Certification Problem

Input (TPDB SRS_Standard/ICFP_2010/96274)

The rewrite relation of the following TRS is considered.

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

1.1 Rule Removal

Using the linear polynomial interpretation over the naturals
[3(x1)] = 1 · x1 + 6
[2(x1)] = 1 · x1 + 4
[1(x1)] = 1 · x1 + 4
[0(x1)] = 1 · x1 + 5
[4(x1)] = 1 · x1 + 4
[5(x1)] = 1 · x1 + 6
all of the following rules can be deleted.
3(2(1(0(x1)))) → 3(2(4(4(x1)))) (26)
4(5(3(3(5(x1))))) → 2(0(1(5(x1)))) (27)
4(5(5(3(3(5(x1)))))) → 2(3(4(2(4(x1))))) (29)
5(1(1(5(4(1(5(x1))))))) → 5(2(3(2(0(4(1(x1))))))) (30)
5(0(3(1(3(4(3(3(x1)))))))) → 2(5(0(5(4(2(5(3(x1)))))))) (31)
3(1(3(1(2(2(2(1(3(x1))))))))) → 2(2(0(5(1(3(4(1(x1)))))))) (32)
3(5(3(2(5(0(2(4(3(x1))))))))) → 1(5(0(2(2(4(4(5(3(x1))))))))) (33)
0(0(4(5(3(3(1(5(5(x1))))))))) → 3(4(2(4(1(0(1(3(x1)))))))) (34)
0(5(1(0(5(1(5(2(0(3(x1)))))))))) → 4(4(3(4(0(0(2(2(1(x1))))))))) (35)
3(0(0(2(4(4(4(5(5(3(x1)))))))))) → 1(4(3(2(3(2(1(1(x1)))))))) (36)
2(4(0(4(4(0(5(3(3(4(0(3(x1)))))))))))) → 2(4(1(5(0(2(3(5(5(4(3(x1))))))))))) (37)
0(2(1(1(1(2(0(5(4(0(2(5(x1)))))))))))) → 0(1(3(1(5(4(2(2(0(0(3(x1))))))))))) (38)
5(4(0(0(5(4(5(4(3(3(4(5(5(x1))))))))))))) → 1(3(2(1(4(1(3(0(1(0(5(x1))))))))))) (39)
1(0(0(2(2(5(4(2(2(0(3(1(2(5(x1)))))))))))))) → 1(0(3(0(5(3(3(4(1(5(4(3(x1)))))))))))) (40)
4(2(3(2(5(4(4(3(3(5(5(2(5(1(3(x1))))))))))))))) → 4(2(3(4(1(0(0(2(4(4(4(5(0(3(x1)))))))))))))) (41)
3(3(3(3(4(2(4(2(4(4(3(4(5(5(4(x1))))))))))))))) → 0(0(0(4(3(4(4(5(4(4(0(0(5(4(x1)))))))))))))) (42)
3(4(2(3(5(2(0(5(1(4(1(1(3(4(2(1(0(x1))))))))))))))))) → 1(4(5(5(2(2(1(5(4(0(3(1(3(1(2(2(4(x1))))))))))))))))) (43)
4(4(0(0(2(0(1(5(2(0(0(2(4(0(3(4(2(x1))))))))))))))))) → 2(0(0(1(3(4(0(2(0(1(2(1(2(1(4(5(x1)))))))))))))))) (44)
2(0(2(2(5(4(0(0(4(2(5(1(1(2(0(1(3(3(3(x1))))))))))))))))))) → 1(3(2(3(4(1(3(0(3(5(5(5(1(3(2(2(3(x1))))))))))))))))) (45)
4(5(0(3(3(5(2(3(4(2(0(3(1(2(5(2(2(3(5(x1))))))))))))))))))) → 2(0(0(1(1(2(0(4(5(5(0(5(5(4(3(3(0(3(1(x1))))))))))))))))))) (46)
1(3(1(3(5(1(2(1(0(1(2(1(0(2(5(5(5(4(5(x1))))))))))))))))))) → 1(4(2(0(3(4(5(0(4(1(0(2(0(3(5(3(0(0(x1)))))))))))))))))) (47)
0(0(5(3(5(2(0(3(1(3(5(2(3(0(1(5(0(4(0(4(x1)))))))))))))))))))) → 1(5(2(3(1(3(4(4(1(0(5(4(4(5(0(2(5(3(5(1(x1)))))))))))))))))))) (48)
0(3(1(5(0(1(5(5(0(0(2(2(4(5(2(3(1(2(4(5(x1)))))))))))))))))))) → 0(2(1(1(1(0(0(4(4(3(1(5(2(3(1(0(4(2(4(4(x1)))))))))))))))))))) (49)
2(3(0(5(3(4(1(0(4(5(3(5(5(3(4(1(4(5(4(0(3(x1))))))))))))))))))))) → 2(1(3(1(2(4(1(5(2(4(0(4(2(4(5(2(1(1(4(2(1(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.
5#(0(5(1(3(2(x1)))))) → 5#(0(5(3(1(2(x1)))))) (51)
5#(0(5(1(3(2(x1)))))) → 5#(3(1(2(x1)))) (52)

1.1.1.1.1 Dependency Graph Processor

The dependency pairs are split into 0 components.