Certification Problem

Input (TPDB SRS_Relative/Mixed_relative_SRS/zr08)

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

b(b(x1)) → c(d(x1)) (1)
c(c(x1)) → d(d(d(x1))) (2)
d(d(d(x1))) → a(c(x1)) (3)

and S is the following TRS.

a(a(x1)) → b(c(x1)) (4)
b(c(x1)) → a(a(x1)) (5)

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
b(b(x1)) → d(c(x1)) (6)
c(c(x1)) → d(d(d(x1))) (2)
d(d(d(x1))) → c(a(x1)) (7)
a(a(x1)) → c(b(x1)) (8)
c(b(x1)) → a(a(x1)) (9)

1.1 Rule Removal

Using the linear polynomial interpretation over the arctic semiring over the integers
[d(x1)] = 6 · x1 + -∞
[a(x1)] = 9 · x1 + -∞
[b(x1)] = 9 · x1 + -∞
[c(x1)] = 9 · x1 + -∞
all of the following rules can be deleted.
b(b(x1)) → d(c(x1)) (6)

1.1.1 Rule Removal

Using the linear polynomial interpretation over (3 x 3)-matrices with strict dimension 1 over the naturals
[d(x1)] =
1 1 0
0 0 0
0 0 0
· x1 +
0 0 0
1 0 0
1 0 0
[a(x1)] =
1 0 0
0 0 0
0 1 0
· x1 +
0 0 0
1 0 0
0 0 0
[b(x1)] =
1 0 0
0 0 0
0 0 0
· x1 +
0 0 0
0 0 0
0 0 0
[c(x1)] =
1 1 1
0 0 0
0 0 0
· x1 +
0 0 0
1 0 0
1 0 0
all of the following rules can be deleted.
d(d(d(x1))) → c(a(x1)) (7)

1.1.1.1 Rule Removal

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

1.1.1.1.1 R is empty

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