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(minus, x), 0) -> x
app2(app2(minus, app2(s, x)), app2(s, y)) -> app2(app2(minus, x), y)
app2(double, 0) -> 0
app2(double, app2(s, x)) -> app2(s, app2(s, app2(double, x)))
app2(app2(plus, 0), y) -> y
app2(app2(plus, app2(s, x)), y) -> app2(s, app2(app2(plus, x), y))
app2(app2(plus, app2(s, x)), y) -> app2(app2(plus, x), app2(s, y))
app2(app2(plus, app2(s, x)), y) -> app2(s, app2(app2(plus, app2(app2(minus, x), y)), app2(double, y)))
Q is empty.
↳ QTRS
↳ DependencyPairsProof
Q restricted rewrite system:
The TRS R consists of the following rules:
app2(app2(minus, x), 0) -> x
app2(app2(minus, app2(s, x)), app2(s, y)) -> app2(app2(minus, x), y)
app2(double, 0) -> 0
app2(double, app2(s, x)) -> app2(s, app2(s, app2(double, x)))
app2(app2(plus, 0), y) -> y
app2(app2(plus, app2(s, x)), y) -> app2(s, app2(app2(plus, x), y))
app2(app2(plus, app2(s, x)), y) -> app2(app2(plus, x), app2(s, y))
app2(app2(plus, app2(s, x)), y) -> app2(s, app2(app2(plus, app2(app2(minus, x), y)), app2(double, y)))
Q is empty.
Q DP problem:
The TRS P consists of the following rules:
APP2(app2(minus, app2(s, x)), app2(s, y)) -> APP2(minus, x)
APP2(app2(plus, app2(s, x)), y) -> APP2(s, app2(app2(plus, x), y))
APP2(app2(plus, app2(s, x)), y) -> APP2(app2(plus, app2(app2(minus, x), y)), app2(double, y))
APP2(app2(plus, app2(s, x)), y) -> APP2(app2(minus, x), y)
APP2(app2(plus, app2(s, x)), y) -> APP2(s, app2(app2(plus, app2(app2(minus, x), y)), app2(double, y)))
APP2(double, app2(s, x)) -> APP2(s, app2(double, x))
APP2(app2(plus, app2(s, x)), y) -> APP2(double, y)
APP2(app2(plus, app2(s, x)), y) -> APP2(minus, x)
APP2(app2(plus, app2(s, x)), y) -> APP2(s, y)
APP2(app2(plus, app2(s, x)), y) -> APP2(app2(plus, x), y)
APP2(app2(plus, app2(s, x)), y) -> APP2(plus, app2(app2(minus, x), y))
APP2(app2(plus, app2(s, x)), y) -> APP2(app2(plus, x), app2(s, y))
APP2(app2(minus, app2(s, x)), app2(s, y)) -> APP2(app2(minus, x), y)
APP2(double, app2(s, x)) -> APP2(s, app2(s, app2(double, x)))
APP2(app2(plus, app2(s, x)), y) -> APP2(plus, x)
APP2(double, app2(s, x)) -> APP2(double, x)
The TRS R consists of the following rules:
app2(app2(minus, x), 0) -> x
app2(app2(minus, app2(s, x)), app2(s, y)) -> app2(app2(minus, x), y)
app2(double, 0) -> 0
app2(double, app2(s, x)) -> app2(s, app2(s, app2(double, x)))
app2(app2(plus, 0), y) -> y
app2(app2(plus, app2(s, x)), y) -> app2(s, app2(app2(plus, x), y))
app2(app2(plus, app2(s, x)), y) -> app2(app2(plus, x), app2(s, y))
app2(app2(plus, app2(s, x)), y) -> app2(s, app2(app2(plus, app2(app2(minus, x), y)), app2(double, 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(minus, app2(s, x)), app2(s, y)) -> APP2(minus, x)
APP2(app2(plus, app2(s, x)), y) -> APP2(s, app2(app2(plus, x), y))
APP2(app2(plus, app2(s, x)), y) -> APP2(app2(plus, app2(app2(minus, x), y)), app2(double, y))
APP2(app2(plus, app2(s, x)), y) -> APP2(app2(minus, x), y)
APP2(app2(plus, app2(s, x)), y) -> APP2(s, app2(app2(plus, app2(app2(minus, x), y)), app2(double, y)))
APP2(double, app2(s, x)) -> APP2(s, app2(double, x))
APP2(app2(plus, app2(s, x)), y) -> APP2(double, y)
APP2(app2(plus, app2(s, x)), y) -> APP2(minus, x)
APP2(app2(plus, app2(s, x)), y) -> APP2(s, y)
APP2(app2(plus, app2(s, x)), y) -> APP2(app2(plus, x), y)
APP2(app2(plus, app2(s, x)), y) -> APP2(plus, app2(app2(minus, x), y))
APP2(app2(plus, app2(s, x)), y) -> APP2(app2(plus, x), app2(s, y))
APP2(app2(minus, app2(s, x)), app2(s, y)) -> APP2(app2(minus, x), y)
APP2(double, app2(s, x)) -> APP2(s, app2(s, app2(double, x)))
APP2(app2(plus, app2(s, x)), y) -> APP2(plus, x)
APP2(double, app2(s, x)) -> APP2(double, x)
The TRS R consists of the following rules:
app2(app2(minus, x), 0) -> x
app2(app2(minus, app2(s, x)), app2(s, y)) -> app2(app2(minus, x), y)
app2(double, 0) -> 0
app2(double, app2(s, x)) -> app2(s, app2(s, app2(double, x)))
app2(app2(plus, 0), y) -> y
app2(app2(plus, app2(s, x)), y) -> app2(s, app2(app2(plus, x), y))
app2(app2(plus, app2(s, x)), y) -> app2(app2(plus, x), app2(s, y))
app2(app2(plus, app2(s, x)), y) -> app2(s, app2(app2(plus, app2(app2(minus, x), y)), app2(double, y)))
Q is empty.
We have to consider all minimal (P,Q,R)-chains.
The approximation of the Dependency Graph contains 3 SCCs with 11 less nodes.
↳ QTRS
↳ DependencyPairsProof
↳ QDP
↳ DependencyGraphProof
↳ AND
↳ QDP
↳ QDPAfsSolverProof
↳ QDP
↳ QDP
Q DP problem:
The TRS P consists of the following rules:
APP2(double, app2(s, x)) -> APP2(double, x)
The TRS R consists of the following rules:
app2(app2(minus, x), 0) -> x
app2(app2(minus, app2(s, x)), app2(s, y)) -> app2(app2(minus, x), y)
app2(double, 0) -> 0
app2(double, app2(s, x)) -> app2(s, app2(s, app2(double, x)))
app2(app2(plus, 0), y) -> y
app2(app2(plus, app2(s, x)), y) -> app2(s, app2(app2(plus, x), y))
app2(app2(plus, app2(s, x)), y) -> app2(app2(plus, x), app2(s, y))
app2(app2(plus, app2(s, x)), y) -> app2(s, app2(app2(plus, app2(app2(minus, x), y)), app2(double, 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(double, app2(s, x)) -> APP2(double, x)
Used argument filtering: APP2(x1, x2) = x2
app2(x1, x2) = app1(x2)
s = s
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(minus, x), 0) -> x
app2(app2(minus, app2(s, x)), app2(s, y)) -> app2(app2(minus, x), y)
app2(double, 0) -> 0
app2(double, app2(s, x)) -> app2(s, app2(s, app2(double, x)))
app2(app2(plus, 0), y) -> y
app2(app2(plus, app2(s, x)), y) -> app2(s, app2(app2(plus, x), y))
app2(app2(plus, app2(s, x)), y) -> app2(app2(plus, x), app2(s, y))
app2(app2(plus, app2(s, x)), y) -> app2(s, app2(app2(plus, app2(app2(minus, x), y)), app2(double, 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(minus, app2(s, x)), app2(s, y)) -> APP2(app2(minus, x), y)
The TRS R consists of the following rules:
app2(app2(minus, x), 0) -> x
app2(app2(minus, app2(s, x)), app2(s, y)) -> app2(app2(minus, x), y)
app2(double, 0) -> 0
app2(double, app2(s, x)) -> app2(s, app2(s, app2(double, x)))
app2(app2(plus, 0), y) -> y
app2(app2(plus, app2(s, x)), y) -> app2(s, app2(app2(plus, x), y))
app2(app2(plus, app2(s, x)), y) -> app2(app2(plus, x), app2(s, y))
app2(app2(plus, app2(s, x)), y) -> app2(s, app2(app2(plus, app2(app2(minus, x), y)), app2(double, 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(minus, app2(s, x)), app2(s, y)) -> APP2(app2(minus, x), y)
Used argument filtering: APP2(x1, x2) = x2
app2(x1, x2) = app1(x2)
s = s
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(minus, x), 0) -> x
app2(app2(minus, app2(s, x)), app2(s, y)) -> app2(app2(minus, x), y)
app2(double, 0) -> 0
app2(double, app2(s, x)) -> app2(s, app2(s, app2(double, x)))
app2(app2(plus, 0), y) -> y
app2(app2(plus, app2(s, x)), y) -> app2(s, app2(app2(plus, x), y))
app2(app2(plus, app2(s, x)), y) -> app2(app2(plus, x), app2(s, y))
app2(app2(plus, app2(s, x)), y) -> app2(s, app2(app2(plus, app2(app2(minus, x), y)), app2(double, 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(plus, app2(s, x)), y) -> APP2(app2(plus, x), y)
APP2(app2(plus, app2(s, x)), y) -> APP2(app2(plus, app2(app2(minus, x), y)), app2(double, y))
APP2(app2(plus, app2(s, x)), y) -> APP2(app2(plus, x), app2(s, y))
The TRS R consists of the following rules:
app2(app2(minus, x), 0) -> x
app2(app2(minus, app2(s, x)), app2(s, y)) -> app2(app2(minus, x), y)
app2(double, 0) -> 0
app2(double, app2(s, x)) -> app2(s, app2(s, app2(double, x)))
app2(app2(plus, 0), y) -> y
app2(app2(plus, app2(s, x)), y) -> app2(s, app2(app2(plus, x), y))
app2(app2(plus, app2(s, x)), y) -> app2(app2(plus, x), app2(s, y))
app2(app2(plus, app2(s, x)), y) -> app2(s, app2(app2(plus, app2(app2(minus, x), y)), app2(double, y)))
Q is empty.
We have to consider all minimal (P,Q,R)-chains.