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(append, nil), ys) -> ys
app2(app2(append, app2(app2(cons, x), xs)), ys) -> app2(app2(cons, x), app2(app2(append, xs), ys))
app2(app2(flatwith, f), app2(leaf, x)) -> app2(app2(cons, app2(f, x)), nil)
app2(app2(flatwith, f), app2(node, xs)) -> app2(app2(flatwithsub, f), xs)
app2(app2(flatwithsub, f), nil) -> nil
app2(app2(flatwithsub, f), app2(app2(cons, x), xs)) -> app2(app2(append, app2(app2(flatwith, f), x)), app2(app2(flatwithsub, f), xs))

Q is empty.


↳ QTRS
  ↳ Non-Overlap Check

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

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(flatwith, f), app2(leaf, x)) -> app2(app2(cons, app2(f, x)), nil)
app2(app2(flatwith, f), app2(node, xs)) -> app2(app2(flatwithsub, f), xs)
app2(app2(flatwithsub, f), nil) -> nil
app2(app2(flatwithsub, f), app2(app2(cons, x), xs)) -> app2(app2(append, app2(app2(flatwith, f), x)), app2(app2(flatwithsub, f), 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(append, nil), ys) -> ys
app2(app2(append, app2(app2(cons, x), xs)), ys) -> app2(app2(cons, x), app2(app2(append, xs), ys))
app2(app2(flatwith, f), app2(leaf, x)) -> app2(app2(cons, app2(f, x)), nil)
app2(app2(flatwith, f), app2(node, xs)) -> app2(app2(flatwithsub, f), xs)
app2(app2(flatwithsub, f), nil) -> nil
app2(app2(flatwithsub, f), app2(app2(cons, x), xs)) -> app2(app2(append, app2(app2(flatwith, f), x)), app2(app2(flatwithsub, f), xs))

The set Q consists of the following terms:

app2(app2(append, nil), x0)
app2(app2(append, app2(app2(cons, x0), x1)), x2)
app2(app2(flatwith, x0), app2(leaf, x1))
app2(app2(flatwith, x0), app2(node, x1))
app2(app2(flatwithsub, x0), nil)
app2(app2(flatwithsub, x0), app2(app2(cons, 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(app2(append, app2(app2(cons, x), xs)), ys) -> APP2(append, xs)
APP2(app2(flatwith, f), app2(leaf, x)) -> APP2(cons, app2(f, x))
APP2(app2(flatwithsub, f), app2(app2(cons, x), xs)) -> APP2(app2(flatwithsub, f), xs)
APP2(app2(flatwithsub, f), app2(app2(cons, x), xs)) -> APP2(app2(append, app2(app2(flatwith, f), x)), app2(app2(flatwithsub, f), xs))
APP2(app2(flatwith, f), app2(node, xs)) -> APP2(flatwithsub, f)
APP2(app2(append, app2(app2(cons, x), xs)), ys) -> APP2(app2(append, xs), ys)
APP2(app2(flatwith, f), app2(leaf, x)) -> APP2(app2(cons, app2(f, x)), nil)
APP2(app2(flatwith, f), app2(node, xs)) -> APP2(app2(flatwithsub, f), xs)
APP2(app2(flatwithsub, f), app2(app2(cons, x), xs)) -> APP2(app2(flatwith, f), x)
APP2(app2(append, app2(app2(cons, x), xs)), ys) -> APP2(app2(cons, x), app2(app2(append, xs), ys))
APP2(app2(flatwithsub, f), app2(app2(cons, x), xs)) -> APP2(append, app2(app2(flatwith, f), x))
APP2(app2(flatwithsub, f), app2(app2(cons, x), xs)) -> APP2(flatwith, f)
APP2(app2(flatwith, f), app2(leaf, x)) -> APP2(f, x)

The TRS R consists of the following rules:

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(flatwith, f), app2(leaf, x)) -> app2(app2(cons, app2(f, x)), nil)
app2(app2(flatwith, f), app2(node, xs)) -> app2(app2(flatwithsub, f), xs)
app2(app2(flatwithsub, f), nil) -> nil
app2(app2(flatwithsub, f), app2(app2(cons, x), xs)) -> app2(app2(append, app2(app2(flatwith, f), x)), app2(app2(flatwithsub, f), xs))

The set Q consists of the following terms:

app2(app2(append, nil), x0)
app2(app2(append, app2(app2(cons, x0), x1)), x2)
app2(app2(flatwith, x0), app2(leaf, x1))
app2(app2(flatwith, x0), app2(node, x1))
app2(app2(flatwithsub, x0), nil)
app2(app2(flatwithsub, x0), app2(app2(cons, 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(app2(append, app2(app2(cons, x), xs)), ys) -> APP2(append, xs)
APP2(app2(flatwith, f), app2(leaf, x)) -> APP2(cons, app2(f, x))
APP2(app2(flatwithsub, f), app2(app2(cons, x), xs)) -> APP2(app2(flatwithsub, f), xs)
APP2(app2(flatwithsub, f), app2(app2(cons, x), xs)) -> APP2(app2(append, app2(app2(flatwith, f), x)), app2(app2(flatwithsub, f), xs))
APP2(app2(flatwith, f), app2(node, xs)) -> APP2(flatwithsub, f)
APP2(app2(append, app2(app2(cons, x), xs)), ys) -> APP2(app2(append, xs), ys)
APP2(app2(flatwith, f), app2(leaf, x)) -> APP2(app2(cons, app2(f, x)), nil)
APP2(app2(flatwith, f), app2(node, xs)) -> APP2(app2(flatwithsub, f), xs)
APP2(app2(flatwithsub, f), app2(app2(cons, x), xs)) -> APP2(app2(flatwith, f), x)
APP2(app2(append, app2(app2(cons, x), xs)), ys) -> APP2(app2(cons, x), app2(app2(append, xs), ys))
APP2(app2(flatwithsub, f), app2(app2(cons, x), xs)) -> APP2(append, app2(app2(flatwith, f), x))
APP2(app2(flatwithsub, f), app2(app2(cons, x), xs)) -> APP2(flatwith, f)
APP2(app2(flatwith, f), app2(leaf, x)) -> APP2(f, x)

The TRS R consists of the following rules:

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(flatwith, f), app2(leaf, x)) -> app2(app2(cons, app2(f, x)), nil)
app2(app2(flatwith, f), app2(node, xs)) -> app2(app2(flatwithsub, f), xs)
app2(app2(flatwithsub, f), nil) -> nil
app2(app2(flatwithsub, f), app2(app2(cons, x), xs)) -> app2(app2(append, app2(app2(flatwith, f), x)), app2(app2(flatwithsub, f), xs))

The set Q consists of the following terms:

app2(app2(append, nil), x0)
app2(app2(append, app2(app2(cons, x0), x1)), x2)
app2(app2(flatwith, x0), app2(leaf, x1))
app2(app2(flatwith, x0), app2(node, x1))
app2(app2(flatwithsub, x0), nil)
app2(app2(flatwithsub, x0), app2(app2(cons, x1), x2))

We have to consider all minimal (P,Q,R)-chains.
The approximation of the Dependency Graph [13,14,18] contains 2 SCCs with 8 less nodes.

↳ QTRS
  ↳ Non-Overlap Check
    ↳ QTRS
      ↳ DependencyPairsProof
        ↳ QDP
          ↳ DependencyGraphProof
            ↳ AND
              ↳ QDP
                ↳ QDPOrderProof
              ↳ 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(append, nil), ys) -> ys
app2(app2(append, app2(app2(cons, x), xs)), ys) -> app2(app2(cons, x), app2(app2(append, xs), ys))
app2(app2(flatwith, f), app2(leaf, x)) -> app2(app2(cons, app2(f, x)), nil)
app2(app2(flatwith, f), app2(node, xs)) -> app2(app2(flatwithsub, f), xs)
app2(app2(flatwithsub, f), nil) -> nil
app2(app2(flatwithsub, f), app2(app2(cons, x), xs)) -> app2(app2(append, app2(app2(flatwith, f), x)), app2(app2(flatwithsub, f), xs))

The set Q consists of the following terms:

app2(app2(append, nil), x0)
app2(app2(append, app2(app2(cons, x0), x1)), x2)
app2(app2(flatwith, x0), app2(leaf, x1))
app2(app2(flatwith, x0), app2(node, x1))
app2(app2(flatwithsub, x0), nil)
app2(app2(flatwithsub, x0), app2(app2(cons, 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)  =  x1
app2(x1, x2)  =  app1(x2)
append  =  append
cons  =  cons

Lexicographic Path Order [19].
Precedence:
trivial

The following usable rules [14] were oriented: none



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

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

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(flatwith, f), app2(leaf, x)) -> app2(app2(cons, app2(f, x)), nil)
app2(app2(flatwith, f), app2(node, xs)) -> app2(app2(flatwithsub, f), xs)
app2(app2(flatwithsub, f), nil) -> nil
app2(app2(flatwithsub, f), app2(app2(cons, x), xs)) -> app2(app2(append, app2(app2(flatwith, f), x)), app2(app2(flatwithsub, f), xs))

The set Q consists of the following terms:

app2(app2(append, nil), x0)
app2(app2(append, app2(app2(cons, x0), x1)), x2)
app2(app2(flatwith, x0), app2(leaf, x1))
app2(app2(flatwith, x0), app2(node, x1))
app2(app2(flatwithsub, x0), nil)
app2(app2(flatwithsub, x0), app2(app2(cons, 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

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

APP2(app2(flatwith, f), app2(node, xs)) -> APP2(app2(flatwithsub, f), xs)
APP2(app2(flatwithsub, f), app2(app2(cons, x), xs)) -> APP2(app2(flatwith, f), x)
APP2(app2(flatwithsub, f), app2(app2(cons, x), xs)) -> APP2(app2(flatwithsub, f), xs)
APP2(app2(flatwith, f), app2(leaf, x)) -> APP2(f, x)

The TRS R consists of the following rules:

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(flatwith, f), app2(leaf, x)) -> app2(app2(cons, app2(f, x)), nil)
app2(app2(flatwith, f), app2(node, xs)) -> app2(app2(flatwithsub, f), xs)
app2(app2(flatwithsub, f), nil) -> nil
app2(app2(flatwithsub, f), app2(app2(cons, x), xs)) -> app2(app2(append, app2(app2(flatwith, f), x)), app2(app2(flatwithsub, f), xs))

The set Q consists of the following terms:

app2(app2(append, nil), x0)
app2(app2(append, app2(app2(cons, x0), x1)), x2)
app2(app2(flatwith, x0), app2(leaf, x1))
app2(app2(flatwith, x0), app2(node, x1))
app2(app2(flatwithsub, x0), nil)
app2(app2(flatwithsub, x0), app2(app2(cons, 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(flatwith, f), app2(node, xs)) -> APP2(app2(flatwithsub, f), xs)
APP2(app2(flatwithsub, f), app2(app2(cons, x), xs)) -> APP2(app2(flatwith, f), x)
APP2(app2(flatwithsub, f), app2(app2(cons, x), xs)) -> APP2(app2(flatwithsub, f), xs)
APP2(app2(flatwith, f), app2(leaf, x)) -> APP2(f, x)
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)
flatwith  =  flatwith
node  =  node
flatwithsub  =  flatwithsub
cons  =  cons
leaf  =  leaf

Lexicographic Path Order [19].
Precedence:
node > app2
node > flatwithsub
cons > app2
cons > flatwith
cons > flatwithsub

The following usable rules [14] were oriented: none



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

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

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(flatwith, f), app2(leaf, x)) -> app2(app2(cons, app2(f, x)), nil)
app2(app2(flatwith, f), app2(node, xs)) -> app2(app2(flatwithsub, f), xs)
app2(app2(flatwithsub, f), nil) -> nil
app2(app2(flatwithsub, f), app2(app2(cons, x), xs)) -> app2(app2(append, app2(app2(flatwith, f), x)), app2(app2(flatwithsub, f), xs))

The set Q consists of the following terms:

app2(app2(append, nil), x0)
app2(app2(append, app2(app2(cons, x0), x1)), x2)
app2(app2(flatwith, x0), app2(leaf, x1))
app2(app2(flatwith, x0), app2(node, x1))
app2(app2(flatwithsub, x0), nil)
app2(app2(flatwithsub, x0), app2(app2(cons, 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.