Certification Problem

Input (TPDB SRS_Relative/Zantema_06_relative/rel01)

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

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

and S is the following TRS.

a(x1) → a(c(a(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(a(a(x1))) → a(b(x1)) (4)
b(c(x1)) → a(c(b(x1))) (5)
a(x1) → a(c(a(x1))) (3)

1.1 Rule Removal

Using the linear polynomial interpretation over (3 x 3)-matrices with strict dimension 1 over the naturals
[a(x1)] =
1 0 0
0 1 0
0 1 1
· x1 +
0 0 0
1 0 0
0 0 0
[b(x1)] =
1 0 1
0 1 0
0 0 1
· x1 +
0 0 0
1 0 0
0 0 0
[c(x1)] =
1 0 0
0 0 0
0 0 1
· x1 +
0 0 0
0 0 0
0 0 0
all of the following rules can be deleted.
b(a(a(x1))) → a(b(x1)) (4)

1.1.1 Rule Removal

Using the linear polynomial interpretation over (3 x 3)-matrices with strict dimension 1 over the naturals
[a(x1)] =
1 0 0
1 0 0
1 0 0
· x1 +
0 0 0
1 0 0
0 0 0
[b(x1)] =
1 0 1
1 0 1
1 0 1
· x1 +
0 0 0
1 0 0
1 0 0
[c(x1)] =
1 0 0
1 0 1
0 0 1
· x1 +
0 0 0
1 0 0
1 0 0
all of the following rules can be deleted.
b(c(x1)) → a(c(b(x1))) (5)

1.1.1.1 R is empty

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