Termination w.r.t. Q proof of /home/nowonder/forschung/aprove/satrung/aprove/lpar16/AG01SLASHHASH3.53.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:

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

  ↳ Non-Overlap Check

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.

The TRS is non-overlapping. Hence, we can switch to innermost.

↳ QTRS

  ↳ Non-Overlap Check

    ↳ 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))

The set Q consists of the following terms:

minus₂(x0, 0)
minus₂(s₁(x0), s₁(x1))
quot₂(0, s₁(x0))
quot₂(s₁(x0), s₁(x1))
app₂(nil, x0)
app₂(add₂(x0, x1), x2)
reverse₁(nil)
reverse₁(add₂(x0, x1))
shuffle₁(nil)
shuffle₁(add₂(x0, x1))
concat₂(leaf, x0)
concat₂(cons₂(x0, x1), x2)
less_leaves₂(x0, leaf)
less_leaves₂(leaf, cons₂(x0, x1))
less_leaves₂(cons₂(x0, x1), cons₂(x2, x3))

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))

The set Q consists of the following terms:

minus₂(x0, 0)
minus₂(s₁(x0), s₁(x1))
quot₂(0, s₁(x0))
quot₂(s₁(x0), s₁(x1))
app₂(nil, x0)
app₂(add₂(x0, x1), x2)
reverse₁(nil)
reverse₁(add₂(x0, x1))
shuffle₁(nil)
shuffle₁(add₂(x0, x1))
concat₂(leaf, x0)
concat₂(cons₂(x0, x1), x2)
less_leaves₂(x0, leaf)
less_leaves₂(leaf, cons₂(x0, x1))
less_leaves₂(cons₂(x0, x1), cons₂(x2, x3))

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

↳ QTRS

  ↳ Non-Overlap Check

    ↳ 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))

The set Q consists of the following terms:

minus₂(x0, 0)
minus₂(s₁(x0), s₁(x1))
quot₂(0, s₁(x0))
quot₂(s₁(x0), s₁(x1))
app₂(nil, x0)
app₂(add₂(x0, x1), x2)
reverse₁(nil)
reverse₁(add₂(x0, x1))
shuffle₁(nil)
shuffle₁(add₂(x0, x1))
concat₂(leaf, x0)
concat₂(cons₂(x0, x1), x2)
less_leaves₂(x0, leaf)
less_leaves₂(leaf, cons₂(x0, x1))
less_leaves₂(cons₂(x0, x1), cons₂(x2, x3))

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

  ↳ Non-Overlap Check

    ↳ 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))

The set Q consists of the following terms:

minus₂(x0, 0)
minus₂(s₁(x0), s₁(x1))
quot₂(0, s₁(x0))
quot₂(s₁(x0), s₁(x1))
app₂(nil, x0)
app₂(add₂(x0, x1), x2)
reverse₁(nil)
reverse₁(add₂(x0, x1))
shuffle₁(nil)
shuffle₁(add₂(x0, x1))
concat₂(leaf, x0)
concat₂(cons₂(x0, x1), x2)
less_leaves₂(x0, leaf)
less_leaves₂(leaf, cons₂(x0, x1))
less_leaves₂(cons₂(x0, x1), cons₂(x2, x3))

We have to consider all minimal (P,Q,R)-chains.
We use the reduction pair processor [13].

The following pairs can be strictly oriented and are deleted.

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

The remaining pairs can at least by weakly be oriented.
none
Used ordering: Combined order from the following AFS and order.
CONCAT₂(x1, x2) = CONCAT₁(x1)
cons₂(x1, x2) = cons₁(x2)

Lexicographic Path Order [19].
Precedence:

[CONCAT₁, cons_1]

The following usable rules [14] were oriented: none

↳ QTRS

  ↳ Non-Overlap Check

    ↳ 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))

The set Q consists of the following terms:

minus₂(x0, 0)
minus₂(s₁(x0), s₁(x1))
quot₂(0, s₁(x0))
quot₂(s₁(x0), s₁(x1))
app₂(nil, x0)
app₂(add₂(x0, x1), x2)
reverse₁(nil)
reverse₁(add₂(x0, x1))
shuffle₁(nil)
shuffle₁(add₂(x0, x1))
concat₂(leaf, x0)
concat₂(cons₂(x0, x1), x2)
less_leaves₂(x0, leaf)
less_leaves₂(leaf, cons₂(x0, x1))
less_leaves₂(cons₂(x0, x1), cons₂(x2, x3))

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

↳ QTRS

  ↳ Non-Overlap Check

    ↳ QTRS

      ↳ DependencyPairsProof

        ↳ QDP

          ↳ DependencyGraphProof

            ↳ AND

              ↳ QDP

              ↳ QDP

              ↳ 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))

The set Q consists of the following terms:

minus₂(x0, 0)
minus₂(s₁(x0), s₁(x1))
quot₂(0, s₁(x0))
quot₂(s₁(x0), s₁(x1))
app₂(nil, x0)
app₂(add₂(x0, x1), x2)
reverse₁(nil)
reverse₁(add₂(x0, x1))
shuffle₁(nil)
shuffle₁(add₂(x0, x1))
concat₂(leaf, x0)
concat₂(cons₂(x0, x1), x2)
less_leaves₂(x0, leaf)
less_leaves₂(leaf, cons₂(x0, x1))
less_leaves₂(cons₂(x0, x1), cons₂(x2, x3))

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

↳ QTRS

  ↳ Non-Overlap Check

    ↳ 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))

The set Q consists of the following terms:

minus₂(x0, 0)
minus₂(s₁(x0), s₁(x1))
quot₂(0, s₁(x0))
quot₂(s₁(x0), s₁(x1))
app₂(nil, x0)
app₂(add₂(x0, x1), x2)
reverse₁(nil)
reverse₁(add₂(x0, x1))
shuffle₁(nil)
shuffle₁(add₂(x0, x1))
concat₂(leaf, x0)
concat₂(cons₂(x0, x1), x2)
less_leaves₂(x0, leaf)
less_leaves₂(leaf, cons₂(x0, x1))
less_leaves₂(cons₂(x0, x1), cons₂(x2, x3))

We have to consider all minimal (P,Q,R)-chains.
We use the reduction pair processor [13].

The following pairs can be strictly oriented and are deleted.

APP₂(add₂(n, x), y) -> APP₂(x, y)

The remaining pairs can at least by weakly be oriented.
none
Used ordering: Combined order from the following AFS and order.
APP₂(x1, x2) = APP₁(x1)
add₂(x1, x2) = add₁(x2)

Lexicographic Path Order [19].
Precedence:

[APP₁, add_1]

The following usable rules [14] were oriented: none

↳ QTRS

  ↳ Non-Overlap Check

    ↳ 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))

The set Q consists of the following terms:

minus₂(x0, 0)
minus₂(s₁(x0), s₁(x1))
quot₂(0, s₁(x0))
quot₂(s₁(x0), s₁(x1))
app₂(nil, x0)
app₂(add₂(x0, x1), x2)
reverse₁(nil)
reverse₁(add₂(x0, x1))
shuffle₁(nil)
shuffle₁(add₂(x0, x1))
concat₂(leaf, x0)
concat₂(cons₂(x0, x1), x2)
less_leaves₂(x0, leaf)
less_leaves₂(leaf, cons₂(x0, x1))
less_leaves₂(cons₂(x0, x1), cons₂(x2, x3))

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

↳ QTRS

  ↳ Non-Overlap Check

    ↳ 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))

The set Q consists of the following terms:

minus₂(x0, 0)
minus₂(s₁(x0), s₁(x1))
quot₂(0, s₁(x0))
quot₂(s₁(x0), s₁(x1))
app₂(nil, x0)
app₂(add₂(x0, x1), x2)
reverse₁(nil)
reverse₁(add₂(x0, x1))
shuffle₁(nil)
shuffle₁(add₂(x0, x1))
concat₂(leaf, x0)
concat₂(cons₂(x0, x1), x2)
less_leaves₂(x0, leaf)
less_leaves₂(leaf, cons₂(x0, x1))
less_leaves₂(cons₂(x0, x1), cons₂(x2, x3))

We have to consider all minimal (P,Q,R)-chains.
We use the reduction pair processor [13].

The following pairs can be strictly oriented and are deleted.

REVERSE₁(add₂(n, x)) -> REVERSE₁(x)

The remaining pairs can at least by weakly be oriented.
none
Used ordering: Combined order from the following AFS and order.
REVERSE₁(x1) = REVERSE₁(x1)
add₂(x1, x2) = add₂(x1, x2)

Lexicographic Path Order [19].
Precedence:

add₂ > REVERSE₁

The following usable rules [14] were oriented: none

↳ QTRS

  ↳ Non-Overlap Check

    ↳ 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))

The set Q consists of the following terms:

minus₂(x0, 0)
minus₂(s₁(x0), s₁(x1))
quot₂(0, s₁(x0))
quot₂(s₁(x0), s₁(x1))
app₂(nil, x0)
app₂(add₂(x0, x1), x2)
reverse₁(nil)
reverse₁(add₂(x0, x1))
shuffle₁(nil)
shuffle₁(add₂(x0, x1))
concat₂(leaf, x0)
concat₂(cons₂(x0, x1), x2)
less_leaves₂(x0, leaf)
less_leaves₂(leaf, cons₂(x0, x1))
less_leaves₂(cons₂(x0, x1), cons₂(x2, x3))

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

↳ QTRS

  ↳ Non-Overlap Check

    ↳ QTRS

      ↳ DependencyPairsProof

        ↳ QDP

          ↳ DependencyGraphProof

            ↳ AND

              ↳ QDP

              ↳ QDP

              ↳ QDP

              ↳ QDP

              ↳ QDP

              ↳ 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))

The set Q consists of the following terms:

minus₂(x0, 0)
minus₂(s₁(x0), s₁(x1))
quot₂(0, s₁(x0))
quot₂(s₁(x0), s₁(x1))
app₂(nil, x0)
app₂(add₂(x0, x1), x2)
reverse₁(nil)
reverse₁(add₂(x0, x1))
shuffle₁(nil)
shuffle₁(add₂(x0, x1))
concat₂(leaf, x0)
concat₂(cons₂(x0, x1), x2)
less_leaves₂(x0, leaf)
less_leaves₂(leaf, cons₂(x0, x1))
less_leaves₂(cons₂(x0, x1), cons₂(x2, x3))

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

↳ QTRS

  ↳ Non-Overlap Check

    ↳ 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))

The set Q consists of the following terms:

minus₂(x0, 0)
minus₂(s₁(x0), s₁(x1))
quot₂(0, s₁(x0))
quot₂(s₁(x0), s₁(x1))
app₂(nil, x0)
app₂(add₂(x0, x1), x2)
reverse₁(nil)
reverse₁(add₂(x0, x1))
shuffle₁(nil)
shuffle₁(add₂(x0, x1))
concat₂(leaf, x0)
concat₂(cons₂(x0, x1), x2)
less_leaves₂(x0, leaf)
less_leaves₂(leaf, cons₂(x0, x1))
less_leaves₂(cons₂(x0, x1), cons₂(x2, x3))

We have to consider all minimal (P,Q,R)-chains.
We use the reduction pair processor [13].

The following pairs can be strictly oriented and are deleted.

MINUS₂(s₁(x), s₁(y)) -> MINUS₂(x, y)

The remaining pairs can at least by weakly be oriented.
none
Used ordering: Combined order from the following AFS and order.
MINUS₂(x1, x2) = MINUS₁(x1)
s₁(x1) = s₁(x1)

Lexicographic Path Order [19].
Precedence:

trivial

The following usable rules [14] were oriented: none

↳ QTRS

  ↳ Non-Overlap Check

    ↳ 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))

The set Q consists of the following terms:

minus₂(x0, 0)
minus₂(s₁(x0), s₁(x1))
quot₂(0, s₁(x0))
quot₂(s₁(x0), s₁(x1))
app₂(nil, x0)
app₂(add₂(x0, x1), x2)
reverse₁(nil)
reverse₁(add₂(x0, x1))
shuffle₁(nil)
shuffle₁(add₂(x0, x1))
concat₂(leaf, x0)
concat₂(cons₂(x0, x1), x2)
less_leaves₂(x0, leaf)
less_leaves₂(leaf, cons₂(x0, x1))
less_leaves₂(cons₂(x0, x1), cons₂(x2, x3))

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

↳ QTRS

  ↳ Non-Overlap Check

    ↳ 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))

The set Q consists of the following terms:

minus₂(x0, 0)
minus₂(s₁(x0), s₁(x1))
quot₂(0, s₁(x0))
quot₂(s₁(x0), s₁(x1))
app₂(nil, x0)
app₂(add₂(x0, x1), x2)
reverse₁(nil)
reverse₁(add₂(x0, x1))
shuffle₁(nil)
shuffle₁(add₂(x0, x1))
concat₂(leaf, x0)
concat₂(cons₂(x0, x1), x2)
less_leaves₂(x0, leaf)
less_leaves₂(leaf, cons₂(x0, x1))
less_leaves₂(cons₂(x0, x1), cons₂(x2, x3))

We have to consider all minimal (P,Q,R)-chains.
We use the reduction pair processor [13].

The following pairs can be strictly oriented and are deleted.

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

The remaining pairs can at least by weakly be oriented.
none
Used ordering: Combined order from the following AFS and order.
QUOT₂(x1, x2) = x1
s₁(x1) = s₁(x1)
minus₂(x1, x2) = x1
0 = 0

Lexicographic Path Order [19].
Precedence:

trivial

The following usable rules [14] were oriented:

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

↳ QTRS

  ↳ Non-Overlap Check

    ↳ 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))

The set Q consists of the following terms:

minus₂(x0, 0)
minus₂(s₁(x0), s₁(x1))
quot₂(0, s₁(x0))
quot₂(s₁(x0), s₁(x1))
app₂(nil, x0)
app₂(add₂(x0, x1), x2)
reverse₁(nil)
reverse₁(add₂(x0, x1))
shuffle₁(nil)
shuffle₁(add₂(x0, x1))
concat₂(leaf, x0)
concat₂(cons₂(x0, x1), x2)
less_leaves₂(x0, leaf)
less_leaves₂(leaf, cons₂(x0, x1))
less_leaves₂(cons₂(x0, x1), cons₂(x2, x3))

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