Certification Problem

Input (TPDB SRS_Standard/Zantema_04/z066)

The rewrite relation of the following TRS is considered.

a(b(c(x1))) → c(b(a(x1))) (1)
C(B(A(x1))) → A(B(C(x1))) (2)
b(a(C(x1))) → C(a(b(x1))) (3)
c(A(B(x1))) → B(A(c(x1))) (4)
A(c(b(x1))) → b(c(A(x1))) (5)
B(C(a(x1))) → a(C(B(x1))) (6)
a(A(x1)) → x1 (7)
A(a(x1)) → x1 (8)
b(B(x1)) → x1 (9)
B(b(x1)) → x1 (10)
c(C(x1)) → x1 (11)
C(c(x1)) → x1 (12)

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
c(b(a(x1))) → a(b(c(x1))) (13)
A(B(C(x1))) → C(B(A(x1))) (14)
C(a(b(x1))) → b(a(C(x1))) (15)
B(A(c(x1))) → c(A(B(x1))) (16)
b(c(A(x1))) → A(c(b(x1))) (17)
a(C(B(x1))) → B(C(a(x1))) (18)
A(a(x1)) → x1 (8)
a(A(x1)) → x1 (7)
B(b(x1)) → x1 (10)
b(B(x1)) → x1 (9)
C(c(x1)) → x1 (12)
c(C(x1)) → x1 (11)

1.1 Rule Removal

Using the linear polynomial interpretation over the arctic semiring over the integers
[B(x1)] = 3 · x1 + -∞
[b(x1)] = 1 · x1 + -∞
[A(x1)] = 1 · x1 + -∞
[C(x1)] = 10 · x1 + -∞
[c(x1)] = 0 · x1 + -∞
[a(x1)] = 1 · x1 + -∞
all of the following rules can be deleted.
A(a(x1)) → x1 (8)
a(A(x1)) → x1 (7)
B(b(x1)) → x1 (10)
b(B(x1)) → x1 (9)
C(c(x1)) → x1 (12)
c(C(x1)) → x1 (11)

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 1
0 0 0
0 1 0
· x1 +
0 0 0
0 0 0
0 0 0
[b(x1)] =
1 0 0
0 1 1
0 0 0
· x1 +
0 0 0
0 0 0
0 0 0
[A(x1)] =
1 1 1
1 1 1
0 0 0
· x1 +
0 0 0
0 0 0
0 0 0
[C(x1)] =
1 0 1
1 1 1
0 1 0
· x1 +
1 0 0
1 0 0
0 0 0
[c(x1)] =
1 0 0
0 0 0
0 1 1
· x1 +
0 0 0
0 0 0
0 0 0
[a(x1)] =
1 1 1
0 0 1
1 1 0
· x1 +
1 0 0
0 0 0
1 0 0
all of the following rules can be deleted.
A(B(C(x1))) → C(B(A(x1))) (14)

1.1.1.1 Rule Removal

Using the linear polynomial interpretation over (3 x 3)-matrices with strict dimension 1 over the naturals
[B(x1)] =
1 0 0
0 0 1
0 1 0
· x1 +
0 0 0
1 0 0
1 0 0
[b(x1)] =
1 0 0
0 1 0
0 0 1
· x1 +
0 0 0
0 0 0
0 0 0
[A(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
1 0 1
1 1 0
· x1 +
0 0 0
1 0 0
1 0 0
[c(x1)] =
1 0 0
0 0 0
0 0 0
· x1 +
0 0 0
0 0 0
0 0 0
[a(x1)] =
1 0 0
0 0 1
0 1 0
· x1 +
0 0 0
0 0 0
0 0 0
all of the following rules can be deleted.
a(C(B(x1))) → B(C(a(x1))) (18)

1.1.1.1.1 Rule Removal

Using the Knuth Bendix order with w0 = 1 and the following precedence and weight functions
prec(C) = 7 weight(C) = 2
prec(B) = 3 weight(B) = 2
prec(A) = 2 weight(A) = 2
prec(a) = 0 weight(a) = 2
prec(b) = 6 weight(b) = 2
prec(c) = 1 weight(c) = 4
all of the following rules can be deleted.
c(b(a(x1))) → a(b(c(x1))) (13)
C(a(b(x1))) → b(a(C(x1))) (15)
B(A(c(x1))) → c(A(B(x1))) (16)
b(c(A(x1))) → A(c(b(x1))) (17)

1.1.1.1.1.1 R is empty

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