Certification Problem

Input (TPDB TRS_Standard/Rubio_04/gmnp)

The rewrite relation of the following TRS is considered.

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

1.1 Rule Removal

Using the linear polynomial interpretation over (5 x 5)-matrices with strict dimension 1 over the naturals
[d(x1)] =
1 1 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 1
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
[b] =
0 0 0 0 0
1 0 0 0 0
0 0 0 0 0
1 0 0 0 0
0 0 0 0 0
[g(x1)] =
1 0 0 0 0
0 0 1 0 0
0 0 0 0 0
0 0 0 0 0
0 1 0 0 1
· x1 +
0 0 0 0 0
0 0 0 0 0
0 0 0 0 0
0 0 0 0 0
1 0 0 0 0
[a] =
0 0 0 0 0
0 0 0 0 0
0 0 0 0 0
0 0 0 0 0
1 0 0 0 0
[e(x1)] =
1 1 1 0 1
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
[c(x1)] =
1 0 0 0 1
0 0 0 0 0
0 0 0 1 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(a) → f(d(a)) (4)
f(c(b)) → f(d(a)) (6)
e(g(X)) → e(X) (7)

1.1.1 Rule Removal

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

1.1.1.1 Rule Removal

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

1.1.1.1.1 R is empty

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