Termination w.r.t. Q proof of TRS/higher-order/AotoYam015.trs

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(app(if, true), x), y) → x
app(app(app(if, false), x), y) → y
app(app(filter, f), nil) → nil
app(app(filter, f), app(app(cons, x), xs)) → app(app(app(if, app(f, x)), app(app(cons, x), app(app(filter, f), xs))), app(app(filter, f), xs))

Q is empty.

↳ QTRS

  ↳ Overlay + Local Confluence

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

app(app(app(if, true), x), y) → x
app(app(app(if, false), x), y) → y
app(app(filter, f), nil) → nil
app(app(filter, f), app(app(cons, x), xs)) → app(app(app(if, app(f, x)), app(app(cons, x), app(app(filter, f), xs))), app(app(filter, f), xs))

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(app(if, true), x), y) → x
app(app(app(if, false), x), y) → y
app(app(filter, f), nil) → nil
app(app(filter, f), app(app(cons, x), xs)) → app(app(app(if, app(f, x)), app(app(cons, x), app(app(filter, f), xs))), app(app(filter, f), xs))

The set Q consists of the following terms:

app(app(app(if, true), x0), x1)
app(app(app(if, false), x0), x1)
app(app(filter, x0), nil)
app(app(filter, x0), app(app(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:

APP(app(filter, f), app(app(cons, x), xs)) → APP(app(app(if, app(f, x)), app(app(cons, x), app(app(filter, f), xs))), app(app(filter, f), xs))
APP(app(filter, f), app(app(cons, x), xs)) → APP(if, app(f, x))
APP(app(filter, f), app(app(cons, x), xs)) → APP(f, x)
APP(app(filter, f), app(app(cons, x), xs)) → APP(app(if, app(f, x)), app(app(cons, x), app(app(filter, f), xs)))
APP(app(filter, f), app(app(cons, x), xs)) → APP(app(filter, f), xs)
APP(app(filter, f), app(app(cons, x), xs)) → APP(app(cons, x), app(app(filter, f), xs))

The TRS R consists of the following rules:

app(app(app(if, true), x), y) → x
app(app(app(if, false), x), y) → y
app(app(filter, f), nil) → nil
app(app(filter, f), app(app(cons, x), xs)) → app(app(app(if, app(f, x)), app(app(cons, x), app(app(filter, f), xs))), app(app(filter, f), xs))

The set Q consists of the following terms:

app(app(app(if, true), x0), x1)
app(app(app(if, false), x0), x1)
app(app(filter, x0), nil)
app(app(filter, x0), app(app(cons, 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(app(filter, f), app(app(cons, x), xs)) → APP(app(app(if, app(f, x)), app(app(cons, x), app(app(filter, f), xs))), app(app(filter, f), xs))
APP(app(filter, f), app(app(cons, x), xs)) → APP(if, app(f, x))
APP(app(filter, f), app(app(cons, x), xs)) → APP(f, x)
APP(app(filter, f), app(app(cons, x), xs)) → APP(app(if, app(f, x)), app(app(cons, x), app(app(filter, f), xs)))
APP(app(filter, f), app(app(cons, x), xs)) → APP(app(filter, f), xs)
APP(app(filter, f), app(app(cons, x), xs)) → APP(app(cons, x), app(app(filter, f), xs))

The TRS R consists of the following rules:

app(app(app(if, true), x), y) → x
app(app(app(if, false), x), y) → y
app(app(filter, f), nil) → nil
app(app(filter, f), app(app(cons, x), xs)) → app(app(app(if, app(f, x)), app(app(cons, x), app(app(filter, f), xs))), app(app(filter, f), xs))

The set Q consists of the following terms:

app(app(app(if, true), x0), x1)
app(app(app(if, false), x0), x1)
app(app(filter, x0), nil)
app(app(filter, x0), app(app(cons, 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(app(filter, f), app(app(cons, x), xs)) → APP(app(app(if, app(f, x)), app(app(cons, x), app(app(filter, f), xs))), app(app(filter, f), xs))
APP(app(filter, f), app(app(cons, x), xs)) → APP(if, app(f, x))
APP(app(filter, f), app(app(cons, x), xs)) → APP(f, x)
APP(app(filter, f), app(app(cons, x), xs)) → APP(app(if, app(f, x)), app(app(cons, x), app(app(filter, f), xs)))
APP(app(filter, f), app(app(cons, x), xs)) → APP(app(filter, f), xs)
APP(app(filter, f), app(app(cons, x), xs)) → APP(app(cons, x), app(app(filter, f), xs))

The TRS R consists of the following rules:

app(app(app(if, true), x), y) → x
app(app(app(if, false), x), y) → y
app(app(filter, f), nil) → nil
app(app(filter, f), app(app(cons, x), xs)) → app(app(app(if, app(f, x)), app(app(cons, x), app(app(filter, f), xs))), app(app(filter, f), xs))

The set Q consists of the following terms:

app(app(app(if, true), x0), x1)
app(app(app(if, false), x0), x1)
app(app(filter, x0), nil)
app(app(filter, x0), app(app(cons, x1), x2))

We have to consider all minimal (P,Q,R)-chains.
The approximation of the Dependency Graph [13,14,18] contains 1 SCC with 4 less nodes.

↳ QTRS

  ↳ Overlay + Local Confluence

    ↳ QTRS

      ↳ DependencyPairsProof

        ↳ QDP

          ↳ EdgeDeletionProof

            ↳ QDP

              ↳ DependencyGraphProof

                ↳ QDP

                  ↳ QDPOrderProof

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

APP(app(filter, f), app(app(cons, x), xs)) → APP(f, x)
APP(app(filter, f), app(app(cons, x), xs)) → APP(app(filter, f), xs)

The TRS R consists of the following rules:

app(app(app(if, true), x), y) → x
app(app(app(if, false), x), y) → y
app(app(filter, f), nil) → nil
app(app(filter, f), app(app(cons, x), xs)) → app(app(app(if, app(f, x)), app(app(cons, x), app(app(filter, f), xs))), app(app(filter, f), xs))

The set Q consists of the following terms:

app(app(app(if, true), x0), x1)
app(app(app(if, false), x0), x1)
app(app(filter, x0), nil)
app(app(filter, x0), app(app(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 oriented strictly and are deleted.

APP(app(filter, f), app(app(cons, x), xs)) → APP(f, x)
APP(app(filter, f), app(app(cons, x), xs)) → APP(app(filter, f), xs)

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

Lexicographic Path Order [19].
Precedence:

filter > app₂
cons > app₂

The following usable rules [14] were oriented: none

↳ QTRS

  ↳ Overlay + Local Confluence

    ↳ QTRS

      ↳ DependencyPairsProof

        ↳ QDP

          ↳ EdgeDeletionProof

            ↳ QDP

              ↳ DependencyGraphProof

                ↳ QDP

                  ↳ QDPOrderProof

                    ↳ QDP

                      ↳ PisEmptyProof

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

app(app(app(if, true), x), y) → x
app(app(app(if, false), x), y) → y
app(app(filter, f), nil) → nil
app(app(filter, f), app(app(cons, x), xs)) → app(app(app(if, app(f, x)), app(app(cons, x), app(app(filter, f), xs))), app(app(filter, f), xs))

The set Q consists of the following terms:

app(app(app(if, true), x0), x1)
app(app(app(if, false), x0), x1)
app(app(filter, x0), nil)
app(app(filter, x0), app(app(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.