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)


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(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 [13,14,18] contains 3 SCCs with 10 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(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.
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
  ↳ 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(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
                ↳ QDPOrderProof
              ↳ 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.
We use the reduction pair processor [13].


The following pairs can be strictly oriented and are deleted.


APP2(concat, app2(app2(cons, x), xs)) -> APP2(concat, 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)
concat  =  concat
app2(x1, x2)  =  app1(x2)
cons  =  cons

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


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

Lexicographic Path Order [19].
Precedence:
app2 > flatten
node > flatten


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(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.