Certification Problem

Input (TPDB SRS_Standard/ICFP_2010/86559)

The rewrite relation of the following TRS is considered.

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

Property / Task

Prove or disprove termination.

Answer / Result

Yes.

Proof (by ttt2 @ termCOMP 2023)

1 String Reversal

Since only unary symbols occur, one can reverse all terms and obtains the TRS
1(0(1(0(x1)))) → 1(3(2(0(x1)))) (31)
1(0(4(1(2(x1))))) → 1(0(0(3(x1)))) (32)
1(3(3(3(2(x1))))) → 1(3(1(1(x1)))) (33)
4(3(2(1(4(x1))))) → 0(3(5(1(x1)))) (34)
4(4(2(3(5(x1))))) → 0(3(1(0(x1)))) (35)
3(3(4(0(5(5(x1)))))) → 3(1(2(3(0(x1))))) (36)
0(1(0(1(4(1(3(0(5(x1))))))))) → 0(3(2(2(3(1(4(2(2(x1))))))))) (37)
1(2(0(4(0(4(4(5(5(5(x1)))))))))) → 3(0(4(0(5(0(4(5(4(5(x1)))))))))) (38)
0(5(4(0(2(3(3(0(2(2(2(x1))))))))))) → 1(5(2(0(0(0(0(4(0(x1))))))))) (39)
4(4(2(5(4(1(5(3(0(2(0(1(x1)))))))))))) → 3(3(0(1(2(3(2(3(0(1(0(1(x1)))))))))))) (40)
0(5(4(4(1(0(5(3(2(4(0(4(x1)))))))))))) → 1(4(1(5(2(1(5(2(0(5(3(x1))))))))))) (41)
1(1(4(4(1(4(5(3(2(0(1(4(x1)))))))))))) → 2(4(4(2(1(5(5(4(4(5(2(x1))))))))))) (42)
4(1(4(3(2(2(2(0(5(5(0(0(2(x1))))))))))))) → 3(4(5(4(1(4(1(3(1(3(3(3(3(x1))))))))))))) (43)
2(5(2(1(2(5(3(0(0(3(1(5(4(x1))))))))))))) → 4(0(4(5(3(5(5(3(5(3(0(5(4(x1))))))))))))) (44)
4(1(4(4(5(0(4(2(0(5(5(4(1(2(x1)))))))))))))) → 1(2(5(2(3(5(2(3(4(1(2(0(0(x1))))))))))))) (45)
1(5(3(1(1(5(4(1(1(4(5(3(1(3(x1)))))))))))))) → 4(3(1(5(5(1(5(3(4(2(1(2(3(x1))))))))))))) (46)
1(2(0(5(3(0(0(0(2(0(3(5(4(4(x1)))))))))))))) → 5(3(1(4(2(4(1(0(0(1(4(0(2(2(x1)))))))))))))) (47)
1(1(3(3(2(1(5(2(4(4(4(4(5(0(5(x1))))))))))))))) → 3(0(1(3(3(0(1(3(1(3(5(0(3(1(x1)))))))))))))) (48)
1(4(1(2(2(3(4(2(2(0(4(5(5(3(5(x1))))))))))))))) → 0(0(1(2(4(4(0(4(5(3(2(4(1(5(x1)))))))))))))) (49)
0(5(0(5(0(1(0(5(2(3(3(1(2(5(5(x1))))))))))))))) → 0(5(0(3(4(5(0(1(4(2(0(5(3(2(x1)))))))))))))) (50)
1(1(2(3(0(0(5(5(5(2(1(2(3(2(1(2(x1)))))))))))))))) → 3(2(3(0(3(1(2(4(1(3(3(1(1(4(4(x1))))))))))))))) (51)
2(1(4(3(1(1(0(3(0(0(2(4(1(3(3(3(x1)))))))))))))))) → 4(0(3(4(4(0(5(3(5(0(5(1(4(0(3(x1))))))))))))))) (52)
1(5(4(5(4(1(5(2(4(1(0(4(5(3(1(1(2(x1))))))))))))))))) → 0(3(4(4(4(2(2(5(5(4(1(2(3(3(3(3(2(x1))))))))))))))))) (53)
2(5(2(2(2(2(2(2(1(3(4(4(4(4(2(0(5(x1))))))))))))))))) → 0(0(5(5(5(0(3(4(5(5(3(4(4(0(4(5(x1)))))))))))))))) (54)
4(1(0(0(5(1(1(5(0(4(1(4(4(0(1(5(5(2(x1)))))))))))))))))) → 0(5(3(4(4(2(0(2(1(2(1(1(1(0(0(1(3(3(x1)))))))))))))))))) (55)
2(4(4(4(5(0(3(4(4(3(5(2(3(5(4(1(1(0(0(x1))))))))))))))))))) → 4(0(4(0(4(3(3(0(4(0(3(5(1(4(1(5(4(0(0(x1))))))))))))))))))) (56)
3(2(0(2(1(4(4(5(4(3(3(4(4(2(5(1(4(1(4(1(x1)))))))))))))))))))) → 3(3(4(1(1(5(1(0(5(2(4(4(4(0(1(0(5(1(2(1(x1)))))))))))))))))))) (57)
2(5(3(3(5(5(3(3(4(3(3(1(5(5(2(4(0(0(1(2(x1)))))))))))))))))))) → 5(4(2(5(2(5(1(3(1(1(2(2(4(0(5(3(3(4(4(x1))))))))))))))))))) (58)
4(2(4(1(2(5(2(0(1(3(5(1(2(5(4(5(4(2(4(2(4(x1))))))))))))))))))))) → 1(2(3(4(5(2(1(2(5(3(4(5(0(2(5(0(5(1(3(5(1(x1))))))))))))))))))))) (59)
5(3(5(5(0(2(4(3(3(3(5(1(4(3(4(3(0(1(2(0(5(x1))))))))))))))))))))) → 5(1(4(4(3(4(4(4(0(5(1(2(5(4(5(0(1(1(5(3(x1)))))))))))))))))))) (60)

1.1 Rule Removal

Using the Knuth Bendix order with w0 = 1 and the following precedence and weight functions
prec(5) = 0 weight(5) = 1
prec(4) = 5 weight(4) = 1
prec(2) = 6 weight(2) = 1
prec(3) = 1 weight(3) = 1
prec(0) = 2 weight(0) = 1
prec(1) = 4 weight(1) = 1
all of the following rules can be deleted.
1(0(1(0(x1)))) → 1(3(2(0(x1)))) (31)
1(0(4(1(2(x1))))) → 1(0(0(3(x1)))) (32)
1(3(3(3(2(x1))))) → 1(3(1(1(x1)))) (33)
4(3(2(1(4(x1))))) → 0(3(5(1(x1)))) (34)
4(4(2(3(5(x1))))) → 0(3(1(0(x1)))) (35)
3(3(4(0(5(5(x1)))))) → 3(1(2(3(0(x1))))) (36)
0(1(0(1(4(1(3(0(5(x1))))))))) → 0(3(2(2(3(1(4(2(2(x1))))))))) (37)
1(2(0(4(0(4(4(5(5(5(x1)))))))))) → 3(0(4(0(5(0(4(5(4(5(x1)))))))))) (38)
0(5(4(0(2(3(3(0(2(2(2(x1))))))))))) → 1(5(2(0(0(0(0(4(0(x1))))))))) (39)
4(4(2(5(4(1(5(3(0(2(0(1(x1)))))))))))) → 3(3(0(1(2(3(2(3(0(1(0(1(x1)))))))))))) (40)
0(5(4(4(1(0(5(3(2(4(0(4(x1)))))))))))) → 1(4(1(5(2(1(5(2(0(5(3(x1))))))))))) (41)
1(1(4(4(1(4(5(3(2(0(1(4(x1)))))))))))) → 2(4(4(2(1(5(5(4(4(5(2(x1))))))))))) (42)
4(1(4(3(2(2(2(0(5(5(0(0(2(x1))))))))))))) → 3(4(5(4(1(4(1(3(1(3(3(3(3(x1))))))))))))) (43)
2(5(2(1(2(5(3(0(0(3(1(5(4(x1))))))))))))) → 4(0(4(5(3(5(5(3(5(3(0(5(4(x1))))))))))))) (44)
4(1(4(4(5(0(4(2(0(5(5(4(1(2(x1)))))))))))))) → 1(2(5(2(3(5(2(3(4(1(2(0(0(x1))))))))))))) (45)
1(5(3(1(1(5(4(1(1(4(5(3(1(3(x1)))))))))))))) → 4(3(1(5(5(1(5(3(4(2(1(2(3(x1))))))))))))) (46)
1(2(0(5(3(0(0(0(2(0(3(5(4(4(x1)))))))))))))) → 5(3(1(4(2(4(1(0(0(1(4(0(2(2(x1)))))))))))))) (47)
1(1(3(3(2(1(5(2(4(4(4(4(5(0(5(x1))))))))))))))) → 3(0(1(3(3(0(1(3(1(3(5(0(3(1(x1)))))))))))))) (48)
1(4(1(2(2(3(4(2(2(0(4(5(5(3(5(x1))))))))))))))) → 0(0(1(2(4(4(0(4(5(3(2(4(1(5(x1)))))))))))))) (49)
0(5(0(5(0(1(0(5(2(3(3(1(2(5(5(x1))))))))))))))) → 0(5(0(3(4(5(0(1(4(2(0(5(3(2(x1)))))))))))))) (50)
1(1(2(3(0(0(5(5(5(2(1(2(3(2(1(2(x1)))))))))))))))) → 3(2(3(0(3(1(2(4(1(3(3(1(1(4(4(x1))))))))))))))) (51)
2(1(4(3(1(1(0(3(0(0(2(4(1(3(3(3(x1)))))))))))))))) → 4(0(3(4(4(0(5(3(5(0(5(1(4(0(3(x1))))))))))))))) (52)
1(5(4(5(4(1(5(2(4(1(0(4(5(3(1(1(2(x1))))))))))))))))) → 0(3(4(4(4(2(2(5(5(4(1(2(3(3(3(3(2(x1))))))))))))))))) (53)
2(5(2(2(2(2(2(2(1(3(4(4(4(4(2(0(5(x1))))))))))))))))) → 0(0(5(5(5(0(3(4(5(5(3(4(4(0(4(5(x1)))))))))))))))) (54)
4(1(0(0(5(1(1(5(0(4(1(4(4(0(1(5(5(2(x1)))))))))))))))))) → 0(5(3(4(4(2(0(2(1(2(1(1(1(0(0(1(3(3(x1)))))))))))))))))) (55)
2(4(4(4(5(0(3(4(4(3(5(2(3(5(4(1(1(0(0(x1))))))))))))))))))) → 4(0(4(0(4(3(3(0(4(0(3(5(1(4(1(5(4(0(0(x1))))))))))))))))))) (56)
3(2(0(2(1(4(4(5(4(3(3(4(4(2(5(1(4(1(4(1(x1)))))))))))))))))))) → 3(3(4(1(1(5(1(0(5(2(4(4(4(0(1(0(5(1(2(1(x1)))))))))))))))))))) (57)
2(5(3(3(5(5(3(3(4(3(3(1(5(5(2(4(0(0(1(2(x1)))))))))))))))))))) → 5(4(2(5(2(5(1(3(1(1(2(2(4(0(5(3(3(4(4(x1))))))))))))))))))) (58)
4(2(4(1(2(5(2(0(1(3(5(1(2(5(4(5(4(2(4(2(4(x1))))))))))))))))))))) → 1(2(3(4(5(2(1(2(5(3(4(5(0(2(5(0(5(1(3(5(1(x1))))))))))))))))))))) (59)
5(3(5(5(0(2(4(3(3(3(5(1(4(3(4(3(0(1(2(0(5(x1))))))))))))))))))))) → 5(1(4(4(3(4(4(4(0(5(1(2(5(4(5(0(1(1(5(3(x1)))))))))))))))))))) (60)

1.1.1 R is empty

There are no rules in the TRS. Hence, it is terminating.