Certification Problem

Input (TPDB TRS_Standard/HirokawaMiddeldorp_04/t004)

The rewrite relation of the following TRS is considered.

f(s(x)) → s(f(f(p(s(x))))) (1)
f(0) → 0 (2)
p(s(x)) → 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 2 0
0 1 0
0 2 0
· x1 +
0 0 0
0 0 0
0 0 0
[0] =
3 0 0
2 0 0
0 0 0
[s(x1)] =
1 0 1
0 3 0
0 1 0
· x1 +
0 0 0
0 0 0
0 0 0
[p(x1)] =
1 0 0
0 0 1
1 0 0
· x1 +
0 0 0
0 0 0
2 0 0
all of the following rules can be deleted.
f(0) → 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 2
0 0 1
0 0 1
· x1 +
0 0 0
0 0 0
0 0 0
[s(x1)] =
1 1 0
0 0 1
0 0 3
· x1 +
1 0 0
0 0 0
2 0 0
[p(x1)] =
1 0 0
2 0 2
0 1 0
· x1 +
3 0 0
1 0 0
0 0 0
all of the following rules can be deleted.
p(s(x)) → x (3)

1.1.1 Rule Removal

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

1.1.1.1 R is empty

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