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:
f(0(x1)) → s(0(x1))
d(0(x1)) → 0(x1)
d(s(x1)) → s(s(d(p(s(x1)))))
f(s(x1)) → d(f(p(s(x1))))
p(s(x1)) → x1
Q is empty.
↳ QTRS
↳ DependencyPairsProof
Q restricted rewrite system:
The TRS R consists of the following rules:
f(0(x1)) → s(0(x1))
d(0(x1)) → 0(x1)
d(s(x1)) → s(s(d(p(s(x1)))))
f(s(x1)) → d(f(p(s(x1))))
p(s(x1)) → 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:
F(s(x1)) → D(f(p(s(x1))))
F(s(x1)) → F(p(s(x1)))
F(s(x1)) → P(s(x1))
D(s(x1)) → P(s(x1))
D(s(x1)) → D(p(s(x1)))
The TRS R consists of the following rules:
f(0(x1)) → s(0(x1))
d(0(x1)) → 0(x1)
d(s(x1)) → s(s(d(p(s(x1)))))
f(s(x1)) → d(f(p(s(x1))))
p(s(x1)) → x1
Q is empty.
We have to consider all minimal (P,Q,R)-chains.
↳ QTRS
↳ DependencyPairsProof
↳ QDP
↳ DependencyGraphProof
Q DP problem:
The TRS P consists of the following rules:
F(s(x1)) → D(f(p(s(x1))))
F(s(x1)) → F(p(s(x1)))
F(s(x1)) → P(s(x1))
D(s(x1)) → P(s(x1))
D(s(x1)) → D(p(s(x1)))
The TRS R consists of the following rules:
f(0(x1)) → s(0(x1))
d(0(x1)) → 0(x1)
d(s(x1)) → s(s(d(p(s(x1)))))
f(s(x1)) → d(f(p(s(x1))))
p(s(x1)) → x1
Q is empty.
We have to consider all minimal (P,Q,R)-chains.
The approximation of the Dependency Graph [15,17,22] contains 2 SCCs with 3 less nodes.
↳ QTRS
↳ DependencyPairsProof
↳ QDP
↳ DependencyGraphProof
↳ AND
↳ QDP
↳ QDP
Q DP problem:
The TRS P consists of the following rules:
D(s(x1)) → D(p(s(x1)))
The TRS R consists of the following rules:
f(0(x1)) → s(0(x1))
d(0(x1)) → 0(x1)
d(s(x1)) → s(s(d(p(s(x1)))))
f(s(x1)) → d(f(p(s(x1))))
p(s(x1)) → x1
Q is empty.
We have to consider all minimal (P,Q,R)-chains.
↳ QTRS
↳ DependencyPairsProof
↳ QDP
↳ DependencyGraphProof
↳ AND
↳ QDP
↳ QDP
Q DP problem:
The TRS P consists of the following rules:
F(s(x1)) → F(p(s(x1)))
The TRS R consists of the following rules:
f(0(x1)) → s(0(x1))
d(0(x1)) → 0(x1)
d(s(x1)) → s(s(d(p(s(x1)))))
f(s(x1)) → d(f(p(s(x1))))
p(s(x1)) → x1
Q is empty.
We have to consider all minimal (P,Q,R)-chains.