Certification Problem

Input (TPDB TRS_Standard/SK90/4.51)

The rewrite relation of the following TRS is considered.

f(a) → g(h(a)) (1)
h(g(x)) → g(h(f(x))) (2)
k(x,h(x),a) → h(x) (3)
k(f(x),y,x) → f(x) (4)

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
[k(x1, x2, x3)] =
1 0 0
1 1 1
1 0 1
· x1 +
1 0 0
1 1 0
0 1 0
· x2 +
1 0 1
0 1 0
0 1 0
· x3 +
1 0 0
0 0 0
1 0 0
[f(x1)] =
1 0 0
0 0 0
0 1 1
· x1 +
0 0 0
0 0 0
0 0 0
[g(x1)] =
1 0 0
0 0 0
0 0 0
· x1 +
0 0 0
0 0 0
0 0 0
[a] =
0 0 0
1 0 0
1 0 0
[h(x1)] =
1 0 0
0 1 1
1 0 0
· x1 +
0 0 0
0 0 0
1 0 0
all of the following rules can be deleted.
k(x,h(x),a) → h(x) (3)
k(f(x),y,x) → f(x) (4)

1.1 Rule Removal

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

1.1.1 Rule Removal

Using the Knuth Bendix order with w0 = 1 and the following precedence and weight functions
prec(g) = 0 weight(g) = 2
prec(h) = 1 weight(h) = 2
prec(f) = 3 weight(f) = 0
all of the following rules can be deleted.
h(g(x)) → g(h(f(x))) (2)

1.1.1.1 R is empty

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