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:

app(app(map, f), nil) → nil
app(app(map, f), app(app(cons, x), xs)) → app(app(cons, app(f, x)), app(app(map, f), xs))
app(flatten, app(app(node, x), xs)) → app(app(cons, x), app(concat, app(app(map, flatten), xs)))
app(concat, nil) → nil
app(concat, app(app(cons, x), xs)) → app(app(append, x), app(concat, xs))
app(app(append, nil), xs) → xs
app(app(append, app(app(cons, x), xs)), ys) → app(app(cons, x), app(app(append, xs), ys))

Q is empty.


QTRS
  ↳ Overlay + Local Confluence

Q restricted rewrite system:
The TRS R consists of the following rules:

app(app(map, f), nil) → nil
app(app(map, f), app(app(cons, x), xs)) → app(app(cons, app(f, x)), app(app(map, f), xs))
app(flatten, app(app(node, x), xs)) → app(app(cons, x), app(concat, app(app(map, flatten), xs)))
app(concat, nil) → nil
app(concat, app(app(cons, x), xs)) → app(app(append, x), app(concat, xs))
app(app(append, nil), xs) → xs
app(app(append, app(app(cons, x), xs)), ys) → app(app(cons, x), app(app(append, xs), ys))

Q is empty.

The TRS is overlay and locally confluent. By [15] we can switch to innermost.

↳ QTRS
  ↳ Overlay + Local Confluence
QTRS
      ↳ DependencyPairsProof

Q restricted rewrite system:
The TRS R consists of the following rules:

app(app(map, f), nil) → nil
app(app(map, f), app(app(cons, x), xs)) → app(app(cons, app(f, x)), app(app(map, f), xs))
app(flatten, app(app(node, x), xs)) → app(app(cons, x), app(concat, app(app(map, flatten), xs)))
app(concat, nil) → nil
app(concat, app(app(cons, x), xs)) → app(app(append, x), app(concat, xs))
app(app(append, nil), xs) → xs
app(app(append, app(app(cons, x), xs)), ys) → app(app(cons, x), app(app(append, xs), ys))

The set Q consists of the following terms:

app(app(map, x0), nil)
app(app(map, x0), app(app(cons, x1), x2))
app(flatten, app(app(node, x0), x1))
app(concat, nil)
app(concat, app(app(cons, x0), x1))
app(app(append, nil), x0)
app(app(append, app(app(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:

APP(concat, app(app(cons, x), xs)) → APP(append, x)
APP(app(map, f), app(app(cons, x), xs)) → APP(app(cons, app(f, x)), app(app(map, f), xs))
APP(app(append, app(app(cons, x), xs)), ys) → APP(append, xs)
APP(app(map, f), app(app(cons, x), xs)) → APP(app(map, f), xs)
APP(concat, app(app(cons, x), xs)) → APP(concat, xs)
APP(app(append, app(app(cons, x), xs)), ys) → APP(app(append, xs), ys)
APP(flatten, app(app(node, x), xs)) → APP(concat, app(app(map, flatten), xs))
APP(flatten, app(app(node, x), xs)) → APP(app(map, flatten), xs)
APP(flatten, app(app(node, x), xs)) → APP(app(cons, x), app(concat, app(app(map, flatten), xs)))
APP(app(map, f), app(app(cons, x), xs)) → APP(f, x)
APP(flatten, app(app(node, x), xs)) → APP(cons, x)
APP(concat, app(app(cons, x), xs)) → APP(app(append, x), app(concat, xs))
APP(app(append, app(app(cons, x), xs)), ys) → APP(app(cons, x), app(app(append, xs), ys))
APP(flatten, app(app(node, x), xs)) → APP(map, flatten)
APP(app(map, f), app(app(cons, x), xs)) → APP(cons, app(f, x))

The TRS R consists of the following rules:

app(app(map, f), nil) → nil
app(app(map, f), app(app(cons, x), xs)) → app(app(cons, app(f, x)), app(app(map, f), xs))
app(flatten, app(app(node, x), xs)) → app(app(cons, x), app(concat, app(app(map, flatten), xs)))
app(concat, nil) → nil
app(concat, app(app(cons, x), xs)) → app(app(append, x), app(concat, xs))
app(app(append, nil), xs) → xs
app(app(append, app(app(cons, x), xs)), ys) → app(app(cons, x), app(app(append, xs), ys))

The set Q consists of the following terms:

app(app(map, x0), nil)
app(app(map, x0), app(app(cons, x1), x2))
app(flatten, app(app(node, x0), x1))
app(concat, nil)
app(concat, app(app(cons, x0), x1))
app(app(append, nil), x0)
app(app(append, app(app(cons, x0), x1)), x2)

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

↳ QTRS
  ↳ Overlay + Local Confluence
    ↳ QTRS
      ↳ DependencyPairsProof
QDP
          ↳ EdgeDeletionProof

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

APP(concat, app(app(cons, x), xs)) → APP(append, x)
APP(app(map, f), app(app(cons, x), xs)) → APP(app(cons, app(f, x)), app(app(map, f), xs))
APP(app(append, app(app(cons, x), xs)), ys) → APP(append, xs)
APP(app(map, f), app(app(cons, x), xs)) → APP(app(map, f), xs)
APP(concat, app(app(cons, x), xs)) → APP(concat, xs)
APP(app(append, app(app(cons, x), xs)), ys) → APP(app(append, xs), ys)
APP(flatten, app(app(node, x), xs)) → APP(concat, app(app(map, flatten), xs))
APP(flatten, app(app(node, x), xs)) → APP(app(map, flatten), xs)
APP(flatten, app(app(node, x), xs)) → APP(app(cons, x), app(concat, app(app(map, flatten), xs)))
APP(app(map, f), app(app(cons, x), xs)) → APP(f, x)
APP(flatten, app(app(node, x), xs)) → APP(cons, x)
APP(concat, app(app(cons, x), xs)) → APP(app(append, x), app(concat, xs))
APP(app(append, app(app(cons, x), xs)), ys) → APP(app(cons, x), app(app(append, xs), ys))
APP(flatten, app(app(node, x), xs)) → APP(map, flatten)
APP(app(map, f), app(app(cons, x), xs)) → APP(cons, app(f, x))

The TRS R consists of the following rules:

app(app(map, f), nil) → nil
app(app(map, f), app(app(cons, x), xs)) → app(app(cons, app(f, x)), app(app(map, f), xs))
app(flatten, app(app(node, x), xs)) → app(app(cons, x), app(concat, app(app(map, flatten), xs)))
app(concat, nil) → nil
app(concat, app(app(cons, x), xs)) → app(app(append, x), app(concat, xs))
app(app(append, nil), xs) → xs
app(app(append, app(app(cons, x), xs)), ys) → app(app(cons, x), app(app(append, xs), ys))

The set Q consists of the following terms:

app(app(map, x0), nil)
app(app(map, x0), app(app(cons, x1), x2))
app(flatten, app(app(node, x0), x1))
app(concat, nil)
app(concat, app(app(cons, x0), x1))
app(app(append, nil), x0)
app(app(append, app(app(cons, x0), x1)), x2)

We have to consider all minimal (P,Q,R)-chains.
We deleted some edges using various graph approximations

↳ QTRS
  ↳ Overlay + Local Confluence
    ↳ QTRS
      ↳ DependencyPairsProof
        ↳ QDP
          ↳ EdgeDeletionProof
QDP
              ↳ DependencyGraphProof

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

APP(concat, app(app(cons, x), xs)) → APP(append, x)
APP(app(map, f), app(app(cons, x), xs)) → APP(app(cons, app(f, x)), app(app(map, f), xs))
APP(app(append, app(app(cons, x), xs)), ys) → APP(append, xs)
APP(concat, app(app(cons, x), xs)) → APP(concat, xs)
APP(app(map, f), app(app(cons, x), xs)) → APP(app(map, f), xs)
APP(app(append, app(app(cons, x), xs)), ys) → APP(app(append, xs), ys)
APP(flatten, app(app(node, x), xs)) → APP(app(map, flatten), xs)
APP(flatten, app(app(node, x), xs)) → APP(concat, app(app(map, flatten), xs))
APP(flatten, app(app(node, x), xs)) → APP(app(cons, x), app(concat, app(app(map, flatten), xs)))
APP(flatten, app(app(node, x), xs)) → APP(cons, x)
APP(app(map, f), app(app(cons, x), xs)) → APP(f, x)
APP(app(append, app(app(cons, x), xs)), ys) → APP(app(cons, x), app(app(append, xs), ys))
APP(concat, app(app(cons, x), xs)) → APP(app(append, x), app(concat, xs))
APP(flatten, app(app(node, x), xs)) → APP(map, flatten)
APP(app(map, f), app(app(cons, x), xs)) → APP(cons, app(f, x))

The TRS R consists of the following rules:

app(app(map, f), nil) → nil
app(app(map, f), app(app(cons, x), xs)) → app(app(cons, app(f, x)), app(app(map, f), xs))
app(flatten, app(app(node, x), xs)) → app(app(cons, x), app(concat, app(app(map, flatten), xs)))
app(concat, nil) → nil
app(concat, app(app(cons, x), xs)) → app(app(append, x), app(concat, xs))
app(app(append, nil), xs) → xs
app(app(append, app(app(cons, x), xs)), ys) → app(app(cons, x), app(app(append, xs), ys))

The set Q consists of the following terms:

app(app(map, x0), nil)
app(app(map, x0), app(app(cons, x1), x2))
app(flatten, app(app(node, x0), x1))
app(concat, nil)
app(concat, app(app(cons, x0), x1))
app(app(append, nil), x0)
app(app(append, app(app(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
  ↳ Overlay + Local Confluence
    ↳ QTRS
      ↳ DependencyPairsProof
        ↳ QDP
          ↳ EdgeDeletionProof
            ↳ QDP
              ↳ DependencyGraphProof
                ↳ AND
QDP
                    ↳ QDPOrderProof
                  ↳ QDP
                  ↳ QDP

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

APP(app(append, app(app(cons, x), xs)), ys) → APP(app(append, xs), ys)

The TRS R consists of the following rules:

app(app(map, f), nil) → nil
app(app(map, f), app(app(cons, x), xs)) → app(app(cons, app(f, x)), app(app(map, f), xs))
app(flatten, app(app(node, x), xs)) → app(app(cons, x), app(concat, app(app(map, flatten), xs)))
app(concat, nil) → nil
app(concat, app(app(cons, x), xs)) → app(app(append, x), app(concat, xs))
app(app(append, nil), xs) → xs
app(app(append, app(app(cons, x), xs)), ys) → app(app(cons, x), app(app(append, xs), ys))

The set Q consists of the following terms:

app(app(map, x0), nil)
app(app(map, x0), app(app(cons, x1), x2))
app(flatten, app(app(node, x0), x1))
app(concat, nil)
app(concat, app(app(cons, x0), x1))
app(app(append, nil), x0)
app(app(append, app(app(cons, x0), x1)), x2)

We have to consider all minimal (P,Q,R)-chains.
We use the reduction pair processor [13]. Here, we combined the reduction pair processor with the A-transformation [14] which results in the following intermediate Q-DP Problem.
Q DP problem:
The TRS P consists of the following rules:

APPEND(cons(x, xs), ys) → APPEND(xs, ys)

R is empty.
The set Q consists of the following terms:

map(x0, nil)
map(x0, cons(x1, x2))
flatten(node(x0, x1))
concat(nil)
concat(cons(x0, x1))
append(nil, x0)
append(cons(x0, x1), x2)

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


The following pairs can be oriented strictly and are deleted.


APP(app(append, app(app(cons, x), xs)), ys) → APP(app(append, xs), ys)
The remaining pairs can at least be oriented weakly.
none
Used ordering: Combined order from the following AFS and order.
APPEND(x1, x2)  =  x1
cons(x1, x2)  =  cons(x1, x2)

Lexicographic Path Order [19].
Precedence:
trivial

The following usable rules [14] were oriented: none



↳ QTRS
  ↳ Overlay + Local Confluence
    ↳ QTRS
      ↳ DependencyPairsProof
        ↳ QDP
          ↳ EdgeDeletionProof
            ↳ QDP
              ↳ DependencyGraphProof
                ↳ AND
                  ↳ QDP
                    ↳ QDPOrderProof
QDP
                        ↳ PisEmptyProof
                  ↳ QDP
                  ↳ QDP

Q DP problem:
P is empty.
The TRS R consists of the following rules:

app(app(map, f), nil) → nil
app(app(map, f), app(app(cons, x), xs)) → app(app(cons, app(f, x)), app(app(map, f), xs))
app(flatten, app(app(node, x), xs)) → app(app(cons, x), app(concat, app(app(map, flatten), xs)))
app(concat, nil) → nil
app(concat, app(app(cons, x), xs)) → app(app(append, x), app(concat, xs))
app(app(append, nil), xs) → xs
app(app(append, app(app(cons, x), xs)), ys) → app(app(cons, x), app(app(append, xs), ys))

The set Q consists of the following terms:

app(app(map, x0), nil)
app(app(map, x0), app(app(cons, x1), x2))
app(flatten, app(app(node, x0), x1))
app(concat, nil)
app(concat, app(app(cons, x0), x1))
app(app(append, nil), x0)
app(app(append, app(app(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
  ↳ Overlay + Local Confluence
    ↳ QTRS
      ↳ DependencyPairsProof
        ↳ QDP
          ↳ EdgeDeletionProof
            ↳ QDP
              ↳ DependencyGraphProof
                ↳ AND
                  ↳ QDP
QDP
                    ↳ QDPOrderProof
                  ↳ QDP

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

APP(concat, app(app(cons, x), xs)) → APP(concat, xs)

The TRS R consists of the following rules:

app(app(map, f), nil) → nil
app(app(map, f), app(app(cons, x), xs)) → app(app(cons, app(f, x)), app(app(map, f), xs))
app(flatten, app(app(node, x), xs)) → app(app(cons, x), app(concat, app(app(map, flatten), xs)))
app(concat, nil) → nil
app(concat, app(app(cons, x), xs)) → app(app(append, x), app(concat, xs))
app(app(append, nil), xs) → xs
app(app(append, app(app(cons, x), xs)), ys) → app(app(cons, x), app(app(append, xs), ys))

The set Q consists of the following terms:

app(app(map, x0), nil)
app(app(map, x0), app(app(cons, x1), x2))
app(flatten, app(app(node, x0), x1))
app(concat, nil)
app(concat, app(app(cons, x0), x1))
app(app(append, nil), x0)
app(app(append, app(app(cons, x0), x1)), x2)

We have to consider all minimal (P,Q,R)-chains.
We use the reduction pair processor [13]. Here, we combined the reduction pair processor with the A-transformation [14] which results in the following intermediate Q-DP Problem.
Q DP problem:
The TRS P consists of the following rules:

CONCAT(cons(x, xs)) → CONCAT(xs)

R is empty.
The set Q consists of the following terms:

map(x0, nil)
map(x0, cons(x1, x2))
flatten(node(x0, x1))
concat(nil)
concat(cons(x0, x1))
append(nil, x0)
append(cons(x0, x1), x2)

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


The following pairs can be oriented strictly and are deleted.


APP(concat, app(app(cons, x), xs)) → APP(concat, xs)
The remaining pairs can at least be oriented weakly.
none
Used ordering: Combined order from the following AFS and order.
CONCAT(x1)  =  CONCAT(x1)
cons(x1, x2)  =  cons(x1, x2)

Lexicographic Path Order [19].
Precedence:
cons2 > CONCAT1

The following usable rules [14] were oriented: none



↳ QTRS
  ↳ Overlay + Local Confluence
    ↳ QTRS
      ↳ DependencyPairsProof
        ↳ QDP
          ↳ EdgeDeletionProof
            ↳ QDP
              ↳ DependencyGraphProof
                ↳ AND
                  ↳ QDP
                  ↳ QDP
                    ↳ QDPOrderProof
QDP
                        ↳ PisEmptyProof
                  ↳ QDP

Q DP problem:
P is empty.
The TRS R consists of the following rules:

app(app(map, f), nil) → nil
app(app(map, f), app(app(cons, x), xs)) → app(app(cons, app(f, x)), app(app(map, f), xs))
app(flatten, app(app(node, x), xs)) → app(app(cons, x), app(concat, app(app(map, flatten), xs)))
app(concat, nil) → nil
app(concat, app(app(cons, x), xs)) → app(app(append, x), app(concat, xs))
app(app(append, nil), xs) → xs
app(app(append, app(app(cons, x), xs)), ys) → app(app(cons, x), app(app(append, xs), ys))

The set Q consists of the following terms:

app(app(map, x0), nil)
app(app(map, x0), app(app(cons, x1), x2))
app(flatten, app(app(node, x0), x1))
app(concat, nil)
app(concat, app(app(cons, x0), x1))
app(app(append, nil), x0)
app(app(append, app(app(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
  ↳ Overlay + Local Confluence
    ↳ QTRS
      ↳ DependencyPairsProof
        ↳ QDP
          ↳ EdgeDeletionProof
            ↳ QDP
              ↳ DependencyGraphProof
                ↳ AND
                  ↳ QDP
                  ↳ QDP
QDP
                    ↳ QDPOrderProof

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

APP(app(map, f), app(app(cons, x), xs)) → APP(f, x)
APP(app(map, f), app(app(cons, x), xs)) → APP(app(map, f), xs)
APP(flatten, app(app(node, x), xs)) → APP(app(map, flatten), xs)

The TRS R consists of the following rules:

app(app(map, f), nil) → nil
app(app(map, f), app(app(cons, x), xs)) → app(app(cons, app(f, x)), app(app(map, f), xs))
app(flatten, app(app(node, x), xs)) → app(app(cons, x), app(concat, app(app(map, flatten), xs)))
app(concat, nil) → nil
app(concat, app(app(cons, x), xs)) → app(app(append, x), app(concat, xs))
app(app(append, nil), xs) → xs
app(app(append, app(app(cons, x), xs)), ys) → app(app(cons, x), app(app(append, xs), ys))

The set Q consists of the following terms:

app(app(map, x0), nil)
app(app(map, x0), app(app(cons, x1), x2))
app(flatten, app(app(node, x0), x1))
app(concat, nil)
app(concat, app(app(cons, x0), x1))
app(app(append, nil), x0)
app(app(append, app(app(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 oriented strictly and are deleted.


APP(app(map, f), app(app(cons, x), xs)) → APP(f, x)
APP(app(map, f), app(app(cons, x), xs)) → APP(app(map, f), xs)
APP(flatten, app(app(node, x), xs)) → APP(app(map, flatten), xs)
The remaining pairs can at least be oriented weakly.
none
Used ordering: Combined order from the following AFS and order.
APP(x1, x2)  =  APP(x1, x2)
app(x1, x2)  =  app(x1, x2)
map  =  map
cons  =  cons
flatten  =  flatten
node  =  node

Lexicographic Path Order [19].
Precedence:
node > app2 > APP2
node > map > APP2
node > flatten > APP2

The following usable rules [14] were oriented: none



↳ QTRS
  ↳ Overlay + Local Confluence
    ↳ QTRS
      ↳ DependencyPairsProof
        ↳ QDP
          ↳ EdgeDeletionProof
            ↳ QDP
              ↳ DependencyGraphProof
                ↳ AND
                  ↳ QDP
                  ↳ QDP
                  ↳ QDP
                    ↳ QDPOrderProof
QDP
                        ↳ PisEmptyProof

Q DP problem:
P is empty.
The TRS R consists of the following rules:

app(app(map, f), nil) → nil
app(app(map, f), app(app(cons, x), xs)) → app(app(cons, app(f, x)), app(app(map, f), xs))
app(flatten, app(app(node, x), xs)) → app(app(cons, x), app(concat, app(app(map, flatten), xs)))
app(concat, nil) → nil
app(concat, app(app(cons, x), xs)) → app(app(append, x), app(concat, xs))
app(app(append, nil), xs) → xs
app(app(append, app(app(cons, x), xs)), ys) → app(app(cons, x), app(app(append, xs), ys))

The set Q consists of the following terms:

app(app(map, x0), nil)
app(app(map, x0), app(app(cons, x1), x2))
app(flatten, app(app(node, x0), x1))
app(concat, nil)
app(concat, app(app(cons, x0), x1))
app(app(append, nil), x0)
app(app(append, app(app(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.