Certification Problem

Input (TPDB SRS_Standard/ICFP_2010/96086)

The rewrite relation of the following TRS is considered.

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

Property / Task

Prove or disprove termination.

Answer / Result

Yes.

Proof (by ttt2 @ termCOMP 2023)

1 Rule Removal

Using the Knuth Bendix order with w0 = 1 and the following precedence and weight functions
prec(4) = 1 weight(4) = 2
prec(5) = 2 weight(5) = 2
prec(3) = 3 weight(3) = 2
prec(0) = 6 weight(0) = 2
prec(1) = 0 weight(1) = 2
prec(2) = 7 weight(2) = 2
all of the following rules can be deleted.
0(0(1(2(x1)))) → 2(3(0(x1))) (1)
4(2(5(2(x1)))) → 1(4(0(x1))) (2)
4(5(3(4(x1)))) → 4(4(2(4(x1)))) (3)
1(3(5(5(2(x1))))) → 1(3(2(3(x1)))) (4)
5(2(2(1(2(x1))))) → 0(0(2(3(x1)))) (5)
2(2(5(3(2(2(x1)))))) → 5(1(1(0(3(x1))))) (6)
2(5(1(2(1(1(x1)))))) → 5(2(1(2(4(1(x1)))))) (7)
3(4(1(4(2(4(x1)))))) → 1(3(3(3(x1)))) (8)
3(5(2(2(4(5(x1)))))) → 3(2(4(3(0(x1))))) (9)
5(2(1(0(1(5(x1)))))) → 5(4(2(4(5(1(x1)))))) (10)
1(3(5(4(1(2(2(x1))))))) → 3(3(3(4(4(0(x1)))))) (11)
4(5(4(3(0(5(1(x1))))))) → 4(3(3(5(4(1(x1)))))) (12)
2(1(5(2(1(3(4(4(x1)))))))) → 4(0(3(4(0(1(2(x1))))))) (13)
5(4(0(2(2(4(0(4(x1)))))))) → 3(1(5(1(3(0(4(x1))))))) (14)
3(4(2(1(1(2(2(5(4(x1))))))))) → 3(3(3(1(3(3(4(x1))))))) (15)
5(4(4(5(0(1(4(5(4(x1))))))))) → 1(5(5(0(4(1(4(5(4(x1))))))))) (16)
5(2(1(3(1(5(2(5(4(4(x1)))))))))) → 5(3(4(5(0(1(4(0(3(x1))))))))) (17)
2(4(1(2(5(2(4(1(3(2(0(3(x1)))))))))))) → 4(3(0(4(2(3(4(3(4(2(0(x1))))))))))) (18)
0(2(3(5(4(2(2(1(0(3(3(5(0(x1))))))))))))) → 3(3(1(2(3(0(4(0(0(0(2(0(x1)))))))))))) (19)
2(1(0(2(1(4(0(0(2(0(0(0(5(2(x1)))))))))))))) → 2(2(4(2(1(4(3(0(5(1(3(3(0(x1))))))))))))) (20)
4(5(0(3(1(3(2(2(5(2(2(4(1(3(2(x1))))))))))))))) → 4(5(3(2(1(4(5(0(0(0(4(5(4(0(0(x1))))))))))))))) (21)
5(2(2(5(2(4(4(1(2(0(1(1(0(1(1(x1))))))))))))))) → 5(0(5(4(3(2(1(0(3(3(5(0(4(1(x1)))))))))))))) (22)
5(4(2(0(3(3(0(0(4(0(3(2(0(5(1(x1))))))))))))))) → 5(0(1(0(0(2(1(1(0(3(2(2(1(3(x1)))))))))))))) (23)
5(4(1(5(1(5(4(4(2(2(0(4(3(1(5(4(4(3(1(x1))))))))))))))))))) → 3(0(4(5(1(1(3(5(3(4(4(4(5(1(4(3(3(1(x1)))))))))))))))))) (24)
2(0(3(0(2(2(2(0(1(4(2(1(0(4(4(3(3(1(4(4(x1)))))))))))))))))))) → 5(2(4(1(1(4(5(1(0(1(2(0(3(0(1(2(3(4(3(1(x1)))))))))))))))))))) (25)

1.1 R is empty

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