Certification Problem

Input (TPDB SRS_Standard/ICFP_2010/85675)

The rewrite relation of the following TRS is considered.

0(1(0(2(x1)))) → 3(4(4(x1))) (1)
2(0(0(3(x1)))) → 3(4(1(x1))) (2)
5(3(5(0(1(x1))))) → 5(2(3(0(x1)))) (3)
0(0(1(2(2(3(x1)))))) → 4(0(2(2(3(x1))))) (4)
0(5(3(1(4(3(x1)))))) → 1(4(0(5(2(x1))))) (5)
2(0(4(3(5(3(3(1(x1)))))))) → 2(1(2(2(2(0(3(1(x1)))))))) (6)
2(4(0(2(3(0(0(2(x1)))))))) → 2(4(0(3(4(4(2(x1))))))) (7)
2(1(4(0(4(1(5(0(2(x1))))))))) → 3(3(2(2(3(0(3(3(4(x1))))))))) (8)
5(0(4(0(0(1(3(5(0(x1))))))))) → 5(0(0(0(3(5(5(3(2(x1))))))))) (9)
5(3(5(4(4(2(2(2(1(x1))))))))) → 5(0(3(0(0(4(5(2(1(x1))))))))) (10)
2(0(4(2(3(3(3(5(4(2(x1)))))))))) → 3(4(4(2(0(2(3(3(5(2(x1)))))))))) (11)
4(4(3(2(0(1(1(4(0(2(x1)))))))))) → 2(3(4(0(0(1(2(1(3(4(x1)))))))))) (12)
1(5(5(2(4(2(4(0(3(3(0(x1))))))))))) → 1(3(5(4(5(5(5(3(2(4(2(x1))))))))))) (13)
2(5(5(0(1(1(5(1(4(2(3(3(x1)))))))))))) → 4(2(3(0(3(0(0(2(3(2(3(x1))))))))))) (14)
3(0(2(4(5(2(1(2(0(2(5(1(x1)))))))))))) → 3(1(3(0(5(0(2(2(4(0(4(x1))))))))))) (15)
3(3(3(4(3(0(0(4(4(5(0(5(x1)))))))))))) → 3(1(5(5(0(3(1(0(5(2(5(x1))))))))))) (16)
0(5(3(5(1(0(5(1(2(4(4(5(0(x1))))))))))))) → 0(3(1(0(4(5(5(2(5(5(2(4(0(x1))))))))))))) (17)
2(5(2(4(1(5(3(3(1(0(2(5(3(x1))))))))))))) → 2(3(1(3(3(5(1(2(0(3(2(3(0(x1))))))))))))) (18)
2(5(5(3(5(3(1(3(1(2(5(0(0(x1))))))))))))) → 2(5(5(5(0(4(5(5(1(1(3(2(2(x1))))))))))))) (19)
3(5(3(3(3(2(3(0(2(4(2(1(3(x1))))))))))))) → 5(1(3(1(4(0(0(1(5(0(4(5(x1)))))))))))) (20)
2(0(3(0(2(3(1(0(0(3(0(3(5(3(x1)))))))))))))) → 3(4(0(2(1(4(4(4(0(5(5(5(2(x1))))))))))))) (21)
3(5(3(3(1(2(4(0(4(3(5(4(2(1(x1)))))))))))))) → 5(2(0(1(4(4(3(3(3(5(3(4(1(2(x1)))))))))))))) (22)
3(5(5(0(2(2(0(0(3(5(1(3(3(5(3(5(x1)))))))))))))))) → 3(1(4(2(0(3(4(0(4(1(5(4(2(5(x1)))))))))))))) (23)
5(5(3(5(0(5(3(5(2(3(1(3(3(4(0(2(x1)))))))))))))))) → 5(2(4(2(4(4(0(5(5(1(3(4(0(4(0(x1))))))))))))))) (24)
2(2(1(3(4(0(1(1(4(2(2(2(0(4(4(2(4(4(x1)))))))))))))))))) → 4(3(4(4(5(5(0(1(0(1(3(2(5(5(1(1(4(4(x1)))))))))))))))))) (25)
0(3(0(2(2(2(5(5(2(4(4(3(5(1(0(0(0(3(3(x1))))))))))))))))))) → 3(3(1(3(4(5(5(2(0(4(2(3(1(1(5(1(2(3(x1)))))))))))))))))) (26)
3(3(4(3(3(4(3(0(2(5(3(1(4(5(2(5(2(4(3(5(x1)))))))))))))))))))) → 5(2(5(0(5(0(1(1(1(3(2(4(4(0(3(2(4(1(5(x1))))))))))))))))))) (27)
2(1(5(5(3(0(1(3(3(3(1(2(0(5(2(0(3(5(2(2(2(x1))))))))))))))))))))) → 3(0(3(3(4(0(1(4(5(1(3(3(3(4(1(2(1(3(2(1(x1)))))))))))))))))))) (28)
5(1(3(4(0(3(0(2(5(3(0(2(2(0(1(2(3(3(4(1(1(x1))))))))))))))))))))) → 5(0(1(1(4(2(0(1(1(2(1(4(0(4(3(1(2(0(0(2(x1)))))))))))))))))))) (29)
5(4(0(0(3(2(5(3(0(3(0(4(5(0(4(4(5(1(1(1(3(x1))))))))))))))))))))) → 5(2(5(1(0(3(5(3(2(1(0(0(5(5(4(5(4(4(2(2(2(x1))))))))))))))))))))) (30)

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
2(0(1(0(x1)))) → 4(4(3(x1))) (31)
3(0(0(2(x1)))) → 1(4(3(x1))) (32)
1(0(5(3(5(x1))))) → 0(3(2(5(x1)))) (33)
3(2(2(1(0(0(x1)))))) → 3(2(2(0(4(x1))))) (34)
3(4(1(3(5(0(x1)))))) → 2(5(0(4(1(x1))))) (35)
1(3(3(5(3(4(0(2(x1)))))))) → 1(3(0(2(2(2(1(2(x1)))))))) (36)
2(0(0(3(2(0(4(2(x1)))))))) → 2(4(4(3(0(4(2(x1))))))) (37)
2(0(5(1(4(0(4(1(2(x1))))))))) → 4(3(3(0(3(2(2(3(3(x1))))))))) (38)
0(5(3(1(0(0(4(0(5(x1))))))))) → 2(3(5(5(3(0(0(0(5(x1))))))))) (39)
1(2(2(2(4(4(5(3(5(x1))))))))) → 1(2(5(4(0(0(3(0(5(x1))))))))) (40)
2(4(5(3(3(3(2(4(0(2(x1)))))))))) → 2(5(3(3(2(0(2(4(4(3(x1)))))))))) (41)
2(0(4(1(1(0(2(3(4(4(x1)))))))))) → 4(3(1(2(1(0(0(4(3(2(x1)))))))))) (42)
0(3(3(0(4(2(4(2(5(5(1(x1))))))))))) → 2(4(2(3(5(5(5(4(5(3(1(x1))))))))))) (43)
3(3(2(4(1(5(1(1(0(5(5(2(x1)))))))))))) → 3(2(3(2(0(0(3(0(3(2(4(x1))))))))))) (44)
1(5(2(0(2(1(2(5(4(2(0(3(x1)))))))))))) → 4(0(4(2(2(0(5(0(3(1(3(x1))))))))))) (45)
5(0(5(4(4(0(0(3(4(3(3(3(x1)))))))))))) → 5(2(5(0(1(3(0(5(5(1(3(x1))))))))))) (46)
0(5(4(4(2(1(5(0(1(5(3(5(0(x1))))))))))))) → 0(4(2(5(5(2(5(5(4(0(1(3(0(x1))))))))))))) (47)
3(5(2(0(1(3(3(5(1(4(2(5(2(x1))))))))))))) → 0(3(2(3(0(2(1(5(3(3(1(3(2(x1))))))))))))) (48)
0(0(5(2(1(3(1(3(5(3(5(5(2(x1))))))))))))) → 2(2(3(1(1(5(5(4(0(5(5(5(2(x1))))))))))))) (49)
3(1(2(4(2(0(3(2(3(3(3(5(3(x1))))))))))))) → 5(4(0(5(1(0(0(4(1(3(1(5(x1)))))))))))) (50)
3(5(3(0(3(0(0(1(3(2(0(3(0(2(x1)))))))))))))) → 2(5(5(5(0(4(4(4(1(2(0(4(3(x1))))))))))))) (51)
1(2(4(5(3(4(0(4(2(1(3(3(5(3(x1)))))))))))))) → 2(1(4(3(5(3(3(3(4(4(1(0(2(5(x1)))))))))))))) (52)
5(3(5(3(3(1(5(3(0(0(2(2(0(5(5(3(x1)))))))))))))))) → 5(2(4(5(1(4(0(4(3(0(2(4(1(3(x1)))))))))))))) (53)
2(0(4(3(3(1(3(2(5(3(5(0(5(3(5(5(x1)))))))))))))))) → 0(4(0(4(3(1(5(5(0(4(4(2(4(2(5(x1))))))))))))))) (54)
4(4(2(4(4(0(2(2(2(4(1(1(0(4(3(1(2(2(x1)))))))))))))))))) → 4(4(1(1(5(5(2(3(1(0(1(0(5(5(4(4(3(4(x1)))))))))))))))))) (55)
3(3(0(0(0(1(5(3(4(4(2(5(5(2(2(2(0(3(0(x1))))))))))))))))))) → 3(2(1(5(1(1(3(2(4(0(2(5(5(4(3(1(3(3(x1)))))))))))))))))) (56)
5(3(4(2(5(2(5(4(1(3(5(2(0(3(4(3(3(4(3(3(x1)))))))))))))))))))) → 5(1(4(2(3(0(4(4(2(3(1(1(1(0(5(0(5(2(5(x1))))))))))))))))))) (57)
2(2(2(5(3(0(2(5(0(2(1(3(3(3(1(0(3(5(5(1(2(x1))))))))))))))))))))) → 1(2(3(1(2(1(4(3(3(3(1(5(4(1(0(4(3(3(0(3(x1)))))))))))))))))))) (58)
1(1(4(3(3(2(1(0(2(2(0(3(5(2(0(3(0(4(3(1(5(x1))))))))))))))))))))) → 2(0(0(2(1(3(4(0(4(1(2(1(1(0(2(4(1(1(0(5(x1)))))))))))))))))))) (59)
3(1(1(1(5(4(4(0(5(4(0(3(0(3(5(2(3(0(0(4(5(x1))))))))))))))))))))) → 2(2(2(4(4(5(4(5(5(0(0(1(2(3(5(3(0(1(5(2(5(x1))))))))))))))))))))) (60)

1.1 Rule Removal

Using the linear polynomial interpretation over the naturals
[2(x1)] = 1 · x1 + 120
[0(x1)] = 1 · x1 + 130
[1(x1)] = 1 · x1 + 129
[4(x1)] = 1 · x1 + 151
[3(x1)] = 1 · x1 + 132
[5(x1)] = 1 · x1 + 122
all of the following rules can be deleted.
2(0(1(0(x1)))) → 4(4(3(x1))) (31)
3(0(0(2(x1)))) → 1(4(3(x1))) (32)
1(0(5(3(5(x1))))) → 0(3(2(5(x1)))) (33)
3(2(2(1(0(0(x1)))))) → 3(2(2(0(4(x1))))) (34)
3(4(1(3(5(0(x1)))))) → 2(5(0(4(1(x1))))) (35)
1(3(3(5(3(4(0(2(x1)))))))) → 1(3(0(2(2(2(1(2(x1)))))))) (36)
2(0(0(3(2(0(4(2(x1)))))))) → 2(4(4(3(0(4(2(x1))))))) (37)
2(0(5(1(4(0(4(1(2(x1))))))))) → 4(3(3(0(3(2(2(3(3(x1))))))))) (38)
0(5(3(1(0(0(4(0(5(x1))))))))) → 2(3(5(5(3(0(0(0(5(x1))))))))) (39)
1(2(2(2(4(4(5(3(5(x1))))))))) → 1(2(5(4(0(0(3(0(5(x1))))))))) (40)
2(0(4(1(1(0(2(3(4(4(x1)))))))))) → 4(3(1(2(1(0(0(4(3(2(x1)))))))))) (42)
0(3(3(0(4(2(4(2(5(5(1(x1))))))))))) → 2(4(2(3(5(5(5(4(5(3(1(x1))))))))))) (43)
3(3(2(4(1(5(1(1(0(5(5(2(x1)))))))))))) → 3(2(3(2(0(0(3(0(3(2(4(x1))))))))))) (44)
1(5(2(0(2(1(2(5(4(2(0(3(x1)))))))))))) → 4(0(4(2(2(0(5(0(3(1(3(x1))))))))))) (45)
5(0(5(4(4(0(0(3(4(3(3(3(x1)))))))))))) → 5(2(5(0(1(3(0(5(5(1(3(x1))))))))))) (46)
0(5(4(4(2(1(5(0(1(5(3(5(0(x1))))))))))))) → 0(4(2(5(5(2(5(5(4(0(1(3(0(x1))))))))))))) (47)
3(5(2(0(1(3(3(5(1(4(2(5(2(x1))))))))))))) → 0(3(2(3(0(2(1(5(3(3(1(3(2(x1))))))))))))) (48)
0(0(5(2(1(3(1(3(5(3(5(5(2(x1))))))))))))) → 2(2(3(1(1(5(5(4(0(5(5(5(2(x1))))))))))))) (49)
3(1(2(4(2(0(3(2(3(3(3(5(3(x1))))))))))))) → 5(4(0(5(1(0(0(4(1(3(1(5(x1)))))))))))) (50)
3(5(3(0(3(0(0(1(3(2(0(3(0(2(x1)))))))))))))) → 2(5(5(5(0(4(4(4(1(2(0(4(3(x1))))))))))))) (51)
5(3(5(3(3(1(5(3(0(0(2(2(0(5(5(3(x1)))))))))))))))) → 5(2(4(5(1(4(0(4(3(0(2(4(1(3(x1)))))))))))))) (53)
2(0(4(3(3(1(3(2(5(3(5(0(5(3(5(5(x1)))))))))))))))) → 0(4(0(4(3(1(5(5(0(4(4(2(4(2(5(x1))))))))))))))) (54)
4(4(2(4(4(0(2(2(2(4(1(1(0(4(3(1(2(2(x1)))))))))))))))))) → 4(4(1(1(5(5(2(3(1(0(1(0(5(5(4(4(3(4(x1)))))))))))))))))) (55)
3(3(0(0(0(1(5(3(4(4(2(5(5(2(2(2(0(3(0(x1))))))))))))))))))) → 3(2(1(5(1(1(3(2(4(0(2(5(5(4(3(1(3(3(x1)))))))))))))))))) (56)
5(3(4(2(5(2(5(4(1(3(5(2(0(3(4(3(3(4(3(3(x1)))))))))))))))))))) → 5(1(4(2(3(0(4(4(2(3(1(1(1(0(5(0(5(2(5(x1))))))))))))))))))) (57)
2(2(2(5(3(0(2(5(0(2(1(3(3(3(1(0(3(5(5(1(2(x1))))))))))))))))))))) → 1(2(3(1(2(1(4(3(3(3(1(5(4(1(0(4(3(3(0(3(x1)))))))))))))))))))) (58)
1(1(4(3(3(2(1(0(2(2(0(3(5(2(0(3(0(4(3(1(5(x1))))))))))))))))))))) → 2(0(0(2(1(3(4(0(4(1(2(1(1(0(2(4(1(1(0(5(x1)))))))))))))))))))) (59)
3(1(1(1(5(4(4(0(5(4(0(3(0(3(5(2(3(0(0(4(5(x1))))))))))))))))))))) → 2(2(2(4(4(5(4(5(5(0(0(1(2(3(5(3(0(1(5(2(5(x1))))))))))))))))))))) (60)

1.1.1 Dependency Pair Transformation

The following set of initial dependency pairs has been identified.
2#(4(5(3(3(3(2(4(0(2(x1)))))))))) → 2#(5(3(3(2(0(2(4(4(3(x1)))))))))) (61)
2#(4(5(3(3(3(2(4(0(2(x1)))))))))) → 2#(0(2(4(4(3(x1)))))) (62)
2#(4(5(3(3(3(2(4(0(2(x1)))))))))) → 2#(4(4(3(x1)))) (63)
1#(2(4(5(3(4(0(4(2(1(3(3(5(3(x1)))))))))))))) → 2#(1(4(3(5(3(3(3(4(4(1(0(2(5(x1)))))))))))))) (64)
1#(2(4(5(3(4(0(4(2(1(3(3(5(3(x1)))))))))))))) → 1#(4(3(5(3(3(3(4(4(1(0(2(5(x1))))))))))))) (65)
1#(2(4(5(3(4(0(4(2(1(3(3(5(3(x1)))))))))))))) → 1#(0(2(5(x1)))) (66)
1#(2(4(5(3(4(0(4(2(1(3(3(5(3(x1)))))))))))))) → 2#(5(x1)) (67)

1.1.1.1 Dependency Graph Processor

The dependency pairs are split into 0 components.