Certification Problem

Input (TPDB SRS_Standard/ICFP_2010/212421)

The rewrite relation of the following TRS is considered.

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

Property / Task

Prove or disprove termination.

Answer / Result

Yes.

Proof (by NaTT @ termCOMP 2023)

1 Dependency Pair Transformation

The following set of initial dependency pairs has been identified.
0#(1(2(x1))) 0#(x1) (49)
5#(0(1(2(x1)))) 0#(2(1(3(3(x1))))) (50)
0#(1(2(2(x1)))) 0#(2(1(0(2(x1))))) (51)
0#(4(5(1(2(x1))))) 0#(2(5(1(1(x1))))) (52)
0#(1(2(x1))) 0#(2(2(1(x1)))) (53)
0#(1(5(2(x1)))) 0#(5(x1)) (54)
0#(1(5(2(x1)))) 5#(0(2(1(3(x1))))) (55)
5#(0(1(2(x1)))) 5#(0(2(1(3(3(x1)))))) (56)
0#(1(2(4(x1)))) 0#(2(2(1(1(x1))))) (57)
0#(1(2(2(x1)))) 0#(2(2(x1))) (58)
0#(2(4(2(x1)))) 5#(4(3(2(2(x1))))) (59)
0#(1(5(2(x1)))) 0#(2(1(3(x1)))) (60)
0#(1(2(x1))) 0#(2(2(1(4(x1))))) (61)
0#(3(5(1(2(x1))))) 0#(x1) (62)
0#(2(4(x1))) 0#(2(1(4(3(x1))))) (63)
0#(1(2(5(x1)))) 5#(2(1(0(x1)))) (64)
0#(1(2(5(x1)))) 5#(5(2(1(0(x1))))) (65)
0#(3(1(2(x1)))) 0#(2(1(3(2(x1))))) (66)
0#(3(1(2(x1)))) 0#(3(2(3(1(3(x1)))))) (67)
0#(1(1(2(5(x1))))) 5#(0(2(5(1(1(x1)))))) (68)
5#(0(1(2(x1)))) 0#(x1) (69)
0#(3(1(5(2(x1))))) 0#(3(2(5(1(2(x1)))))) (70)
0#(4(5(1(2(x1))))) 5#(5(x1)) (71)
0#(1(5(2(x1)))) 5#(x1) (72)
0#(1(1(2(5(x1))))) 5#(1(1(x1))) (73)
0#(0(4(2(x1)))) 0#(2(2(3(4(x1))))) (74)
0#(4(0(4(2(x1))))) 0#(0(2(2(x1)))) (75)
5#(0(1(2(2(x1))))) 5#(0(2(2(1(2(x1)))))) (76)
0#(4(2(1(4(x1))))) 0#(2(1(4(4(4(x1)))))) (77)
0#(3(4(2(x1)))) 0#(2(2(3(4(x1))))) (78)
0#(1(2(5(x1)))) 0#(x1) (79)
0#(4(1(2(5(x1))))) 0#(2(5(x1))) (80)
0#(3(5(1(2(x1))))) 5#(3(2(1(0(x1))))) (81)
0#(1(2(x1))) 5#(2(3(x1))) (82)
0#(1(2(4(x1)))) 5#(5(2(1(x1)))) (83)
0#(1(2(x1))) 0#(1(3(2(x1)))) (84)
0#(1(5(2(x1)))) 5#(5(0(2(1(3(x1)))))) (85)
0#(3(5(1(2(x1))))) 5#(5(3(2(1(0(x1)))))) (86)
0#(4(2(1(2(x1))))) 0#(2(x1)) (87)
0#(3(1(2(x1)))) 5#(0(2(3(x1)))) (88)
0#(4(2(x1))) 0#(2(3(x1))) (89)
0#(4(5(1(2(x1))))) 5#(x1) (90)
0#(1(2(4(x1)))) 5#(2(1(x1))) (91)
0#(4(0(4(2(x1))))) 0#(2(2(x1))) (92)
0#(1(4(2(x1)))) 0#(5(2(1(4(x1))))) (93)
0#(1(2(2(x1)))) 0#(2(x1)) (94)
0#(3(1(2(5(x1))))) 0#(5(x1)) (95)
0#(4(1(2(2(x1))))) 0#(2(2(3(x1)))) (96)
0#(1(2(4(x1)))) 0#(1(4(2(3(x1))))) (97)
0#(4(1(1(2(x1))))) 0#(2(1(x1))) (98)
0#(3(4(1(4(x1))))) 0#(5(3(1(4(4(x1)))))) (99)
0#(1(1(2(5(x1))))) 0#(2(5(1(1(x1))))) (100)
5#(0(1(2(2(x1))))) 0#(2(2(1(2(x1))))) (101)
0#(2(3(4(2(x1))))) 0#(x1) (102)
0#(3(1(2(x1)))) 0#(2(5(3(x1)))) (103)
0#(2(4(2(x1)))) 0#(5(4(3(2(2(x1)))))) (104)
5#(0(1(2(x1)))) 5#(0(x1)) (105)
0#(3(1(2(x1)))) 0#(2(2(1(x1)))) (106)
0#(1(5(2(x1)))) 0#(2(2(1(0(5(x1)))))) (107)
0#(4(2(x1))) 5#(2(x1)) (108)
0#(1(2(x1))) 0#(5(2(3(x1)))) (109)
0#(3(1(2(x1)))) 5#(3(x1)) (110)
0#(1(5(2(x1)))) 5#(0(2(3(x1)))) (111)
0#(4(2(x1))) 5#(5(2(x1))) (112)
0#(3(1(5(2(x1))))) 5#(1(2(x1))) (113)
5#(0(4(4(2(x1))))) 0#(5(2(5(4(4(x1)))))) (114)
0#(3(1(2(x1)))) 0#(2(3(x1))) (115)
5#(0(4(4(2(x1))))) 5#(4(4(x1))) (116)
0#(1(4(2(x1)))) 5#(2(1(4(x1)))) (117)
0#(4(2(x1))) 0#(5(5(2(x1)))) (118)
0#(1(5(2(x1)))) 0#(2(3(x1))) (119)
0#(3(1(2(x1)))) 0#(x1) (120)
0#(4(5(1(2(x1))))) 5#(1(1(x1))) (121)
0#(1(2(x1))) 5#(1(0(5(2(3(x1)))))) (122)
5#(0(2(4(2(x1))))) 5#(1(4(x1))) (123)
0#(4(2(5(2(x1))))) 5#(4(3(2(2(0(x1)))))) (124)
0#(1(2(x1))) 0#(2(1(3(x1)))) (125)
0#(1(2(x1))) 0#(2(1(0(x1)))) (126)
0#(4(5(1(2(x1))))) 0#(5(5(x1))) (127)
0#(3(4(1(4(x1))))) 5#(3(1(4(4(x1))))) (128)
5#(0(2(4(2(x1))))) 0#(2(2(5(1(4(x1)))))) (129)
0#(4(2(5(2(x1))))) 0#(x1) (130)
0#(1(2(4(x1)))) 0#(5(5(2(1(x1))))) (131)
0#(0(4(2(x1)))) 0#(0(2(2(3(4(x1)))))) (132)
5#(0(4(4(2(x1))))) 5#(2(5(4(4(x1))))) (133)

1.1 Dependency Graph Processor

The dependency pairs are split into 1 component.