Certification Problem

Input (TPDB TRS_Standard/SK90/2.28)

The rewrite relation of the following TRS is considered.

dfib(s(s(x)),y) → dfib(s(x),dfib(x,y)) (1)

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
[dfib(x1, x2)] =
1 0 0
0 0 0
0 0 0
· x1 +
1 0 1
0 0 0
0 0 0
· x2 +
0 0 0
0 0 0
1 0 0
[s(x1)] =
1 0 1
1 0 1
1 0 0
· x1 +
1 0 0
0 0 0
1 0 0
all of the following rules can be deleted.
dfib(s(s(x)),y) → dfib(s(x),dfib(x,y)) (1)

1.1 R is empty

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