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))


Using Dependency Pairs [1,13] we result in the following initial DP problem:
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 [13,14,18] contains 3 SCCs with 12 less nodes.

↳ QTRS
  ↳ Non-Overlap Check
    ↳ QTRS
      ↳ DependencyPairsProof
        ↳ QDP
          ↳ DependencyGraphProof
            ↳ AND
              ↳ QDP
                ↳ QDPOrderProof
              ↳ 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.
We use the reduction pair processor [13].


The following pairs can be strictly oriented and are deleted.


APP2(app2(le, app2(s, x)), app2(s, y)) -> APP2(app2(le, x), y)
The remaining pairs can at least by weakly be oriented.
none
Used ordering: Combined order from the following AFS and order.
APP2(x1, x2)  =  APP1(x2)
app2(x1, x2)  =  app1(x2)
le  =  le
s  =  s

Lexicographic Path Order [19].
Precedence:
APP1 > app1
APP1 > le
s > app1
s > le

The following usable rules [14] were oriented: none



↳ QTRS
  ↳ Non-Overlap Check
    ↳ QTRS
      ↳ DependencyPairsProof
        ↳ QDP
          ↳ DependencyGraphProof
            ↳ AND
              ↳ QDP
                ↳ QDPOrderProof
                  ↳ 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
                ↳ QDPOrderProof
              ↳ 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.
We use the reduction pair processor [13].


The following pairs can be strictly oriented and are deleted.


APP2(app2(maxlist, x), app2(app2(cons, y), ys)) -> APP2(app2(maxlist, y), ys)
The remaining pairs can at least by weakly be oriented.
none
Used ordering: Combined order from the following AFS and order.
APP2(x1, x2)  =  APP1(x2)
app2(x1, x2)  =  app1(x2)
maxlist  =  maxlist
cons  =  cons

Lexicographic Path Order [19].
Precedence:
APP1 > app1
cons > app1

The following usable rules [14] were oriented: none



↳ QTRS
  ↳ Non-Overlap Check
    ↳ QTRS
      ↳ DependencyPairsProof
        ↳ QDP
          ↳ DependencyGraphProof
            ↳ AND
              ↳ QDP
              ↳ QDP
                ↳ QDPOrderProof
                  ↳ 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
                ↳ QDPOrderProof

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.
We use the reduction pair processor [13].


The following pairs can be strictly oriented and are deleted.


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 remaining pairs can at least by weakly be oriented.
none
Used ordering: Combined order from the following AFS and order.
APP2(x1, x2)  =  APP1(x2)
height  =  height
app2(x1, x2)  =  app2(x1, x2)
node  =  node
map  =  map
cons  =  cons

Lexicographic Path Order [19].
Precedence:
APP1 > app2
height > app2
cons > app2
cons > map

The following usable rules [14] were oriented: none



↳ QTRS
  ↳ Non-Overlap Check
    ↳ QTRS
      ↳ DependencyPairsProof
        ↳ QDP
          ↳ DependencyGraphProof
            ↳ AND
              ↳ QDP
              ↳ QDP
              ↳ QDP
                ↳ QDPOrderProof
                  ↳ 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.