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:
app2(g, app2(h, app2(g, x))) -> app2(g, x)
app2(g, app2(g, x)) -> app2(g, app2(h, app2(g, x)))
app2(h, app2(h, x)) -> app2(h, app2(app2(f, app2(h, x)), x))
Q is empty.
↳ QTRS
↳ DependencyPairsProof
Q restricted rewrite system:
The TRS R consists of the following rules:
app2(g, app2(h, app2(g, x))) -> app2(g, x)
app2(g, app2(g, x)) -> app2(g, app2(h, app2(g, x)))
app2(h, app2(h, x)) -> app2(h, app2(app2(f, app2(h, x)), x))
Q is empty.
Q DP problem:
The TRS P consists of the following rules:
APP2(h, app2(h, x)) -> APP2(f, app2(h, x))
APP2(h, app2(h, x)) -> APP2(app2(f, app2(h, x)), x)
APP2(h, app2(h, x)) -> APP2(h, app2(app2(f, app2(h, x)), x))
APP2(g, app2(g, x)) -> APP2(h, app2(g, x))
APP2(g, app2(g, x)) -> APP2(g, app2(h, app2(g, x)))
The TRS R consists of the following rules:
app2(g, app2(h, app2(g, x))) -> app2(g, x)
app2(g, app2(g, x)) -> app2(g, app2(h, app2(g, x)))
app2(h, app2(h, x)) -> app2(h, app2(app2(f, app2(h, 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:
APP2(h, app2(h, x)) -> APP2(f, app2(h, x))
APP2(h, app2(h, x)) -> APP2(app2(f, app2(h, x)), x)
APP2(h, app2(h, x)) -> APP2(h, app2(app2(f, app2(h, x)), x))
APP2(g, app2(g, x)) -> APP2(h, app2(g, x))
APP2(g, app2(g, x)) -> APP2(g, app2(h, app2(g, x)))
The TRS R consists of the following rules:
app2(g, app2(h, app2(g, x))) -> app2(g, x)
app2(g, app2(g, x)) -> app2(g, app2(h, app2(g, x)))
app2(h, app2(h, x)) -> app2(h, app2(app2(f, app2(h, x)), x))
Q is empty.
We have to consider all minimal (P,Q,R)-chains.
The approximation of the Dependency Graph contains 1 SCC with 4 less nodes.
↳ QTRS
↳ DependencyPairsProof
↳ QDP
↳ DependencyGraphProof
↳ QDP
↳ QDPAfsSolverProof
Q DP problem:
The TRS P consists of the following rules:
APP2(g, app2(g, x)) -> APP2(g, app2(h, app2(g, x)))
The TRS R consists of the following rules:
app2(g, app2(h, app2(g, x))) -> app2(g, x)
app2(g, app2(g, x)) -> app2(g, app2(h, app2(g, x)))
app2(h, app2(h, x)) -> app2(h, app2(app2(f, app2(h, x)), x))
Q is empty.
We have to consider all minimal (P,Q,R)-chains.
By using an argument filtering and a montonic ordering, at least one Dependency Pair of this SCC can be strictly oriented.
APP2(g, app2(g, x)) -> APP2(g, app2(h, app2(g, x)))
Used argument filtering: APP2(x1, x2) = x2
app2(x1, x2) = x1
g = g
h = h
Used ordering: Precedence:
g > h
↳ QTRS
↳ DependencyPairsProof
↳ QDP
↳ DependencyGraphProof
↳ QDP
↳ QDPAfsSolverProof
↳ QDP
↳ PisEmptyProof
Q DP problem:
P is empty.
The TRS R consists of the following rules:
app2(g, app2(h, app2(g, x))) -> app2(g, x)
app2(g, app2(g, x)) -> app2(g, app2(h, app2(g, x)))
app2(h, app2(h, x)) -> app2(h, app2(app2(f, app2(h, x)), x))
Q is empty.
We have to consider all minimal (P,Q,R)-chains.
The TRS P is empty. Hence, there is no (P,Q,R) chain.