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) → rand(s(x)) (4)
rand(x) → x (5)

Property / Task

Prove or disprove termination.

Answer / Result

Yes.

Proof (by AProVE @ termCOMP 2023)

1 Rule Removal

Using the linear polynomial interpretation over the naturals
[f(x1)] = 1 · x1
[0] = 0
[s(x1)] = 1 · x1
[rand(x1)] = 1 + 1 · x1
all of the following rules can be deleted.
rand(x) → x (5)

1.1 Rule Removal

Using the matrix interpretations of dimension 2 with strict dimension 1 over the integers
[f(x1)] =
0
0
+
1 2
0 0
· x1
[0] =
0
2
[s(x1)] =
0
0
+
1 0
0 0
· x1
[rand(x1)] =
2
2
+
1 0
0 0
· x1
all of the following rules can be deleted.
f(0) → s(0) (1)

1.1.1 Rule Removal

Using the matrix interpretations of dimension 2 with strict dimension 1 over the integers
[f(x1)] =
0
2
+
2 2
0 0
· x1
[s(x1)] =
0
2
+
1 0
1 2
· x1
[0] =
0
0
[rand(x1)] =
1
2
+
1 0
0 0
· x1
all of the following rules can be deleted.
f(s(0)) → s(0) (2)

1.1.1.1 Rule Removal

Using the matrix interpretations of dimension 2 with strict dimension 1 over the integers
[f(x1)] =
0
0
+
1 2
0 1
· x1
[s(x1)] =
0
1
+
1 0
0 2
· x1
[rand(x1)] =
1
2
+
2 0
0 0
· x1
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.