Certification Problem

Input (TPDB TRS_Relative/INVY_15/ex4)

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

a → b (1)

and S is the following TRS.

f(s(x)) → c(x,f(x)) (2)
c(x,c(y,z)) → c(y,c(x,z)) (3)

Property / Task

Prove or disprove termination.

Answer / Result

Yes.

Proof (by ttt2 @ termCOMP 2023)

1 Rule Removal

Using the linear polynomial interpretation over (3 x 3)-matrices with strict dimension 1 over the naturals
[c(x1, x2)] =
1 1 0
0 0 0
0 0 0
· x1 +
1 0 0
0 0 0
0 0 0
· x2 +
0 0 0
1 0 0
0 0 0
[b] =
0 0 0
0 0 0
0 0 0
[f(x1)] =
1 1 1
0 0 0
0 0 0
· x1 +
0 0 0
1 0 0
0 0 0
[y] =
1 0 0
1 0 0
0 0 0
[a] =
1 0 0
0 0 0
0 0 0
[z] =
1 0 0
0 0 0
0 0 0
[s(x1)] =
1 1 0
1 1 0
1 0 1
· x1 +
1 0 0
1 0 0
1 0 0
all of the following rules can be deleted.
a → b (1)
f(s(x)) → c(x,f(x)) (2)

1.1 R is empty

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