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:
app2(app2(h, z), app2(e, x)) -> app2(app2(h, app2(c, z)), app2(app2(d, z), x))
app2(app2(d, z), app2(app2(g, 0), 0)) -> app2(e, 0)
app2(app2(d, z), app2(app2(g, x), y)) -> app2(app2(g, app2(e, x)), app2(app2(d, z), y))
app2(app2(d, app2(c, z)), app2(app2(g, app2(app2(g, x), y)), 0)) -> app2(app2(g, app2(app2(d, app2(c, z)), app2(app2(g, x), y))), app2(app2(d, z), app2(app2(g, x), y)))
app2(app2(g, app2(e, x)), app2(e, y)) -> app2(e, app2(app2(g, x), y))
Q is empty.
↳ QTRS
↳ DependencyPairsProof
Q restricted rewrite system:
The TRS R consists of the following rules:
app2(app2(h, z), app2(e, x)) -> app2(app2(h, app2(c, z)), app2(app2(d, z), x))
app2(app2(d, z), app2(app2(g, 0), 0)) -> app2(e, 0)
app2(app2(d, z), app2(app2(g, x), y)) -> app2(app2(g, app2(e, x)), app2(app2(d, z), y))
app2(app2(d, app2(c, z)), app2(app2(g, app2(app2(g, x), y)), 0)) -> app2(app2(g, app2(app2(d, app2(c, z)), app2(app2(g, x), y))), app2(app2(d, z), app2(app2(g, x), y)))
app2(app2(g, app2(e, x)), app2(e, y)) -> app2(e, app2(app2(g, x), y))
Q is empty.
Q DP problem:
The TRS P consists of the following rules:
APP2(app2(d, z), app2(app2(g, x), y)) -> APP2(app2(d, z), y)
APP2(app2(g, app2(e, x)), app2(e, y)) -> APP2(g, x)
APP2(app2(h, z), app2(e, x)) -> APP2(c, z)
APP2(app2(d, app2(c, z)), app2(app2(g, app2(app2(g, x), y)), 0)) -> APP2(app2(d, z), app2(app2(g, x), y))
APP2(app2(g, app2(e, x)), app2(e, y)) -> APP2(e, app2(app2(g, x), y))
APP2(app2(d, z), app2(app2(g, 0), 0)) -> APP2(e, 0)
APP2(app2(d, z), app2(app2(g, x), y)) -> APP2(g, app2(e, x))
APP2(app2(d, app2(c, z)), app2(app2(g, app2(app2(g, x), y)), 0)) -> APP2(g, app2(app2(d, app2(c, z)), app2(app2(g, x), y)))
APP2(app2(d, app2(c, z)), app2(app2(g, app2(app2(g, x), y)), 0)) -> APP2(app2(d, app2(c, z)), app2(app2(g, x), y))
APP2(app2(d, app2(c, z)), app2(app2(g, app2(app2(g, x), y)), 0)) -> APP2(app2(g, app2(app2(d, app2(c, z)), app2(app2(g, x), y))), app2(app2(d, z), app2(app2(g, x), y)))
APP2(app2(h, z), app2(e, x)) -> APP2(app2(d, z), x)
APP2(app2(h, z), app2(e, x)) -> APP2(app2(h, app2(c, z)), app2(app2(d, z), x))
APP2(app2(h, z), app2(e, x)) -> APP2(h, app2(c, z))
APP2(app2(h, z), app2(e, x)) -> APP2(d, z)
APP2(app2(d, z), app2(app2(g, x), y)) -> APP2(app2(g, app2(e, x)), app2(app2(d, z), y))
APP2(app2(g, app2(e, x)), app2(e, y)) -> APP2(app2(g, x), y)
APP2(app2(d, z), app2(app2(g, x), y)) -> APP2(e, x)
APP2(app2(d, app2(c, z)), app2(app2(g, app2(app2(g, x), y)), 0)) -> APP2(d, z)
The TRS R consists of the following rules:
app2(app2(h, z), app2(e, x)) -> app2(app2(h, app2(c, z)), app2(app2(d, z), x))
app2(app2(d, z), app2(app2(g, 0), 0)) -> app2(e, 0)
app2(app2(d, z), app2(app2(g, x), y)) -> app2(app2(g, app2(e, x)), app2(app2(d, z), y))
app2(app2(d, app2(c, z)), app2(app2(g, app2(app2(g, x), y)), 0)) -> app2(app2(g, app2(app2(d, app2(c, z)), app2(app2(g, x), y))), app2(app2(d, z), app2(app2(g, x), y)))
app2(app2(g, app2(e, x)), app2(e, y)) -> app2(e, app2(app2(g, x), y))
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(app2(d, z), app2(app2(g, x), y)) -> APP2(app2(d, z), y)
APP2(app2(g, app2(e, x)), app2(e, y)) -> APP2(g, x)
APP2(app2(h, z), app2(e, x)) -> APP2(c, z)
APP2(app2(d, app2(c, z)), app2(app2(g, app2(app2(g, x), y)), 0)) -> APP2(app2(d, z), app2(app2(g, x), y))
APP2(app2(g, app2(e, x)), app2(e, y)) -> APP2(e, app2(app2(g, x), y))
APP2(app2(d, z), app2(app2(g, 0), 0)) -> APP2(e, 0)
APP2(app2(d, z), app2(app2(g, x), y)) -> APP2(g, app2(e, x))
APP2(app2(d, app2(c, z)), app2(app2(g, app2(app2(g, x), y)), 0)) -> APP2(g, app2(app2(d, app2(c, z)), app2(app2(g, x), y)))
APP2(app2(d, app2(c, z)), app2(app2(g, app2(app2(g, x), y)), 0)) -> APP2(app2(d, app2(c, z)), app2(app2(g, x), y))
APP2(app2(d, app2(c, z)), app2(app2(g, app2(app2(g, x), y)), 0)) -> APP2(app2(g, app2(app2(d, app2(c, z)), app2(app2(g, x), y))), app2(app2(d, z), app2(app2(g, x), y)))
APP2(app2(h, z), app2(e, x)) -> APP2(app2(d, z), x)
APP2(app2(h, z), app2(e, x)) -> APP2(app2(h, app2(c, z)), app2(app2(d, z), x))
APP2(app2(h, z), app2(e, x)) -> APP2(h, app2(c, z))
APP2(app2(h, z), app2(e, x)) -> APP2(d, z)
APP2(app2(d, z), app2(app2(g, x), y)) -> APP2(app2(g, app2(e, x)), app2(app2(d, z), y))
APP2(app2(g, app2(e, x)), app2(e, y)) -> APP2(app2(g, x), y)
APP2(app2(d, z), app2(app2(g, x), y)) -> APP2(e, x)
APP2(app2(d, app2(c, z)), app2(app2(g, app2(app2(g, x), y)), 0)) -> APP2(d, z)
The TRS R consists of the following rules:
app2(app2(h, z), app2(e, x)) -> app2(app2(h, app2(c, z)), app2(app2(d, z), x))
app2(app2(d, z), app2(app2(g, 0), 0)) -> app2(e, 0)
app2(app2(d, z), app2(app2(g, x), y)) -> app2(app2(g, app2(e, x)), app2(app2(d, z), y))
app2(app2(d, app2(c, z)), app2(app2(g, app2(app2(g, x), y)), 0)) -> app2(app2(g, app2(app2(d, app2(c, z)), app2(app2(g, x), y))), app2(app2(d, z), app2(app2(g, x), y)))
app2(app2(g, app2(e, x)), app2(e, y)) -> app2(e, app2(app2(g, x), y))
Q is empty.
We have to consider all minimal (P,Q,R)-chains.
The approximation of the Dependency Graph contains 3 SCCs with 13 less nodes.
↳ QTRS
↳ DependencyPairsProof
↳ QDP
↳ DependencyGraphProof
↳ AND
↳ QDP
↳ QDPAfsSolverProof
↳ QDP
↳ QDP
Q DP problem:
The TRS P consists of the following rules:
APP2(app2(g, app2(e, x)), app2(e, y)) -> APP2(app2(g, x), y)
The TRS R consists of the following rules:
app2(app2(h, z), app2(e, x)) -> app2(app2(h, app2(c, z)), app2(app2(d, z), x))
app2(app2(d, z), app2(app2(g, 0), 0)) -> app2(e, 0)
app2(app2(d, z), app2(app2(g, x), y)) -> app2(app2(g, app2(e, x)), app2(app2(d, z), y))
app2(app2(d, app2(c, z)), app2(app2(g, app2(app2(g, x), y)), 0)) -> app2(app2(g, app2(app2(d, app2(c, z)), app2(app2(g, x), y))), app2(app2(d, z), app2(app2(g, x), y)))
app2(app2(g, app2(e, x)), app2(e, y)) -> app2(e, app2(app2(g, x), y))
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(app2(g, app2(e, x)), app2(e, y)) -> APP2(app2(g, x), y)
Used argument filtering: APP2(x1, x2) = x2
app2(x1, x2) = app1(x2)
e = e
Used ordering: Quasi Precedence:
trivial
↳ QTRS
↳ DependencyPairsProof
↳ QDP
↳ DependencyGraphProof
↳ AND
↳ QDP
↳ QDPAfsSolverProof
↳ QDP
↳ PisEmptyProof
↳ QDP
↳ QDP
Q DP problem:
P is empty.
The TRS R consists of the following rules:
app2(app2(h, z), app2(e, x)) -> app2(app2(h, app2(c, z)), app2(app2(d, z), x))
app2(app2(d, z), app2(app2(g, 0), 0)) -> app2(e, 0)
app2(app2(d, z), app2(app2(g, x), y)) -> app2(app2(g, app2(e, x)), app2(app2(d, z), y))
app2(app2(d, app2(c, z)), app2(app2(g, app2(app2(g, x), y)), 0)) -> app2(app2(g, app2(app2(d, app2(c, z)), app2(app2(g, x), y))), app2(app2(d, z), app2(app2(g, x), y)))
app2(app2(g, app2(e, x)), app2(e, y)) -> app2(e, app2(app2(g, x), y))
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.
↳ QTRS
↳ DependencyPairsProof
↳ QDP
↳ DependencyGraphProof
↳ AND
↳ QDP
↳ QDP
↳ QDPAfsSolverProof
↳ QDP
Q DP problem:
The TRS P consists of the following rules:
APP2(app2(d, z), app2(app2(g, x), y)) -> APP2(app2(d, z), y)
APP2(app2(d, app2(c, z)), app2(app2(g, app2(app2(g, x), y)), 0)) -> APP2(app2(d, z), app2(app2(g, x), y))
APP2(app2(d, app2(c, z)), app2(app2(g, app2(app2(g, x), y)), 0)) -> APP2(app2(d, app2(c, z)), app2(app2(g, x), y))
The TRS R consists of the following rules:
app2(app2(h, z), app2(e, x)) -> app2(app2(h, app2(c, z)), app2(app2(d, z), x))
app2(app2(d, z), app2(app2(g, 0), 0)) -> app2(e, 0)
app2(app2(d, z), app2(app2(g, x), y)) -> app2(app2(g, app2(e, x)), app2(app2(d, z), y))
app2(app2(d, app2(c, z)), app2(app2(g, app2(app2(g, x), y)), 0)) -> app2(app2(g, app2(app2(d, app2(c, z)), app2(app2(g, x), y))), app2(app2(d, z), app2(app2(g, x), y)))
app2(app2(g, app2(e, x)), app2(e, y)) -> app2(e, app2(app2(g, x), y))
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(app2(d, z), app2(app2(g, x), y)) -> APP2(app2(d, z), y)
APP2(app2(d, app2(c, z)), app2(app2(g, app2(app2(g, x), y)), 0)) -> APP2(app2(d, z), app2(app2(g, x), y))
APP2(app2(d, app2(c, z)), app2(app2(g, app2(app2(g, x), y)), 0)) -> APP2(app2(d, app2(c, z)), app2(app2(g, x), y))
Used argument filtering: APP2(x1, x2) = x2
app2(x1, x2) = app2(x1, x2)
g = g
0 = 0
e = e
Used ordering: Quasi Precedence:
trivial
↳ QTRS
↳ DependencyPairsProof
↳ QDP
↳ DependencyGraphProof
↳ AND
↳ QDP
↳ QDP
↳ QDPAfsSolverProof
↳ QDP
↳ PisEmptyProof
↳ QDP
Q DP problem:
P is empty.
The TRS R consists of the following rules:
app2(app2(h, z), app2(e, x)) -> app2(app2(h, app2(c, z)), app2(app2(d, z), x))
app2(app2(d, z), app2(app2(g, 0), 0)) -> app2(e, 0)
app2(app2(d, z), app2(app2(g, x), y)) -> app2(app2(g, app2(e, x)), app2(app2(d, z), y))
app2(app2(d, app2(c, z)), app2(app2(g, app2(app2(g, x), y)), 0)) -> app2(app2(g, app2(app2(d, app2(c, z)), app2(app2(g, x), y))), app2(app2(d, z), app2(app2(g, x), y)))
app2(app2(g, app2(e, x)), app2(e, y)) -> app2(e, app2(app2(g, x), y))
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.
↳ QTRS
↳ DependencyPairsProof
↳ QDP
↳ DependencyGraphProof
↳ AND
↳ QDP
↳ QDP
↳ QDP
Q DP problem:
The TRS P consists of the following rules:
APP2(app2(h, z), app2(e, x)) -> APP2(app2(h, app2(c, z)), app2(app2(d, z), x))
The TRS R consists of the following rules:
app2(app2(h, z), app2(e, x)) -> app2(app2(h, app2(c, z)), app2(app2(d, z), x))
app2(app2(d, z), app2(app2(g, 0), 0)) -> app2(e, 0)
app2(app2(d, z), app2(app2(g, x), y)) -> app2(app2(g, app2(e, x)), app2(app2(d, z), y))
app2(app2(d, app2(c, z)), app2(app2(g, app2(app2(g, x), y)), 0)) -> app2(app2(g, app2(app2(d, app2(c, z)), app2(app2(g, x), y))), app2(app2(d, z), app2(app2(g, x), y)))
app2(app2(g, app2(e, x)), app2(e, y)) -> app2(e, app2(app2(g, x), y))
Q is empty.
We have to consider all minimal (P,Q,R)-chains.