YES Termination Proof

Termination Proof

by ttt2 (version ttt2 1.15)

Input

The rewrite relation of the following TRS is considered.

P(x0) → Q(Q(p(x0)))
p(p(x0)) → q(q(x0))
p(Q(Q(x0))) → Q(Q(p(x0)))
Q(p(q(x0))) → q(p(Q(x0)))
q(q(p(x0))) → p(q(q(x0)))
q(Q(x0)) → x0
Q(q(x0)) → x0
p(P(x0)) → x0
P(p(x0)) → x0

Proof

1 Rule Removal

Using the linear polynomial interpretation over the arctic semiring over the integers
[p(x1)] = 0 · x1 + -∞
[q(x1)] = 0 · x1 + -∞
[P(x1)] = 9 · x1 + -∞
[Q(x1)] = 0 · x1 + -∞
the rules
p(p(x0)) → q(q(x0))
p(Q(Q(x0))) → Q(Q(p(x0)))
Q(p(q(x0))) → q(p(Q(x0)))
q(q(p(x0))) → p(q(q(x0)))
q(Q(x0)) → x0
Q(q(x0)) → x0
remain.

1.1 Rule Removal

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

1.1.1 Rule Removal

Using the linear polynomial interpretation over the arctic semiring over the integers
[p(x1)] = 2 · x1 + -∞
[q(x1)] = 1 · x1 + -∞
[Q(x1)] = 4 · x1 + -∞
the rules
p(Q(Q(x0))) → Q(Q(p(x0)))
Q(p(q(x0))) → q(p(Q(x0)))
q(q(p(x0))) → p(q(q(x0)))
remain.

1.1.1.1 String Reversal

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

1.1.1.1.1 Bounds

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