Certification Problem

Input (TPDB TRS_Standard/AG01/#3.51)

The rewrite relation of the following TRS is considered.

f(f(x)) f(c(f(x))) (1)
f(f(x)) f(d(f(x))) (2)
g(c(x)) x (3)
g(d(x)) x (4)
g(c(h(0))) g(d(1)) (5)
g(c(1)) g(d(h(0))) (6)
g(h(x)) g(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
[0] =
0 0 0
0 0 0
1 0 0
[c(x1)] =
1 0 0
1 0 0
1 1 1
· x1 +
0 0 0
1 0 0
0 0 0
[g(x1)] =
1 1 0
1 0 1
1 1 1
· x1 +
1 0 0
0 0 0
0 0 0
[h(x1)] =
1 0 0
0 1 0
0 0 1
· x1 +
0 0 0
0 0 0
1 0 0
[f(x1)] =
1 0 0
0 0 0
0 0 0
· x1 +
0 0 0
0 0 0
0 0 0
[1] =
0 0 0
1 0 0
1 0 0
[d(x1)] =
1 0 0
0 0 0
0 1 1
· x1 +
0 0 0
0 0 0
0 0 0
all of the following rules can be deleted.
g(c(x)) x (3)
g(d(x)) x (4)
g(c(h(0))) g(d(1)) (5)
g(c(1)) g(d(h(0))) (6)

1.1 Rule Removal

Using the linear polynomial interpretation over (3 x 3)-matrices with strict dimension 1 over the naturals
[c(x1)] =
1 1 0
0 0 0
0 0 0
· x1 +
0 0 0
0 0 0
0 0 0
[g(x1)] =
1 0 1
0 0 0
0 0 0
· x1 +
1 0 0
1 0 0
0 0 0
[h(x1)] =
1 0 0
0 0 0
0 1 1
· x1 +
0 0 0
0 0 0
1 0 0
[f(x1)] =
1 1 0
1 1 0
0 0 0
· x1 +
0 0 0
0 0 0
0 0 0
[d(x1)] =
1 1 0
0 0 0
1 1 0
· x1 +
0 0 0
0 0 0
0 0 0
all of the following rules can be deleted.
g(h(x)) g(x) (7)

1.1.1 Rule Removal

Using the linear polynomial interpretation over (5 x 5)-matrices with strict dimension 1 over the naturals
[c(x1)] =
1 0 0 0 0
0 0 0 0 0
0 0 0 0 0
0 0 0 0 0
0 0 0 0 0
· x1 +
0 0 0 0 0
0 0 0 0 0
0 0 0 0 0
0 0 0 0 0
0 0 0 0 0
[f(x1)] =
1 0 1 0 0
0 0 0 0 0
0 0 0 0 0
0 0 0 0 0
0 0 0 0 0
· x1 +
0 0 0 0 0
0 0 0 0 0
1 0 0 0 0
0 0 0 0 0
0 0 0 0 0
[d(x1)] =
1 0 0 0 0
0 0 0 0 0
0 0 0 0 0
0 0 0 0 0
0 0 0 0 0
· x1 +
0 0 0 0 0
0 0 0 0 0
0 0 0 0 0
0 0 0 0 0
0 0 0 0 0
all of the following rules can be deleted.
f(f(x)) f(c(f(x))) (1)
f(f(x)) f(d(f(x))) (2)

1.1.1.1 R is empty

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