Termination w.r.t. Q proof of TRS/AG01#3.17a.trs

Termination w.r.t. Q of the following Term Rewriting System could not be shown:

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

app(nil, k) → k
app(l, nil) → l
app(cons(x, l), k) → cons(x, app(l, k))
sum(cons(x, nil)) → cons(x, nil)
sum(cons(x, cons(y, l))) → sum(cons(plus(x, y), l))
sum(app(l, cons(x, cons(y, k)))) → sum(app(l, sum(cons(x, cons(y, k)))))
plus(0, y) → y
plus(s(x), y) → s(plus(x, y))
sum(plus(cons(0, x), cons(y, l))) → pred(sum(cons(s(x), cons(y, l))))
pred(cons(s(x), nil)) → cons(x, nil)

Q is empty.

↳ QTRS

  ↳ DependencyPairsProof

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

app(nil, k) → k
app(l, nil) → l
app(cons(x, l), k) → cons(x, app(l, k))
sum(cons(x, nil)) → cons(x, nil)
sum(cons(x, cons(y, l))) → sum(cons(plus(x, y), l))
sum(app(l, cons(x, cons(y, k)))) → sum(app(l, sum(cons(x, cons(y, k)))))
plus(0, y) → y
plus(s(x), y) → s(plus(x, y))
sum(plus(cons(0, x), cons(y, l))) → pred(sum(cons(s(x), cons(y, l))))
pred(cons(s(x), nil)) → cons(x, nil)

Q is empty.

Using Dependency Pairs [1,13] we result in the following initial DP problem:
Q DP problem:
The TRS P consists of the following rules:

PLUS(s(x), y) → PLUS(x, y)
SUM(app(l, cons(x, cons(y, k)))) → APP(l, sum(cons(x, cons(y, k))))
SUM(cons(x, cons(y, l))) → PLUS(x, y)
SUM(plus(cons(0, x), cons(y, l))) → SUM(cons(s(x), cons(y, l)))
SUM(cons(x, cons(y, l))) → SUM(cons(plus(x, y), l))
SUM(plus(cons(0, x), cons(y, l))) → PRED(sum(cons(s(x), cons(y, l))))
SUM(app(l, cons(x, cons(y, k)))) → SUM(cons(x, cons(y, k)))
SUM(app(l, cons(x, cons(y, k)))) → SUM(app(l, sum(cons(x, cons(y, k)))))
APP(cons(x, l), k) → APP(l, k)

The TRS R consists of the following rules:

app(nil, k) → k
app(l, nil) → l
app(cons(x, l), k) → cons(x, app(l, k))
sum(cons(x, nil)) → cons(x, nil)
sum(cons(x, cons(y, l))) → sum(cons(plus(x, y), l))
sum(app(l, cons(x, cons(y, k)))) → sum(app(l, sum(cons(x, cons(y, k)))))
plus(0, y) → y
plus(s(x), y) → s(plus(x, y))
sum(plus(cons(0, x), cons(y, l))) → pred(sum(cons(s(x), cons(y, l))))
pred(cons(s(x), nil)) → cons(x, nil)

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

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ EdgeDeletionProof

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

PLUS(s(x), y) → PLUS(x, y)
SUM(app(l, cons(x, cons(y, k)))) → APP(l, sum(cons(x, cons(y, k))))
SUM(cons(x, cons(y, l))) → PLUS(x, y)
SUM(plus(cons(0, x), cons(y, l))) → SUM(cons(s(x), cons(y, l)))
SUM(cons(x, cons(y, l))) → SUM(cons(plus(x, y), l))
SUM(plus(cons(0, x), cons(y, l))) → PRED(sum(cons(s(x), cons(y, l))))
SUM(app(l, cons(x, cons(y, k)))) → SUM(cons(x, cons(y, k)))
SUM(app(l, cons(x, cons(y, k)))) → SUM(app(l, sum(cons(x, cons(y, k)))))
APP(cons(x, l), k) → APP(l, k)

The TRS R consists of the following rules:

app(nil, k) → k
app(l, nil) → l
app(cons(x, l), k) → cons(x, app(l, k))
sum(cons(x, nil)) → cons(x, nil)
sum(cons(x, cons(y, l))) → sum(cons(plus(x, y), l))
sum(app(l, cons(x, cons(y, k)))) → sum(app(l, sum(cons(x, cons(y, k)))))
plus(0, y) → y
plus(s(x), y) → s(plus(x, y))
sum(plus(cons(0, x), cons(y, l))) → pred(sum(cons(s(x), cons(y, l))))
pred(cons(s(x), nil)) → cons(x, nil)

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

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ EdgeDeletionProof

        ↳ QDP

          ↳ DependencyGraphProof

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

PLUS(s(x), y) → PLUS(x, y)
SUM(app(l, cons(x, cons(y, k)))) → APP(l, sum(cons(x, cons(y, k))))
SUM(cons(x, cons(y, l))) → PLUS(x, y)
SUM(plus(cons(0, x), cons(y, l))) → SUM(cons(s(x), cons(y, l)))
SUM(cons(x, cons(y, l))) → SUM(cons(plus(x, y), l))
SUM(app(l, cons(x, cons(y, k)))) → SUM(cons(x, cons(y, k)))
SUM(plus(cons(0, x), cons(y, l))) → PRED(sum(cons(s(x), cons(y, l))))
APP(cons(x, l), k) → APP(l, k)
SUM(app(l, cons(x, cons(y, k)))) → SUM(app(l, sum(cons(x, cons(y, k)))))

The TRS R consists of the following rules:

app(nil, k) → k
app(l, nil) → l
app(cons(x, l), k) → cons(x, app(l, k))
sum(cons(x, nil)) → cons(x, nil)
sum(cons(x, cons(y, l))) → sum(cons(plus(x, y), l))
sum(app(l, cons(x, cons(y, k)))) → sum(app(l, sum(cons(x, cons(y, k)))))
plus(0, y) → y
plus(s(x), y) → s(plus(x, y))
sum(plus(cons(0, x), cons(y, l))) → pred(sum(cons(s(x), cons(y, l))))
pred(cons(s(x), nil)) → cons(x, nil)

Q is empty.
We have to consider all minimal (P,Q,R)-chains.
The approximation of the Dependency Graph [13,14,18] contains 4 SCCs with 5 less nodes.

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ EdgeDeletionProof

        ↳ QDP

          ↳ DependencyGraphProof

            ↳ AND

              ↳ QDP

                ↳ QDPOrderProof

              ↳ QDP

              ↳ QDP

              ↳ QDP

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

PLUS(s(x), y) → PLUS(x, y)

The TRS R consists of the following rules:

app(nil, k) → k
app(l, nil) → l
app(cons(x, l), k) → cons(x, app(l, k))
sum(cons(x, nil)) → cons(x, nil)
sum(cons(x, cons(y, l))) → sum(cons(plus(x, y), l))
sum(app(l, cons(x, cons(y, k)))) → sum(app(l, sum(cons(x, cons(y, k)))))
plus(0, y) → y
plus(s(x), y) → s(plus(x, y))
sum(plus(cons(0, x), cons(y, l))) → pred(sum(cons(s(x), cons(y, l))))
pred(cons(s(x), nil)) → cons(x, nil)

Q is empty.
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.

PLUS(s(x), y) → 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)

Recursive Path Order [2].
Precedence:

s₁ > PLUS₁

The following usable rules [14] were oriented: none

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ EdgeDeletionProof

        ↳ QDP

          ↳ DependencyGraphProof

            ↳ AND

              ↳ QDP

                ↳ QDPOrderProof

                  ↳ QDP

                    ↳ PisEmptyProof

              ↳ QDP

              ↳ QDP

              ↳ QDP

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

app(nil, k) → k
app(l, nil) → l
app(cons(x, l), k) → cons(x, app(l, k))
sum(cons(x, nil)) → cons(x, nil)
sum(cons(x, cons(y, l))) → sum(cons(plus(x, y), l))
sum(app(l, cons(x, cons(y, k)))) → sum(app(l, sum(cons(x, cons(y, k)))))
plus(0, y) → y
plus(s(x), y) → s(plus(x, y))
sum(plus(cons(0, x), cons(y, l))) → pred(sum(cons(s(x), cons(y, l))))
pred(cons(s(x), nil)) → cons(x, nil)

Q is empty.
We have to consider all minimal (P,Q,R)-chains.
The TRS P is empty. Hence, there is no (P,Q,R) chain.

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ EdgeDeletionProof

        ↳ QDP

          ↳ DependencyGraphProof

            ↳ AND

              ↳ QDP

              ↳ QDP

                ↳ QDPOrderProof

              ↳ QDP

              ↳ QDP

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

APP(cons(x, l), k) → APP(l, k)

The TRS R consists of the following rules:

app(nil, k) → k
app(l, nil) → l
app(cons(x, l), k) → cons(x, app(l, k))
sum(cons(x, nil)) → cons(x, nil)
sum(cons(x, cons(y, l))) → sum(cons(plus(x, y), l))
sum(app(l, cons(x, cons(y, k)))) → sum(app(l, sum(cons(x, cons(y, k)))))
plus(0, y) → y
plus(s(x), y) → s(plus(x, y))
sum(plus(cons(0, x), cons(y, l))) → pred(sum(cons(s(x), cons(y, l))))
pred(cons(s(x), nil)) → cons(x, nil)

Q is empty.
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(cons(x, l), k) → APP(l, k)

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

Recursive Path Order [2].
Precedence:

trivial

The following usable rules [14] were oriented: none

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ EdgeDeletionProof

        ↳ QDP

          ↳ DependencyGraphProof

            ↳ AND

              ↳ QDP

              ↳ QDP

                ↳ QDPOrderProof

                  ↳ QDP

                    ↳ PisEmptyProof

              ↳ QDP

              ↳ QDP

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

app(nil, k) → k
app(l, nil) → l
app(cons(x, l), k) → cons(x, app(l, k))
sum(cons(x, nil)) → cons(x, nil)
sum(cons(x, cons(y, l))) → sum(cons(plus(x, y), l))
sum(app(l, cons(x, cons(y, k)))) → sum(app(l, sum(cons(x, cons(y, k)))))
plus(0, y) → y
plus(s(x), y) → s(plus(x, y))
sum(plus(cons(0, x), cons(y, l))) → pred(sum(cons(s(x), cons(y, l))))
pred(cons(s(x), nil)) → cons(x, nil)

Q is empty.
We have to consider all minimal (P,Q,R)-chains.
The TRS P is empty. Hence, there is no (P,Q,R) chain.

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ EdgeDeletionProof

        ↳ QDP

          ↳ DependencyGraphProof

            ↳ AND

              ↳ QDP

              ↳ QDP

              ↳ QDP

                ↳ QDPOrderProof

              ↳ QDP

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

SUM(cons(x, cons(y, l))) → SUM(cons(plus(x, y), l))

The TRS R consists of the following rules:

app(nil, k) → k
app(l, nil) → l
app(cons(x, l), k) → cons(x, app(l, k))
sum(cons(x, nil)) → cons(x, nil)
sum(cons(x, cons(y, l))) → sum(cons(plus(x, y), l))
sum(app(l, cons(x, cons(y, k)))) → sum(app(l, sum(cons(x, cons(y, k)))))
plus(0, y) → y
plus(s(x), y) → s(plus(x, y))
sum(plus(cons(0, x), cons(y, l))) → pred(sum(cons(s(x), cons(y, l))))
pred(cons(s(x), nil)) → cons(x, nil)

Q is empty.
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.

SUM(cons(x, cons(y, l))) → SUM(cons(plus(x, y), l))

The remaining pairs can at least be oriented weakly.
none
Used ordering: Combined order from the following AFS and order.
SUM(x1) = x1
cons(x1, x2) = cons(x1, x2)
plus(x1, x2) = x2
0 = 0
s(x1) = x1

Recursive Path Order [2].
Precedence:

trivial

The following usable rules [14] were oriented:

plus(0, y) → y
plus(s(x), y) → s(plus(x, y))

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ EdgeDeletionProof

        ↳ QDP

          ↳ DependencyGraphProof

            ↳ AND

              ↳ QDP

              ↳ QDP

              ↳ QDP

                ↳ QDPOrderProof

                  ↳ QDP

                    ↳ PisEmptyProof

              ↳ QDP

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

app(nil, k) → k
app(l, nil) → l
app(cons(x, l), k) → cons(x, app(l, k))
sum(cons(x, nil)) → cons(x, nil)
sum(cons(x, cons(y, l))) → sum(cons(plus(x, y), l))
sum(app(l, cons(x, cons(y, k)))) → sum(app(l, sum(cons(x, cons(y, k)))))
plus(0, y) → y
plus(s(x), y) → s(plus(x, y))
sum(plus(cons(0, x), cons(y, l))) → pred(sum(cons(s(x), cons(y, l))))
pred(cons(s(x), nil)) → cons(x, nil)

Q is empty.
We have to consider all minimal (P,Q,R)-chains.
The TRS P is empty. Hence, there is no (P,Q,R) chain.

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ EdgeDeletionProof

        ↳ QDP

          ↳ DependencyGraphProof

            ↳ AND

              ↳ QDP

              ↳ QDP

              ↳ QDP

              ↳ QDP

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

SUM(app(l, cons(x, cons(y, k)))) → SUM(app(l, sum(cons(x, cons(y, k)))))

The TRS R consists of the following rules:

app(nil, k) → k
app(l, nil) → l
app(cons(x, l), k) → cons(x, app(l, k))
sum(cons(x, nil)) → cons(x, nil)
sum(cons(x, cons(y, l))) → sum(cons(plus(x, y), l))
sum(app(l, cons(x, cons(y, k)))) → sum(app(l, sum(cons(x, cons(y, k)))))
plus(0, y) → y
plus(s(x), y) → s(plus(x, y))
sum(plus(cons(0, x), cons(y, l))) → pred(sum(cons(s(x), cons(y, l))))
pred(cons(s(x), nil)) → cons(x, nil)

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