Termination w.r.t. Q proof of TRS/AG01#3.53.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:

minus₂(x, 0) -> x
minus₂(s₁(x), s₁(y)) -> minus₂(x, y)
quot₂(0, s₁(y)) -> 0
quot₂(s₁(x), s₁(y)) -> s₁(quot₂(minus₂(x, y), s₁(y)))
app₂(nil, y) -> y
app₂(add₂(n, x), y) -> add₂(n, app₂(x, y))
reverse₁(nil) -> nil
reverse₁(add₂(n, x)) -> app₂(reverse₁(x), add₂(n, nil))
shuffle₁(nil) -> nil
shuffle₁(add₂(n, x)) -> add₂(n, shuffle₁(reverse₁(x)))
concat₂(leaf, y) -> y
concat₂(cons₂(u, v), y) -> cons₂(u, concat₂(v, y))
less_leaves₂(x, leaf) -> false
less_leaves₂(leaf, cons₂(w, z)) -> true
less_leaves₂(cons₂(u, v), cons₂(w, z)) -> less_leaves₂(concat₂(u, v), concat₂(w, z))

Q is empty.

↳ QTRS

  ↳ DependencyPairsProof

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

minus₂(x, 0) -> x
minus₂(s₁(x), s₁(y)) -> minus₂(x, y)
quot₂(0, s₁(y)) -> 0
quot₂(s₁(x), s₁(y)) -> s₁(quot₂(minus₂(x, y), s₁(y)))
app₂(nil, y) -> y
app₂(add₂(n, x), y) -> add₂(n, app₂(x, y))
reverse₁(nil) -> nil
reverse₁(add₂(n, x)) -> app₂(reverse₁(x), add₂(n, nil))
shuffle₁(nil) -> nil
shuffle₁(add₂(n, x)) -> add₂(n, shuffle₁(reverse₁(x)))
concat₂(leaf, y) -> y
concat₂(cons₂(u, v), y) -> cons₂(u, concat₂(v, y))
less_leaves₂(x, leaf) -> false
less_leaves₂(leaf, cons₂(w, z)) -> true
less_leaves₂(cons₂(u, v), cons₂(w, z)) -> less_leaves₂(concat₂(u, v), concat₂(w, z))

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:

LESS_LEAVES₂(cons₂(u, v), cons₂(w, z)) -> CONCAT₂(u, v)
LESS_LEAVES₂(cons₂(u, v), cons₂(w, z)) -> LESS_LEAVES₂(concat₂(u, v), concat₂(w, z))
QUOT₂(s₁(x), s₁(y)) -> QUOT₂(minus₂(x, y), s₁(y))
REVERSE₁(add₂(n, x)) -> APP₂(reverse₁(x), add₂(n, nil))
REVERSE₁(add₂(n, x)) -> REVERSE₁(x)
LESS_LEAVES₂(cons₂(u, v), cons₂(w, z)) -> CONCAT₂(w, z)
SHUFFLE₁(add₂(n, x)) -> SHUFFLE₁(reverse₁(x))
MINUS₂(s₁(x), s₁(y)) -> MINUS₂(x, y)
SHUFFLE₁(add₂(n, x)) -> REVERSE₁(x)
QUOT₂(s₁(x), s₁(y)) -> MINUS₂(x, y)
CONCAT₂(cons₂(u, v), y) -> CONCAT₂(v, y)
APP₂(add₂(n, x), y) -> APP₂(x, y)

The TRS R consists of the following rules:

minus₂(x, 0) -> x
minus₂(s₁(x), s₁(y)) -> minus₂(x, y)
quot₂(0, s₁(y)) -> 0
quot₂(s₁(x), s₁(y)) -> s₁(quot₂(minus₂(x, y), s₁(y)))
app₂(nil, y) -> y
app₂(add₂(n, x), y) -> add₂(n, app₂(x, y))
reverse₁(nil) -> nil
reverse₁(add₂(n, x)) -> app₂(reverse₁(x), add₂(n, nil))
shuffle₁(nil) -> nil
shuffle₁(add₂(n, x)) -> add₂(n, shuffle₁(reverse₁(x)))
concat₂(leaf, y) -> y
concat₂(cons₂(u, v), y) -> cons₂(u, concat₂(v, y))
less_leaves₂(x, leaf) -> false
less_leaves₂(leaf, cons₂(w, z)) -> true
less_leaves₂(cons₂(u, v), cons₂(w, z)) -> less_leaves₂(concat₂(u, v), concat₂(w, z))

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

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ DependencyGraphProof

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

LESS_LEAVES₂(cons₂(u, v), cons₂(w, z)) -> CONCAT₂(u, v)
LESS_LEAVES₂(cons₂(u, v), cons₂(w, z)) -> LESS_LEAVES₂(concat₂(u, v), concat₂(w, z))
QUOT₂(s₁(x), s₁(y)) -> QUOT₂(minus₂(x, y), s₁(y))
REVERSE₁(add₂(n, x)) -> APP₂(reverse₁(x), add₂(n, nil))
REVERSE₁(add₂(n, x)) -> REVERSE₁(x)
LESS_LEAVES₂(cons₂(u, v), cons₂(w, z)) -> CONCAT₂(w, z)
SHUFFLE₁(add₂(n, x)) -> SHUFFLE₁(reverse₁(x))
MINUS₂(s₁(x), s₁(y)) -> MINUS₂(x, y)
SHUFFLE₁(add₂(n, x)) -> REVERSE₁(x)
QUOT₂(s₁(x), s₁(y)) -> MINUS₂(x, y)
CONCAT₂(cons₂(u, v), y) -> CONCAT₂(v, y)
APP₂(add₂(n, x), y) -> APP₂(x, y)

The TRS R consists of the following rules:

minus₂(x, 0) -> x
minus₂(s₁(x), s₁(y)) -> minus₂(x, y)
quot₂(0, s₁(y)) -> 0
quot₂(s₁(x), s₁(y)) -> s₁(quot₂(minus₂(x, y), s₁(y)))
app₂(nil, y) -> y
app₂(add₂(n, x), y) -> add₂(n, app₂(x, y))
reverse₁(nil) -> nil
reverse₁(add₂(n, x)) -> app₂(reverse₁(x), add₂(n, nil))
shuffle₁(nil) -> nil
shuffle₁(add₂(n, x)) -> add₂(n, shuffle₁(reverse₁(x)))
concat₂(leaf, y) -> y
concat₂(cons₂(u, v), y) -> cons₂(u, concat₂(v, y))
less_leaves₂(x, leaf) -> false
less_leaves₂(leaf, cons₂(w, z)) -> true
less_leaves₂(cons₂(u, v), cons₂(w, z)) -> less_leaves₂(concat₂(u, v), concat₂(w, z))

Q is empty.
We have to consider all minimal (P,Q,R)-chains.
The approximation of the Dependency Graph [13,14,18] contains 7 SCCs with 5 less nodes.

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ DependencyGraphProof

        ↳ AND

          ↳ QDP

            ↳ QDPOrderProof

          ↳ QDP

          ↳ QDP

          ↳ QDP

          ↳ QDP

          ↳ QDP

          ↳ QDP

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

CONCAT₂(cons₂(u, v), y) -> CONCAT₂(v, y)

The TRS R consists of the following rules:

minus₂(x, 0) -> x
minus₂(s₁(x), s₁(y)) -> minus₂(x, y)
quot₂(0, s₁(y)) -> 0
quot₂(s₁(x), s₁(y)) -> s₁(quot₂(minus₂(x, y), s₁(y)))
app₂(nil, y) -> y
app₂(add₂(n, x), y) -> add₂(n, app₂(x, y))
reverse₁(nil) -> nil
reverse₁(add₂(n, x)) -> app₂(reverse₁(x), add₂(n, nil))
shuffle₁(nil) -> nil
shuffle₁(add₂(n, x)) -> add₂(n, shuffle₁(reverse₁(x)))
concat₂(leaf, y) -> y
concat₂(cons₂(u, v), y) -> cons₂(u, concat₂(v, y))
less_leaves₂(x, leaf) -> false
less_leaves₂(leaf, cons₂(w, z)) -> true
less_leaves₂(cons₂(u, v), cons₂(w, z)) -> less_leaves₂(concat₂(u, v), concat₂(w, z))

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.

CONCAT₂(cons₂(u, v), y) -> CONCAT₂(v, y)

The remaining pairs can at least be oriented weakly.
none
Used ordering: Polynomial interpretation [21]:

POL(CONCAT₂(x₁, x₂)) = 2·x₁
POL(cons₂(x₁, x₂)) = 1 + 2·x₂

The following usable rules [14] were oriented: none

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ DependencyGraphProof

        ↳ AND

          ↳ QDP

            ↳ QDPOrderProof

              ↳ QDP

                ↳ PisEmptyProof

          ↳ QDP

          ↳ QDP

          ↳ QDP

          ↳ QDP

          ↳ QDP

          ↳ QDP

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

minus₂(x, 0) -> x
minus₂(s₁(x), s₁(y)) -> minus₂(x, y)
quot₂(0, s₁(y)) -> 0
quot₂(s₁(x), s₁(y)) -> s₁(quot₂(minus₂(x, y), s₁(y)))
app₂(nil, y) -> y
app₂(add₂(n, x), y) -> add₂(n, app₂(x, y))
reverse₁(nil) -> nil
reverse₁(add₂(n, x)) -> app₂(reverse₁(x), add₂(n, nil))
shuffle₁(nil) -> nil
shuffle₁(add₂(n, x)) -> add₂(n, shuffle₁(reverse₁(x)))
concat₂(leaf, y) -> y
concat₂(cons₂(u, v), y) -> cons₂(u, concat₂(v, y))
less_leaves₂(x, leaf) -> false
less_leaves₂(leaf, cons₂(w, z)) -> true
less_leaves₂(cons₂(u, v), cons₂(w, z)) -> less_leaves₂(concat₂(u, v), concat₂(w, z))

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

      ↳ DependencyGraphProof

        ↳ AND

          ↳ QDP

          ↳ QDP

            ↳ QDPOrderProof

          ↳ QDP

          ↳ QDP

          ↳ QDP

          ↳ QDP

          ↳ QDP

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

LESS_LEAVES₂(cons₂(u, v), cons₂(w, z)) -> LESS_LEAVES₂(concat₂(u, v), concat₂(w, z))

The TRS R consists of the following rules:

minus₂(x, 0) -> x
minus₂(s₁(x), s₁(y)) -> minus₂(x, y)
quot₂(0, s₁(y)) -> 0
quot₂(s₁(x), s₁(y)) -> s₁(quot₂(minus₂(x, y), s₁(y)))
app₂(nil, y) -> y
app₂(add₂(n, x), y) -> add₂(n, app₂(x, y))
reverse₁(nil) -> nil
reverse₁(add₂(n, x)) -> app₂(reverse₁(x), add₂(n, nil))
shuffle₁(nil) -> nil
shuffle₁(add₂(n, x)) -> add₂(n, shuffle₁(reverse₁(x)))
concat₂(leaf, y) -> y
concat₂(cons₂(u, v), y) -> cons₂(u, concat₂(v, y))
less_leaves₂(x, leaf) -> false
less_leaves₂(leaf, cons₂(w, z)) -> true
less_leaves₂(cons₂(u, v), cons₂(w, z)) -> less_leaves₂(concat₂(u, v), concat₂(w, z))

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.

LESS_LEAVES₂(cons₂(u, v), cons₂(w, z)) -> LESS_LEAVES₂(concat₂(u, v), concat₂(w, z))

The remaining pairs can at least be oriented weakly.
none
Used ordering: Polynomial interpretation [21]:

POL(LESS_LEAVES₂(x₁, x₂)) = 2·x₂
POL(concat₂(x₁, x₂)) = x₁ + x₂
POL(cons₂(x₁, x₂)) = 2 + x₁ + x₂
POL(leaf) = 0

The following usable rules [14] were oriented:

concat₂(leaf, y) -> y
concat₂(cons₂(u, v), y) -> cons₂(u, concat₂(v, y))

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ DependencyGraphProof

        ↳ AND

          ↳ QDP

          ↳ QDP

            ↳ QDPOrderProof

              ↳ QDP

                ↳ PisEmptyProof

          ↳ QDP

          ↳ QDP

          ↳ QDP

          ↳ QDP

          ↳ QDP

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

minus₂(x, 0) -> x
minus₂(s₁(x), s₁(y)) -> minus₂(x, y)
quot₂(0, s₁(y)) -> 0
quot₂(s₁(x), s₁(y)) -> s₁(quot₂(minus₂(x, y), s₁(y)))
app₂(nil, y) -> y
app₂(add₂(n, x), y) -> add₂(n, app₂(x, y))
reverse₁(nil) -> nil
reverse₁(add₂(n, x)) -> app₂(reverse₁(x), add₂(n, nil))
shuffle₁(nil) -> nil
shuffle₁(add₂(n, x)) -> add₂(n, shuffle₁(reverse₁(x)))
concat₂(leaf, y) -> y
concat₂(cons₂(u, v), y) -> cons₂(u, concat₂(v, y))
less_leaves₂(x, leaf) -> false
less_leaves₂(leaf, cons₂(w, z)) -> true
less_leaves₂(cons₂(u, v), cons₂(w, z)) -> less_leaves₂(concat₂(u, v), concat₂(w, z))

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

      ↳ DependencyGraphProof

        ↳ AND

          ↳ QDP

          ↳ QDP

          ↳ QDP

            ↳ QDPOrderProof

          ↳ QDP

          ↳ QDP

          ↳ QDP

          ↳ QDP

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

APP₂(add₂(n, x), y) -> APP₂(x, y)

The TRS R consists of the following rules:

minus₂(x, 0) -> x
minus₂(s₁(x), s₁(y)) -> minus₂(x, y)
quot₂(0, s₁(y)) -> 0
quot₂(s₁(x), s₁(y)) -> s₁(quot₂(minus₂(x, y), s₁(y)))
app₂(nil, y) -> y
app₂(add₂(n, x), y) -> add₂(n, app₂(x, y))
reverse₁(nil) -> nil
reverse₁(add₂(n, x)) -> app₂(reverse₁(x), add₂(n, nil))
shuffle₁(nil) -> nil
shuffle₁(add₂(n, x)) -> add₂(n, shuffle₁(reverse₁(x)))
concat₂(leaf, y) -> y
concat₂(cons₂(u, v), y) -> cons₂(u, concat₂(v, y))
less_leaves₂(x, leaf) -> false
less_leaves₂(leaf, cons₂(w, z)) -> true
less_leaves₂(cons₂(u, v), cons₂(w, z)) -> less_leaves₂(concat₂(u, v), concat₂(w, z))

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.

APP₂(add₂(n, x), y) -> APP₂(x, y)

The remaining pairs can at least be oriented weakly.
none
Used ordering: Polynomial interpretation [21]:

POL(APP₂(x₁, x₂)) = 2·x₁
POL(add₂(x₁, x₂)) = 1 + 2·x₂

The following usable rules [14] were oriented: none

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ DependencyGraphProof

        ↳ AND

          ↳ QDP

          ↳ QDP

          ↳ QDP

            ↳ QDPOrderProof

              ↳ QDP

                ↳ PisEmptyProof

          ↳ QDP

          ↳ QDP

          ↳ QDP

          ↳ QDP

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

minus₂(x, 0) -> x
minus₂(s₁(x), s₁(y)) -> minus₂(x, y)
quot₂(0, s₁(y)) -> 0
quot₂(s₁(x), s₁(y)) -> s₁(quot₂(minus₂(x, y), s₁(y)))
app₂(nil, y) -> y
app₂(add₂(n, x), y) -> add₂(n, app₂(x, y))
reverse₁(nil) -> nil
reverse₁(add₂(n, x)) -> app₂(reverse₁(x), add₂(n, nil))
shuffle₁(nil) -> nil
shuffle₁(add₂(n, x)) -> add₂(n, shuffle₁(reverse₁(x)))
concat₂(leaf, y) -> y
concat₂(cons₂(u, v), y) -> cons₂(u, concat₂(v, y))
less_leaves₂(x, leaf) -> false
less_leaves₂(leaf, cons₂(w, z)) -> true
less_leaves₂(cons₂(u, v), cons₂(w, z)) -> less_leaves₂(concat₂(u, v), concat₂(w, z))

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

      ↳ DependencyGraphProof

        ↳ AND

          ↳ QDP

          ↳ QDP

          ↳ QDP

          ↳ QDP

            ↳ QDPOrderProof

          ↳ QDP

          ↳ QDP

          ↳ QDP

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

REVERSE₁(add₂(n, x)) -> REVERSE₁(x)

The TRS R consists of the following rules:

minus₂(x, 0) -> x
minus₂(s₁(x), s₁(y)) -> minus₂(x, y)
quot₂(0, s₁(y)) -> 0
quot₂(s₁(x), s₁(y)) -> s₁(quot₂(minus₂(x, y), s₁(y)))
app₂(nil, y) -> y
app₂(add₂(n, x), y) -> add₂(n, app₂(x, y))
reverse₁(nil) -> nil
reverse₁(add₂(n, x)) -> app₂(reverse₁(x), add₂(n, nil))
shuffle₁(nil) -> nil
shuffle₁(add₂(n, x)) -> add₂(n, shuffle₁(reverse₁(x)))
concat₂(leaf, y) -> y
concat₂(cons₂(u, v), y) -> cons₂(u, concat₂(v, y))
less_leaves₂(x, leaf) -> false
less_leaves₂(leaf, cons₂(w, z)) -> true
less_leaves₂(cons₂(u, v), cons₂(w, z)) -> less_leaves₂(concat₂(u, v), concat₂(w, z))

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.

REVERSE₁(add₂(n, x)) -> REVERSE₁(x)

The remaining pairs can at least be oriented weakly.
none
Used ordering: Polynomial interpretation [21]:

POL(REVERSE₁(x₁)) = 2·x₁
POL(add₂(x₁, x₂)) = 1 + 2·x₂

The following usable rules [14] were oriented: none

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ DependencyGraphProof

        ↳ AND

          ↳ QDP

          ↳ QDP

          ↳ QDP

          ↳ QDP

            ↳ QDPOrderProof

              ↳ QDP

                ↳ PisEmptyProof

          ↳ QDP

          ↳ QDP

          ↳ QDP

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

minus₂(x, 0) -> x
minus₂(s₁(x), s₁(y)) -> minus₂(x, y)
quot₂(0, s₁(y)) -> 0
quot₂(s₁(x), s₁(y)) -> s₁(quot₂(minus₂(x, y), s₁(y)))
app₂(nil, y) -> y
app₂(add₂(n, x), y) -> add₂(n, app₂(x, y))
reverse₁(nil) -> nil
reverse₁(add₂(n, x)) -> app₂(reverse₁(x), add₂(n, nil))
shuffle₁(nil) -> nil
shuffle₁(add₂(n, x)) -> add₂(n, shuffle₁(reverse₁(x)))
concat₂(leaf, y) -> y
concat₂(cons₂(u, v), y) -> cons₂(u, concat₂(v, y))
less_leaves₂(x, leaf) -> false
less_leaves₂(leaf, cons₂(w, z)) -> true
less_leaves₂(cons₂(u, v), cons₂(w, z)) -> less_leaves₂(concat₂(u, v), concat₂(w, z))

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

      ↳ DependencyGraphProof

        ↳ AND

          ↳ QDP

          ↳ QDP

          ↳ QDP

          ↳ QDP

          ↳ QDP

            ↳ QDPOrderProof

          ↳ QDP

          ↳ QDP

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

SHUFFLE₁(add₂(n, x)) -> SHUFFLE₁(reverse₁(x))

The TRS R consists of the following rules:

minus₂(x, 0) -> x
minus₂(s₁(x), s₁(y)) -> minus₂(x, y)
quot₂(0, s₁(y)) -> 0
quot₂(s₁(x), s₁(y)) -> s₁(quot₂(minus₂(x, y), s₁(y)))
app₂(nil, y) -> y
app₂(add₂(n, x), y) -> add₂(n, app₂(x, y))
reverse₁(nil) -> nil
reverse₁(add₂(n, x)) -> app₂(reverse₁(x), add₂(n, nil))
shuffle₁(nil) -> nil
shuffle₁(add₂(n, x)) -> add₂(n, shuffle₁(reverse₁(x)))
concat₂(leaf, y) -> y
concat₂(cons₂(u, v), y) -> cons₂(u, concat₂(v, y))
less_leaves₂(x, leaf) -> false
less_leaves₂(leaf, cons₂(w, z)) -> true
less_leaves₂(cons₂(u, v), cons₂(w, z)) -> less_leaves₂(concat₂(u, v), concat₂(w, z))

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.

SHUFFLE₁(add₂(n, x)) -> SHUFFLE₁(reverse₁(x))

The remaining pairs can at least be oriented weakly.
none
Used ordering: Polynomial interpretation [21]:

POL(SHUFFLE₁(x₁)) = 2·x₁
POL(add₂(x₁, x₂)) = 1 + x₂
POL(app₂(x₁, x₂)) = x₁ + x₂
POL(nil) = 0
POL(reverse₁(x₁)) = x₁

The following usable rules [14] were oriented:

app₂(nil, y) -> y
reverse₁(nil) -> nil
reverse₁(add₂(n, x)) -> app₂(reverse₁(x), add₂(n, nil))
app₂(add₂(n, x), y) -> add₂(n, app₂(x, y))

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ DependencyGraphProof

        ↳ AND

          ↳ QDP

          ↳ QDP

          ↳ QDP

          ↳ QDP

          ↳ QDP

            ↳ QDPOrderProof

              ↳ QDP

                ↳ PisEmptyProof

          ↳ QDP

          ↳ QDP

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

minus₂(x, 0) -> x
minus₂(s₁(x), s₁(y)) -> minus₂(x, y)
quot₂(0, s₁(y)) -> 0
quot₂(s₁(x), s₁(y)) -> s₁(quot₂(minus₂(x, y), s₁(y)))
app₂(nil, y) -> y
app₂(add₂(n, x), y) -> add₂(n, app₂(x, y))
reverse₁(nil) -> nil
reverse₁(add₂(n, x)) -> app₂(reverse₁(x), add₂(n, nil))
shuffle₁(nil) -> nil
shuffle₁(add₂(n, x)) -> add₂(n, shuffle₁(reverse₁(x)))
concat₂(leaf, y) -> y
concat₂(cons₂(u, v), y) -> cons₂(u, concat₂(v, y))
less_leaves₂(x, leaf) -> false
less_leaves₂(leaf, cons₂(w, z)) -> true
less_leaves₂(cons₂(u, v), cons₂(w, z)) -> less_leaves₂(concat₂(u, v), concat₂(w, z))

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

      ↳ DependencyGraphProof

        ↳ AND

          ↳ QDP

          ↳ QDP

          ↳ QDP

          ↳ QDP

          ↳ QDP

          ↳ QDP

            ↳ QDPOrderProof

          ↳ QDP

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

MINUS₂(s₁(x), s₁(y)) -> MINUS₂(x, y)

The TRS R consists of the following rules:

minus₂(x, 0) -> x
minus₂(s₁(x), s₁(y)) -> minus₂(x, y)
quot₂(0, s₁(y)) -> 0
quot₂(s₁(x), s₁(y)) -> s₁(quot₂(minus₂(x, y), s₁(y)))
app₂(nil, y) -> y
app₂(add₂(n, x), y) -> add₂(n, app₂(x, y))
reverse₁(nil) -> nil
reverse₁(add₂(n, x)) -> app₂(reverse₁(x), add₂(n, nil))
shuffle₁(nil) -> nil
shuffle₁(add₂(n, x)) -> add₂(n, shuffle₁(reverse₁(x)))
concat₂(leaf, y) -> y
concat₂(cons₂(u, v), y) -> cons₂(u, concat₂(v, y))
less_leaves₂(x, leaf) -> false
less_leaves₂(leaf, cons₂(w, z)) -> true
less_leaves₂(cons₂(u, v), cons₂(w, z)) -> less_leaves₂(concat₂(u, v), concat₂(w, z))

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.

MINUS₂(s₁(x), s₁(y)) -> MINUS₂(x, y)

The remaining pairs can at least be oriented weakly.
none
Used ordering: Polynomial interpretation [21]:

POL(MINUS₂(x₁, x₂)) = x₂
POL(s₁(x₁)) = 1 + 2·x₁

The following usable rules [14] were oriented: none

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ DependencyGraphProof

        ↳ AND

          ↳ QDP

          ↳ QDP

          ↳ QDP

          ↳ QDP

          ↳ QDP

          ↳ QDP

            ↳ QDPOrderProof

              ↳ QDP

                ↳ PisEmptyProof

          ↳ QDP

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

minus₂(x, 0) -> x
minus₂(s₁(x), s₁(y)) -> minus₂(x, y)
quot₂(0, s₁(y)) -> 0
quot₂(s₁(x), s₁(y)) -> s₁(quot₂(minus₂(x, y), s₁(y)))
app₂(nil, y) -> y
app₂(add₂(n, x), y) -> add₂(n, app₂(x, y))
reverse₁(nil) -> nil
reverse₁(add₂(n, x)) -> app₂(reverse₁(x), add₂(n, nil))
shuffle₁(nil) -> nil
shuffle₁(add₂(n, x)) -> add₂(n, shuffle₁(reverse₁(x)))
concat₂(leaf, y) -> y
concat₂(cons₂(u, v), y) -> cons₂(u, concat₂(v, y))
less_leaves₂(x, leaf) -> false
less_leaves₂(leaf, cons₂(w, z)) -> true
less_leaves₂(cons₂(u, v), cons₂(w, z)) -> less_leaves₂(concat₂(u, v), concat₂(w, z))

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

      ↳ DependencyGraphProof

        ↳ AND

          ↳ QDP

          ↳ QDP

          ↳ QDP

          ↳ QDP

          ↳ QDP

          ↳ QDP

          ↳ QDP

            ↳ QDPOrderProof

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

QUOT₂(s₁(x), s₁(y)) -> QUOT₂(minus₂(x, y), s₁(y))

The TRS R consists of the following rules:

minus₂(x, 0) -> x
minus₂(s₁(x), s₁(y)) -> minus₂(x, y)
quot₂(0, s₁(y)) -> 0
quot₂(s₁(x), s₁(y)) -> s₁(quot₂(minus₂(x, y), s₁(y)))
app₂(nil, y) -> y
app₂(add₂(n, x), y) -> add₂(n, app₂(x, y))
reverse₁(nil) -> nil
reverse₁(add₂(n, x)) -> app₂(reverse₁(x), add₂(n, nil))
shuffle₁(nil) -> nil
shuffle₁(add₂(n, x)) -> add₂(n, shuffle₁(reverse₁(x)))
concat₂(leaf, y) -> y
concat₂(cons₂(u, v), y) -> cons₂(u, concat₂(v, y))
less_leaves₂(x, leaf) -> false
less_leaves₂(leaf, cons₂(w, z)) -> true
less_leaves₂(cons₂(u, v), cons₂(w, z)) -> less_leaves₂(concat₂(u, v), concat₂(w, z))

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.

QUOT₂(s₁(x), s₁(y)) -> QUOT₂(minus₂(x, y), s₁(y))

The remaining pairs can at least be oriented weakly.
none
Used ordering: Polynomial interpretation [21]:

POL(0) = 0
POL(QUOT₂(x₁, x₂)) = 2·x₁
POL(minus₂(x₁, x₂)) = 2·x₁
POL(s₁(x₁)) = 1 + 2·x₁

The following usable rules [14] were oriented:

minus₂(s₁(x), s₁(y)) -> minus₂(x, y)
minus₂(x, 0) -> x

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ DependencyGraphProof

        ↳ AND

          ↳ QDP

          ↳ QDP

          ↳ QDP

          ↳ QDP

          ↳ QDP

          ↳ QDP

          ↳ QDP

            ↳ QDPOrderProof

              ↳ QDP

                ↳ PisEmptyProof

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

minus₂(x, 0) -> x
minus₂(s₁(x), s₁(y)) -> minus₂(x, y)
quot₂(0, s₁(y)) -> 0
quot₂(s₁(x), s₁(y)) -> s₁(quot₂(minus₂(x, y), s₁(y)))
app₂(nil, y) -> y
app₂(add₂(n, x), y) -> add₂(n, app₂(x, y))
reverse₁(nil) -> nil
reverse₁(add₂(n, x)) -> app₂(reverse₁(x), add₂(n, nil))
shuffle₁(nil) -> nil
shuffle₁(add₂(n, x)) -> add₂(n, shuffle₁(reverse₁(x)))
concat₂(leaf, y) -> y
concat₂(cons₂(u, v), y) -> cons₂(u, concat₂(v, y))
less_leaves₂(x, leaf) -> false
less_leaves₂(leaf, cons₂(w, z)) -> true
less_leaves₂(cons₂(u, v), cons₂(w, z)) -> less_leaves₂(concat₂(u, v), concat₂(w, z))

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.