Certification Problem

Input (TPDB SRS_Standard/ICFP_2010/212026)

The rewrite relation of the following TRS is considered.

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

Property / Task

Prove or disprove termination.

Answer / Result

Yes.

Proof (by matchbox @ termCOMP 2023)

1 Dependency Pair Transformation

The following set of initial dependency pairs has been identified.

There are 132 ruless (increase limit for explicit display).

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 +
1
[1(x1)] = x1 +
0
[0(x1)] = x1 +
0
[5#(x1)] = x1 +
1
[4#(x1)] = x1 +
0
[3#(x1)] = x1 +
0
[0#(x1)] = x1 +
0
together with the usable rules
0(0(0(x1))) → 0(0(1(0(1(x1))))) (1)
0(2(0(x1))) → 0(0(2(1(1(x1))))) (2)
0(3(2(x1))) → 2(1(3(0(x1)))) (3)
0(4(2(x1))) → 4(1(1(1(0(2(x1)))))) (4)
3(0(4(x1))) → 0(1(3(4(1(x1))))) (5)
3(2(0(x1))) → 0(2(1(3(x1)))) (6)
3(2(0(x1))) → 3(0(2(1(x1)))) (7)
3(2(0(x1))) → 0(1(2(1(3(x1))))) (8)
3(2(0(x1))) → 3(0(1(2(1(4(x1)))))) (9)
3(2(4(x1))) → 1(2(1(3(3(4(x1)))))) (10)
3(2(4(x1))) → 4(1(2(1(1(3(x1)))))) (11)
3(2(5(x1))) → 1(3(5(2(x1)))) (12)
3(2(5(x1))) → 3(1(4(5(2(x1))))) (13)
3(2(5(x1))) → 1(1(3(3(5(2(x1)))))) (14)
3(2(5(x1))) → 1(3(4(5(4(2(x1)))))) (15)
3(5(0(x1))) → 3(1(0(5(1(1(x1)))))) (16)
5(0(2(x1))) → 1(1(0(2(5(x1))))) (17)
5(0(4(x1))) → 0(5(4(1(1(x1))))) (18)
5(2(4(x1))) → 5(2(1(1(3(4(x1)))))) (19)
5(3(2(x1))) → 1(1(3(5(1(2(x1)))))) (20)
5(5(0(x1))) → 5(1(1(5(0(x1))))) (21)
0(0(2(4(x1)))) → 0(4(0(2(1(1(x1)))))) (22)
0(3(1(2(x1)))) → 1(1(3(0(2(x1))))) (23)
0(3(2(4(x1)))) → 2(1(3(0(4(4(x1)))))) (24)
0(4(3(2(x1)))) → 2(3(4(0(1(x1))))) (25)
3(0(2(4(x1)))) → 3(0(4(2(1(1(x1)))))) (26)
3(0(5(4(x1)))) → 3(4(0(5(1(3(x1)))))) (27)
3(1(0(0(x1)))) → 0(1(1(3(0(1(x1)))))) (28)
3(2(4(0(x1)))) → 3(1(0(4(2(x1))))) (29)
3(2(4(5(x1)))) → 3(1(4(2(1(5(x1)))))) (30)
3(2(5(4(x1)))) → 3(4(3(5(2(x1))))) (31)
3(5(3(2(x1)))) → 3(3(0(2(5(x1))))) (32)
4(0(4(2(x1)))) → 4(4(2(1(3(0(x1)))))) (33)
4(3(2(0(x1)))) → 0(2(1(3(4(x1))))) (34)
5(4(1(0(x1)))) → 5(1(1(4(0(1(x1)))))) (35)
5(5(0(0(x1)))) → 1(1(0(5(5(0(x1)))))) (36)
5(5(2(2(x1)))) → 5(5(1(1(2(2(x1)))))) (37)
0(0(1(2(4(x1))))) → 0(0(2(1(1(4(x1)))))) (38)
0(5(3(1(2(x1))))) → 5(2(1(3(1(0(x1)))))) (39)
0(5(5(1(0(x1))))) → 5(0(0(5(1(1(x1)))))) (40)
3(0(1(4(5(x1))))) → 3(4(0(5(1(1(x1)))))) (41)
3(0(3(1(2(x1))))) → 0(2(1(1(3(3(x1)))))) (42)
3(2(2(0(4(x1))))) → 3(2(2(0(1(4(x1)))))) (43)
3(5(1(4(0(x1))))) → 1(3(3(4(5(0(x1)))))) (44)
3(5(3(0(4(x1))))) → 3(0(1(5(3(4(x1)))))) (45)
5(1(0(2(4(x1))))) → 2(1(5(4(0(3(x1)))))) (46)
5(1(3(2(0(x1))))) → 2(3(5(0(1(1(x1)))))) (47)
5(4(2(0(2(x1))))) → 5(0(4(2(1(2(x1)))))) (48)
(w.r.t. the implicit argument filter of the reduction pair), the pairs
5#(5(2(2(x1)))) → 5#(1(1(2(2(x1))))) (50)
5#(5(0(0(x1)))) → 5#(0(x1)) (53)
5#(4(2(0(2(x1))))) → 4#(2(1(2(x1)))) (56)
5#(4(2(0(2(x1))))) → 0#(4(2(1(2(x1))))) (57)
5#(4(1(0(x1)))) → 4#(0(1(x1))) (59)
5#(4(1(0(x1)))) → 0#(1(x1)) (60)
5#(2(4(x1))) → 3#(4(x1)) (64)
5#(1(3(2(0(x1))))) → 5#(0(1(1(x1)))) (65)
5#(1(3(2(0(x1))))) → 3#(5(0(1(1(x1))))) (66)
5#(1(3(2(0(x1))))) → 0#(1(1(x1))) (67)
5#(1(0(2(4(x1))))) → 5#(4(0(3(x1)))) (68)
5#(1(0(2(4(x1))))) → 4#(0(3(x1))) (69)
5#(1(0(2(4(x1))))) → 3#(x1) (70)
5#(1(0(2(4(x1))))) → 0#(3(x1)) (71)
5#(0(4(x1))) → 4#(1(1(x1))) (73)
5#(0(2(x1))) → 5#(x1) (75)
4#(3(2(0(x1)))) → 4#(x1) (77)
4#(3(2(0(x1)))) → 3#(4(x1)) (78)
4#(0(4(2(x1)))) → 3#(0(x1)) (82)
4#(0(4(2(x1)))) → 0#(x1) (83)
3#(5(3(2(x1)))) → 5#(x1) (84)
3#(5(3(0(4(x1))))) → 3#(4(x1)) (89)
3#(2(5(x1))) → 4#(2(x1)) (103)
3#(2(4(x1))) → 3#(x1) (113)
3#(2(4(x1))) → 3#(4(x1)) (114)
3#(2(4(x1))) → 3#(3(4(x1))) (115)
3#(2(2(0(4(x1))))) → 0#(1(4(x1))) (122)
3#(2(0(x1))) → 4#(x1) (123)
3#(2(0(x1))) → 3#(x1) (124)
3#(0(5(4(x1)))) → 3#(x1) (136)
3#(0(3(1(2(x1))))) → 3#(x1) (142)
3#(0(3(1(2(x1))))) → 3#(3(x1)) (143)
0#(5(5(1(0(x1))))) → 5#(1(1(x1))) (152)
0#(5(5(1(0(x1))))) → 0#(5(1(1(x1)))) (154)
0#(5(5(1(0(x1))))) → 0#(0(5(1(1(x1))))) (155)
0#(5(3(1(2(x1))))) → 3#(1(0(x1))) (157)
0#(5(3(1(2(x1))))) → 0#(x1) (158)
0#(4(3(2(x1)))) → 4#(0(1(x1))) (159)
0#(4(3(2(x1)))) → 3#(4(0(1(x1)))) (160)
0#(4(3(2(x1)))) → 0#(1(x1)) (161)
0#(3(2(x1))) → 3#(0(x1)) (164)
0#(3(2(x1))) → 0#(x1) (165)
0#(3(2(4(x1)))) → 4#(4(x1)) (166)
0#(3(2(4(x1)))) → 3#(0(4(4(x1)))) (167)
0#(3(2(4(x1)))) → 0#(4(4(x1))) (168)
and no rules could be deleted.

1.1.1 Dependency Graph Processor

The dependency pairs are split into 1 component.