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) → x (5)
rand(x) → rand(s(x)) (6)

Property / Task

Prove or disprove termination.

Answer / Result

Yes.

Proof (by ttt2 @ termCOMP 2023)

1 Rule Removal

Using the linear polynomial interpretation over the naturals
[rand(x1)] = 1 · x1 + 8
[c(x1, x2)] = 8 · x1 + 12 · x2 + 0
[g(x1)] = 16 · x1 + 12
[s(x1)] = 1 · x1 + 0
[f(x1)] = 16 · x1 + 12
all of the following rules can be deleted.
rand(x) → x (5)

1.1 Rule Removal

Using the linear polynomial interpretation over (3 x 3)-matrices with strict dimension 1 over the naturals
[rand(x1)] =
1 0 0
1 1 0
0 0 0
· x1 +
0 0 0
0 0 0
0 0 0
[c(x1, x2)] =
1 0 1
0 0 0
0 0 0
· x1 +
1 0 0
1 0 1
1 0 1
· x2 +
1 0 0
0 0 0
0 0 0
[g(x1)] =
1 1 0
0 0 0
0 0 0
· x1 +
0 0 0
0 0 0
0 0 0
[s(x1)] =
1 0 0
0 0 0
0 0 1
· x1 +
0 0 0
0 0 0
1 0 0
[f(x1)] =
1 0 1
0 0 0
0 0 0
· x1 +
0 0 0
0 0 0
0 0 0
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 linear polynomial interpretation over (3 x 3)-matrices with strict dimension 1 over the naturals
[rand(x1)] =
1 0 1
0 0 1
1 0 0
· x1 +
1 0 0
0 0 0
0 0 0
[c(x1, x2)] =
1 0 0
0 1 0
0 0 0
· x1 +
1 0 0
0 1 1
1 1 0
· x2 +
0 0 0
1 0 0
0 0 0
[g(x1)] =
1 0 1
0 0 0
0 0 0
· x1 +
0 0 0
0 0 0
0 0 0
[s(x1)] =
1 0 0
0 1 0
0 0 0
· x1 +
0 0 0
1 0 0
0 0 0
[f(x1)] =
1 0 0
0 1 0
0 1 0
· x1 +
0 0 0
0 0 0
0 0 0
all of the following rules can be deleted.
g(c(x,s(y))) → g(c(s(x),y)) (3)

1.1.1.1 Rule Removal

Using the linear polynomial interpretation over (3 x 3)-matrices with strict dimension 1 over the naturals
[rand(x1)] =
1 0 0
0 0 0
0 0 0
· x1 +
0 0 0
0 0 0
1 0 0
[c(x1, x2)] =
1 1 0
0 1 1
0 0 0
· x1 +
1 0 0
0 0 0
0 0 0
· x2 +
0 0 0
0 0 0
0 0 0
[s(x1)] =
1 0 0
0 1 0
0 0 1
· x1 +
0 0 0
1 0 0
1 0 0
[f(x1)] =
1 0 0
0 1 0
0 0 0
· x1 +
1 0 0
0 0 0
0 0 0
all of the following rules can be deleted.
f(c(s(x),y)) → f(c(x,s(y))) (1)

1.1.1.1.1 R is empty

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