Certification Problem

Input (TPDB TRS_Standard/Transformed_CSR_04/ExProp7_Luc06_Z)

The rewrite relation of the following TRS is considered.

f(0) → cons(0,n__f(s(0))) (1)
f(s(0)) → f(p(s(0))) (2)
p(s(X)) → X (3)
f(X) → n__f(X) (4)
activate(n__f(X)) → f(X) (5)
activate(X) → X (6)

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
[cons(x1, x2)] =
1 0 0
0 0 0
0 0 0
· x1 +
1 0 0
0 0 0
0 0 0
· x2 +
0 0 0
0 0 0
0 0 0
[f(x1)] =
1 0 0
1 0 0
0 0 0
· x1 +
0 0 0
0 0 0
0 0 0
[n__f(x1)] =
1 0 0
1 0 0
0 0 0
· x1 +
0 0 0
0 0 0
0 0 0
[p(x1)] =
1 0 0
1 1 1
0 1 0
· x1 +
0 0 0
1 0 0
0 0 0
[0] =
0 0 0
0 0 0
0 0 0
[activate(x1)] =
1 1 0
1 1 0
0 0 1
· x1 +
1 0 0
0 0 0
0 0 0
[s(x1)] =
1 0 0
0 1 1
0 1 0
· x1 +
0 0 0
1 0 0
1 0 0
all of the following rules can be deleted.
activate(n__f(X)) → f(X) (5)
activate(X) → X (6)

1.1 Rule Removal

Using the linear polynomial interpretation over (5 x 5)-matrices with strict dimension 1 over the naturals
[cons(x1, x2)] =
1 0 0 0 0
0 0 0 0 0
0 0 0 0 0
0 0 0 0 0
0 0 0 0 0
· x1 +
1 0 0 0 0
0 0 0 0 0
0 0 0 0 0
0 0 0 0 0
0 0 0 0 0
· x2 +
0 0 0 0 0
0 0 0 0 0
0 0 0 0 0
0 0 0 0 0
0 0 0 0 0
[f(x1)] =
1 0 0 0 0
0 0 0 0 0
0 0 0 0 0
0 0 1 0 1
0 0 0 0 0
· x1 +
1 0 0 0 0
0 0 0 0 0
0 0 0 0 0
0 0 0 0 0
0 0 0 0 0
[n__f(x1)] =
1 0 0 0 0
0 0 0 0 0
0 0 0 0 0
0 0 0 0 0
0 0 0 0 0
· x1 +
0 0 0 0 0
0 0 0 0 0
0 0 0 0 0
0 0 0 0 0
0 0 0 0 0
[p(x1)] =
1 1 0 0 0
0 0 1 0 0
0 0 0 0 1
0 0 1 1 0
0 1 1 0 0
· x1 +
0 0 0 0 0
0 0 0 0 0
0 0 0 0 0
0 0 0 0 0
0 0 0 0 0
[0] =
0 0 0 0 0
0 0 0 0 0
0 0 0 0 0
0 0 0 0 0
0 0 0 0 0
[s(x1)] =
1 0 0 0 0
0 0 0 0 1
1 1 0 0 0
0 0 0 1 0
0 0 1 0 0
· x1 +
0 0 0 0 0
0 0 0 0 0
1 0 0 0 0
0 0 0 0 0
0 0 0 0 0
all of the following rules can be deleted.
f(0) → cons(0,n__f(s(0))) (1)
f(X) → n__f(X) (4)

1.1.1 Rule Removal

Using the linear polynomial interpretation over (5 x 5)-matrices with strict dimension 1 over the naturals
[f(x1)] =
1 0 0 1 0
0 0 0 0 0
0 0 0 0 0
0 0 0 0 0
0 0 0 0 0
· x1 +
0 0 0 0 0
0 0 0 0 0
0 0 0 0 0
1 0 0 0 0
0 0 0 0 0
[p(x1)] =
1 0 0 1 0
1 0 0 1 0
0 0 0 1 0
0 0 0 0 1
0 0 0 0 1
· x1 +
0 0 0 0 0
0 0 0 0 0
0 0 0 0 0
0 0 0 0 0
1 0 0 0 0
[0] =
0 0 0 0 0
0 0 0 0 0
0 0 0 0 0
0 0 0 0 0
0 0 0 0 0
[s(x1)] =
1 1 0 0 0
0 0 0 0 0
0 0 0 0 0
0 0 1 0 0
0 0 0 1 1
· x1 +
0 0 0 0 0
0 0 0 0 0
0 0 0 0 0
1 0 0 0 0
0 0 0 0 0
all of the following rules can be deleted.
p(s(X)) → X (3)

1.1.1.1 Rule Removal

Using the linear polynomial interpretation over (4 x 4)-matrices with strict dimension 1 over the naturals
[f(x1)] =
1 0 0 1
0 0 1 1
0 0 1 1
1 1 0 0
· x1 +
0 0 0 0
0 0 0 0
0 0 0 0
0 0 0 0
[p(x1)] =
1 1 0 0
0 0 0 0
1 1 0 0
0 0 0 0
· x1 +
0 0 0 0
0 0 0 0
0 0 0 0
0 0 0 0
[0] =
0 0 0 0
1 0 0 0
0 0 0 0
1 0 0 0
[s(x1)] =
1 0 0 1
0 0 0 0
0 0 0 0
0 1 0 1
· x1 +
0 0 0 0
1 0 0 0
0 0 0 0
0 0 0 0
all of the following rules can be deleted.
f(s(0)) → f(p(s(0))) (2)

1.1.1.1.1 R is empty

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