Certification Problem

Input (TPDB SRS_Standard/ICFP_2010/212693)

The rewrite relation of the following TRS is considered.

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

Property / Task

Prove or disprove termination.

Answer / Result

Yes.

Proof (by ttt2 @ termCOMP 2023)

1 String Reversal

Since only unary symbols occur, one can reverse all terms and obtains the TRS
2(1(0(x1))) 1(2(0(0(x1)))) (50)
2(1(0(x1))) 3(1(2(0(x1)))) (51)
2(1(0(x1))) 4(4(1(2(0(0(x1)))))) (52)
1(3(0(x1))) 0(4(3(1(0(x1))))) (53)
1(3(0(x1))) 4(4(3(1(0(x1))))) (54)
1(3(0(x1))) 0(4(4(4(3(1(x1)))))) (55)
2(3(0(x1))) 3(1(2(0(x1)))) (56)
2(3(0(x1))) 4(3(2(0(x1)))) (57)
2(3(0(x1))) 3(4(2(0(0(x1))))) (58)
2(3(0(x1))) 3(4(1(2(0(x1))))) (59)
2(3(0(x1))) 3(3(4(2(0(x1))))) (60)
2(3(0(x1))) 4(3(3(1(2(0(x1)))))) (61)
2(3(0(x1))) 5(5(4(3(2(0(x1)))))) (62)
2(3(0(x1))) 0(4(3(4(4(2(x1)))))) (63)
1(4(0(x1))) 4(4(1(0(x1)))) (64)
1(4(0(x1))) 4(1(2(0(x1)))) (65)
2(4(0(x1))) 4(1(2(0(x1)))) (66)
2(4(0(x1))) 4(3(2(0(x1)))) (67)
2(4(0(x1))) 3(4(2(0(x1)))) (68)
1(0(2(x1))) 1(2(0(5(x1)))) (69)
1(3(2(x1))) 2(5(3(1(x1)))) (70)
1(3(2(x1))) 5(3(1(2(0(x1))))) (71)
1(3(2(x1))) 2(5(3(4(1(x1))))) (72)
1(0(2(0(x1)))) 1(2(0(0(5(x1))))) (73)
1(1(3(0(x1)))) 4(3(1(4(1(0(x1)))))) (74)
1(2(3(0(x1)))) 1(2(4(3(0(0(x1)))))) (75)
2(2(3(0(x1)))) 2(2(0(4(3(1(x1)))))) (76)
2(1(4(0(x1)))) 5(2(0(4(1(x1))))) (77)
2(3(4(0(x1)))) 0(0(4(4(3(2(x1)))))) (78)
1(3(5(0(x1)))) 4(5(3(4(1(0(x1)))))) (79)
1(3(5(0(x1)))) 0(4(3(5(1(0(x1)))))) (80)
2(3(5(0(x1)))) 3(5(4(2(0(x1))))) (81)
2(3(5(0(x1)))) 3(3(5(2(0(x1))))) (82)
1(3(0(2(x1)))) 2(5(3(1(0(2(x1)))))) (83)
1(4(0(2(x1)))) 5(4(1(0(2(x1))))) (84)
2(3(5(2(x1)))) 3(3(2(5(2(x1))))) (85)
2(4(5(2(x1)))) 4(2(5(2(0(x1))))) (86)
1(2(3(0(0(x1))))) 2(5(3(1(0(0(x1)))))) (87)
2(3(0(1(0(x1))))) 0(2(3(4(1(0(x1)))))) (88)
2(3(0(1(0(x1))))) 5(0(0(1(3(2(x1)))))) (89)
1(5(2(3(0(x1))))) 3(3(1(5(2(0(x1)))))) (90)
2(1(1(5(0(x1))))) 5(1(1(4(2(0(x1)))))) (91)
2(2(1(5(0(x1))))) 2(1(2(5(2(0(x1)))))) (92)
1(2(3(5(0(x1))))) 5(2(4(3(1(0(x1)))))) (93)
2(3(5(5(0(x1))))) 5(3(1(5(2(0(x1)))))) (94)
1(1(3(0(2(x1))))) 4(3(1(0(1(2(x1)))))) (95)
1(3(0(2(2(x1))))) 2(5(2(0(3(1(x1)))))) (96)
1(5(0(2(2(x1))))) 1(5(1(2(0(2(x1)))))) (97)
1(4(5(5(2(x1))))) 4(3(1(2(5(5(x1)))))) (98)

1.1 Dependency Pair Transformation

The following set of initial dependency pairs has been identified.
2#(1(0(x1))) 2#(0(0(x1))) (99)
2#(1(0(x1))) 1#(2(0(0(x1)))) (100)
2#(1(0(x1))) 2#(0(x1)) (101)
2#(1(0(x1))) 1#(2(0(x1))) (102)
1#(3(0(x1))) 1#(0(x1)) (103)
1#(3(0(x1))) 1#(x1) (104)
2#(3(0(x1))) 2#(0(x1)) (105)
2#(3(0(x1))) 1#(2(0(x1))) (106)
2#(3(0(x1))) 2#(0(0(x1))) (107)
2#(3(0(x1))) 2#(x1) (108)
1#(4(0(x1))) 1#(0(x1)) (109)
1#(4(0(x1))) 2#(0(x1)) (110)
1#(4(0(x1))) 1#(2(0(x1))) (111)
2#(4(0(x1))) 2#(0(x1)) (112)
2#(4(0(x1))) 1#(2(0(x1))) (113)
1#(0(2(x1))) 2#(0(5(x1))) (114)
1#(0(2(x1))) 1#(2(0(5(x1)))) (115)
1#(3(2(x1))) 1#(x1) (116)
1#(3(2(x1))) 2#(5(3(1(x1)))) (117)
1#(3(2(x1))) 2#(0(x1)) (118)
1#(3(2(x1))) 1#(2(0(x1))) (119)
1#(3(2(x1))) 2#(5(3(4(1(x1))))) (120)
1#(0(2(0(x1)))) 2#(0(0(5(x1)))) (121)
1#(0(2(0(x1)))) 1#(2(0(0(5(x1))))) (122)
1#(1(3(0(x1)))) 1#(0(x1)) (123)
1#(1(3(0(x1)))) 1#(4(1(0(x1)))) (124)
1#(2(3(0(x1)))) 2#(4(3(0(0(x1))))) (125)
1#(2(3(0(x1)))) 1#(2(4(3(0(0(x1)))))) (126)
2#(2(3(0(x1)))) 1#(x1) (127)
2#(2(3(0(x1)))) 2#(0(4(3(1(x1))))) (128)
2#(2(3(0(x1)))) 2#(2(0(4(3(1(x1)))))) (129)
2#(1(4(0(x1)))) 1#(x1) (130)
2#(1(4(0(x1)))) 2#(0(4(1(x1)))) (131)
2#(3(4(0(x1)))) 2#(x1) (132)
1#(3(5(0(x1)))) 1#(0(x1)) (133)
2#(3(5(0(x1)))) 2#(0(x1)) (134)
1#(3(0(2(x1)))) 1#(0(2(x1))) (135)
1#(3(0(2(x1)))) 2#(5(3(1(0(2(x1)))))) (136)
1#(4(0(2(x1)))) 1#(0(2(x1))) (137)
2#(3(5(2(x1)))) 2#(5(2(x1))) (138)
2#(4(5(2(x1)))) 2#(0(x1)) (139)
2#(4(5(2(x1)))) 2#(5(2(0(x1)))) (140)
1#(2(3(0(0(x1))))) 1#(0(0(x1))) (141)
1#(2(3(0(0(x1))))) 2#(5(3(1(0(0(x1)))))) (142)
2#(3(0(1(0(x1))))) 2#(3(4(1(0(x1))))) (143)
2#(3(0(1(0(x1))))) 2#(x1) (144)
2#(3(0(1(0(x1))))) 1#(3(2(x1))) (145)
1#(5(2(3(0(x1))))) 2#(0(x1)) (146)
1#(5(2(3(0(x1))))) 1#(5(2(0(x1)))) (147)
2#(1(1(5(0(x1))))) 2#(0(x1)) (148)
2#(1(1(5(0(x1))))) 1#(4(2(0(x1)))) (149)
2#(1(1(5(0(x1))))) 1#(1(4(2(0(x1))))) (150)
2#(2(1(5(0(x1))))) 2#(0(x1)) (151)
2#(2(1(5(0(x1))))) 2#(5(2(0(x1)))) (152)
2#(2(1(5(0(x1))))) 1#(2(5(2(0(x1))))) (153)
2#(2(1(5(0(x1))))) 2#(1(2(5(2(0(x1)))))) (154)
1#(2(3(5(0(x1))))) 1#(0(x1)) (155)
1#(2(3(5(0(x1))))) 2#(4(3(1(0(x1))))) (156)
2#(3(5(5(0(x1))))) 2#(0(x1)) (157)
2#(3(5(5(0(x1))))) 1#(5(2(0(x1)))) (158)
1#(1(3(0(2(x1))))) 1#(2(x1)) (159)
1#(1(3(0(2(x1))))) 1#(0(1(2(x1)))) (160)
1#(3(0(2(2(x1))))) 1#(x1) (161)
1#(3(0(2(2(x1))))) 2#(0(3(1(x1)))) (162)
1#(3(0(2(2(x1))))) 2#(5(2(0(3(1(x1)))))) (163)
1#(5(0(2(2(x1))))) 2#(0(2(x1))) (164)
1#(5(0(2(2(x1))))) 1#(2(0(2(x1)))) (165)
1#(5(0(2(2(x1))))) 1#(5(1(2(0(2(x1)))))) (166)
1#(4(5(5(2(x1))))) 2#(5(5(x1))) (167)
1#(4(5(5(2(x1))))) 1#(2(5(5(x1)))) (168)

1.1.1 Dependency Graph Processor

The dependency pairs are split into 2 components.