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: Quasi 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: Quasi 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: Quasi 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.