Certification Problem

Input (TPDB TRS_Relative/INVY_15/#3.15_rand)

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

average(s(x),y) → average(x,s(y)) (1)
average(x,s(s(s(y)))) → s(average(s(x),y)) (2)
average(0,0) → 0 (3)
average(0,s(0)) → 0 (4)
average(0,s(s(0))) → s(0) (5)

and S is the following TRS.

rand(x) → rand(s(x)) (6)
rand(x) → x (7)

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
[average(x1, x2)] = 1 · x1 + 1 · x2
[s(x1)] = 1 · x1
[0] = 0
[rand(x1)] = 1 + 1 · x1
all of the following rules can be deleted.
rand(x) → x (7)

1.1 Rule Removal

Using the linear polynomial interpretation over the naturals
[average(x1, x2)] = 1 + 1 · x1 + 1 · x2
[s(x1)] = 1 · x1
[0] = 0
[rand(x1)] = 1 · x1
all of the following rules can be deleted.
average(0,0) → 0 (3)
average(0,s(0)) → 0 (4)
average(0,s(s(0))) → s(0) (5)

1.1.1 Rule Removal

Using the matrix interpretations of dimension 2 with strict dimension 1 over the integers
[average(x1, x2)] =
0
0
+
1 1
1 1
· x1 +
1 1
0 1
· x2
[s(x1)] =
0
1
+
1 0
0 1
· x1
[rand(x1)] =
1
1
+
1 0
1 0
· x1
all of the following rules can be deleted.
average(x,s(s(s(y)))) → s(average(s(x),y)) (2)

1.1.1.1 Rule Removal

Using the matrix interpretations of dimension 2 with strict dimension 1 over the integers
[average(x1, x2)] =
1
1
+
1 1
1 1
· x1 +
1 0
1 1
· x2
[s(x1)] =
0
1
+
1 0
0 1
· x1
[rand(x1)] =
1
1
+
1 0
1 0
· x1
all of the following rules can be deleted.
average(s(x),y) → average(x,s(y)) (1)

1.1.1.1.1 R is empty

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