Certification Problem

Input (TPDB TRS_Standard/Transformed_CSR_04/ExConc_Zan97_FR)

The rewrite relation of the following TRS is considered.

f(X) → g(n__h(n__f(X))) (1)
h(X) → n__h(X) (2)
f(X) → n__f(X) (3)
activate(n__h(X)) → h(activate(X)) (4)
activate(n__f(X)) → f(activate(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
[h(x1)] =
1 1 0
0 0 0
0 0 1
· x1 +
0 0 0
0 0 0
0 0 0
[n__f(x1)] =
1 0 0
0 0 0
0 1 1
· x1 +
0 0 0
0 0 0
1 0 0
[g(x1)] =
1 0 0
0 0 0
0 0 0
· x1 +
0 0 0
0 0 0
0 0 0
[activate(x1)] =
1 0 1
0 1 0
0 0 1
· x1 +
0 0 0
0 0 0
0 0 0
[f(x1)] =
1 1 0
0 0 0
0 1 1
· x1 +
1 0 0
0 0 0
1 0 0
[n__h(x1)] =
1 1 0
0 0 0
0 0 1
· x1 +
0 0 0
0 0 0
0 0 0
all of the following rules can be deleted.
f(X) → g(n__h(n__f(X))) (1)
f(X) → n__f(X) (3)

1.1 Rule Removal

Using the Knuth Bendix order with w0 = 1 and the following precedence and weight functions
prec(activate) = 3 weight(activate) = 2
prec(h) = 1 weight(h) = 4
prec(n__h) = 0 weight(n__h) = 4
prec(n__f) = 7 weight(n__f) = 2
prec(f) = 2 weight(f) = 2
all of the following rules can be deleted.
h(X) → n__h(X) (2)
activate(n__h(X)) → h(activate(X)) (4)
activate(n__f(X)) → f(activate(X)) (5)
activate(X) → X (6)

1.1.1 R is empty

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