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(append, xs), nil) -> xs
app2(app2(append, nil), ys) -> ys
app2(app2(append, app2(app2(cons, x), xs)), ys) -> app2(app2(cons, x), app2(app2(append, xs), ys))
app2(app2(zip, nil), yss) -> yss
app2(app2(zip, xss), nil) -> xss
app2(app2(zip, app2(app2(cons, xs), xss)), app2(app2(cons, ys), yss)) -> app2(app2(cons, app2(app2(append, xs), ys)), app2(app2(zip, xss), yss))
app2(app2(combine, xs), nil) -> xs
app2(app2(combine, xs), app2(app2(cons, ys), yss)) -> app2(app2(combine, app2(app2(zip, xs), ys)), yss)
app2(levels, app2(app2(node, x), xs)) -> app2(app2(cons, app2(app2(cons, x), nil)), app2(app2(combine, nil), app2(app2(map, levels), xs)))

Q is empty.


↳ 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(append, xs), nil) -> xs
app2(app2(append, nil), ys) -> ys
app2(app2(append, app2(app2(cons, x), xs)), ys) -> app2(app2(cons, x), app2(app2(append, xs), ys))
app2(app2(zip, nil), yss) -> yss
app2(app2(zip, xss), nil) -> xss
app2(app2(zip, app2(app2(cons, xs), xss)), app2(app2(cons, ys), yss)) -> app2(app2(cons, app2(app2(append, xs), ys)), app2(app2(zip, xss), yss))
app2(app2(combine, xs), nil) -> xs
app2(app2(combine, xs), app2(app2(cons, ys), yss)) -> app2(app2(combine, app2(app2(zip, xs), ys)), yss)
app2(levels, app2(app2(node, x), xs)) -> app2(app2(cons, app2(app2(cons, x), nil)), app2(app2(combine, nil), app2(app2(map, levels), xs)))

Q is empty.

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(levels, app2(app2(node, x), xs)) -> APP2(combine, nil)
APP2(app2(map, f), app2(app2(cons, x), xs)) -> APP2(app2(cons, app2(f, x)), app2(app2(map, f), xs))
APP2(levels, app2(app2(node, x), xs)) -> APP2(map, levels)
APP2(levels, app2(app2(node, x), xs)) -> APP2(app2(cons, x), nil)
APP2(app2(zip, app2(app2(cons, xs), xss)), app2(app2(cons, ys), yss)) -> APP2(app2(cons, app2(app2(append, xs), ys)), app2(app2(zip, xss), yss))
APP2(levels, app2(app2(node, x), xs)) -> APP2(cons, x)
APP2(app2(zip, app2(app2(cons, xs), xss)), app2(app2(cons, ys), yss)) -> APP2(app2(append, xs), ys)
APP2(app2(combine, xs), app2(app2(cons, ys), yss)) -> APP2(app2(zip, xs), ys)
APP2(app2(map, f), app2(app2(cons, x), xs)) -> APP2(f, x)
APP2(app2(zip, app2(app2(cons, xs), xss)), app2(app2(cons, ys), yss)) -> APP2(zip, xss)
APP2(app2(combine, xs), app2(app2(cons, ys), yss)) -> APP2(zip, xs)
APP2(levels, app2(app2(node, x), xs)) -> APP2(app2(cons, app2(app2(cons, x), nil)), app2(app2(combine, nil), app2(app2(map, levels), xs)))
APP2(app2(zip, app2(app2(cons, xs), xss)), app2(app2(cons, ys), yss)) -> APP2(append, xs)
APP2(app2(append, app2(app2(cons, x), xs)), ys) -> APP2(append, xs)
APP2(levels, app2(app2(node, x), xs)) -> APP2(cons, app2(app2(cons, x), nil))
APP2(app2(zip, app2(app2(cons, xs), xss)), app2(app2(cons, ys), yss)) -> APP2(cons, app2(app2(append, xs), ys))
APP2(app2(zip, app2(app2(cons, xs), xss)), app2(app2(cons, ys), yss)) -> APP2(app2(zip, xss), yss)
APP2(app2(map, f), app2(app2(cons, x), xs)) -> APP2(app2(map, f), xs)
APP2(app2(combine, xs), app2(app2(cons, ys), yss)) -> APP2(app2(combine, app2(app2(zip, xs), ys)), yss)
APP2(app2(append, app2(app2(cons, x), xs)), ys) -> APP2(app2(append, xs), ys)
APP2(app2(combine, xs), app2(app2(cons, ys), yss)) -> APP2(combine, app2(app2(zip, xs), ys))
APP2(levels, app2(app2(node, x), xs)) -> APP2(app2(combine, nil), app2(app2(map, levels), xs))
APP2(app2(append, app2(app2(cons, x), xs)), ys) -> APP2(app2(cons, x), app2(app2(append, xs), ys))
APP2(levels, app2(app2(node, x), xs)) -> APP2(app2(map, levels), xs)
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(append, xs), nil) -> xs
app2(app2(append, nil), ys) -> ys
app2(app2(append, app2(app2(cons, x), xs)), ys) -> app2(app2(cons, x), app2(app2(append, xs), ys))
app2(app2(zip, nil), yss) -> yss
app2(app2(zip, xss), nil) -> xss
app2(app2(zip, app2(app2(cons, xs), xss)), app2(app2(cons, ys), yss)) -> app2(app2(cons, app2(app2(append, xs), ys)), app2(app2(zip, xss), yss))
app2(app2(combine, xs), nil) -> xs
app2(app2(combine, xs), app2(app2(cons, ys), yss)) -> app2(app2(combine, app2(app2(zip, xs), ys)), yss)
app2(levels, app2(app2(node, x), xs)) -> app2(app2(cons, app2(app2(cons, x), nil)), app2(app2(combine, nil), app2(app2(map, levels), xs)))

Q is empty.
We have to consider all minimal (P,Q,R)-chains.

↳ QTRS
  ↳ DependencyPairsProof
    ↳ QDP
      ↳ DependencyGraphProof

Q DP problem:
The TRS P consists of the following rules:

APP2(levels, app2(app2(node, x), xs)) -> APP2(combine, nil)
APP2(app2(map, f), app2(app2(cons, x), xs)) -> APP2(app2(cons, app2(f, x)), app2(app2(map, f), xs))
APP2(levels, app2(app2(node, x), xs)) -> APP2(map, levels)
APP2(levels, app2(app2(node, x), xs)) -> APP2(app2(cons, x), nil)
APP2(app2(zip, app2(app2(cons, xs), xss)), app2(app2(cons, ys), yss)) -> APP2(app2(cons, app2(app2(append, xs), ys)), app2(app2(zip, xss), yss))
APP2(levels, app2(app2(node, x), xs)) -> APP2(cons, x)
APP2(app2(zip, app2(app2(cons, xs), xss)), app2(app2(cons, ys), yss)) -> APP2(app2(append, xs), ys)
APP2(app2(combine, xs), app2(app2(cons, ys), yss)) -> APP2(app2(zip, xs), ys)
APP2(app2(map, f), app2(app2(cons, x), xs)) -> APP2(f, x)
APP2(app2(zip, app2(app2(cons, xs), xss)), app2(app2(cons, ys), yss)) -> APP2(zip, xss)
APP2(app2(combine, xs), app2(app2(cons, ys), yss)) -> APP2(zip, xs)
APP2(levels, app2(app2(node, x), xs)) -> APP2(app2(cons, app2(app2(cons, x), nil)), app2(app2(combine, nil), app2(app2(map, levels), xs)))
APP2(app2(zip, app2(app2(cons, xs), xss)), app2(app2(cons, ys), yss)) -> APP2(append, xs)
APP2(app2(append, app2(app2(cons, x), xs)), ys) -> APP2(append, xs)
APP2(levels, app2(app2(node, x), xs)) -> APP2(cons, app2(app2(cons, x), nil))
APP2(app2(zip, app2(app2(cons, xs), xss)), app2(app2(cons, ys), yss)) -> APP2(cons, app2(app2(append, xs), ys))
APP2(app2(zip, app2(app2(cons, xs), xss)), app2(app2(cons, ys), yss)) -> APP2(app2(zip, xss), yss)
APP2(app2(map, f), app2(app2(cons, x), xs)) -> APP2(app2(map, f), xs)
APP2(app2(combine, xs), app2(app2(cons, ys), yss)) -> APP2(app2(combine, app2(app2(zip, xs), ys)), yss)
APP2(app2(append, app2(app2(cons, x), xs)), ys) -> APP2(app2(append, xs), ys)
APP2(app2(combine, xs), app2(app2(cons, ys), yss)) -> APP2(combine, app2(app2(zip, xs), ys))
APP2(levels, app2(app2(node, x), xs)) -> APP2(app2(combine, nil), app2(app2(map, levels), xs))
APP2(app2(append, app2(app2(cons, x), xs)), ys) -> APP2(app2(cons, x), app2(app2(append, xs), ys))
APP2(levels, app2(app2(node, x), xs)) -> APP2(app2(map, levels), xs)
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(append, xs), nil) -> xs
app2(app2(append, nil), ys) -> ys
app2(app2(append, app2(app2(cons, x), xs)), ys) -> app2(app2(cons, x), app2(app2(append, xs), ys))
app2(app2(zip, nil), yss) -> yss
app2(app2(zip, xss), nil) -> xss
app2(app2(zip, app2(app2(cons, xs), xss)), app2(app2(cons, ys), yss)) -> app2(app2(cons, app2(app2(append, xs), ys)), app2(app2(zip, xss), yss))
app2(app2(combine, xs), nil) -> xs
app2(app2(combine, xs), app2(app2(cons, ys), yss)) -> app2(app2(combine, app2(app2(zip, xs), ys)), yss)
app2(levels, app2(app2(node, x), xs)) -> app2(app2(cons, app2(app2(cons, x), nil)), app2(app2(combine, nil), app2(app2(map, levels), xs)))

Q is empty.
We have to consider all minimal (P,Q,R)-chains.
The approximation of the Dependency Graph [13,14,18] contains 4 SCCs with 19 less nodes.

↳ QTRS
  ↳ DependencyPairsProof
    ↳ QDP
      ↳ DependencyGraphProof
        ↳ AND
          ↳ QDP
            ↳ QDPOrderProof
          ↳ QDP
          ↳ 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(app2(append, xs), nil) -> xs
app2(app2(append, nil), ys) -> ys
app2(app2(append, app2(app2(cons, x), xs)), ys) -> app2(app2(cons, x), app2(app2(append, xs), ys))
app2(app2(zip, nil), yss) -> yss
app2(app2(zip, xss), nil) -> xss
app2(app2(zip, app2(app2(cons, xs), xss)), app2(app2(cons, ys), yss)) -> app2(app2(cons, app2(app2(append, xs), ys)), app2(app2(zip, xss), yss))
app2(app2(combine, xs), nil) -> xs
app2(app2(combine, xs), app2(app2(cons, ys), yss)) -> app2(app2(combine, app2(app2(zip, xs), ys)), yss)
app2(levels, app2(app2(node, x), xs)) -> app2(app2(cons, app2(app2(cons, x), nil)), app2(app2(combine, nil), app2(app2(map, levels), xs)))

Q is empty.
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(append, app2(app2(cons, x), xs)), ys) -> APP2(app2(append, xs), 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(x1)
app2(x1, x2)  =  app1(x2)
append  =  append
cons  =  cons

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


The following usable rules [14] were oriented: none



↳ QTRS
  ↳ DependencyPairsProof
    ↳ QDP
      ↳ DependencyGraphProof
        ↳ AND
          ↳ QDP
            ↳ QDPOrderProof
              ↳ QDP
                ↳ PisEmptyProof
          ↳ QDP
          ↳ 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(append, xs), nil) -> xs
app2(app2(append, nil), ys) -> ys
app2(app2(append, app2(app2(cons, x), xs)), ys) -> app2(app2(cons, x), app2(app2(append, xs), ys))
app2(app2(zip, nil), yss) -> yss
app2(app2(zip, xss), nil) -> xss
app2(app2(zip, app2(app2(cons, xs), xss)), app2(app2(cons, ys), yss)) -> app2(app2(cons, app2(app2(append, xs), ys)), app2(app2(zip, xss), yss))
app2(app2(combine, xs), nil) -> xs
app2(app2(combine, xs), app2(app2(cons, ys), yss)) -> app2(app2(combine, app2(app2(zip, xs), ys)), yss)
app2(levels, app2(app2(node, x), xs)) -> app2(app2(cons, app2(app2(cons, x), nil)), app2(app2(combine, nil), app2(app2(map, levels), xs)))

Q is empty.
We have to consider all minimal (P,Q,R)-chains.
The TRS P is empty. Hence, there is no (P,Q,R) chain.

↳ QTRS
  ↳ DependencyPairsProof
    ↳ QDP
      ↳ DependencyGraphProof
        ↳ AND
          ↳ QDP
          ↳ QDP
            ↳ QDPOrderProof
          ↳ QDP
          ↳ QDP

Q DP problem:
The TRS P consists of the following rules:

APP2(app2(zip, app2(app2(cons, xs), xss)), app2(app2(cons, ys), yss)) -> APP2(app2(zip, xss), yss)

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(append, xs), nil) -> xs
app2(app2(append, nil), ys) -> ys
app2(app2(append, app2(app2(cons, x), xs)), ys) -> app2(app2(cons, x), app2(app2(append, xs), ys))
app2(app2(zip, nil), yss) -> yss
app2(app2(zip, xss), nil) -> xss
app2(app2(zip, app2(app2(cons, xs), xss)), app2(app2(cons, ys), yss)) -> app2(app2(cons, app2(app2(append, xs), ys)), app2(app2(zip, xss), yss))
app2(app2(combine, xs), nil) -> xs
app2(app2(combine, xs), app2(app2(cons, ys), yss)) -> app2(app2(combine, app2(app2(zip, xs), ys)), yss)
app2(levels, app2(app2(node, x), xs)) -> app2(app2(cons, app2(app2(cons, x), nil)), app2(app2(combine, nil), app2(app2(map, levels), xs)))

Q is empty.
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(zip, app2(app2(cons, xs), xss)), app2(app2(cons, ys), yss)) -> APP2(app2(zip, xss), yss)
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)
zip  =  zip
cons  =  cons

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


The following usable rules [14] were oriented: none



↳ QTRS
  ↳ DependencyPairsProof
    ↳ QDP
      ↳ DependencyGraphProof
        ↳ AND
          ↳ QDP
          ↳ 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(append, xs), nil) -> xs
app2(app2(append, nil), ys) -> ys
app2(app2(append, app2(app2(cons, x), xs)), ys) -> app2(app2(cons, x), app2(app2(append, xs), ys))
app2(app2(zip, nil), yss) -> yss
app2(app2(zip, xss), nil) -> xss
app2(app2(zip, app2(app2(cons, xs), xss)), app2(app2(cons, ys), yss)) -> app2(app2(cons, app2(app2(append, xs), ys)), app2(app2(zip, xss), yss))
app2(app2(combine, xs), nil) -> xs
app2(app2(combine, xs), app2(app2(cons, ys), yss)) -> app2(app2(combine, app2(app2(zip, xs), ys)), yss)
app2(levels, app2(app2(node, x), xs)) -> app2(app2(cons, app2(app2(cons, x), nil)), app2(app2(combine, nil), app2(app2(map, levels), xs)))

Q is empty.
We have to consider all minimal (P,Q,R)-chains.
The TRS P is empty. Hence, there is no (P,Q,R) chain.

↳ QTRS
  ↳ DependencyPairsProof
    ↳ QDP
      ↳ DependencyGraphProof
        ↳ AND
          ↳ QDP
          ↳ QDP
          ↳ QDP
            ↳ QDPOrderProof
          ↳ QDP

Q DP problem:
The TRS P consists of the following rules:

APP2(app2(combine, xs), app2(app2(cons, ys), yss)) -> APP2(app2(combine, app2(app2(zip, xs), ys)), yss)

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(append, xs), nil) -> xs
app2(app2(append, nil), ys) -> ys
app2(app2(append, app2(app2(cons, x), xs)), ys) -> app2(app2(cons, x), app2(app2(append, xs), ys))
app2(app2(zip, nil), yss) -> yss
app2(app2(zip, xss), nil) -> xss
app2(app2(zip, app2(app2(cons, xs), xss)), app2(app2(cons, ys), yss)) -> app2(app2(cons, app2(app2(append, xs), ys)), app2(app2(zip, xss), yss))
app2(app2(combine, xs), nil) -> xs
app2(app2(combine, xs), app2(app2(cons, ys), yss)) -> app2(app2(combine, app2(app2(zip, xs), ys)), yss)
app2(levels, app2(app2(node, x), xs)) -> app2(app2(cons, app2(app2(cons, x), nil)), app2(app2(combine, nil), app2(app2(map, levels), xs)))

Q is empty.
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(combine, xs), app2(app2(cons, ys), yss)) -> APP2(app2(combine, app2(app2(zip, xs), ys)), yss)
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)
combine  =  combine
cons  =  cons
zip  =  zip
append  =  append
nil  =  nil

Lexicographic Path Order [19].
Precedence:
combine > [APP1, zip, append]
cons > app1 > [APP1, zip, append]
nil > [APP1, zip, append]


The following usable rules [14] were oriented: none



↳ QTRS
  ↳ DependencyPairsProof
    ↳ QDP
      ↳ DependencyGraphProof
        ↳ AND
          ↳ QDP
          ↳ 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(append, xs), nil) -> xs
app2(app2(append, nil), ys) -> ys
app2(app2(append, app2(app2(cons, x), xs)), ys) -> app2(app2(cons, x), app2(app2(append, xs), ys))
app2(app2(zip, nil), yss) -> yss
app2(app2(zip, xss), nil) -> xss
app2(app2(zip, app2(app2(cons, xs), xss)), app2(app2(cons, ys), yss)) -> app2(app2(cons, app2(app2(append, xs), ys)), app2(app2(zip, xss), yss))
app2(app2(combine, xs), nil) -> xs
app2(app2(combine, xs), app2(app2(cons, ys), yss)) -> app2(app2(combine, app2(app2(zip, xs), ys)), yss)
app2(levels, app2(app2(node, x), xs)) -> app2(app2(cons, app2(app2(cons, x), nil)), app2(app2(combine, nil), app2(app2(map, levels), xs)))

Q is empty.
We have to consider all minimal (P,Q,R)-chains.
The TRS P is empty. Hence, there is no (P,Q,R) chain.

↳ QTRS
  ↳ DependencyPairsProof
    ↳ QDP
      ↳ DependencyGraphProof
        ↳ AND
          ↳ QDP
          ↳ QDP
          ↳ QDP
          ↳ QDP
            ↳ QDPOrderProof

Q DP problem:
The TRS P consists of the following rules:

APP2(app2(map, f), app2(app2(cons, x), xs)) -> APP2(f, x)
APP2(levels, app2(app2(node, x), xs)) -> APP2(app2(map, levels), xs)
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(append, xs), nil) -> xs
app2(app2(append, nil), ys) -> ys
app2(app2(append, app2(app2(cons, x), xs)), ys) -> app2(app2(cons, x), app2(app2(append, xs), ys))
app2(app2(zip, nil), yss) -> yss
app2(app2(zip, xss), nil) -> xss
app2(app2(zip, app2(app2(cons, xs), xss)), app2(app2(cons, ys), yss)) -> app2(app2(cons, app2(app2(append, xs), ys)), app2(app2(zip, xss), yss))
app2(app2(combine, xs), nil) -> xs
app2(app2(combine, xs), app2(app2(cons, ys), yss)) -> app2(app2(combine, app2(app2(zip, xs), ys)), yss)
app2(levels, app2(app2(node, x), xs)) -> app2(app2(cons, app2(app2(cons, x), nil)), app2(app2(combine, nil), app2(app2(map, levels), xs)))

Q is empty.
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(map, f), app2(app2(cons, x), xs)) -> APP2(f, x)
APP2(levels, app2(app2(node, x), xs)) -> APP2(app2(map, levels), xs)
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)  =  APP2(x1, x2)
app2(x1, x2)  =  app2(x1, x2)
map  =  map
cons  =  cons
levels  =  levels
node  =  node

Lexicographic Path Order [19].
Precedence:
node > [map, cons] > [app2, levels] > APP2


The following usable rules [14] were oriented: none



↳ QTRS
  ↳ DependencyPairsProof
    ↳ QDP
      ↳ DependencyGraphProof
        ↳ AND
          ↳ QDP
          ↳ 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(append, xs), nil) -> xs
app2(app2(append, nil), ys) -> ys
app2(app2(append, app2(app2(cons, x), xs)), ys) -> app2(app2(cons, x), app2(app2(append, xs), ys))
app2(app2(zip, nil), yss) -> yss
app2(app2(zip, xss), nil) -> xss
app2(app2(zip, app2(app2(cons, xs), xss)), app2(app2(cons, ys), yss)) -> app2(app2(cons, app2(app2(append, xs), ys)), app2(app2(zip, xss), yss))
app2(app2(combine, xs), nil) -> xs
app2(app2(combine, xs), app2(app2(cons, ys), yss)) -> app2(app2(combine, app2(app2(zip, xs), ys)), yss)
app2(levels, app2(app2(node, x), xs)) -> app2(app2(cons, app2(app2(cons, x), nil)), app2(app2(combine, nil), app2(app2(map, levels), xs)))

Q is empty.
We have to consider all minimal (P,Q,R)-chains.
The TRS P is empty. Hence, there is no (P,Q,R) chain.