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(app2(map, f), nil) -> nil
app2(app2(map, f), app2(app2(cons, x), xs)) -> app2(app2(cons, app2(f, x)), app2(app2(map, f), xs))
app2(flatten, app2(app2(node, x), xs)) -> app2(app2(cons, x), app2(concat, app2(app2(map, flatten), xs)))
app2(concat, nil) -> nil
app2(concat, app2(app2(cons, x), xs)) -> app2(app2(append, x), app2(concat, xs))
app2(app2(append, nil), xs) -> xs
app2(app2(append, app2(app2(cons, x), xs)), ys) -> app2(app2(cons, x), app2(app2(append, xs), ys))
Q is empty.
↳ QTRS
↳ Non-Overlap Check
Q restricted rewrite system:
The TRS R consists of the following rules:
app2(app2(map, f), nil) -> nil
app2(app2(map, f), app2(app2(cons, x), xs)) -> app2(app2(cons, app2(f, x)), app2(app2(map, f), xs))
app2(flatten, app2(app2(node, x), xs)) -> app2(app2(cons, x), app2(concat, app2(app2(map, flatten), xs)))
app2(concat, nil) -> nil
app2(concat, app2(app2(cons, x), xs)) -> app2(app2(append, x), app2(concat, xs))
app2(app2(append, nil), xs) -> xs
app2(app2(append, app2(app2(cons, x), xs)), ys) -> app2(app2(cons, x), app2(app2(append, xs), ys))
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(app2(map, f), nil) -> nil
app2(app2(map, f), app2(app2(cons, x), xs)) -> app2(app2(cons, app2(f, x)), app2(app2(map, f), xs))
app2(flatten, app2(app2(node, x), xs)) -> app2(app2(cons, x), app2(concat, app2(app2(map, flatten), xs)))
app2(concat, nil) -> nil
app2(concat, app2(app2(cons, x), xs)) -> app2(app2(append, x), app2(concat, xs))
app2(app2(append, nil), xs) -> xs
app2(app2(append, app2(app2(cons, x), xs)), ys) -> app2(app2(cons, x), app2(app2(append, xs), ys))
The set Q consists of the following terms:
app2(app2(map, x0), nil)
app2(app2(map, x0), app2(app2(cons, x1), x2))
app2(flatten, app2(app2(node, x0), x1))
app2(concat, nil)
app2(concat, app2(app2(cons, x0), x1))
app2(app2(append, nil), x0)
app2(app2(append, app2(app2(cons, x0), x1)), x2)
Q DP problem:
The TRS P consists of the following rules:
APP2(concat, app2(app2(cons, x), xs)) -> APP2(append, x)
APP2(app2(map, f), app2(app2(cons, x), xs)) -> APP2(app2(cons, app2(f, x)), app2(app2(map, f), xs))
APP2(app2(append, app2(app2(cons, x), xs)), ys) -> APP2(append, xs)
APP2(app2(map, f), app2(app2(cons, x), xs)) -> APP2(app2(map, f), xs)
APP2(concat, app2(app2(cons, x), xs)) -> APP2(concat, xs)
APP2(app2(append, app2(app2(cons, x), xs)), ys) -> APP2(app2(append, xs), ys)
APP2(flatten, app2(app2(node, x), xs)) -> APP2(concat, app2(app2(map, flatten), xs))
APP2(flatten, app2(app2(node, x), xs)) -> APP2(app2(map, flatten), xs)
APP2(flatten, app2(app2(node, x), xs)) -> APP2(app2(cons, x), app2(concat, app2(app2(map, flatten), xs)))
APP2(app2(map, f), app2(app2(cons, x), xs)) -> APP2(f, x)
APP2(flatten, app2(app2(node, x), xs)) -> APP2(cons, x)
APP2(concat, app2(app2(cons, x), xs)) -> APP2(app2(append, x), app2(concat, xs))
APP2(app2(append, app2(app2(cons, x), xs)), ys) -> APP2(app2(cons, x), app2(app2(append, xs), ys))
APP2(flatten, app2(app2(node, x), xs)) -> APP2(map, flatten)
APP2(app2(map, f), app2(app2(cons, x), xs)) -> APP2(cons, app2(f, x))
The TRS R consists of the following rules:
app2(app2(map, f), nil) -> nil
app2(app2(map, f), app2(app2(cons, x), xs)) -> app2(app2(cons, app2(f, x)), app2(app2(map, f), xs))
app2(flatten, app2(app2(node, x), xs)) -> app2(app2(cons, x), app2(concat, app2(app2(map, flatten), xs)))
app2(concat, nil) -> nil
app2(concat, app2(app2(cons, x), xs)) -> app2(app2(append, x), app2(concat, xs))
app2(app2(append, nil), xs) -> xs
app2(app2(append, app2(app2(cons, x), xs)), ys) -> app2(app2(cons, x), app2(app2(append, xs), ys))
The set Q consists of the following terms:
app2(app2(map, x0), nil)
app2(app2(map, x0), app2(app2(cons, x1), x2))
app2(flatten, app2(app2(node, x0), x1))
app2(concat, nil)
app2(concat, app2(app2(cons, x0), x1))
app2(app2(append, nil), x0)
app2(app2(append, app2(app2(cons, x0), x1)), x2)
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(concat, app2(app2(cons, x), xs)) -> APP2(append, x)
APP2(app2(map, f), app2(app2(cons, x), xs)) -> APP2(app2(cons, app2(f, x)), app2(app2(map, f), xs))
APP2(app2(append, app2(app2(cons, x), xs)), ys) -> APP2(append, xs)
APP2(app2(map, f), app2(app2(cons, x), xs)) -> APP2(app2(map, f), xs)
APP2(concat, app2(app2(cons, x), xs)) -> APP2(concat, xs)
APP2(app2(append, app2(app2(cons, x), xs)), ys) -> APP2(app2(append, xs), ys)
APP2(flatten, app2(app2(node, x), xs)) -> APP2(concat, app2(app2(map, flatten), xs))
APP2(flatten, app2(app2(node, x), xs)) -> APP2(app2(map, flatten), xs)
APP2(flatten, app2(app2(node, x), xs)) -> APP2(app2(cons, x), app2(concat, app2(app2(map, flatten), xs)))
APP2(app2(map, f), app2(app2(cons, x), xs)) -> APP2(f, x)
APP2(flatten, app2(app2(node, x), xs)) -> APP2(cons, x)
APP2(concat, app2(app2(cons, x), xs)) -> APP2(app2(append, x), app2(concat, xs))
APP2(app2(append, app2(app2(cons, x), xs)), ys) -> APP2(app2(cons, x), app2(app2(append, xs), ys))
APP2(flatten, app2(app2(node, x), xs)) -> APP2(map, flatten)
APP2(app2(map, f), app2(app2(cons, x), xs)) -> APP2(cons, app2(f, x))
The TRS R consists of the following rules:
app2(app2(map, f), nil) -> nil
app2(app2(map, f), app2(app2(cons, x), xs)) -> app2(app2(cons, app2(f, x)), app2(app2(map, f), xs))
app2(flatten, app2(app2(node, x), xs)) -> app2(app2(cons, x), app2(concat, app2(app2(map, flatten), xs)))
app2(concat, nil) -> nil
app2(concat, app2(app2(cons, x), xs)) -> app2(app2(append, x), app2(concat, xs))
app2(app2(append, nil), xs) -> xs
app2(app2(append, app2(app2(cons, x), xs)), ys) -> app2(app2(cons, x), app2(app2(append, xs), ys))
The set Q consists of the following terms:
app2(app2(map, x0), nil)
app2(app2(map, x0), app2(app2(cons, x1), x2))
app2(flatten, app2(app2(node, x0), x1))
app2(concat, nil)
app2(concat, app2(app2(cons, x0), x1))
app2(app2(append, nil), x0)
app2(app2(append, app2(app2(cons, x0), x1)), x2)
We have to consider all minimal (P,Q,R)-chains.
The approximation of the Dependency Graph contains 3 SCCs with 10 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(append, app2(app2(cons, x), xs)), ys) -> APP2(app2(append, xs), ys)
The TRS R consists of the following rules:
app2(app2(map, f), nil) -> nil
app2(app2(map, f), app2(app2(cons, x), xs)) -> app2(app2(cons, app2(f, x)), app2(app2(map, f), xs))
app2(flatten, app2(app2(node, x), xs)) -> app2(app2(cons, x), app2(concat, app2(app2(map, flatten), xs)))
app2(concat, nil) -> nil
app2(concat, app2(app2(cons, x), xs)) -> app2(app2(append, x), app2(concat, xs))
app2(app2(append, nil), xs) -> xs
app2(app2(append, app2(app2(cons, x), xs)), ys) -> app2(app2(cons, x), app2(app2(append, xs), ys))
The set Q consists of the following terms:
app2(app2(map, x0), nil)
app2(app2(map, x0), app2(app2(cons, x1), x2))
app2(flatten, app2(app2(node, x0), x1))
app2(concat, nil)
app2(concat, app2(app2(cons, x0), x1))
app2(app2(append, nil), x0)
app2(app2(append, app2(app2(cons, x0), x1)), x2)
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(append, app2(app2(cons, x), xs)), ys) -> APP2(app2(append, xs), ys)
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
↳ QDPAfsSolverProof
↳ QDP
↳ PisEmptyProof
↳ QDP
↳ QDP
Q DP problem:
P is empty.
The TRS R consists of the following rules:
app2(app2(map, f), nil) -> nil
app2(app2(map, f), app2(app2(cons, x), xs)) -> app2(app2(cons, app2(f, x)), app2(app2(map, f), xs))
app2(flatten, app2(app2(node, x), xs)) -> app2(app2(cons, x), app2(concat, app2(app2(map, flatten), xs)))
app2(concat, nil) -> nil
app2(concat, app2(app2(cons, x), xs)) -> app2(app2(append, x), app2(concat, xs))
app2(app2(append, nil), xs) -> xs
app2(app2(append, app2(app2(cons, x), xs)), ys) -> app2(app2(cons, x), app2(app2(append, xs), ys))
The set Q consists of the following terms:
app2(app2(map, x0), nil)
app2(app2(map, x0), app2(app2(cons, x1), x2))
app2(flatten, app2(app2(node, x0), x1))
app2(concat, nil)
app2(concat, app2(app2(cons, x0), x1))
app2(app2(append, nil), x0)
app2(app2(append, app2(app2(cons, x0), x1)), x2)
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(concat, app2(app2(cons, x), xs)) -> APP2(concat, xs)
The TRS R consists of the following rules:
app2(app2(map, f), nil) -> nil
app2(app2(map, f), app2(app2(cons, x), xs)) -> app2(app2(cons, app2(f, x)), app2(app2(map, f), xs))
app2(flatten, app2(app2(node, x), xs)) -> app2(app2(cons, x), app2(concat, app2(app2(map, flatten), xs)))
app2(concat, nil) -> nil
app2(concat, app2(app2(cons, x), xs)) -> app2(app2(append, x), app2(concat, xs))
app2(app2(append, nil), xs) -> xs
app2(app2(append, app2(app2(cons, x), xs)), ys) -> app2(app2(cons, x), app2(app2(append, xs), ys))
The set Q consists of the following terms:
app2(app2(map, x0), nil)
app2(app2(map, x0), app2(app2(cons, x1), x2))
app2(flatten, app2(app2(node, x0), x1))
app2(concat, nil)
app2(concat, app2(app2(cons, x0), x1))
app2(app2(append, nil), x0)
app2(app2(append, app2(app2(cons, x0), x1)), x2)
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(concat, app2(app2(cons, x), xs)) -> APP2(concat, xs)
Used argument filtering: APP2(x1, x2) = x2
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(app2(map, f), nil) -> nil
app2(app2(map, f), app2(app2(cons, x), xs)) -> app2(app2(cons, app2(f, x)), app2(app2(map, f), xs))
app2(flatten, app2(app2(node, x), xs)) -> app2(app2(cons, x), app2(concat, app2(app2(map, flatten), xs)))
app2(concat, nil) -> nil
app2(concat, app2(app2(cons, x), xs)) -> app2(app2(append, x), app2(concat, xs))
app2(app2(append, nil), xs) -> xs
app2(app2(append, app2(app2(cons, x), xs)), ys) -> app2(app2(cons, x), app2(app2(append, xs), ys))
The set Q consists of the following terms:
app2(app2(map, x0), nil)
app2(app2(map, x0), app2(app2(cons, x1), x2))
app2(flatten, app2(app2(node, x0), x1))
app2(concat, nil)
app2(concat, app2(app2(cons, x0), x1))
app2(app2(append, nil), x0)
app2(app2(append, app2(app2(cons, x0), x1)), x2)
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
↳ QDPAfsSolverProof
Q DP problem:
The TRS P consists of the following rules:
APP2(app2(map, f), app2(app2(cons, x), xs)) -> APP2(f, x)
APP2(app2(map, f), app2(app2(cons, x), xs)) -> APP2(app2(map, f), xs)
APP2(flatten, app2(app2(node, x), xs)) -> APP2(app2(map, flatten), xs)
The TRS R consists of the following rules:
app2(app2(map, f), nil) -> nil
app2(app2(map, f), app2(app2(cons, x), xs)) -> app2(app2(cons, app2(f, x)), app2(app2(map, f), xs))
app2(flatten, app2(app2(node, x), xs)) -> app2(app2(cons, x), app2(concat, app2(app2(map, flatten), xs)))
app2(concat, nil) -> nil
app2(concat, app2(app2(cons, x), xs)) -> app2(app2(append, x), app2(concat, xs))
app2(app2(append, nil), xs) -> xs
app2(app2(append, app2(app2(cons, x), xs)), ys) -> app2(app2(cons, x), app2(app2(append, xs), ys))
The set Q consists of the following terms:
app2(app2(map, x0), nil)
app2(app2(map, x0), app2(app2(cons, x1), x2))
app2(flatten, app2(app2(node, x0), x1))
app2(concat, nil)
app2(concat, app2(app2(cons, x0), x1))
app2(app2(append, nil), x0)
app2(app2(append, app2(app2(cons, x0), x1)), x2)
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(map, f), app2(app2(cons, x), xs)) -> APP2(f, x)
APP2(app2(map, f), app2(app2(cons, x), xs)) -> APP2(app2(map, f), xs)
APP2(flatten, app2(app2(node, x), xs)) -> APP2(app2(map, flatten), xs)
Used argument filtering: APP2(x1, x2) = x2
app2(x1, x2) = app2(x1, x2)
cons = cons
node = node
Used ordering: Precedence:
trivial
↳ QTRS
↳ Non-Overlap Check
↳ QTRS
↳ DependencyPairsProof
↳ QDP
↳ DependencyGraphProof
↳ AND
↳ QDP
↳ QDP
↳ QDP
↳ QDPAfsSolverProof
↳ QDP
↳ PisEmptyProof
Q DP problem:
P is empty.
The TRS R consists of the following rules:
app2(app2(map, f), nil) -> nil
app2(app2(map, f), app2(app2(cons, x), xs)) -> app2(app2(cons, app2(f, x)), app2(app2(map, f), xs))
app2(flatten, app2(app2(node, x), xs)) -> app2(app2(cons, x), app2(concat, app2(app2(map, flatten), xs)))
app2(concat, nil) -> nil
app2(concat, app2(app2(cons, x), xs)) -> app2(app2(append, x), app2(concat, xs))
app2(app2(append, nil), xs) -> xs
app2(app2(append, app2(app2(cons, x), xs)), ys) -> app2(app2(cons, x), app2(app2(append, xs), ys))
The set Q consists of the following terms:
app2(app2(map, x0), nil)
app2(app2(map, x0), app2(app2(cons, x1), x2))
app2(flatten, app2(app2(node, x0), x1))
app2(concat, nil)
app2(concat, app2(app2(cons, x0), x1))
app2(app2(append, nil), x0)
app2(app2(append, app2(app2(cons, x0), x1)), x2)
We have to consider all minimal (P,Q,R)-chains.
The TRS P is empty. Hence, there is no (P,Q,R) chain.