Certification Problem

Input (TPDB TRS_Standard/SK90/2.04)

The rewrite relation of the following TRS is considered.

f(+(x,0)) → f(x) (1)
+(x,+(y,z)) → +(+(x,y),z) (2)

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
[+(x1, x2)] =
1 0 0
0 0 0
0 0 0
· x1 +
1 0 0
0 0 0
1 0 0
· x2 +
0 0 0
0 0 0
1 0 0
[0] =
1 0 0
0 0 0
0 0 0
[f(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(+(x,0)) → f(x) (1)

1.1 Rule Removal

Using the linear polynomial interpretation over (3 x 3)-matrices with strict dimension 1 over the naturals
[+(x1, x2)] =
1 0 0
0 1 0
1 1 0
· x1 +
1 1 0
0 1 0
0 1 1
· x2 +
0 0 0
1 0 0
0 0 0
all of the following rules can be deleted.
+(x,+(y,z)) → +(+(x,y),z) (2)

1.1.1 R is empty

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