Certification Problem

Input (TPDB TRS_Standard/Mixed_TRS/jones5)

The rewrite relation of the following TRS is considered.

f(x,empty) → x (1)
f(empty,cons(a,k)) → f(cons(a,k),k) (2)
f(cons(a,k),y) → f(y,k) (3)

Property / Task

Prove or disprove termination.

Answer / Result

Yes.

Proof (by ttt2 @ termCOMP 2023)

1 Rule Removal

Using the linear polynomial interpretation over (3 x 3)-matrices with strict dimension 1 over the naturals
[f(x1, x2)] =
1 0 1
0 1 0
1 0 1
· x1 +
1 1 1
0 1 1
1 1 1
· x2 +
0 0 0
0 0 0
0 0 0
[empty] =
1 0 0
0 0 0
0 0 0
[cons(x1, x2)] =
1 0 0
1 0 0
1 0 0
· x1 +
1 0 0
1 1 1
0 1 1
· x2 +
0 0 0
0 0 0
0 0 0
all of the following rules can be deleted.
f(x,empty) → x (1)
f(empty,cons(a,k)) → f(cons(a,k),k) (2)

1.1 Rule Removal

Using the Knuth Bendix order with w0 = 1 and the following precedence and weight functions
prec(cons) = 1 weight(cons) = 0
prec(f) = 0 weight(f) = 0
all of the following rules can be deleted.
f(cons(a,k),y) → f(y,k) (3)

1.1.1 R is empty

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