YES Termination Proof

Termination Proof

by ttt2 (version ttt2 1.15)

Input

The rewrite relation of the following TRS is considered.

a(b(x0)) → C(x0)
b(c(x0)) → A(x0)
c(a(x0)) → B(x0)
A(C(x0)) → b(x0)
C(B(x0)) → a(x0)
B(A(x0)) → c(x0)
a(a(a(a(x0)))) → A(A(A(x0)))
A(A(A(A(x0)))) → a(a(a(x0)))
b(b(b(b(x0)))) → B(B(B(x0)))
B(B(B(B(x0)))) → b(b(b(x0)))
c(c(c(c(x0)))) → C(C(C(x0)))
C(C(C(C(x0)))) → c(c(c(x0)))
B(a(a(a(x0)))) → c(A(A(A(x0))))
A(A(A(b(x0)))) → a(a(a(C(x0))))
C(b(b(b(x0)))) → a(B(B(B(x0))))
B(B(B(c(x0)))) → b(b(b(A(x0))))
A(c(c(c(x0)))) → b(C(C(C(x0))))
C(C(C(a(x0)))) → c(c(c(B(x0))))
a(A(x0)) → x0
A(a(x0)) → x0
b(B(x0)) → x0
B(b(x0)) → x0
c(C(x0)) → x0
C(c(x0)) → x0

Proof

1 Rule Removal

Using the Knuth Bendix order with w0 = 1 and the following precedence and weight function
prec(B) = 4 weight(B) = 1
prec(A) = 1 weight(A) = 1
prec(c) = 0 weight(c) = 1
prec(C) = 7 weight(C) = 1
prec(a) = 0 weight(a) = 1
prec(b) = 0 weight(b) = 1
all rules could be removed.

1.1 R is empty

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