YES Termination Proof

Termination Proof

by ttt2 (version ttt2 1.15)

Input

The rewrite relation of the following TRS is considered.

C(x0) → c(x0)
c(c(x0)) → x0
b(b(x0)) → B(x0)
B(B(x0)) → b(x0)
c(B(c(b(c(x0))))) → B(c(b(c(B(c(b(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
[c(x1)] = 0 · x1 + -∞
[B(x1)] = 0 · x1 + -∞
[C(x1)] = 12 · x1 + -∞
[b(x1)] = 0 · x1 + -∞
the rules
c(c(x0)) → x0
b(b(x0)) → B(x0)
B(B(x0)) → b(x0)
c(B(c(b(c(x0))))) → B(c(b(c(B(c(b(x0)))))))
b(B(x0)) → x0
B(b(x0)) → x0
remain.

1.1 Rule Removal

Using the linear polynomial interpretation over the arctic semiring over the integers
[c(x1)] = 1 · x1 + -∞
[B(x1)] = 0 · x1 + -∞
[b(x1)] = 0 · x1 + -∞
the rules
b(b(x0)) → B(x0)
B(B(x0)) → b(x0)
c(B(c(b(c(x0))))) → B(c(b(c(B(c(b(x0)))))))
b(B(x0)) → x0
B(b(x0)) → x0
remain.

1.1.1 String Reversal

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

1.1.1.1 Bounds

The given TRS is match-bounded by 1. This is shown by the following automaton.