Certification Problem

Input (TPDB SRS_Standard/ICFP_2010/68989)

The rewrite relation of the following TRS is considered.

0(0(1(2(1(x1))))) 1(0(1(1(0(x1))))) (1)
0(3(1(2(4(x1))))) 0(5(1(4(x1)))) (2)
4(3(0(1(1(x1))))) 5(4(5(2(x1)))) (3)
4(3(4(2(2(x1))))) 0(5(0(2(x1)))) (4)
2(1(1(4(0(2(x1)))))) 1(1(5(3(5(x1))))) (5)
2(2(4(1(3(4(2(x1))))))) 2(3(2(4(3(5(x1)))))) (6)
4(0(5(4(2(4(0(x1))))))) 0(1(2(2(2(2(0(x1))))))) (7)
4(2(4(2(5(1(0(1(5(x1))))))))) 5(0(3(1(0(2(2(5(x1)))))))) (8)
5(2(4(5(0(3(0(2(3(x1))))))))) 5(3(2(1(0(1(3(4(0(x1))))))))) (9)
1(4(1(3(2(3(3(1(2(1(x1)))))))))) 1(5(3(2(4(5(1(3(4(x1))))))))) (10)
4(0(4(2(3(4(5(1(1(5(1(x1))))))))))) 5(4(3(2(2(3(2(1(2(4(0(x1))))))))))) (11)
4(5(2(2(5(4(4(3(4(5(4(x1))))))))))) 5(2(3(2(5(0(0(0(5(4(x1)))))))))) (12)
4(3(4(4(0(3(0(3(2(3(2(1(x1)))))))))))) 5(5(5(5(5(0(2(2(4(4(2(0(x1)))))))))))) (13)
5(3(4(4(3(3(5(2(5(2(1(1(4(2(x1)))))))))))))) 5(4(0(5(2(5(5(3(1(0(3(3(5(x1))))))))))))) (14)
0(5(1(0(3(3(2(5(5(4(0(5(5(5(2(x1))))))))))))))) 1(0(1(3(4(5(3(3(3(2(2(3(3(5(2(x1))))))))))))))) (15)
4(5(3(2(1(1(5(2(2(3(4(3(2(3(1(x1))))))))))))))) 0(1(3(5(0(1(3(4(0(3(5(4(3(1(x1)))))))))))))) (16)
5(0(1(0(1(1(5(1(1(5(5(2(1(1(0(x1))))))))))))))) 5(1(5(1(1(1(3(0(3(3(3(3(1(0(x1)))))))))))))) (17)
5(3(0(4(4(1(1(5(3(4(1(1(2(3(2(x1))))))))))))))) 5(4(1(4(0(2(1(2(2(5(3(5(3(4(4(x1))))))))))))))) (18)
2(4(1(0(2(3(2(3(5(3(1(2(3(1(1(4(x1)))))))))))))))) 2(2(2(1(4(5(0(1(0(3(1(3(5(1(2(x1))))))))))))))) (19)
0(1(1(3(2(2(0(0(0(5(0(2(4(3(3(0(1(x1))))))))))))))))) 5(4(1(1(0(5(2(0(2(3(3(3(0(5(0(1(x1)))))))))))))))) (20)
2(1(1(5(3(1(3(4(3(5(3(3(2(4(3(1(4(x1))))))))))))))))) 1(0(1(0(0(2(1(3(2(2(0(3(0(5(2(4(x1)))))))))))))))) (21)
2(4(3(5(0(2(5(5(1(5(0(4(4(4(1(4(3(x1))))))))))))))))) 2(0(5(2(2(0(5(4(1(3(2(4(1(4(1(1(0(x1))))))))))))))))) (22)
0(4(5(4(5(0(2(3(1(2(4(5(3(5(0(4(3(3(2(x1))))))))))))))))))) 1(0(2(4(5(5(2(2(4(2(1(1(4(0(1(2(3(0(2(5(x1)))))))))))))))))))) (23)
4(5(1(0(2(0(5(4(5(4(4(2(5(5(2(3(5(4(2(3(x1)))))))))))))))))))) 0(5(5(2(0(5(2(4(2(5(2(5(2(0(1(5(2(3(3(0(x1)))))))))))))))))))) (24)
3(1(2(4(3(4(3(2(0(3(2(3(4(3(4(5(4(3(4(1(1(x1))))))))))))))))))))) 3(4(0(0(2(4(5(0(0(4(3(5(4(3(0(3(2(2(1(1(x1)))))))))))))))))))) (25)

Property / Task

Prove or disprove termination.

Answer / Result

Yes.

Proof (by AProVE @ termCOMP 2023)

1 String Reversal

Since only unary symbols occur, one can reverse all terms and obtains the TRS
1(2(1(0(0(x1))))) 0(1(1(0(1(x1))))) (26)
4(2(1(3(0(x1))))) 4(1(5(0(x1)))) (27)
1(1(0(3(4(x1))))) 2(5(4(5(x1)))) (28)
2(2(4(3(4(x1))))) 2(0(5(0(x1)))) (29)
2(0(4(1(1(2(x1)))))) 5(3(5(1(1(x1))))) (30)
2(4(3(1(4(2(2(x1))))))) 5(3(4(2(3(2(x1)))))) (31)
0(4(2(4(5(0(4(x1))))))) 0(2(2(2(2(1(0(x1))))))) (32)
5(1(0(1(5(2(4(2(4(x1))))))))) 5(2(2(0(1(3(0(5(x1)))))))) (33)
3(2(0(3(0(5(4(2(5(x1))))))))) 0(4(3(1(0(1(2(3(5(x1))))))))) (34)
1(2(1(3(3(2(3(1(4(1(x1)))))))))) 4(3(1(5(4(2(3(5(1(x1))))))))) (35)
1(5(1(1(5(4(3(2(4(0(4(x1))))))))))) 0(4(2(1(2(3(2(2(3(4(5(x1))))))))))) (36)
4(5(4(3(4(4(5(2(2(5(4(x1))))))))))) 4(5(0(0(0(5(2(3(2(5(x1)))))))))) (37)
1(2(3(2(3(0(3(0(4(4(3(4(x1)))))))))))) 0(2(4(4(2(2(0(5(5(5(5(5(x1)))))))))))) (38)
2(4(1(1(2(5(2(5(3(3(4(4(3(5(x1)))))))))))))) 5(3(3(0(1(3(5(5(2(5(0(4(5(x1))))))))))))) (39)
2(5(5(5(0(4(5(5(2(3(3(0(1(5(0(x1))))))))))))))) 2(5(3(3(2(2(3(3(3(5(4(3(1(0(1(x1))))))))))))))) (40)
1(3(2(3(4(3(2(2(5(1(1(2(3(5(4(x1))))))))))))))) 1(3(4(5(3(0(4(3(1(0(5(3(1(0(x1)))))))))))))) (41)
0(1(1(2(5(5(1(1(5(1(1(0(1(0(5(x1))))))))))))))) 0(1(3(3(3(3(0(3(1(1(1(5(1(5(x1)))))))))))))) (42)
2(3(2(1(1(4(3(5(1(1(4(4(0(3(5(x1))))))))))))))) 4(4(3(5(3(5(2(2(1(2(0(4(1(4(5(x1))))))))))))))) (43)
4(1(1(3(2(1(3(5(3(2(3(2(0(1(4(2(x1)))))))))))))))) 2(1(5(3(1(3(0(1(0(5(4(1(2(2(2(x1))))))))))))))) (44)
1(0(3(3(4(2(0(5(0(0(0(2(2(3(1(1(0(x1))))))))))))))))) 1(0(5(0(3(3(3(2(0(2(5(0(1(1(4(5(x1)))))))))))))))) (45)
4(1(3(4(2(3(3(5(3(4(3(1(3(5(1(1(2(x1))))))))))))))))) 4(2(5(0(3(0(2(2(3(1(2(0(0(1(0(1(x1)))))))))))))))) (46)
3(4(1(4(4(4(0(5(1(5(5(2(0(5(3(4(2(x1))))))))))))))))) 0(1(1(4(1(4(2(3(1(4(5(0(2(2(5(0(2(x1))))))))))))))))) (47)
2(3(3(4(0(5(3(5(4(2(1(3(2(0(5(4(5(4(0(x1))))))))))))))))))) 5(2(0(3(2(1(0(4(1(1(2(4(2(2(5(5(4(2(0(1(x1)))))))))))))))))))) (48)
3(2(4(5(3(2(5(5(2(4(4(5(4(5(0(2(0(1(5(4(x1)))))))))))))))))))) 0(3(3(2(5(1(0(2(5(2(5(2(4(2(5(0(2(5(5(0(x1)))))))))))))))))))) (49)
1(1(4(3(4(5(4(3(4(3(2(3(0(2(3(4(3(4(2(1(3(x1))))))))))))))))))))) 1(1(2(2(3(0(3(4(5(3(4(0(0(5(4(2(0(0(4(3(x1)))))))))))))))))))) (50)

1.1 Rule Removal

Using the linear polynomial interpretation over the naturals
[1(x1)] = 1 · x1 + 7
[2(x1)] = 1 · x1 + 7
[0(x1)] = 1 · x1 + 9
[4(x1)] = 1 · x1 + 8
[3(x1)] = 1 · x1 + 9
[5(x1)] = 1 · x1 + 8
all of the following rules can be deleted.
4(2(1(3(0(x1))))) 4(1(5(0(x1)))) (27)
1(1(0(3(4(x1))))) 2(5(4(5(x1)))) (28)
2(2(4(3(4(x1))))) 2(0(5(0(x1)))) (29)
2(0(4(1(1(2(x1)))))) 5(3(5(1(1(x1))))) (30)
2(4(3(1(4(2(2(x1))))))) 5(3(4(2(3(2(x1)))))) (31)
0(4(2(4(5(0(4(x1))))))) 0(2(2(2(2(1(0(x1))))))) (32)
5(1(0(1(5(2(4(2(4(x1))))))))) 5(2(2(0(1(3(0(5(x1)))))))) (33)
3(2(0(3(0(5(4(2(5(x1))))))))) 0(4(3(1(0(1(2(3(5(x1))))))))) (34)
1(2(1(3(3(2(3(1(4(1(x1)))))))))) 4(3(1(5(4(2(3(5(1(x1))))))))) (35)
4(5(4(3(4(4(5(2(2(5(4(x1))))))))))) 4(5(0(0(0(5(2(3(2(5(x1)))))))))) (37)
1(2(3(2(3(0(3(0(4(4(3(4(x1)))))))))))) 0(2(4(4(2(2(0(5(5(5(5(5(x1)))))))))))) (38)
2(4(1(1(2(5(2(5(3(3(4(4(3(5(x1)))))))))))))) 5(3(3(0(1(3(5(5(2(5(0(4(5(x1))))))))))))) (39)
1(3(2(3(4(3(2(2(5(1(1(2(3(5(4(x1))))))))))))))) 1(3(4(5(3(0(4(3(1(0(5(3(1(0(x1)))))))))))))) (41)
0(1(1(2(5(5(1(1(5(1(1(0(1(0(5(x1))))))))))))))) 0(1(3(3(3(3(0(3(1(1(1(5(1(5(x1)))))))))))))) (42)
4(1(1(3(2(1(3(5(3(2(3(2(0(1(4(2(x1)))))))))))))))) 2(1(5(3(1(3(0(1(0(5(4(1(2(2(2(x1))))))))))))))) (44)
1(0(3(3(4(2(0(5(0(0(0(2(2(3(1(1(0(x1))))))))))))))))) 1(0(5(0(3(3(3(2(0(2(5(0(1(1(4(5(x1)))))))))))))))) (45)
4(1(3(4(2(3(3(5(3(4(3(1(3(5(1(1(2(x1))))))))))))))))) 4(2(5(0(3(0(2(2(3(1(2(0(0(1(0(1(x1)))))))))))))))) (46)
3(4(1(4(4(4(0(5(1(5(5(2(0(5(3(4(2(x1))))))))))))))))) 0(1(1(4(1(4(2(3(1(4(5(0(2(2(5(0(2(x1))))))))))))))))) (47)
2(3(3(4(0(5(3(5(4(2(1(3(2(0(5(4(5(4(0(x1))))))))))))))))))) 5(2(0(3(2(1(0(4(1(1(2(4(2(2(5(5(4(2(0(1(x1)))))))))))))))))))) (48)
1(1(4(3(4(5(4(3(4(3(2(3(0(2(3(4(3(4(2(1(3(x1))))))))))))))))))))) 1(1(2(2(3(0(3(4(5(3(4(0(0(5(4(2(0(0(4(3(x1)))))))))))))))))))) (50)

1.1.1 Switch to Innermost Termination

The TRS is overlay and locally confluent:

10

Hence, it suffices to show innermost termination in the following.

1.1.1.1 Dependency Pair Transformation

The following set of initial dependency pairs has been identified.
1#(2(1(0(0(x1))))) 1#(1(0(1(x1)))) (51)
1#(2(1(0(0(x1))))) 1#(0(1(x1))) (52)
1#(2(1(0(0(x1))))) 1#(x1) (53)
1#(5(1(1(5(4(3(2(4(0(4(x1))))))))))) 2#(1(2(3(2(2(3(4(5(x1))))))))) (54)
1#(5(1(1(5(4(3(2(4(0(4(x1))))))))))) 1#(2(3(2(2(3(4(5(x1)))))))) (55)
1#(5(1(1(5(4(3(2(4(0(4(x1))))))))))) 2#(3(2(2(3(4(5(x1))))))) (56)
1#(5(1(1(5(4(3(2(4(0(4(x1))))))))))) 3#(2(2(3(4(5(x1)))))) (57)
1#(5(1(1(5(4(3(2(4(0(4(x1))))))))))) 2#(2(3(4(5(x1))))) (58)
1#(5(1(1(5(4(3(2(4(0(4(x1))))))))))) 2#(3(4(5(x1)))) (59)
1#(5(1(1(5(4(3(2(4(0(4(x1))))))))))) 3#(4(5(x1))) (60)
2#(5(5(5(0(4(5(5(2(3(3(0(1(5(0(x1))))))))))))))) 2#(5(3(3(2(2(3(3(3(5(4(3(1(0(1(x1))))))))))))))) (61)
2#(5(5(5(0(4(5(5(2(3(3(0(1(5(0(x1))))))))))))))) 3#(3(2(2(3(3(3(5(4(3(1(0(1(x1))))))))))))) (62)
2#(5(5(5(0(4(5(5(2(3(3(0(1(5(0(x1))))))))))))))) 3#(2(2(3(3(3(5(4(3(1(0(1(x1)))))))))))) (63)
2#(5(5(5(0(4(5(5(2(3(3(0(1(5(0(x1))))))))))))))) 2#(2(3(3(3(5(4(3(1(0(1(x1))))))))))) (64)
2#(5(5(5(0(4(5(5(2(3(3(0(1(5(0(x1))))))))))))))) 2#(3(3(3(5(4(3(1(0(1(x1)))))))))) (65)
2#(5(5(5(0(4(5(5(2(3(3(0(1(5(0(x1))))))))))))))) 3#(3(3(5(4(3(1(0(1(x1))))))))) (66)
2#(5(5(5(0(4(5(5(2(3(3(0(1(5(0(x1))))))))))))))) 3#(3(5(4(3(1(0(1(x1)))))))) (67)
2#(5(5(5(0(4(5(5(2(3(3(0(1(5(0(x1))))))))))))))) 3#(5(4(3(1(0(1(x1))))))) (68)
2#(5(5(5(0(4(5(5(2(3(3(0(1(5(0(x1))))))))))))))) 3#(1(0(1(x1)))) (69)
2#(5(5(5(0(4(5(5(2(3(3(0(1(5(0(x1))))))))))))))) 1#(0(1(x1))) (70)
2#(5(5(5(0(4(5(5(2(3(3(0(1(5(0(x1))))))))))))))) 1#(x1) (71)
2#(3(2(1(1(4(3(5(1(1(4(4(0(3(5(x1))))))))))))))) 3#(5(3(5(2(2(1(2(0(4(1(4(5(x1))))))))))))) (72)
2#(3(2(1(1(4(3(5(1(1(4(4(0(3(5(x1))))))))))))))) 3#(5(2(2(1(2(0(4(1(4(5(x1))))))))))) (73)
2#(3(2(1(1(4(3(5(1(1(4(4(0(3(5(x1))))))))))))))) 2#(2(1(2(0(4(1(4(5(x1))))))))) (74)
2#(3(2(1(1(4(3(5(1(1(4(4(0(3(5(x1))))))))))))))) 2#(1(2(0(4(1(4(5(x1)))))))) (75)
2#(3(2(1(1(4(3(5(1(1(4(4(0(3(5(x1))))))))))))))) 1#(2(0(4(1(4(5(x1))))))) (76)
2#(3(2(1(1(4(3(5(1(1(4(4(0(3(5(x1))))))))))))))) 2#(0(4(1(4(5(x1)))))) (77)
2#(3(2(1(1(4(3(5(1(1(4(4(0(3(5(x1))))))))))))))) 1#(4(5(x1))) (78)
3#(2(4(5(3(2(5(5(2(4(4(5(4(5(0(2(0(1(5(4(x1)))))))))))))))))))) 3#(3(2(5(1(0(2(5(2(5(2(4(2(5(0(2(5(5(0(x1))))))))))))))))))) (79)
3#(2(4(5(3(2(5(5(2(4(4(5(4(5(0(2(0(1(5(4(x1)))))))))))))))))))) 3#(2(5(1(0(2(5(2(5(2(4(2(5(0(2(5(5(0(x1)))))))))))))))))) (80)
3#(2(4(5(3(2(5(5(2(4(4(5(4(5(0(2(0(1(5(4(x1)))))))))))))))))))) 2#(5(1(0(2(5(2(5(2(4(2(5(0(2(5(5(0(x1))))))))))))))))) (81)
3#(2(4(5(3(2(5(5(2(4(4(5(4(5(0(2(0(1(5(4(x1)))))))))))))))))))) 1#(0(2(5(2(5(2(4(2(5(0(2(5(5(0(x1))))))))))))))) (82)
3#(2(4(5(3(2(5(5(2(4(4(5(4(5(0(2(0(1(5(4(x1)))))))))))))))))))) 2#(5(2(5(2(4(2(5(0(2(5(5(0(x1))))))))))))) (83)
3#(2(4(5(3(2(5(5(2(4(4(5(4(5(0(2(0(1(5(4(x1)))))))))))))))))))) 2#(5(2(4(2(5(0(2(5(5(0(x1))))))))))) (84)
3#(2(4(5(3(2(5(5(2(4(4(5(4(5(0(2(0(1(5(4(x1)))))))))))))))))))) 2#(4(2(5(0(2(5(5(0(x1))))))))) (85)
3#(2(4(5(3(2(5(5(2(4(4(5(4(5(0(2(0(1(5(4(x1)))))))))))))))))))) 2#(5(0(2(5(5(0(x1))))))) (86)
3#(2(4(5(3(2(5(5(2(4(4(5(4(5(0(2(0(1(5(4(x1)))))))))))))))))))) 2#(5(5(0(x1)))) (87)

1.1.1.1.1 Dependency Graph Processor

The dependency pairs are split into 1 component.