Certification Problem

Input (TPDB TRS_Relative/Relative_05/rt1-3)

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

t(f(x),g(y),f(z)) → t(z,g(x),g(y)) (1)
t(g(x),g(y),f(z)) → t(f(y),f(z),x) (2)

and S is the following TRS.

f(g(x)) → g(f(x)) (3)
g(f(x)) → f(g(x)) (4)
f(f(x)) → g(g(x)) (5)
g(g(x)) → f(f(x)) (6)

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
[g(x1)] =
1 0 0
0 0 0
0 0 0
· x1 +
1 0 0
0 0 0
0 0 0
[f(x1)] =
1 0 0
0 0 0
0 0 0
· x1 +
1 0 0
0 0 0
0 0 0
[t(x1, x2, x3)] =
1 0 0
0 0 0
0 0 0
· x1 +
1 0 0
0 0 0
0 0 0
· x2 +
1 0 0
0 0 0
0 0 0
· x3 +
0 0 0
1 0 0
1 0 0
all of the following rules can be deleted.
t(f(x),g(y),f(z)) → t(z,g(x),g(y)) (1)
t(g(x),g(y),f(z)) → t(f(y),f(z),x) (2)

1.1 R is empty

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