Certification Problem

Input (TPDB SRS_Standard/ICFP_2010/96224)

The rewrite relation of the following TRS is considered.

0(1(2(2(3(0(4(4(x1)))))))) → 0(4(5(0(3(4(0(4(x1)))))))) (1)
0(2(2(2(0(5(2(5(4(x1))))))))) → 0(0(0(3(4(3(0(1(0(x1))))))))) (2)
2(1(1(3(4(3(1(1(5(x1))))))))) → 1(0(2(3(1(0(5(1(5(x1))))))))) (3)
5(4(0(4(3(3(1(2(5(3(0(x1))))))))))) → 5(5(2(5(3(1(5(0(3(5(2(x1))))))))))) (4)
2(4(5(0(1(1(3(3(5(3(0(0(x1)))))))))))) → 4(4(2(2(1(0(4(0(1(3(2(0(x1)))))))))))) (5)
4(4(3(0(3(1(5(3(5(1(3(1(5(3(x1)))))))))))))) → 0(2(4(5(0(0(0(5(1(0(5(4(4(x1))))))))))))) (6)
2(2(0(0(2(1(0(5(3(2(2(1(4(0(5(x1))))))))))))))) → 0(0(3(5(3(0(4(3(1(3(0(2(5(5(x1)))))))))))))) (7)
3(0(0(4(2(5(5(1(3(0(2(3(3(5(1(4(5(x1))))))))))))))))) → 3(4(0(1(0(5(5(3(1(4(0(3(5(3(2(2(5(x1))))))))))))))))) (8)
5(1(1(4(1(5(3(0(4(3(2(5(4(1(3(3(5(x1))))))))))))))))) → 5(0(1(0(4(0(2(4(5(1(5(4(1(5(3(3(5(x1))))))))))))))))) (9)
0(1(4(4(3(2(0(4(1(4(3(4(4(1(5(3(4(4(x1)))))))))))))))))) → 0(2(4(2(3(1(0(1(1(1(3(0(2(4(4(1(1(2(x1)))))))))))))))))) (10)
1(1(4(1(0(1(0(3(3(4(4(1(5(4(0(4(4(5(5(3(x1)))))))))))))))))))) → 1(3(0(3(2(2(4(4(2(0(3(3(4(0(3(0(4(3(4(0(x1)))))))))))))))))))) (11)
2(0(3(3(3(4(1(1(0(4(4(0(3(3(3(0(0(1(5(3(x1)))))))))))))))))))) → 5(3(3(1(4(0(4(5(4(4(4(2(4(3(1(1(1(5(4(x1))))))))))))))))))) (12)
2(3(3(2(1(5(0(5(0(1(3(3(2(5(1(5(0(3(0(5(x1)))))))))))))))))))) → 3(4(0(2(5(5(2(4(2(4(3(1(1(4(4(5(5(3(5(x1))))))))))))))))))) (13)
3(3(2(2(3(3(4(0(0(0(2(5(0(5(3(0(0(1(1(4(x1)))))))))))))))))))) → 3(1(1(0(3(3(5(4(2(2(1(0(1(1(0(5(0(3(4(0(x1)))))))))))))))))))) (14)
4(2(4(1(0(5(0(4(1(0(3(0(2(5(4(3(5(3(5(3(x1)))))))))))))))))))) → 4(2(5(4(4(2(4(5(0(0(3(1(5(0(2(0(2(2(1(x1))))))))))))))))))) (15)

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
4(4(0(3(2(2(1(0(x1)))))))) → 4(0(4(3(0(5(4(0(x1)))))))) (16)
4(5(2(5(0(2(2(2(0(x1))))))))) → 0(1(0(3(4(3(0(0(0(x1))))))))) (17)
5(1(1(3(4(3(1(1(2(x1))))))))) → 5(1(5(0(1(3(2(0(1(x1))))))))) (18)
0(3(5(2(1(3(3(4(0(4(5(x1))))))))))) → 2(5(3(0(5(1(3(5(2(5(5(x1))))))))))) (19)
0(0(3(5(3(3(1(1(0(5(4(2(x1)))))))))))) → 0(2(3(1(0(4(0(1(2(2(4(4(x1)))))))))))) (20)
3(5(1(3(1(5(3(5(1(3(0(3(4(4(x1)))))))))))))) → 4(4(5(0(1(5(0(0(0(5(4(2(0(x1))))))))))))) (21)
5(0(4(1(2(2(3(5(0(1(2(0(0(2(2(x1))))))))))))))) → 5(5(2(0(3(1(3(4(0(3(5(3(0(0(x1)))))))))))))) (22)
5(4(1(5(3(3(2(0(3(1(5(5(2(4(0(0(3(x1))))))))))))))))) → 5(2(2(3(5(3(0(4(1(3(5(5(0(1(0(4(3(x1))))))))))))))))) (23)
5(3(3(1(4(5(2(3(4(0(3(5(1(4(1(1(5(x1))))))))))))))))) → 5(3(3(5(1(4(5(1(5(4(2(0(4(0(1(0(5(x1))))))))))))))))) (24)
4(4(3(5(1(4(4(3(4(1(4(0(2(3(4(4(1(0(x1)))))))))))))))))) → 2(1(1(4(4(2(0(3(1(1(1(0(1(3(2(4(2(0(x1)))))))))))))))))) (25)
3(5(5(4(4(0(4(5(1(4(4(3(3(0(1(0(1(4(1(1(x1)))))))))))))))))))) → 0(4(3(4(0(3(0(4(3(3(0(2(4(4(2(2(3(0(3(1(x1)))))))))))))))))))) (26)
3(5(1(0(0(3(3(3(0(4(4(0(1(1(4(3(3(3(0(2(x1)))))))))))))))))))) → 4(5(1(1(1(3(4(2(4(4(4(5(4(0(4(1(3(3(5(x1))))))))))))))))))) (27)
5(0(3(0(5(1(5(2(3(3(1(0(5(0(5(1(2(3(3(2(x1)))))))))))))))))))) → 5(3(5(5(4(4(1(1(3(4(2(4(2(5(5(2(0(4(3(x1))))))))))))))))))) (28)
4(1(1(0(0(3(5(0(5(2(0(0(0(4(3(3(2(2(3(3(x1)))))))))))))))))))) → 0(4(3(0(5(0(1(1(0(1(2(2(4(5(3(3(0(1(1(3(x1)))))))))))))))))))) (29)
3(5(3(5(3(4(5(2(0(3(0(1(4(0(5(0(1(4(2(4(x1)))))))))))))))))))) → 1(2(2(0(2(0(5(1(3(0(0(5(4(2(4(4(5(2(4(x1))))))))))))))))))) (30)

1.1 Rule Removal

Using the Knuth Bendix order with w0 = 1 and the following precedence and weight functions
prec(5) = 2 weight(5) = 1
prec(1) = 6 weight(1) = 1
prec(2) = 0 weight(2) = 1
prec(3) = 3 weight(3) = 1
prec(0) = 1 weight(0) = 1
prec(4) = 4 weight(4) = 1
all of the following rules can be deleted.
4(4(0(3(2(2(1(0(x1)))))))) → 4(0(4(3(0(5(4(0(x1)))))))) (16)
4(5(2(5(0(2(2(2(0(x1))))))))) → 0(1(0(3(4(3(0(0(0(x1))))))))) (17)
5(1(1(3(4(3(1(1(2(x1))))))))) → 5(1(5(0(1(3(2(0(1(x1))))))))) (18)
0(3(5(2(1(3(3(4(0(4(5(x1))))))))))) → 2(5(3(0(5(1(3(5(2(5(5(x1))))))))))) (19)
0(0(3(5(3(3(1(1(0(5(4(2(x1)))))))))))) → 0(2(3(1(0(4(0(1(2(2(4(4(x1)))))))))))) (20)
3(5(1(3(1(5(3(5(1(3(0(3(4(4(x1)))))))))))))) → 4(4(5(0(1(5(0(0(0(5(4(2(0(x1))))))))))))) (21)
5(0(4(1(2(2(3(5(0(1(2(0(0(2(2(x1))))))))))))))) → 5(5(2(0(3(1(3(4(0(3(5(3(0(0(x1)))))))))))))) (22)
5(4(1(5(3(3(2(0(3(1(5(5(2(4(0(0(3(x1))))))))))))))))) → 5(2(2(3(5(3(0(4(1(3(5(5(0(1(0(4(3(x1))))))))))))))))) (23)
5(3(3(1(4(5(2(3(4(0(3(5(1(4(1(1(5(x1))))))))))))))))) → 5(3(3(5(1(4(5(1(5(4(2(0(4(0(1(0(5(x1))))))))))))))))) (24)
4(4(3(5(1(4(4(3(4(1(4(0(2(3(4(4(1(0(x1)))))))))))))))))) → 2(1(1(4(4(2(0(3(1(1(1(0(1(3(2(4(2(0(x1)))))))))))))))))) (25)
3(5(5(4(4(0(4(5(1(4(4(3(3(0(1(0(1(4(1(1(x1)))))))))))))))))))) → 0(4(3(4(0(3(0(4(3(3(0(2(4(4(2(2(3(0(3(1(x1)))))))))))))))))))) (26)
3(5(1(0(0(3(3(3(0(4(4(0(1(1(4(3(3(3(0(2(x1)))))))))))))))))))) → 4(5(1(1(1(3(4(2(4(4(4(5(4(0(4(1(3(3(5(x1))))))))))))))))))) (27)
5(0(3(0(5(1(5(2(3(3(1(0(5(0(5(1(2(3(3(2(x1)))))))))))))))))))) → 5(3(5(5(4(4(1(1(3(4(2(4(2(5(5(2(0(4(3(x1))))))))))))))))))) (28)
4(1(1(0(0(3(5(0(5(2(0(0(0(4(3(3(2(2(3(3(x1)))))))))))))))))))) → 0(4(3(0(5(0(1(1(0(1(2(2(4(5(3(3(0(1(1(3(x1)))))))))))))))))))) (29)
3(5(3(5(3(4(5(2(0(3(0(1(4(0(5(0(1(4(2(4(x1)))))))))))))))))))) → 1(2(2(0(2(0(5(1(3(0(0(5(4(2(4(4(5(2(4(x1))))))))))))))))))) (30)

1.1.1 R is empty

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