Certification Problem

Input (TPDB SRS_Relative/Zantema_06_relative/rel08)

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

a(x1) → x1 (1)
b(x1) → x1 (2)
a(c(x1)) → b(b(c(a(x1)))) (3)
d(b(b(b(x1)))) → a(d(a(x1))) (4)

and S is the following TRS.

a(b(x1)) → b(a(x1)) (5)
b(a(x1)) → a(b(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
a(x1) → x1 (1)
b(x1) → x1 (2)
c(a(x1)) → a(c(b(b(x1)))) (7)
b(b(b(d(x1)))) → a(d(a(x1))) (8)
b(a(x1)) → a(b(x1)) (6)
a(b(x1)) → b(a(x1)) (5)

1.1 Rule Removal

Using the linear polynomial interpretation over (2 x 2)-matrices with strict dimension 1 over the naturals
[b(x1)] =
1 0
0 1
· x1 +
0 0
1 0
[d(x1)] =
4 0
4 0
· x1 +
0 0
0 0
[a(x1)] =
1 0
0 1
· x1 +
0 0
3 0
[c(x1)] =
2 1
0 3
· x1 +
0 0
3 0
all of the following rules can be deleted.
c(a(x1)) → a(c(b(b(x1)))) (7)

1.1.1 Rule Removal

Using the linear polynomial interpretation over (3 x 3)-matrices with strict dimension 1 over the naturals
[b(x1)] =
1 0 0
1 1 0
0 1 1
· x1 +
1 0 0
0 0 0
0 0 0
[d(x1)] =
1 0 1
0 0 0
0 0 0
· x1 +
0 0 0
0 0 0
0 0 0
[a(x1)] =
1 0 0
0 1 0
0 0 1
· x1 +
0 0 0
0 0 0
1 0 0
all of the following rules can be deleted.
b(x1) → x1 (2)
b(b(b(d(x1)))) → a(d(a(x1))) (8)

1.1.1.1 Rule Removal

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

1.1.1.1.1 R is empty

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