Termination w.r.t. Q proof of /tmp/tmpT

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(append, nil), ys) → ys
app(app(append, app(app(cons, x), xs)), ys) → app(app(cons, x), app(app(append, xs), ys))
app(app(flatwith, f), app(leaf, x)) → app(app(cons, app(f, x)), nil)
app(app(flatwith, f), app(node, xs)) → app(app(flatwithsub, f), xs)
app(app(flatwithsub, f), nil) → nil
app(app(flatwithsub, f), app(app(cons, x), xs)) → app(app(append, app(app(flatwith, f), x)), app(app(flatwithsub, f), xs))

Q is empty.

↳ QTRS

  ↳ Overlay + Local Confluence

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

app(app(append, nil), ys) → ys
app(app(append, app(app(cons, x), xs)), ys) → app(app(cons, x), app(app(append, xs), ys))
app(app(flatwith, f), app(leaf, x)) → app(app(cons, app(f, x)), nil)
app(app(flatwith, f), app(node, xs)) → app(app(flatwithsub, f), xs)
app(app(flatwithsub, f), nil) → nil
app(app(flatwithsub, f), app(app(cons, x), xs)) → app(app(append, app(app(flatwith, f), x)), app(app(flatwithsub, f), xs))

Q is empty.

The TRS is overlay and locally confluent. By [19] 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(append, nil), ys) → ys
app(app(append, app(app(cons, x), xs)), ys) → app(app(cons, x), app(app(append, xs), ys))
app(app(flatwith, f), app(leaf, x)) → app(app(cons, app(f, x)), nil)
app(app(flatwith, f), app(node, xs)) → app(app(flatwithsub, f), xs)
app(app(flatwithsub, f), nil) → nil
app(app(flatwithsub, f), app(app(cons, x), xs)) → app(app(append, app(app(flatwith, f), x)), app(app(flatwithsub, f), xs))

The set Q consists of the following terms:

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

Using Dependency Pairs [1,15] we result in the following initial DP problem:
Q DP problem:
The TRS P consists of the following rules:

APP(app(flatwith, f), app(node, xs)) → APP(app(flatwithsub, f), xs)
APP(app(append, app(app(cons, x), xs)), ys) → APP(append, xs)
APP(app(flatwith, f), app(leaf, x)) → APP(f, x)
APP(app(append, app(app(cons, x), xs)), ys) → APP(app(append, xs), ys)
APP(app(append, app(app(cons, x), xs)), ys) → APP(app(cons, x), app(app(append, xs), ys))
APP(app(flatwith, f), app(leaf, x)) → APP(app(cons, app(f, x)), nil)
APP(app(flatwith, f), app(leaf, x)) → APP(cons, app(f, x))
APP(app(flatwithsub, f), app(app(cons, x), xs)) → APP(app(flatwith, f), x)
APP(app(flatwithsub, f), app(app(cons, x), xs)) → APP(append, app(app(flatwith, f), x))
APP(app(flatwithsub, f), app(app(cons, x), xs)) → APP(app(flatwithsub, f), xs)
APP(app(flatwithsub, f), app(app(cons, x), xs)) → APP(app(append, app(app(flatwith, f), x)), app(app(flatwithsub, f), xs))
APP(app(flatwithsub, f), app(app(cons, x), xs)) → APP(flatwith, f)
APP(app(flatwith, f), app(node, xs)) → APP(flatwithsub, f)

The TRS R consists of the following rules:

app(app(append, nil), ys) → ys
app(app(append, app(app(cons, x), xs)), ys) → app(app(cons, x), app(app(append, xs), ys))
app(app(flatwith, f), app(leaf, x)) → app(app(cons, app(f, x)), nil)
app(app(flatwith, f), app(node, xs)) → app(app(flatwithsub, f), xs)
app(app(flatwithsub, f), nil) → nil
app(app(flatwithsub, f), app(app(cons, x), xs)) → app(app(append, app(app(flatwith, f), x)), app(app(flatwithsub, f), xs))

The set Q consists of the following terms:

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

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

↳ QTRS

  ↳ Overlay + Local Confluence

    ↳ QTRS

      ↳ DependencyPairsProof

        ↳ QDP

          ↳ DependencyGraphProof

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

APP(app(flatwith, f), app(node, xs)) → APP(app(flatwithsub, f), xs)
APP(app(append, app(app(cons, x), xs)), ys) → APP(append, xs)
APP(app(flatwith, f), app(leaf, x)) → APP(f, x)
APP(app(append, app(app(cons, x), xs)), ys) → APP(app(append, xs), ys)
APP(app(append, app(app(cons, x), xs)), ys) → APP(app(cons, x), app(app(append, xs), ys))
APP(app(flatwith, f), app(leaf, x)) → APP(app(cons, app(f, x)), nil)
APP(app(flatwith, f), app(leaf, x)) → APP(cons, app(f, x))
APP(app(flatwithsub, f), app(app(cons, x), xs)) → APP(app(flatwith, f), x)
APP(app(flatwithsub, f), app(app(cons, x), xs)) → APP(append, app(app(flatwith, f), x))
APP(app(flatwithsub, f), app(app(cons, x), xs)) → APP(app(flatwithsub, f), xs)
APP(app(flatwithsub, f), app(app(cons, x), xs)) → APP(app(append, app(app(flatwith, f), x)), app(app(flatwithsub, f), xs))
APP(app(flatwithsub, f), app(app(cons, x), xs)) → APP(flatwith, f)
APP(app(flatwith, f), app(node, xs)) → APP(flatwithsub, f)

The TRS R consists of the following rules:

app(app(append, nil), ys) → ys
app(app(append, app(app(cons, x), xs)), ys) → app(app(cons, x), app(app(append, xs), ys))
app(app(flatwith, f), app(leaf, x)) → app(app(cons, app(f, x)), nil)
app(app(flatwith, f), app(node, xs)) → app(app(flatwithsub, f), xs)
app(app(flatwithsub, f), nil) → nil
app(app(flatwithsub, f), app(app(cons, x), xs)) → app(app(append, app(app(flatwith, f), x)), app(app(flatwithsub, f), xs))

The set Q consists of the following terms:

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

We have to consider all minimal (P,Q,R)-chains.
The approximation of the Dependency Graph [15,17,22] contains 2 SCCs with 8 less nodes.

↳ QTRS

  ↳ Overlay + Local Confluence

    ↳ QTRS

      ↳ DependencyPairsProof

        ↳ QDP

          ↳ DependencyGraphProof

            ↳ AND

              ↳ QDP

                ↳ UsableRulesProof

              ↳ 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(append, nil), ys) → ys
app(app(append, app(app(cons, x), xs)), ys) → app(app(cons, x), app(app(append, xs), ys))
app(app(flatwith, f), app(leaf, x)) → app(app(cons, app(f, x)), nil)
app(app(flatwith, f), app(node, xs)) → app(app(flatwithsub, f), xs)
app(app(flatwithsub, f), nil) → nil
app(app(flatwithsub, f), app(app(cons, x), xs)) → app(app(append, app(app(flatwith, f), x)), app(app(flatwithsub, f), xs))

The set Q consists of the following terms:

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

We have to consider all minimal (P,Q,R)-chains.
As all Q-normal forms are R-normal forms we are in the innermost case. Hence, by the usable rules processor [15] we can delete all non-usable rules [17] from R.

↳ QTRS

  ↳ Overlay + Local Confluence

    ↳ QTRS

      ↳ DependencyPairsProof

        ↳ QDP

          ↳ DependencyGraphProof

            ↳ AND

              ↳ QDP

                ↳ UsableRulesProof

                  ↳ QDP

                    ↳ ATransformationProof

              ↳ 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)

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

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

We have to consider all minimal (P,Q,R)-chains.
We have applied the A-Transformation [17] to get from an applicative problem to a standard problem.

↳ QTRS

  ↳ Overlay + Local Confluence

    ↳ QTRS

      ↳ DependencyPairsProof

        ↳ QDP

          ↳ DependencyGraphProof

            ↳ AND

              ↳ QDP

                ↳ UsableRulesProof

                  ↳ QDP

                    ↳ ATransformationProof

                      ↳ QDP

                        ↳ QReductionProof

              ↳ QDP

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

append1(cons(x, xs), ys) → append1(xs, ys)

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

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

We have to consider all minimal (P,Q,R)-chains.
We deleted the following terms from Q as each root-symbol of these terms does neither occur in P nor in R.

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

↳ QTRS

  ↳ Overlay + Local Confluence

    ↳ QTRS

      ↳ DependencyPairsProof

        ↳ QDP

          ↳ DependencyGraphProof

            ↳ AND

              ↳ QDP

                ↳ UsableRulesProof

                  ↳ QDP

                    ↳ ATransformationProof

                      ↳ QDP

                        ↳ QReductionProof

                          ↳ QDP

                            ↳ QDPSizeChangeProof

              ↳ QDP

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

append1(cons(x, xs), ys) → append1(xs, ys)

R is empty.
Q is empty.
We have to consider all minimal (P,Q,R)-chains.
By using the subterm criterion [20] together with the size-change analysis [32] we have proven that there are no infinite chains for this DP problem.

From the DPs we obtained the following set of size-change graphs:

append1(cons(x, xs), ys) → append1(xs, ys)
The graph contains the following edges 1 > 1, 2 >= 2

↳ QTRS

  ↳ Overlay + Local Confluence

    ↳ QTRS

      ↳ DependencyPairsProof

        ↳ QDP

          ↳ DependencyGraphProof

            ↳ AND

              ↳ QDP

              ↳ QDP

                ↳ UsableRulesProof

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

APP(app(flatwith, f), app(node, xs)) → APP(app(flatwithsub, f), xs)
APP(app(flatwithsub, f), app(app(cons, x), xs)) → APP(app(flatwithsub, f), xs)
APP(app(flatwith, f), app(leaf, x)) → APP(f, x)
APP(app(flatwithsub, f), app(app(cons, x), xs)) → APP(app(flatwith, f), x)

The TRS R consists of the following rules:

app(app(append, nil), ys) → ys
app(app(append, app(app(cons, x), xs)), ys) → app(app(cons, x), app(app(append, xs), ys))
app(app(flatwith, f), app(leaf, x)) → app(app(cons, app(f, x)), nil)
app(app(flatwith, f), app(node, xs)) → app(app(flatwithsub, f), xs)
app(app(flatwithsub, f), nil) → nil
app(app(flatwithsub, f), app(app(cons, x), xs)) → app(app(append, app(app(flatwith, f), x)), app(app(flatwithsub, f), xs))

The set Q consists of the following terms:

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

↳ QTRS

  ↳ Overlay + Local Confluence

    ↳ QTRS

      ↳ DependencyPairsProof

        ↳ QDP

          ↳ DependencyGraphProof

            ↳ AND

              ↳ QDP

              ↳ QDP

                ↳ UsableRulesProof

                  ↳ QDP

                    ↳ QDPSizeChangeProof

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

APP(app(flatwith, f), app(node, xs)) → APP(app(flatwithsub, f), xs)
APP(app(flatwithsub, f), app(app(cons, x), xs)) → APP(app(flatwithsub, f), xs)
APP(app(flatwith, f), app(leaf, x)) → APP(f, x)
APP(app(flatwithsub, f), app(app(cons, x), xs)) → APP(app(flatwith, f), x)

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

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

We have to consider all minimal (P,Q,R)-chains.
By using the subterm criterion [20] together with the size-change analysis [32] we have proven that there are no infinite chains for this DP problem.

From the DPs we obtained the following set of size-change graphs:

APP(app(flatwithsub, f), app(app(cons, x), xs)) → APP(app(flatwith, f), x)
The graph contains the following edges 2 > 2
APP(app(flatwithsub, f), app(app(cons, x), xs)) → APP(app(flatwithsub, f), xs)
The graph contains the following edges 1 >= 1, 2 > 2
APP(app(flatwith, f), app(leaf, x)) → APP(f, x)
The graph contains the following edges 1 > 1, 2 > 2
APP(app(flatwith, f), app(node, xs)) → APP(app(flatwithsub, f), xs)
The graph contains the following edges 2 > 2