Certification Problem

Input (TPDB SRS_Relative/Zantema_06_relative/rel09)

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

b(q(b(x1))) → b(p(b(x1))) (1)

and S is the following TRS.

0(p(0(x1))) → q(x1) (2)
1(p(1(x1))) → q(x1) (3)
0(q(0(x1))) → q(x1) (4)
1(q(1(x1))) → q(x1) (5)
p(x1) → 1(p(1(0(1(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
b(q(b(x1))) → b(p(b(x1))) (1)
0(p(0(x1))) → q(x1) (2)
1(p(1(x1))) → q(x1) (3)
0(q(0(x1))) → q(x1) (4)
1(q(1(x1))) → q(x1) (5)
p(x1) → 1(0(1(p(1(x1))))) (7)

1.1 Rule Removal

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

1.1.1 Rule Removal

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

1.1.1.1 R is empty

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