Certification Problem

Input (TPDB TRS_Standard/Secret_06_TRS/gen-1)

The rewrite relation of the following TRS is considered.

f(c(a,z,x)) → b(a,z) (1)
b(x,b(z,y)) → f(b(f(f(z)),c(x,z,y))) (2)
b(y,z) → z (3)

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
[c(x1, x2, x3)] =
1 0 0
0 0 0
0 0 0
· x1 +
1 1 1
0 0 0
0 0 0
· x2 +
1 1 0
0 1 0
0 0 0
· x3 +
1 0 0
1 0 0
0 0 0
[b(x1, x2)] =
1 0 0
1 1 0
1 0 1
· x1 +
1 1 1
0 1 1
1 1 1
· x2 +
0 0 0
1 0 0
1 0 0
[a] =
1 0 0
0 0 0
0 0 0
[f(x1)] =
1 0 0
1 0 0
1 0 0
· x1 +
0 0 0
0 0 0
0 0 0
all of the following rules can be deleted.
f(c(a,z,x)) → b(a,z) (1)

1.1 Rule Removal

Using the linear polynomial interpretation over (3 x 3)-matrices with strict dimension 1 over the naturals
[c(x1, x2, x3)] =
1 0 1
0 0 0
0 0 0
· x1 +
1 0 1
1 0 1
0 0 1
· x2 +
1 0 0
0 0 1
0 0 0
· x3 +
0 0 0
0 0 0
1 0 0
[b(x1, x2)] =
1 0 1
0 0 0
1 0 1
· x1 +
1 0 1
0 1 0
0 1 1
· x2 +
1 0 0
0 0 0
0 0 0
[f(x1)] =
1 0 0
0 0 0
0 1 0
· x1 +
0 0 0
0 0 0
0 0 0
all of the following rules can be deleted.
b(y,z) → z (3)

1.1.1 Rule Removal

Using the linear polynomial interpretation over (3 x 3)-matrices with strict dimension 1 over the naturals
[c(x1, x2, x3)] =
1 0 0
0 0 0
0 1 1
· x1 +
1 0 1
1 0 0
0 0 1
· x2 +
1 0 0
0 0 0
0 0 0
· x3 +
0 0 0
1 0 0
0 0 0
[b(x1, x2)] =
1 1 1
0 0 0
1 0 1
· x1 +
1 0 1
0 1 0
1 1 0
· x2 +
1 0 0
1 0 0
1 0 0
[f(x1)] =
1 0 0
0 0 0
0 0 0
· x1 +
0 0 0
1 0 0
0 0 0
all of the following rules can be deleted.
b(x,b(z,y)) → f(b(f(f(z)),c(x,z,y))) (2)

1.1.1.1 R is empty

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