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(app2(le, 0), y) -> true
app2(app2(le, app2(s, x)), 0) -> false
app2(app2(le, app2(s, x)), app2(s, y)) -> app2(app2(le, x), y)
app2(app2(maxlist, x), app2(app2(cons, y), ys)) -> app2(app2(if, app2(app2(le, x), y)), app2(app2(maxlist, y), ys))
app2(app2(maxlist, x), nil) -> x
app2(height, app2(app2(node, x), xs)) -> app2(s, app2(app2(maxlist, 0), app2(app2(map, height), xs)))
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(app2(le, 0), y) -> true
app2(app2(le, app2(s, x)), 0) -> false
app2(app2(le, app2(s, x)), app2(s, y)) -> app2(app2(le, x), y)
app2(app2(maxlist, x), app2(app2(cons, y), ys)) -> app2(app2(if, app2(app2(le, x), y)), app2(app2(maxlist, y), ys))
app2(app2(maxlist, x), nil) -> x
app2(height, app2(app2(node, x), xs)) -> app2(s, app2(app2(maxlist, 0), app2(app2(map, height), xs)))
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(app2(le, 0), y) -> true
app2(app2(le, app2(s, x)), 0) -> false
app2(app2(le, app2(s, x)), app2(s, y)) -> app2(app2(le, x), y)
app2(app2(maxlist, x), app2(app2(cons, y), ys)) -> app2(app2(if, app2(app2(le, x), y)), app2(app2(maxlist, y), ys))
app2(app2(maxlist, x), nil) -> x
app2(height, app2(app2(node, x), xs)) -> app2(s, app2(app2(maxlist, 0), app2(app2(map, height), xs)))
The set Q consists of the following terms:
app2(app2(map, x0), nil)
app2(app2(map, x0), app2(app2(cons, x1), x2))
app2(app2(le, 0), x0)
app2(app2(le, app2(s, x0)), 0)
app2(app2(le, app2(s, x0)), app2(s, x1))
app2(app2(maxlist, x0), app2(app2(cons, x1), x2))
app2(app2(maxlist, x0), nil)
app2(height, app2(app2(node, x0), x1))
Q DP problem:
The TRS P consists of the following rules:
APP2(app2(maxlist, x), app2(app2(cons, y), ys)) -> APP2(maxlist, y)
APP2(height, app2(app2(node, x), xs)) -> APP2(maxlist, 0)
APP2(app2(map, f), app2(app2(cons, x), xs)) -> APP2(app2(cons, app2(f, x)), app2(app2(map, f), xs))
APP2(app2(le, app2(s, x)), app2(s, y)) -> APP2(app2(le, x), y)
APP2(app2(maxlist, x), app2(app2(cons, y), ys)) -> APP2(app2(maxlist, y), ys)
APP2(app2(map, f), app2(app2(cons, x), xs)) -> APP2(app2(map, f), xs)
APP2(app2(le, app2(s, x)), app2(s, y)) -> APP2(le, x)
APP2(app2(maxlist, x), app2(app2(cons, y), ys)) -> APP2(le, x)
APP2(height, app2(app2(node, x), xs)) -> APP2(s, app2(app2(maxlist, 0), app2(app2(map, height), xs)))
APP2(app2(maxlist, x), app2(app2(cons, y), ys)) -> APP2(if, app2(app2(le, x), y))
APP2(height, app2(app2(node, x), xs)) -> APP2(app2(map, height), xs)
APP2(height, app2(app2(node, x), xs)) -> APP2(map, height)
APP2(height, app2(app2(node, x), xs)) -> APP2(app2(maxlist, 0), app2(app2(map, height), xs))
APP2(app2(map, f), app2(app2(cons, x), xs)) -> APP2(f, x)
APP2(app2(maxlist, x), app2(app2(cons, y), ys)) -> APP2(app2(if, app2(app2(le, x), y)), app2(app2(maxlist, y), ys))
APP2(app2(maxlist, x), app2(app2(cons, y), ys)) -> APP2(app2(le, x), y)
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(app2(le, 0), y) -> true
app2(app2(le, app2(s, x)), 0) -> false
app2(app2(le, app2(s, x)), app2(s, y)) -> app2(app2(le, x), y)
app2(app2(maxlist, x), app2(app2(cons, y), ys)) -> app2(app2(if, app2(app2(le, x), y)), app2(app2(maxlist, y), ys))
app2(app2(maxlist, x), nil) -> x
app2(height, app2(app2(node, x), xs)) -> app2(s, app2(app2(maxlist, 0), app2(app2(map, height), xs)))
The set Q consists of the following terms:
app2(app2(map, x0), nil)
app2(app2(map, x0), app2(app2(cons, x1), x2))
app2(app2(le, 0), x0)
app2(app2(le, app2(s, x0)), 0)
app2(app2(le, app2(s, x0)), app2(s, x1))
app2(app2(maxlist, x0), app2(app2(cons, x1), x2))
app2(app2(maxlist, x0), nil)
app2(height, app2(app2(node, x0), 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(maxlist, x), app2(app2(cons, y), ys)) -> APP2(maxlist, y)
APP2(height, app2(app2(node, x), xs)) -> APP2(maxlist, 0)
APP2(app2(map, f), app2(app2(cons, x), xs)) -> APP2(app2(cons, app2(f, x)), app2(app2(map, f), xs))
APP2(app2(le, app2(s, x)), app2(s, y)) -> APP2(app2(le, x), y)
APP2(app2(maxlist, x), app2(app2(cons, y), ys)) -> APP2(app2(maxlist, y), ys)
APP2(app2(map, f), app2(app2(cons, x), xs)) -> APP2(app2(map, f), xs)
APP2(app2(le, app2(s, x)), app2(s, y)) -> APP2(le, x)
APP2(app2(maxlist, x), app2(app2(cons, y), ys)) -> APP2(le, x)
APP2(height, app2(app2(node, x), xs)) -> APP2(s, app2(app2(maxlist, 0), app2(app2(map, height), xs)))
APP2(app2(maxlist, x), app2(app2(cons, y), ys)) -> APP2(if, app2(app2(le, x), y))
APP2(height, app2(app2(node, x), xs)) -> APP2(app2(map, height), xs)
APP2(height, app2(app2(node, x), xs)) -> APP2(map, height)
APP2(height, app2(app2(node, x), xs)) -> APP2(app2(maxlist, 0), app2(app2(map, height), xs))
APP2(app2(map, f), app2(app2(cons, x), xs)) -> APP2(f, x)
APP2(app2(maxlist, x), app2(app2(cons, y), ys)) -> APP2(app2(if, app2(app2(le, x), y)), app2(app2(maxlist, y), ys))
APP2(app2(maxlist, x), app2(app2(cons, y), ys)) -> APP2(app2(le, x), y)
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(app2(le, 0), y) -> true
app2(app2(le, app2(s, x)), 0) -> false
app2(app2(le, app2(s, x)), app2(s, y)) -> app2(app2(le, x), y)
app2(app2(maxlist, x), app2(app2(cons, y), ys)) -> app2(app2(if, app2(app2(le, x), y)), app2(app2(maxlist, y), ys))
app2(app2(maxlist, x), nil) -> x
app2(height, app2(app2(node, x), xs)) -> app2(s, app2(app2(maxlist, 0), app2(app2(map, height), xs)))
The set Q consists of the following terms:
app2(app2(map, x0), nil)
app2(app2(map, x0), app2(app2(cons, x1), x2))
app2(app2(le, 0), x0)
app2(app2(le, app2(s, x0)), 0)
app2(app2(le, app2(s, x0)), app2(s, x1))
app2(app2(maxlist, x0), app2(app2(cons, x1), x2))
app2(app2(maxlist, x0), nil)
app2(height, app2(app2(node, x0), x1))
We have to consider all minimal (P,Q,R)-chains.
The approximation of the Dependency Graph contains 3 SCCs with 12 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(le, app2(s, x)), app2(s, y)) -> APP2(app2(le, x), y)
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(app2(le, 0), y) -> true
app2(app2(le, app2(s, x)), 0) -> false
app2(app2(le, app2(s, x)), app2(s, y)) -> app2(app2(le, x), y)
app2(app2(maxlist, x), app2(app2(cons, y), ys)) -> app2(app2(if, app2(app2(le, x), y)), app2(app2(maxlist, y), ys))
app2(app2(maxlist, x), nil) -> x
app2(height, app2(app2(node, x), xs)) -> app2(s, app2(app2(maxlist, 0), app2(app2(map, height), xs)))
The set Q consists of the following terms:
app2(app2(map, x0), nil)
app2(app2(map, x0), app2(app2(cons, x1), x2))
app2(app2(le, 0), x0)
app2(app2(le, app2(s, x0)), 0)
app2(app2(le, app2(s, x0)), app2(s, x1))
app2(app2(maxlist, x0), app2(app2(cons, x1), x2))
app2(app2(maxlist, x0), nil)
app2(height, app2(app2(node, x0), 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(le, app2(s, x)), app2(s, y)) -> APP2(app2(le, 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(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(app2(le, 0), y) -> true
app2(app2(le, app2(s, x)), 0) -> false
app2(app2(le, app2(s, x)), app2(s, y)) -> app2(app2(le, x), y)
app2(app2(maxlist, x), app2(app2(cons, y), ys)) -> app2(app2(if, app2(app2(le, x), y)), app2(app2(maxlist, y), ys))
app2(app2(maxlist, x), nil) -> x
app2(height, app2(app2(node, x), xs)) -> app2(s, app2(app2(maxlist, 0), app2(app2(map, height), xs)))
The set Q consists of the following terms:
app2(app2(map, x0), nil)
app2(app2(map, x0), app2(app2(cons, x1), x2))
app2(app2(le, 0), x0)
app2(app2(le, app2(s, x0)), 0)
app2(app2(le, app2(s, x0)), app2(s, x1))
app2(app2(maxlist, x0), app2(app2(cons, x1), x2))
app2(app2(maxlist, x0), nil)
app2(height, app2(app2(node, x0), 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(maxlist, x), app2(app2(cons, y), ys)) -> APP2(app2(maxlist, y), 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(app2(le, 0), y) -> true
app2(app2(le, app2(s, x)), 0) -> false
app2(app2(le, app2(s, x)), app2(s, y)) -> app2(app2(le, x), y)
app2(app2(maxlist, x), app2(app2(cons, y), ys)) -> app2(app2(if, app2(app2(le, x), y)), app2(app2(maxlist, y), ys))
app2(app2(maxlist, x), nil) -> x
app2(height, app2(app2(node, x), xs)) -> app2(s, app2(app2(maxlist, 0), app2(app2(map, height), xs)))
The set Q consists of the following terms:
app2(app2(map, x0), nil)
app2(app2(map, x0), app2(app2(cons, x1), x2))
app2(app2(le, 0), x0)
app2(app2(le, app2(s, x0)), 0)
app2(app2(le, app2(s, x0)), app2(s, x1))
app2(app2(maxlist, x0), app2(app2(cons, x1), x2))
app2(app2(maxlist, x0), nil)
app2(height, app2(app2(node, x0), 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(maxlist, x), app2(app2(cons, y), ys)) -> APP2(app2(maxlist, y), ys)
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(app2(le, 0), y) -> true
app2(app2(le, app2(s, x)), 0) -> false
app2(app2(le, app2(s, x)), app2(s, y)) -> app2(app2(le, x), y)
app2(app2(maxlist, x), app2(app2(cons, y), ys)) -> app2(app2(if, app2(app2(le, x), y)), app2(app2(maxlist, y), ys))
app2(app2(maxlist, x), nil) -> x
app2(height, app2(app2(node, x), xs)) -> app2(s, app2(app2(maxlist, 0), app2(app2(map, height), xs)))
The set Q consists of the following terms:
app2(app2(map, x0), nil)
app2(app2(map, x0), app2(app2(cons, x1), x2))
app2(app2(le, 0), x0)
app2(app2(le, app2(s, x0)), 0)
app2(app2(le, app2(s, x0)), app2(s, x1))
app2(app2(maxlist, x0), app2(app2(cons, x1), x2))
app2(app2(maxlist, x0), nil)
app2(height, app2(app2(node, x0), 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
↳ QDPAfsSolverProof
Q DP problem:
The TRS P consists of the following rules:
APP2(height, app2(app2(node, x), xs)) -> APP2(app2(map, height), xs)
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)
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(app2(le, 0), y) -> true
app2(app2(le, app2(s, x)), 0) -> false
app2(app2(le, app2(s, x)), app2(s, y)) -> app2(app2(le, x), y)
app2(app2(maxlist, x), app2(app2(cons, y), ys)) -> app2(app2(if, app2(app2(le, x), y)), app2(app2(maxlist, y), ys))
app2(app2(maxlist, x), nil) -> x
app2(height, app2(app2(node, x), xs)) -> app2(s, app2(app2(maxlist, 0), app2(app2(map, height), xs)))
The set Q consists of the following terms:
app2(app2(map, x0), nil)
app2(app2(map, x0), app2(app2(cons, x1), x2))
app2(app2(le, 0), x0)
app2(app2(le, app2(s, x0)), 0)
app2(app2(le, app2(s, x0)), app2(s, x1))
app2(app2(maxlist, x0), app2(app2(cons, x1), x2))
app2(app2(maxlist, x0), nil)
app2(height, app2(app2(node, x0), 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(height, app2(app2(node, x), xs)) -> APP2(app2(map, height), xs)
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)
Used argument filtering: APP2(x1, x2) = x2
app2(x1, x2) = app2(x1, x2)
node = node
cons = cons
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(app2(le, 0), y) -> true
app2(app2(le, app2(s, x)), 0) -> false
app2(app2(le, app2(s, x)), app2(s, y)) -> app2(app2(le, x), y)
app2(app2(maxlist, x), app2(app2(cons, y), ys)) -> app2(app2(if, app2(app2(le, x), y)), app2(app2(maxlist, y), ys))
app2(app2(maxlist, x), nil) -> x
app2(height, app2(app2(node, x), xs)) -> app2(s, app2(app2(maxlist, 0), app2(app2(map, height), xs)))
The set Q consists of the following terms:
app2(app2(map, x0), nil)
app2(app2(map, x0), app2(app2(cons, x1), x2))
app2(app2(le, 0), x0)
app2(app2(le, app2(s, x0)), 0)
app2(app2(le, app2(s, x0)), app2(s, x1))
app2(app2(maxlist, x0), app2(app2(cons, x1), x2))
app2(app2(maxlist, x0), nil)
app2(height, app2(app2(node, x0), x1))
We have to consider all minimal (P,Q,R)-chains.
The TRS P is empty. Hence, there is no (P,Q,R) chain.