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.