Certification Problem

Input (TPDB TRS_Relative/INVY_15/#3.49_rand)

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

f(c(s(x),y)) → f(c(x,s(y))) (1)
f(c(s(x),s(y))) → g(c(x,y)) (2)
g(c(x,s(y))) → g(c(s(x),y)) (3)
g(c(s(x),s(y))) → f(c(x,y)) (4)

and S is the following TRS.

rand(x) → rand(s(x)) (5)
rand(x) → x (6)

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
[c(x1, x2)] = 1 · x1 + 1 · x2
[s(x1)] = 1 · x1
[g(x1)] = 1 · x1
[rand(x1)] = 1 + 1 · x1
all of the following rules can be deleted.
rand(x) → x (6)

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
[c(x1, x2)] =
0
0
+
1 1
0 0
· x1 +
1 1
0 0
· x2
[s(x1)] =
0
1
+
1 0
0 1
· x1
[g(x1)] =
0
0
+
1 0
0 0
· x1
[rand(x1)] =
1
1
+
1 0
1 0
· x1
all of the following rules can be deleted.
f(c(s(x),s(y))) → g(c(x,y)) (2)
g(c(s(x),s(y))) → f(c(x,y)) (4)

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

1.1.1.1 Rule Removal

Using the matrix interpretations of dimension 2 with strict dimension 1 over the integers
[g(x1)] =
0
0
+
1 1
0 0
· x1
[c(x1, x2)] =
0
0
+
1 0
1 0
· 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.
g(c(x,s(y))) → g(c(s(x),y)) (3)

1.1.1.1.1 R is empty

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