Certification Problem

Input (TPDB TRS_Standard/Transformed_CSR_04/Ex4_7_15_Bor03_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(0)) → 0 (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 (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 1 1 0
0 0 1 0 0
0 0 0 0 0
1 0 0 1 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 1 1 0
0 0 0 0 0
0 0 0 0 0
1 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 0 0 0 0
0 0 0 0 0
0 0 0 0 0
0 0 0 0 0
0 0 0 0 1
· 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
[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
[activate(x1)] =
1 0 0 0 0
0 1 0 0 0
0 1 1 0 1
0 0 0 1 0
0 1 0 0 1
· 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
[s(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
1 0 0 0 0
1 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)
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
[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 1
· 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__f(x1)] =
1 0 0 0 0
0 0 0 0 1
0 0 0 0 0
0 0 0 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 0 0 0 0
[p(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
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
1 0 0 0 0
0 0 0 0 0
0 0 0 0 0
[activate(x1)] =
1 0 0 1 0
0 0 0 0 0
0 0 0 0 0
0 0 0 0 0
0 1 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
[s(x1)] =
1 0 0 0 0
0 0 0 0 0
0 0 1 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
all of the following rules can be deleted.
f(s(0)) → f(p(s(0))) (2)
activate(n__f(X)) → f(X) (5)

1.1.1 Rule Removal

Using the Weighted Path Order with the following precedence and status
prec(p) = 0 status(p) = [1] list-extension(p) = Lex
prec(s) = 0 status(s) = [1] list-extension(s) = Lex
prec(0) = 0 status(0) = [] list-extension(0) = Lex
and the following Max-polynomial interpretation
[p(x1)] = max(0, 1 + 1 · x1)
[s(x1)] = max(1, 0 + 1 · x1)
[0] = max(6)
all of the following rules can be deleted.
p(s(0)) → 0 (3)

1.1.1.1 R is empty

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