Certification Problem

Input (TPDB TRS_Relative/INVY_15/#3.24_rand)

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

f(0) → s(0) (1)
f(s(0)) → s(0) (2)
f(s(s(x))) → f(f(s(x))) (3)

and S is the following TRS.

rand(x) → x (4)
rand(x) → rand(s(x)) (5)

Property / Task

Prove or disprove termination.

Answer / Result

Yes.

Proof (by ttt2 @ termCOMP 2023)

1 Rule Removal

Using the linear polynomial interpretation over (4 x 4)-matrices with strict dimension 1 over the naturals
[f(x1)] =
1 0 0 0
0 0 0 0
0 0 0 0
0 1 1 1
· x1 +
0 0 0 0
0 0 0 0
0 0 0 0
0 0 0 0
[rand(x1)] =
1 1 0 0
1 1 0 0
0 1 1 1
1 1 1 1
· x1 +
1 0 0 0
0 0 0 0
1 0 0 0
0 0 0 0
[0] =
0 0 0 0
1 0 0 0
0 0 0 0
0 0 0 0
[s(x1)] =
1 0 0 0
0 0 0 0
0 0 0 1
0 1 1 0
· x1 +
0 0 0 0
0 0 0 0
0 0 0 0
0 0 0 0
all of the following rules can be deleted.
rand(x) → 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 1 0
0 1 0
0 0 0
· x1 +
0 0 0
0 0 0
1 0 0
[rand(x1)] =
1 0 0
0 0 0
1 0 0
· x1 +
1 0 0
0 0 0
0 0 0
[0] =
0 0 0
1 0 0
0 0 0
[s(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.
f(0) → s(0) (1)

1.1.1 Rule Removal

Using the linear polynomial interpretation over (3 x 3)-matrices with strict dimension 1 over the naturals
[f(x1)] =
1 1 0
0 0 0
0 0 0
· x1 +
0 0 0
1 0 0
1 0 0
[rand(x1)] =
1 0 0
1 0 1
0 0 0
· x1 +
1 0 0
0 0 0
0 0 0
[0] =
0 0 0
0 0 0
1 0 0
[s(x1)] =
1 0 0
1 1 0
0 0 1
· x1 +
0 0 0
1 0 0
0 0 0
all of the following rules can be deleted.
f(s(0)) → s(0) (2)

1.1.1.1 Rule Removal

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

1.1.1.1.1 R is empty

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