Certification Problem

Input (TPDB TRS_Relative/INVY_15/#3.54_rand)

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

f(g(x)) → g(f(f(x))) (1)
f(h(x)) → h(g(x)) (2)
f'(s(x),y,y) → f'(y,x,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
[g(x1)] = 1 · x1
[h(x1)] = 2 · x1
[f'(x1, x2, x3)] = 3 · x1 + 1 · x2 + 2 · x3
[s(x1)] = 1 · x1
[rand(x1)] = 2 + 7 · 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 0
0 0
· x1
[g(x1)] =
0
0
+
1 0
0 0
· x1
[h(x1)] =
0
0
+
1 0
0 0
· x1
[f'(x1, x2, x3)] =
1
1
+
1 1
1 0
· x1 +
1 1
0 0
· x2 +
1 0
1 0
· x3
[s(x1)] =
0
1
+
1 0
1 1
· x1
[rand(x1)] =
1
1
+
1 0
1 0
· x1
all of the following rules can be deleted.
f'(s(x),y,y) → f'(y,x,s(x)) (3)

1.1.1 Rule Removal

Using the matrix interpretations of dimension 2 with strict dimension 1 over the integers
[f(x1)] =
0
0
+
1 1
0 1
· x1
[g(x1)] =
0
0
+
1 0
0 2
· x1
[h(x1)] =
0
1
+
1 0
0 0
· x1
[rand(x1)] =
3
3
+
3 3
3 3
· x1
[s(x1)] =
0
0
+
1 1
0 0
· x1
all of the following rules can be deleted.
f(h(x)) → h(g(x)) (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 1
0 1
· x1
[g(x1)] =
0
1
+
1 0
0 2
· x1
[rand(x1)] =
3
3
+
3 3
3 3
· x1
[s(x1)] =
0
0
+
1 1
0 0
· x1
all of the following rules can be deleted.
f(g(x)) → g(f(f(x))) (1)

1.1.1.1.1 R is empty

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