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:
a(f, 0) → a(s, 0)
a(d, 0) → 0
a(d, a(s, x)) → a(s, a(s, a(d, a(p, a(s, x)))))
a(f, a(s, x)) → a(d, a(f, a(p, a(s, x))))
a(p, a(s, x)) → x
Q is empty.
↳ QTRS
↳ DependencyPairsProof
Q restricted rewrite system:
The TRS R consists of the following rules:
a(f, 0) → a(s, 0)
a(d, 0) → 0
a(d, a(s, x)) → a(s, a(s, a(d, a(p, a(s, x)))))
a(f, a(s, x)) → a(d, a(f, a(p, a(s, x))))
a(p, a(s, x)) → x
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:
A(d, a(s, x)) → A(p, a(s, x))
A(f, 0) → A(s, 0)
A(f, a(s, x)) → A(d, a(f, a(p, a(s, x))))
A(d, a(s, x)) → A(s, a(d, a(p, a(s, x))))
A(f, a(s, x)) → A(p, a(s, x))
A(f, a(s, x)) → A(f, a(p, a(s, x)))
A(d, a(s, x)) → A(d, a(p, a(s, x)))
A(d, a(s, x)) → A(s, a(s, a(d, a(p, a(s, x)))))
The TRS R consists of the following rules:
a(f, 0) → a(s, 0)
a(d, 0) → 0
a(d, a(s, x)) → a(s, a(s, a(d, a(p, a(s, x)))))
a(f, a(s, x)) → a(d, a(f, a(p, a(s, x))))
a(p, a(s, x)) → x
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:
A(d, a(s, x)) → A(p, a(s, x))
A(f, 0) → A(s, 0)
A(f, a(s, x)) → A(d, a(f, a(p, a(s, x))))
A(d, a(s, x)) → A(s, a(d, a(p, a(s, x))))
A(f, a(s, x)) → A(p, a(s, x))
A(f, a(s, x)) → A(f, a(p, a(s, x)))
A(d, a(s, x)) → A(d, a(p, a(s, x)))
A(d, a(s, x)) → A(s, a(s, a(d, a(p, a(s, x)))))
The TRS R consists of the following rules:
a(f, 0) → a(s, 0)
a(d, 0) → 0
a(d, a(s, x)) → a(s, a(s, a(d, a(p, a(s, x)))))
a(f, a(s, x)) → a(d, a(f, a(p, a(s, x))))
a(p, a(s, x)) → x
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 6 less nodes.
↳ QTRS
↳ DependencyPairsProof
↳ QDP
↳ DependencyGraphProof
↳ AND
↳ QDP
↳ QDP
Q DP problem:
The TRS P consists of the following rules:
A(d, a(s, x)) → A(d, a(p, a(s, x)))
The TRS R consists of the following rules:
a(f, 0) → a(s, 0)
a(d, 0) → 0
a(d, a(s, x)) → a(s, a(s, a(d, a(p, a(s, x)))))
a(f, a(s, x)) → a(d, a(f, a(p, a(s, x))))
a(p, a(s, x)) → x
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:
A(f, a(s, x)) → A(f, a(p, a(s, x)))
The TRS R consists of the following rules:
a(f, 0) → a(s, 0)
a(d, 0) → 0
a(d, a(s, x)) → a(s, a(s, a(d, a(p, a(s, x)))))
a(f, a(s, x)) → a(d, a(f, a(p, a(s, x))))
a(p, a(s, x)) → x
Q is empty.
We have to consider all minimal (P,Q,R)-chains.