Certification Problem

Input (TPDB TRS_Standard/Transformed_CSR_04/Ex15_Luc06_Z)

The rewrite relation of the following TRS is considered.

f(n__f(n__a)) → f(n__g(f(n__a))) (1)
f(X) → n__f(X) (2)
a → n__a (3)
g(X) → n__g(X) (4)
activate(n__f(X)) → f(X) (5)
activate(n__a) → a (6)
activate(n__g(X)) → g(X) (7)
activate(X) → X (8)

Property / Task

Prove or disprove termination.

Answer / Result

Yes.

Proof (by ttt2 @ termCOMP 2023)

1 Rule Removal

Using the linear polynomial interpretation over (4 x 4)-matrices with strict dimension 1 over the naturals
[a] =
0 0 0 0
0 0 0 0
0 0 0 0
1 0 0 0
[n__f(x1)] =
1 0 0 0
1 0 0 0
0 0 0 0
0 0 0 1
· x1 +
1 0 0 0
0 0 0 0
0 0 0 0
0 0 0 0
[n__g(x1)] =
1 0 0 0
1 0 0 0
0 1 0 0
0 1 0 0
· x1 +
0 0 0 0
0 0 0 0
1 0 0 0
1 0 0 0
[g(x1)] =
1 0 0 0
1 0 0 0
0 1 0 0
1 1 0 0
· x1 +
1 0 0 0
0 0 0 0
1 0 0 0
1 0 0 0
[n__a] =
0 0 0 0
0 0 0 0
0 0 0 0
1 0 0 0
[activate(x1)] =
1 1 1 1
0 1 1 0
1 0 1 1
0 1 0 1
· x1 +
0 0 0 0
0 0 0 0
1 0 0 0
0 0 0 0
[f(x1)] =
1 0 0 0
1 0 0 0
0 0 0 1
1 0 0 1
· x1 +
1 0 0 0
0 0 0 0
0 0 0 0
0 0 0 0
all of the following rules can be deleted.
g(X) → n__g(X) (4)
activate(n__a) → a (6)
activate(n__g(X)) → g(X) (7)

1.1 Rule Removal

Using the linear polynomial interpretation over (5 x 5)-matrices with strict dimension 1 over the naturals
[a] =
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
[n__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
[n__g(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
[n__a] =
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
[activate(x1)] =
1 0 0 0 0
0 1 0 0 0
0 0 1 0 0
0 0 0 1 0
0 0 0 0 1
· x1 +
1 0 0 0 0
1 0 0 0 0
0 0 0 0 0
0 0 0 0 0
0 0 0 0 0
[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
1 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.
f(n__f(n__a)) → f(n__g(f(n__a))) (1)
activate(n__f(X)) → f(X) (5)
activate(X) → X (8)

1.1.1 Rule Removal

Using the Weighted Path Order with the following precedence and status
prec(a) = 0 status(a) = [] list-extension(a) = Lex
prec(f) = 0 status(f) = [1] list-extension(f) = Lex
prec(n__f) = 0 status(n__f) = [1] list-extension(n__f) = Lex
prec(n__a) = 0 status(n__a) = [] list-extension(n__a) = Lex
and the following Max-polynomial interpretation
[a] = max(5)
[f(x1)] = 1 + 1 · x1
[n__f(x1)] = max(0, 0 + 1 · x1)
[n__a] = max(1)
all of the following rules can be deleted.
f(X) → n__f(X) (2)
a → n__a (3)

1.1.1.1 R is empty

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