YES Termination Proof

Termination Proof

by ttt2 (version ttt2 1.15)

Input

The rewrite relation of the following TRS is considered.

R(x0) → r(x0)
r(p(x0)) → p(p(r(P(x0))))
r(r(x0)) → x0
r(P(P(x0))) → P(P(r(x0)))
p(P(x0)) → x0
P(p(x0)) → x0
r(R(x0)) → x0
R(r(x0)) → x0

Proof

1 Rule Removal

Using the linear polynomial interpretation over the arctic semiring over the integers
[r(x1)] = 8 · x1 + -∞
[P(x1)] = 0 · x1 + -∞
[R(x1)] = 8 · x1 + -∞
[p(x1)] = 0 · x1 + -∞
the rules
R(x0) → r(x0)
r(p(x0)) → p(p(r(P(x0))))
r(P(P(x0))) → P(P(r(x0)))
p(P(x0)) → x0
P(p(x0)) → x0
remain.

1.1 Rule Removal

Using the linear polynomial interpretation over the arctic semiring over the integers
[r(x1)] = 1 · x1 + -∞
[P(x1)] = 0 · x1 + -∞
[R(x1)] = 10 · x1 + -∞
[p(x1)] = 0 · x1 + -∞
the rules
r(p(x0)) → p(p(r(P(x0))))
r(P(P(x0))) → P(P(r(x0)))
p(P(x0)) → x0
P(p(x0)) → x0
remain.

1.1.1 String Reversal

Since only unary symbols occur, one can reverse all terms and obtains the TRS
p(r(x0)) → P(r(p(p(x0))))
P(P(r(x0))) → r(P(P(x0)))
P(p(x0)) → x0
p(P(x0)) → x0

1.1.1.1 Bounds

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