Certification Problem

Input (TPDB SRS_Standard/ICFP_2010/212094)

The rewrite relation of the following TRS is considered.

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

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

1.1 Dependency Graph Processor

The dependency pairs are split into 1 component.