Termination w.r.t. Q of the following Term Rewriting System could be proven:
Q restricted rewrite system:
The TRS R consists of the following rules:
f(0) → 0
f(s(0)) → s(0)
f(s(s(x))) → p(h(g(x)))
g(0) → pair(s(0), s(0))
g(s(x)) → h(g(x))
h(x) → pair(+(p(x), q(x)), p(x))
p(pair(x, y)) → x
q(pair(x, y)) → y
+(x, 0) → x
+(x, s(y)) → s(+(x, y))
f(s(s(x))) → +(p(g(x)), q(g(x)))
g(s(x)) → pair(+(p(g(x)), q(g(x))), p(g(x)))
Q is empty.
↳ QTRS
↳ DirectTerminationProof
Q restricted rewrite system:
The TRS R consists of the following rules:
f(0) → 0
f(s(0)) → s(0)
f(s(s(x))) → p(h(g(x)))
g(0) → pair(s(0), s(0))
g(s(x)) → h(g(x))
h(x) → pair(+(p(x), q(x)), p(x))
p(pair(x, y)) → x
q(pair(x, y)) → y
+(x, 0) → x
+(x, s(y)) → s(+(x, y))
f(s(s(x))) → +(p(g(x)), q(g(x)))
g(s(x)) → pair(+(p(g(x)), q(g(x))), p(g(x)))
Q is empty.
We use [23] with the following order to prove termination.
Lexicographic path order with status [19].
Quasi-Precedence:
f1 > g1 > h1 > p1
f1 > g1 > h1 > pair2
f1 > g1 > h1 > +2 > s1
f1 > g1 > h1 > q1
Status: trivial