Certification Problem

Input (TPDB SRS_Relative/Waldmann_06_relative/r7)

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

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

and S is the following TRS.

a(x1) → a(c(b(x1))) (3)

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(b(x1))) → x1 (1)
c(c(c(x1))) → a(a(x1)) (2)
a(x1) → b(c(a(x1))) (4)

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 1
0 0 1
0 0 1
· x1 +
0 0 0
0 0 0
1 0 0
[b(x1)] =
1 0 0
0 1 1
0 1 0
· x1 +
0 0 0
0 0 0
0 0 0
[a(x1)] =
1 0 1
0 0 0
0 0 0
· x1 +
0 0 0
1 0 0
0 0 0
all of the following rules can be deleted.
c(c(c(x1))) → a(a(x1)) (2)

1.1.1 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 0 0
· x1 +
0 0 0
0 0 0
0 0 0
[b(x1)] =
1 0 1
0 1 0
0 0 1
· x1 +
0 0 0
0 0 0
1 0 0
[a(x1)] =
1 0 0
0 1 1
1 1 0
· x1 +
1 0 0
1 0 0
1 0 0
all of the following rules can be deleted.
b(b(b(x1))) → x1 (1)

1.1.1.1 R is empty

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