Certification Problem

Input (TPDB TRS_Standard/Strategy_removed_mixed_05/test830)

The rewrite relation of the following TRS is considered.

f(s(X)) → f(X) (1)
g(cons(0,Y)) → g(Y) (2)
g(cons(s(X),Y)) → s(X) (3)
h(cons(X,Y)) → h(g(cons(X,Y))) (4)

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
[g(x1)] =
1 0 0
0 0 0
0 0 0
· x1 +
0 0 0
1 0 0
0 0 0
[f(x1)] =
1 0 0
0 0 0
0 0 0
· x1 +
1 0 0
0 0 0
1 0 0
[cons(x1, x2)] =
1 1 0
1 0 0
0 0 0
· x1 +
1 0 0
0 0 0
1 0 0
· x2 +
0 0 0
1 0 0
0 0 0
[h(x1)] =
1 0 0
0 0 0
0 0 1
· x1 +
1 0 0
0 0 0
0 0 0
[s(x1)] =
1 0 0
0 0 0
0 0 0
· x1 +
1 0 0
1 0 0
0 0 0
[0] =
1 0 0
1 0 0
0 0 0
all of the following rules can be deleted.
f(s(X)) → f(X) (1)
g(cons(0,Y)) → g(Y) (2)
g(cons(s(X),Y)) → s(X) (3)

1.1 Rule Removal

Using the linear polynomial interpretation over (3 x 3)-matrices with strict dimension 1 over the naturals
[g(x1)] =
1 1 0
0 0 0
0 0 0
· x1 +
0 0 0
0 0 0
0 0 0
[cons(x1, x2)] =
1 0 0
1 0 0
0 0 0
· x1 +
1 0 0
1 0 1
0 0 0
· x2 +
1 0 0
1 0 0
1 0 0
[h(x1)] =
1 1 1
1 1 0
0 0 0
· x1 +
0 0 0
0 0 0
0 0 0
all of the following rules can be deleted.
h(cons(X,Y)) → h(g(cons(X,Y))) (4)

1.1.1 R is empty

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