Termination w.r.t. Q proof of TRS/higher-order/AotoYam004.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(plus, 0), y) → y
app(app(plus, app(s, x)), y) → app(s, app(app(plus, x), y))
app(app(sumwith, f), nil) → nil
app(app(sumwith, f), app(app(cons, x), xs)) → app(app(plus, app(f, x)), app(app(sumwith, f), xs))

Q is empty.

↳ QTRS

  ↳ Overlay + Local Confluence

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

app(app(plus, 0), y) → y
app(app(plus, app(s, x)), y) → app(s, app(app(plus, x), y))
app(app(sumwith, f), nil) → nil
app(app(sumwith, f), app(app(cons, x), xs)) → app(app(plus, app(f, x)), app(app(sumwith, 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(plus, 0), y) → y
app(app(plus, app(s, x)), y) → app(s, app(app(plus, x), y))
app(app(sumwith, f), nil) → nil
app(app(sumwith, f), app(app(cons, x), xs)) → app(app(plus, app(f, x)), app(app(sumwith, f), xs))

The set Q consists of the following terms:

app(app(plus, 0), x0)
app(app(plus, app(s, x0)), x1)
app(app(sumwith, x0), nil)
app(app(sumwith, 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(sumwith, f), app(app(cons, x), xs)) → APP(app(plus, app(f, x)), app(app(sumwith, f), xs))
APP(app(plus, app(s, x)), y) → APP(s, app(app(plus, x), y))
APP(app(plus, app(s, x)), y) → APP(app(plus, x), y)
APP(app(sumwith, f), app(app(cons, x), xs)) → APP(f, x)
APP(app(plus, app(s, x)), y) → APP(plus, x)
APP(app(sumwith, f), app(app(cons, x), xs)) → APP(plus, app(f, x))
APP(app(sumwith, f), app(app(cons, x), xs)) → APP(app(sumwith, f), xs)

The TRS R consists of the following rules:

app(app(plus, 0), y) → y
app(app(plus, app(s, x)), y) → app(s, app(app(plus, x), y))
app(app(sumwith, f), nil) → nil
app(app(sumwith, f), app(app(cons, x), xs)) → app(app(plus, app(f, x)), app(app(sumwith, f), xs))

The set Q consists of the following terms:

app(app(plus, 0), x0)
app(app(plus, app(s, x0)), x1)
app(app(sumwith, x0), nil)
app(app(sumwith, 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(sumwith, f), app(app(cons, x), xs)) → APP(app(plus, app(f, x)), app(app(sumwith, f), xs))
APP(app(plus, app(s, x)), y) → APP(s, app(app(plus, x), y))
APP(app(plus, app(s, x)), y) → APP(app(plus, x), y)
APP(app(sumwith, f), app(app(cons, x), xs)) → APP(f, x)
APP(app(plus, app(s, x)), y) → APP(plus, x)
APP(app(sumwith, f), app(app(cons, x), xs)) → APP(plus, app(f, x))
APP(app(sumwith, f), app(app(cons, x), xs)) → APP(app(sumwith, f), xs)

The TRS R consists of the following rules:

app(app(plus, 0), y) → y
app(app(plus, app(s, x)), y) → app(s, app(app(plus, x), y))
app(app(sumwith, f), nil) → nil
app(app(sumwith, f), app(app(cons, x), xs)) → app(app(plus, app(f, x)), app(app(sumwith, f), xs))

The set Q consists of the following terms:

app(app(plus, 0), x0)
app(app(plus, app(s, x0)), x1)
app(app(sumwith, x0), nil)
app(app(sumwith, 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(plus, app(s, x)), y) → APP(s, app(app(plus, x), y))
APP(app(sumwith, f), app(app(cons, x), xs)) → APP(app(plus, app(f, x)), app(app(sumwith, f), xs))
APP(app(plus, app(s, x)), y) → APP(app(plus, x), y)
APP(app(sumwith, f), app(app(cons, x), xs)) → APP(f, x)
APP(app(plus, app(s, x)), y) → APP(plus, x)
APP(app(sumwith, f), app(app(cons, x), xs)) → APP(plus, app(f, x))
APP(app(sumwith, f), app(app(cons, x), xs)) → APP(app(sumwith, f), xs)

The TRS R consists of the following rules:

app(app(plus, 0), y) → y
app(app(plus, app(s, x)), y) → app(s, app(app(plus, x), y))
app(app(sumwith, f), nil) → nil
app(app(sumwith, f), app(app(cons, x), xs)) → app(app(plus, app(f, x)), app(app(sumwith, f), xs))

The set Q consists of the following terms:

app(app(plus, 0), x0)
app(app(plus, app(s, x0)), x1)
app(app(sumwith, x0), nil)
app(app(sumwith, 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 2 SCCs with 4 less nodes.

↳ QTRS

  ↳ Overlay + Local Confluence

    ↳ QTRS

      ↳ DependencyPairsProof

        ↳ QDP

          ↳ EdgeDeletionProof

            ↳ QDP

              ↳ DependencyGraphProof

                ↳ AND

                  ↳ QDP

                    ↳ QDPOrderProof

                  ↳ QDP

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

APP(app(plus, app(s, x)), y) → APP(app(plus, x), y)

The TRS R consists of the following rules:

app(app(plus, 0), y) → y
app(app(plus, app(s, x)), y) → app(s, app(app(plus, x), y))
app(app(sumwith, f), nil) → nil
app(app(sumwith, f), app(app(cons, x), xs)) → app(app(plus, app(f, x)), app(app(sumwith, f), xs))

The set Q consists of the following terms:

app(app(plus, 0), x0)
app(app(plus, app(s, x0)), x1)
app(app(sumwith, x0), nil)
app(app(sumwith, x0), app(app(cons, 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:

PLUS(s(x), y) → PLUS(x, y)

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

plus(0, x0)
plus(s(x0), x1)
sumwith(x0, nil)
sumwith(x0, cons(x1, x2))

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

The following pairs can be oriented strictly and are deleted.

APP(app(plus, app(s, x)), y) → APP(app(plus, x), y)

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

Lexicographic Path Order [19].
Precedence:

s₁ > PLUS₁

The following usable rules [14] were oriented: none

↳ QTRS

  ↳ Overlay + Local Confluence

    ↳ QTRS

      ↳ DependencyPairsProof

        ↳ QDP

          ↳ EdgeDeletionProof

            ↳ QDP

              ↳ DependencyGraphProof

                ↳ AND

                  ↳ QDP

                    ↳ QDPOrderProof

                      ↳ QDP

                        ↳ PisEmptyProof

                  ↳ QDP

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

app(app(plus, 0), y) → y
app(app(plus, app(s, x)), y) → app(s, app(app(plus, x), y))
app(app(sumwith, f), nil) → nil
app(app(sumwith, f), app(app(cons, x), xs)) → app(app(plus, app(f, x)), app(app(sumwith, f), xs))

The set Q consists of the following terms:

app(app(plus, 0), x0)
app(app(plus, app(s, x0)), x1)
app(app(sumwith, x0), nil)
app(app(sumwith, 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.

↳ QTRS

  ↳ Overlay + Local Confluence

    ↳ QTRS

      ↳ DependencyPairsProof

        ↳ QDP

          ↳ EdgeDeletionProof

            ↳ QDP

              ↳ DependencyGraphProof

                ↳ AND

                  ↳ QDP

                  ↳ QDP

                    ↳ QDPOrderProof

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

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

The TRS R consists of the following rules:

app(app(plus, 0), y) → y
app(app(plus, app(s, x)), y) → app(s, app(app(plus, x), y))
app(app(sumwith, f), nil) → nil
app(app(sumwith, f), app(app(cons, x), xs)) → app(app(plus, app(f, x)), app(app(sumwith, f), xs))

The set Q consists of the following terms:

app(app(plus, 0), x0)
app(app(plus, app(s, x0)), x1)
app(app(sumwith, x0), nil)
app(app(sumwith, 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(sumwith, f), app(app(cons, x), xs)) → APP(f, x)
APP(app(sumwith, f), app(app(cons, x), xs)) → APP(app(sumwith, 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)
sumwith = sumwith
cons = cons

Lexicographic Path Order [19].
Precedence:

sumwith > app₂
cons > app₂

The following usable rules [14] were oriented: none

↳ QTRS

  ↳ Overlay + Local Confluence

    ↳ QTRS

      ↳ DependencyPairsProof

        ↳ QDP

          ↳ EdgeDeletionProof

            ↳ QDP

              ↳ DependencyGraphProof

                ↳ AND

                  ↳ QDP

                  ↳ QDP

                    ↳ QDPOrderProof

                      ↳ QDP

                        ↳ PisEmptyProof

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

app(app(plus, 0), y) → y
app(app(plus, app(s, x)), y) → app(s, app(app(plus, x), y))
app(app(sumwith, f), nil) → nil
app(app(sumwith, f), app(app(cons, x), xs)) → app(app(plus, app(f, x)), app(app(sumwith, f), xs))

The set Q consists of the following terms:

app(app(plus, 0), x0)
app(app(plus, app(s, x0)), x1)
app(app(sumwith, x0), nil)
app(app(sumwith, 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.