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(p, app2(s, x)) -> x
app2(fact, 0) -> app2(s, 0)
app2(fact, app2(s, x)) -> app2(app2(*, app2(s, x)), app2(fact, app2(p, app2(s, x))))
app2(app2(*, 0), y) -> 0
app2(app2(*, app2(s, x)), y) -> app2(app2(+, app2(app2(*, x), y)), y)
app2(app2(+, x), 0) -> x
app2(app2(+, x), app2(s, y)) -> app2(s, app2(app2(+, x), y))
Q is empty.
↳ QTRS
↳ Non-Overlap Check
Q restricted rewrite system:
The TRS R consists of the following rules:
app2(p, app2(s, x)) -> x
app2(fact, 0) -> app2(s, 0)
app2(fact, app2(s, x)) -> app2(app2(*, app2(s, x)), app2(fact, app2(p, app2(s, x))))
app2(app2(*, 0), y) -> 0
app2(app2(*, app2(s, x)), y) -> app2(app2(+, app2(app2(*, x), y)), y)
app2(app2(+, x), 0) -> x
app2(app2(+, x), app2(s, y)) -> app2(s, app2(app2(+, x), y))
Q is empty.
The TRS is non-overlapping. Hence, we can switch to innermost.
↳ QTRS
↳ Non-Overlap Check
↳ QTRS
↳ DependencyPairsProof
Q restricted rewrite system:
The TRS R consists of the following rules:
app2(p, app2(s, x)) -> x
app2(fact, 0) -> app2(s, 0)
app2(fact, app2(s, x)) -> app2(app2(*, app2(s, x)), app2(fact, app2(p, app2(s, x))))
app2(app2(*, 0), y) -> 0
app2(app2(*, app2(s, x)), y) -> app2(app2(+, app2(app2(*, x), y)), y)
app2(app2(+, x), 0) -> x
app2(app2(+, x), app2(s, y)) -> app2(s, app2(app2(+, x), y))
The set Q consists of the following terms:
app2(p, app2(s, x0))
app2(fact, 0)
app2(fact, app2(s, x0))
app2(app2(*, 0), x0)
app2(app2(*, app2(s, x0)), x1)
app2(app2(+, x0), 0)
app2(app2(+, x0), app2(s, x1))
Q DP problem:
The TRS P consists of the following rules:
APP2(app2(*, app2(s, x)), y) -> APP2(+, app2(app2(*, x), y))
APP2(app2(+, x), app2(s, y)) -> APP2(app2(+, x), y)
APP2(fact, app2(s, x)) -> APP2(app2(*, app2(s, x)), app2(fact, app2(p, app2(s, x))))
APP2(fact, app2(s, x)) -> APP2(*, app2(s, x))
APP2(app2(*, app2(s, x)), y) -> APP2(app2(*, x), y)
APP2(app2(*, app2(s, x)), y) -> APP2(app2(+, app2(app2(*, x), y)), y)
APP2(fact, 0) -> APP2(s, 0)
APP2(fact, app2(s, x)) -> APP2(fact, app2(p, app2(s, x)))
APP2(fact, app2(s, x)) -> APP2(p, app2(s, x))
APP2(app2(+, x), app2(s, y)) -> APP2(s, app2(app2(+, x), y))
APP2(app2(*, app2(s, x)), y) -> APP2(*, x)
The TRS R consists of the following rules:
app2(p, app2(s, x)) -> x
app2(fact, 0) -> app2(s, 0)
app2(fact, app2(s, x)) -> app2(app2(*, app2(s, x)), app2(fact, app2(p, app2(s, x))))
app2(app2(*, 0), y) -> 0
app2(app2(*, app2(s, x)), y) -> app2(app2(+, app2(app2(*, x), y)), y)
app2(app2(+, x), 0) -> x
app2(app2(+, x), app2(s, y)) -> app2(s, app2(app2(+, x), y))
The set Q consists of the following terms:
app2(p, app2(s, x0))
app2(fact, 0)
app2(fact, app2(s, x0))
app2(app2(*, 0), x0)
app2(app2(*, app2(s, x0)), x1)
app2(app2(+, x0), 0)
app2(app2(+, x0), app2(s, x1))
We have to consider all minimal (P,Q,R)-chains.
↳ QTRS
↳ Non-Overlap Check
↳ QTRS
↳ DependencyPairsProof
↳ QDP
↳ DependencyGraphProof
Q DP problem:
The TRS P consists of the following rules:
APP2(app2(*, app2(s, x)), y) -> APP2(+, app2(app2(*, x), y))
APP2(app2(+, x), app2(s, y)) -> APP2(app2(+, x), y)
APP2(fact, app2(s, x)) -> APP2(app2(*, app2(s, x)), app2(fact, app2(p, app2(s, x))))
APP2(fact, app2(s, x)) -> APP2(*, app2(s, x))
APP2(app2(*, app2(s, x)), y) -> APP2(app2(*, x), y)
APP2(app2(*, app2(s, x)), y) -> APP2(app2(+, app2(app2(*, x), y)), y)
APP2(fact, 0) -> APP2(s, 0)
APP2(fact, app2(s, x)) -> APP2(fact, app2(p, app2(s, x)))
APP2(fact, app2(s, x)) -> APP2(p, app2(s, x))
APP2(app2(+, x), app2(s, y)) -> APP2(s, app2(app2(+, x), y))
APP2(app2(*, app2(s, x)), y) -> APP2(*, x)
The TRS R consists of the following rules:
app2(p, app2(s, x)) -> x
app2(fact, 0) -> app2(s, 0)
app2(fact, app2(s, x)) -> app2(app2(*, app2(s, x)), app2(fact, app2(p, app2(s, x))))
app2(app2(*, 0), y) -> 0
app2(app2(*, app2(s, x)), y) -> app2(app2(+, app2(app2(*, x), y)), y)
app2(app2(+, x), 0) -> x
app2(app2(+, x), app2(s, y)) -> app2(s, app2(app2(+, x), y))
The set Q consists of the following terms:
app2(p, app2(s, x0))
app2(fact, 0)
app2(fact, app2(s, x0))
app2(app2(*, 0), x0)
app2(app2(*, app2(s, x0)), x1)
app2(app2(+, x0), 0)
app2(app2(+, x0), app2(s, x1))
We have to consider all minimal (P,Q,R)-chains.
The approximation of the Dependency Graph contains 3 SCCs with 8 less nodes.
↳ QTRS
↳ Non-Overlap Check
↳ QTRS
↳ DependencyPairsProof
↳ QDP
↳ DependencyGraphProof
↳ AND
↳ QDP
↳ QDPAfsSolverProof
↳ QDP
↳ QDP
Q DP problem:
The TRS P consists of the following rules:
APP2(app2(+, x), app2(s, y)) -> APP2(app2(+, x), y)
The TRS R consists of the following rules:
app2(p, app2(s, x)) -> x
app2(fact, 0) -> app2(s, 0)
app2(fact, app2(s, x)) -> app2(app2(*, app2(s, x)), app2(fact, app2(p, app2(s, x))))
app2(app2(*, 0), y) -> 0
app2(app2(*, app2(s, x)), y) -> app2(app2(+, app2(app2(*, x), y)), y)
app2(app2(+, x), 0) -> x
app2(app2(+, x), app2(s, y)) -> app2(s, app2(app2(+, x), y))
The set Q consists of the following terms:
app2(p, app2(s, x0))
app2(fact, 0)
app2(fact, app2(s, x0))
app2(app2(*, 0), x0)
app2(app2(*, app2(s, x0)), x1)
app2(app2(+, x0), 0)
app2(app2(+, x0), app2(s, x1))
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(+, x), app2(s, y)) -> APP2(app2(+, x), y)
Used argument filtering: APP2(x1, x2) = x2
app2(x1, x2) = app1(x2)
s = s
Used ordering: Precedence:
trivial
↳ QTRS
↳ Non-Overlap Check
↳ 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(p, app2(s, x)) -> x
app2(fact, 0) -> app2(s, 0)
app2(fact, app2(s, x)) -> app2(app2(*, app2(s, x)), app2(fact, app2(p, app2(s, x))))
app2(app2(*, 0), y) -> 0
app2(app2(*, app2(s, x)), y) -> app2(app2(+, app2(app2(*, x), y)), y)
app2(app2(+, x), 0) -> x
app2(app2(+, x), app2(s, y)) -> app2(s, app2(app2(+, x), y))
The set Q consists of the following terms:
app2(p, app2(s, x0))
app2(fact, 0)
app2(fact, app2(s, x0))
app2(app2(*, 0), x0)
app2(app2(*, app2(s, x0)), x1)
app2(app2(+, x0), 0)
app2(app2(+, x0), app2(s, x1))
We have to consider all minimal (P,Q,R)-chains.
The TRS P is empty. Hence, there is no (P,Q,R) chain.
↳ QTRS
↳ Non-Overlap Check
↳ QTRS
↳ DependencyPairsProof
↳ QDP
↳ DependencyGraphProof
↳ AND
↳ QDP
↳ QDP
↳ QDPAfsSolverProof
↳ QDP
Q DP problem:
The TRS P consists of the following rules:
APP2(app2(*, app2(s, x)), y) -> APP2(app2(*, x), y)
The TRS R consists of the following rules:
app2(p, app2(s, x)) -> x
app2(fact, 0) -> app2(s, 0)
app2(fact, app2(s, x)) -> app2(app2(*, app2(s, x)), app2(fact, app2(p, app2(s, x))))
app2(app2(*, 0), y) -> 0
app2(app2(*, app2(s, x)), y) -> app2(app2(+, app2(app2(*, x), y)), y)
app2(app2(+, x), 0) -> x
app2(app2(+, x), app2(s, y)) -> app2(s, app2(app2(+, x), y))
The set Q consists of the following terms:
app2(p, app2(s, x0))
app2(fact, 0)
app2(fact, app2(s, x0))
app2(app2(*, 0), x0)
app2(app2(*, app2(s, x0)), x1)
app2(app2(+, x0), 0)
app2(app2(+, x0), app2(s, x1))
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(*, app2(s, x)), y) -> APP2(app2(*, x), y)
Used argument filtering: APP2(x1, x2) = x1
app2(x1, x2) = app1(x2)
Used ordering: Precedence:
trivial
↳ QTRS
↳ Non-Overlap Check
↳ 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(p, app2(s, x)) -> x
app2(fact, 0) -> app2(s, 0)
app2(fact, app2(s, x)) -> app2(app2(*, app2(s, x)), app2(fact, app2(p, app2(s, x))))
app2(app2(*, 0), y) -> 0
app2(app2(*, app2(s, x)), y) -> app2(app2(+, app2(app2(*, x), y)), y)
app2(app2(+, x), 0) -> x
app2(app2(+, x), app2(s, y)) -> app2(s, app2(app2(+, x), y))
The set Q consists of the following terms:
app2(p, app2(s, x0))
app2(fact, 0)
app2(fact, app2(s, x0))
app2(app2(*, 0), x0)
app2(app2(*, app2(s, x0)), x1)
app2(app2(+, x0), 0)
app2(app2(+, x0), app2(s, x1))
We have to consider all minimal (P,Q,R)-chains.
The TRS P is empty. Hence, there is no (P,Q,R) chain.
↳ QTRS
↳ Non-Overlap Check
↳ QTRS
↳ DependencyPairsProof
↳ QDP
↳ DependencyGraphProof
↳ AND
↳ QDP
↳ QDP
↳ QDP
Q DP problem:
The TRS P consists of the following rules:
APP2(fact, app2(s, x)) -> APP2(fact, app2(p, app2(s, x)))
The TRS R consists of the following rules:
app2(p, app2(s, x)) -> x
app2(fact, 0) -> app2(s, 0)
app2(fact, app2(s, x)) -> app2(app2(*, app2(s, x)), app2(fact, app2(p, app2(s, x))))
app2(app2(*, 0), y) -> 0
app2(app2(*, app2(s, x)), y) -> app2(app2(+, app2(app2(*, x), y)), y)
app2(app2(+, x), 0) -> x
app2(app2(+, x), app2(s, y)) -> app2(s, app2(app2(+, x), y))
The set Q consists of the following terms:
app2(p, app2(s, x0))
app2(fact, 0)
app2(fact, app2(s, x0))
app2(app2(*, 0), x0)
app2(app2(*, app2(s, x0)), x1)
app2(app2(+, x0), 0)
app2(app2(+, x0), app2(s, x1))
We have to consider all minimal (P,Q,R)-chains.