Certification Problem

Input (TPDB SRS_Standard/ICFP_2010/96485)

The rewrite relation of the following TRS is considered.

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

1.1 Rule Removal

Using the linear polynomial interpretation over the naturals
[2(x1)] = 1 · x1 + 14
[0(x1)] = 1 · x1 + 16
[1(x1)] = 1 · x1 + 14
[3(x1)] = 1 · x1 + 27
[4(x1)] = 1 · x1 + 18
[5(x1)] = 1 · x1 + 10
all of the following rules can be deleted.
2(0(1(0(x1)))) → 2(2(2(x1))) (26)
1(1(1(0(x1)))) → 0(1(3(x1))) (27)
1(4(0(2(1(x1))))) → 2(2(5(2(x1)))) (29)
3(0(1(5(1(x1))))) → 3(0(5(5(x1)))) (30)
4(0(1(5(3(1(x1)))))) → 4(1(0(0(0(4(x1)))))) (31)
5(5(4(5(4(1(3(3(x1)))))))) → 2(5(0(2(3(5(3(x1))))))) (32)
2(5(0(0(1(3(5(2(4(x1))))))))) → 2(3(5(3(5(1(4(4(x1)))))))) (33)
3(2(1(0(2(0(1(1(2(2(x1)))))))))) → 3(2(5(5(2(5(4(4(5(x1))))))))) (34)
5(4(5(3(3(4(3(3(2(2(2(x1))))))))))) → 5(0(2(3(5(3(5(2(4(5(5(x1))))))))))) (35)
5(2(0(0(0(1(5(1(1(2(4(x1))))))))))) → 5(2(5(4(1(0(3(0(5(5(x1)))))))))) (36)
2(0(0(3(2(2(0(2(4(5(4(5(x1)))))))))))) → 2(2(1(1(2(3(0(5(4(3(5(x1))))))))))) (37)
4(1(3(2(5(0(5(1(3(1(1(2(0(x1))))))))))))) → 4(1(1(5(0(5(4(4(1(5(1(0(3(x1))))))))))))) (38)
2(1(4(3(1(1(2(2(2(4(2(1(1(x1))))))))))))) → 3(2(2(2(5(0(1(4(3(0(5(1(x1)))))))))))) (39)
1(3(0(2(3(3(0(4(2(0(0(4(1(x1))))))))))))) → 5(5(2(1(0(1(5(0(2(0(5(0(5(1(x1)))))))))))))) (40)
0(0(1(4(5(1(2(0(0(3(1(4(1(4(x1)))))))))))))) → 0(2(1(4(0(3(4(4(5(2(0(3(2(x1))))))))))))) (41)
3(3(0(5(3(0(4(0(0(0(4(4(2(4(x1)))))))))))))) → 3(0(0(5(5(3(5(3(1(0(1(0(1(5(x1)))))))))))))) (42)
1(2(3(3(3(2(3(5(2(2(0(3(3(5(2(1(x1)))))))))))))))) → 1(4(0(5(3(0(0(5(0(1(2(1(4(0(0(1(1(x1))))))))))))))))) (43)
5(0(2(5(1(4(5(1(3(4(3(0(5(2(1(3(x1)))))))))))))))) → 5(1(1(1(2(2(4(4(2(5(0(2(4(0(1(3(x1)))))))))))))))) (44)
1(4(0(2(5(3(2(4(2(1(3(0(0(1(5(4(3(1(x1)))))))))))))))))) → 1(4(1(0(5(1(2(1(0(1(3(5(2(2(4(0(0(3(1(x1))))))))))))))))))) (45)
2(4(2(0(4(1(4(2(3(0(3(0(0(5(2(3(3(0(1(x1))))))))))))))))))) → 2(3(5(1(1(1(2(2(2(3(3(4(2(4(1(2(4(3(4(x1))))))))))))))))))) (46)
0(3(0(3(5(2(3(0(0(3(3(1(3(1(0(1(2(3(5(x1))))))))))))))))))) → 1(3(1(5(0(3(1(5(1(2(3(0(0(4(2(3(0(1(5(x1))))))))))))))))))) (47)
1(5(5(1(2(1(0(3(1(3(0(5(0(0(5(3(3(2(2(0(x1)))))))))))))))))))) → 4(0(5(4(3(4(5(2(1(2(3(2(5(2(5(2(3(0(0(x1))))))))))))))))))) (48)
4(2(0(2(0(5(4(0(2(0(5(3(4(5(3(3(2(4(0(1(x1)))))))))))))))))))) → 4(0(4(5(1(3(4(5(1(3(4(0(5(2(1(1(4(5(4(5(4(x1))))))))))))))))))))) (49)
4(3(0(0(2(5(4(0(4(1(5(1(4(2(4(3(1(3(2(4(1(x1))))))))))))))))))))) → 3(1(2(4(2(1(3(1(5(1(3(4(1(1(5(3(1(4(3(4(x1)))))))))))))))))))) (50)

1.1.1 Dependency Pair Transformation

The following set of initial dependency pairs has been identified.
4#(1(5(4(x1)))) → 4#(4(5(1(x1)))) (51)
4#(1(5(4(x1)))) → 4#(5(1(x1))) (52)

1.1.1.1 Dependency Graph Processor

The dependency pairs are split into 0 components.