Termination w.r.t. Q of the following Term Rewriting System could not be shown:
Q restricted rewrite system:
The TRS R consists of the following rules:
t(o(x1)) → m(a(x1))
t(e(x1)) → n(s(x1))
a(l(x1)) → a(t(x1))
o(m(a(x1))) → t(e(n(x1)))
s(a(x1)) → l(a(t(o(m(a(t(e(x1))))))))
n(s(x1)) → a(l(a(t(x1))))
Q is empty.
↳ QTRS
↳ DependencyPairsProof
Q restricted rewrite system:
The TRS R consists of the following rules:
t(o(x1)) → m(a(x1))
t(e(x1)) → n(s(x1))
a(l(x1)) → a(t(x1))
o(m(a(x1))) → t(e(n(x1)))
s(a(x1)) → l(a(t(o(m(a(t(e(x1))))))))
n(s(x1)) → a(l(a(t(x1))))
Q is empty.
Using Dependency Pairs [1,15] we result in the following initial DP problem:
Q DP problem:
The TRS P consists of the following rules:
N(s(x1)) → T(x1)
T(e(x1)) → N(s(x1))
T(e(x1)) → S(x1)
A(l(x1)) → A(t(x1))
S(a(x1)) → A(t(e(x1)))
S(a(x1)) → O(m(a(t(e(x1)))))
A(l(x1)) → T(x1)
N(s(x1)) → A(l(a(t(x1))))
O(m(a(x1))) → T(e(n(x1)))
S(a(x1)) → T(o(m(a(t(e(x1))))))
T(o(x1)) → A(x1)
S(a(x1)) → T(e(x1))
N(s(x1)) → A(t(x1))
S(a(x1)) → A(t(o(m(a(t(e(x1)))))))
O(m(a(x1))) → N(x1)
The TRS R consists of the following rules:
t(o(x1)) → m(a(x1))
t(e(x1)) → n(s(x1))
a(l(x1)) → a(t(x1))
o(m(a(x1))) → t(e(n(x1)))
s(a(x1)) → l(a(t(o(m(a(t(e(x1))))))))
n(s(x1)) → a(l(a(t(x1))))
Q is empty.
We have to consider all minimal (P,Q,R)-chains.
↳ QTRS
↳ DependencyPairsProof
↳ QDP
Q DP problem:
The TRS P consists of the following rules:
N(s(x1)) → T(x1)
T(e(x1)) → N(s(x1))
T(e(x1)) → S(x1)
A(l(x1)) → A(t(x1))
S(a(x1)) → A(t(e(x1)))
S(a(x1)) → O(m(a(t(e(x1)))))
A(l(x1)) → T(x1)
N(s(x1)) → A(l(a(t(x1))))
O(m(a(x1))) → T(e(n(x1)))
S(a(x1)) → T(o(m(a(t(e(x1))))))
T(o(x1)) → A(x1)
S(a(x1)) → T(e(x1))
N(s(x1)) → A(t(x1))
S(a(x1)) → A(t(o(m(a(t(e(x1)))))))
O(m(a(x1))) → N(x1)
The TRS R consists of the following rules:
t(o(x1)) → m(a(x1))
t(e(x1)) → n(s(x1))
a(l(x1)) → a(t(x1))
o(m(a(x1))) → t(e(n(x1)))
s(a(x1)) → l(a(t(o(m(a(t(e(x1))))))))
n(s(x1)) → a(l(a(t(x1))))
Q is empty.
We have to consider all minimal (P,Q,R)-chains.