Certification Problem

Input (TPDB SRS_Standard/ICFP_2010/96464)

The rewrite relation of the following TRS is considered.

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

1.1 Rule Removal

Using the linear polynomial interpretation over the naturals
[1(x1)] = 1 · x1 + 1
[0(x1)] = 1 · x1 + 1
[2(x1)] = 1 · x1 + 1
[4(x1)] = 1 · x1 + 1
[3(x1)] = 1 · x1 + 1
[5(x1)] = 1 · x1 + 1
all of the following rules can be deleted.
4(2(2(2(0(5(1(2(2(x1))))))))) → 0(5(5(4(0(0(2(x1))))))) (31)
4(2(5(3(0(2(0(3(2(0(x1)))))))))) → 0(4(5(4(4(5(4(0(2(x1))))))))) (32)
0(2(2(5(5(3(1(3(2(3(x1)))))))))) → 0(2(2(2(0(0(0(0(4(x1))))))))) (33)
1(2(1(3(1(1(1(4(4(0(3(3(x1)))))))))))) → 1(0(0(4(4(1(0(3(5(3(3(x1))))))))))) (39)
4(2(5(4(1(3(1(4(1(1(2(0(2(3(x1)))))))))))))) → 0(0(5(0(0(1(4(5(4(4(2(2(0(x1))))))))))))) (40)
3(5(3(2(0(1(1(4(0(1(1(2(0(4(3(x1))))))))))))))) → 3(1(1(4(4(0(3(0(3(0(0(5(4(2(x1)))))))))))))) (42)
5(2(1(5(2(1(1(1(3(2(3(3(3(0(5(0(x1)))))))))))))))) → 4(1(0(1(4(1(1(4(4(3(3(4(1(5(0(x1))))))))))))))) (44)
3(1(3(1(3(3(2(4(2(1(4(0(1(4(5(3(1(3(x1)))))))))))))))))) → 3(4(2(2(5(1(4(1(4(4(3(3(0(0(3(3(0(x1))))))))))))))))) (46)
3(4(2(3(0(3(2(0(0(5(5(4(1(3(1(2(5(0(5(2(x1)))))))))))))))))))) → 3(4(0(1(1(4(4(5(0(2(5(2(4(1(3(0(0(2(0(x1))))))))))))))))))) (47)
5(1(3(2(1(0(4(1(2(1(2(1(5(2(1(5(3(5(0(3(3(x1))))))))))))))))))))) → 5(1(3(5(5(5(0(2(4(0(0(5(2(5(4(5(5(0(2(0(x1)))))))))))))))))))) (50)

1.1.1 Dependency Pair Transformation

The following set of initial dependency pairs has been identified.

There are 134 ruless (increase limit for explicit display).

1.1.1.1 Dependency Graph Processor

The dependency pairs are split into 1 component.