Certification Problem

Input (TPDB TRS_Standard/AG01/#3.24)

The rewrite relation of the following TRS is considered.

f(0) → s(0) (1)
f(s(0)) → s(0) (2)
f(s(s(x))) → f(f(s(x))) (3)

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
[f(x1)] =
1 0 0
0 1 0
0 0 1
· x1 +
1 0 0
1 0 0
0 0 0
[0] =
1 0 0
1 0 0
1 0 0
[s(x1)] =
1 0 0
0 1 0
0 0 0
· x1 +
1 0 0
1 0 0
1 0 0
all of the following rules can be deleted.
f(s(0)) → s(0) (2)

1.1 Rule Removal

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

1.1.1 Rule Removal

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

1.1.1.1 R is empty

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