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(a(x0))))) → A(A(A(x0)))
A(A(A(A(x0)))) → a(a(a(a(x0))))
b(b(b(b(b(x0))))) → B(B(B(x0)))
B(B(B(B(x0)))) → b(b(b(b(x0))))
c(c(c(c(c(x0))))) → C(C(C(x0)))
C(C(C(C(x0)))) → c(c(c(c(x0))))
B(a(a(a(a(x0))))) → c(A(A(A(x0))))
A(A(A(b(x0)))) → a(a(a(a(C(x0)))))
C(b(b(b(b(x0))))) → a(B(B(B(x0))))
B(B(B(c(x0)))) → b(b(b(b(A(x0)))))
A(c(c(c(c(x0))))) → b(C(C(C(x0))))
C(C(C(a(x0)))) → c(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 linear polynomial interpretation over the arctic semiring over the integers
[A(x1)] = 3 · x1 + -∞
[a(x1)] = 2 · x1 + -∞
[c(x1)] = 2 · x1 + -∞
[B(x1)] = 3 · x1 + -∞
[b(x1)] = 2 · x1 + -∞
[C(x1)] = 3 · x1 + -∞
the rules
B(a(a(a(a(x0))))) → c(A(A(A(x0))))
A(A(A(b(x0)))) → a(a(a(a(C(x0)))))
C(b(b(b(b(x0))))) → a(B(B(B(x0))))
B(B(B(c(x0)))) → b(b(b(b(A(x0)))))
A(c(c(c(c(x0))))) → b(C(C(C(x0))))
C(C(C(a(x0)))) → c(c(c(c(B(x0)))))
remain.

1.1 String Reversal

Since only unary symbols occur, one can reverse all terms and obtains the TRS
a(a(a(a(B(x0))))) → A(A(A(c(x0))))
b(A(A(A(x0)))) → C(a(a(a(a(x0)))))
b(b(b(b(C(x0))))) → B(B(B(a(x0))))
c(B(B(B(x0)))) → A(b(b(b(b(x0)))))
c(c(c(c(A(x0))))) → C(C(C(b(x0))))
a(C(C(C(x0)))) → B(c(c(c(c(x0)))))

1.1.1 Bounds

The given TRS is match-bounded by 0. This is shown by the following automaton.