(0(2(x1))))))) → 3(0(3(2(3(2(2(4(5(2(x1)))))))))) (49)
4(3(5(5(4(1(0(x1))))))) → 3(5(1(3(4(5(2(3(3(4(x1)))))))))) (50)
5(0(0(0(1(4(0(x1))))))) → 5(2(5(3(2(2(3(0(5(4(x1)))))))))) (51)
5(0(2(1(5(1(5(x1))))))) → 5(3(2(3(3(3(4(3(3(2(x1)))))))))) (52)
5(0(2(5(4(4(0(x1))))))) → 5(4(0(3(2(2(1(1(3(4(x1)))))))))) (53)
5(0(5(0(1(5(2(x1))))))) → 5(3(2(3(2(4(3(2(0(2(x1)))))))))) (54)
5(0(5(5(5(4(5(x1))))))) → 5(2(3(3(2(3(3(0(3(2(x1)))))))))) (55)
5(3(5(0(1(0(1(x1))))))) → 5(3(2(2(1(5(5(3(5(1(x1)))))))))) (57)
5(4(0(4(1(0(3(x1))))))) → 5(3(2(2(4(4(3(2(4(3(x1)))))))))) (58)

1.1 Dependency Pair Transformation

The following set of initial dependency pairs has been identified.
4#(1(2(5(4(0(0(x1))))))) → 5#(1(3(0(3(2(0(4(x1)))))))) (59)
4#(1(2(5(4(0(0(x1))))))) → 4#(x1) (60)
4#(1(2(5(4(0(0(x1))))))) → 4#(5(1(3(0(3(2(0(4(x1))))))))) (61)
4#(0(0(1(3(1(x1)))))) → 5#(4(4(1(x1)))) (62)
4#(0(0(1(3(1(x1)))))) → 4#(4(1(x1))) (63)
4#(0(0(1(3(1(x1)))))) → 4#(1(x1)) (64)

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 +
0
[3(x1)] = x1 +
0
[2(x1)] = x1 +
0
[1(x1)] = x1 +
0
[0(x1)] = x1 +
1
[5#(x1)] = x1 +
0
[4#(x1)] = x1 +
1
together with the usable rules
4(0(0(1(3(1(x1)))))) → 3(2(2(1(2(0(5(4(4(1(x1)))))))))) (15)
4(1(2(5(4(0(0(x1))))))) → 2(4(5(1(3(0(3(2(0(4(x1)))))))))) (47)
5(3(0(1(4(3(1(x1))))))) → 3(2(2(1(2(1(0(0(3(1(x1)))))))))) (56)
(w.r.t. the implicit argument filter of the reduction pair), the pairs
4#(1(2(5(4(0(0(x1))))))) → 5#(1(3(0(3(2(0(4(x1)))))))) (59)
4#(1(2(5(4(0(0(x1))))))) → 4#(x1) (60)
4#(0(0(1(3(1(x1)))))) → 5#(4(4(1(x1)))) (62)
4#(0(0(1(3(1(x1)))))) → 4#(4(1(x1))) (63)
4#(0(0(1(3(1(x1)))))) → 4#(1(x1)) (64)
and no rules could be deleted.

1.1.1.1 Dependency Graph Processor

The dependency pairs are split into 0 components.