Certification Problem

Input (TPDB TRS_Relative/Relative_05/rt3-2)

The relative rewrite relation R/S is considered where R is the following TRS

p(0,y) → y (1)
p(s(x),y) → s(p(x,y)) (2)

and S is the following TRS.

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

1.1 Rule Removal

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

1.1.1 R is empty

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