Certification Problem

Input (TPDB SRS_Relative/Mixed_relative_SRS/zr01)

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

n(s(x1)) → s(x1) (1)
o(s(x1)) → s(x1) (2)

and S is the following TRS.

t(x1) → t(c(n(x1))) (3)
c(n(x1)) → n(c(x1)) (4)
c(o(x1)) → o(c(x1)) (5)
c(o(x1)) → o(x1) (6)

Property / Task

Prove or disprove termination.

Answer / Result

Yes.

Proof (by ttt2 @ termCOMP 2023)

1 String Reversal

Since only unary symbols occur, one can reverse all terms and obtains the TRS
s(n(x1)) → s(x1) (7)
s(o(x1)) → s(x1) (8)
t(x1) → n(c(t(x1))) (9)
n(c(x1)) → c(n(x1)) (10)
o(c(x1)) → c(o(x1)) (11)
o(c(x1)) → o(x1) (12)

1.1 Rule Removal

Using the linear polynomial interpretation over (3 x 3)-matrices with strict dimension 1 over the naturals
[c(x1)] =
1 0 0
0 0 0
0 1 0
· x1 +
0 0 0
0 0 0
0 0 0
[n(x1)] =
1 0 0
0 0 0
0 0 0
· x1 +
0 0 0
0 0 0
0 0 0
[t(x1)] =
1 0 1
0 1 0
0 0 0
· x1 +
0 0 0
1 0 0
1 0 0
[s(x1)] =
1 0 0
0 0 0
1 0 0
· x1 +
0 0 0
0 0 0
1 0 0
[o(x1)] =
1 0 0
0 0 0
0 0 0
· x1 +
1 0 0
0 0 0
0 0 0
all of the following rules can be deleted.
s(o(x1)) → s(x1) (8)

1.1.1 Rule Removal

Using the linear polynomial interpretation over (5 x 5)-matrices with strict dimension 1 over the naturals
[c(x1)] =
1 0 0 0 0
1 0 0 0 0
0 0 0 0 0
0 0 1 0 0
1 0 0 0 0
· x1 +
0 0 0 0 0
0 0 0 0 0
0 0 0 0 0
0 0 0 0 0
0 0 0 0 0
[n(x1)] =
1 0 0 0 0
0 1 1 0 0
0 0 0 0 0
0 0 0 0 0
0 1 0 0 0
· x1 +
0 0 0 0 0
0 0 0 0 0
1 0 0 0 0
1 0 0 0 0
0 0 0 0 0
[t(x1)] =
1 0 0 0 0
1 0 0 0 1
0 0 0 0 0
1 1 0 0 0
1 1 0 1 0
· x1 +
1 0 0 0 0
1 0 0 0 0
1 0 0 0 0
1 0 0 0 0
1 0 0 0 0
[s(x1)] =
1 1 1 0 0
0 0 0 0 0
0 0 0 0 0
0 0 0 0 0
0 0 0 0 0
· x1 +
0 0 0 0 0
0 0 0 0 0
0 0 0 0 0
0 0 0 0 0
1 0 0 0 0
[o(x1)] =
1 0 0 0 0
1 0 0 0 0
0 0 0 0 0
0 0 0 0 0
1 0 0 0 0
· x1 +
0 0 0 0 0
0 0 0 0 0
0 0 0 0 0
1 0 0 0 0
0 0 0 0 0
all of the following rules can be deleted.
s(n(x1)) → s(x1) (7)

1.1.1.1 R is empty

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