Certification Problem

Input (TPDB TRS_Relative/Relative_05/rt1-5)

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

s(a(x)) → s(b(x)) (1)
b(b(x)) → a(x) (2)

and S is the following TRS.

f(s(x),y) → f(x,s(y)) (3)
s(a(x)) → a(s(x)) (4)
s(b(x)) → b(s(x)) (5)
a(s(x)) → s(a(x)) (6)
b(s(x)) → s(b(x)) (7)

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
[s(x1)] =
1 0 0
0 0 1
0 1 0
· x1 +
0 0 0
0 0 0
0 0 0
[f(x1, x2)] =
1 0 0
0 0 0
0 1 1
· x1 +
1 1 1
0 0 0
0 0 0
· x2 +
0 0 0
0 0 0
0 0 0
[a(x1)] =
1 0 0
0 1 0
0 0 1
· x1 +
1 0 0
0 0 0
0 0 0
[b(x1)] =
1 0 0
0 1 0
0 0 1
· x1 +
1 0 0
0 0 0
0 0 0
all of the following rules can be deleted.
b(b(x)) → a(x) (2)

1.1 Rule Removal

Using the linear polynomial interpretation over (3 x 3)-matrices with strict dimension 1 over the naturals
[s(x1)] =
1 0 0
0 1 0
0 0 0
· x1 +
0 0 0
1 0 0
0 0 0
[f(x1, x2)] =
1 1 0
0 1 0
0 0 0
· x1 +
1 0 0
0 1 0
0 0 0
· x2 +
0 0 0
0 0 0
0 0 0
[a(x1)] =
1 0 0
1 1 0
0 0 0
· x1 +
0 0 0
0 0 0
0 0 0
[b(x1)] =
1 0 0
0 1 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),y) → f(x,s(y)) (3)

1.1.1 Rule Removal

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

1.1.1.1 R is empty

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