Termination w.r.t. Q proof of /tmp/tmpAph3iK/#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:

minus(x, 0) → x
minus(s(x), s(y)) → minus(x, y)
quot(0, s(y)) → 0
quot(s(x), s(y)) → s(quot(minus(x, y), s(y)))
app(nil, y) → y
app(add(n, x), y) → add(n, app(x, y))
reverse(nil) → nil
reverse(add(n, x)) → app(reverse(x), add(n, nil))
shuffle(nil) → nil
shuffle(add(n, x)) → add(n, shuffle(reverse(x)))
concat(leaf, y) → y
concat(cons(u, v), y) → cons(u, concat(v, y))
less_leaves(x, leaf) → false
less_leaves(leaf, cons(w, z)) → true
less_leaves(cons(u, v), cons(w, z)) → less_leaves(concat(u, v), concat(w, z))

Q is empty.

↳ QTRS

  ↳ DependencyPairsProof

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

minus(x, 0) → x
minus(s(x), s(y)) → minus(x, y)
quot(0, s(y)) → 0
quot(s(x), s(y)) → s(quot(minus(x, y), s(y)))
app(nil, y) → y
app(add(n, x), y) → add(n, app(x, y))
reverse(nil) → nil
reverse(add(n, x)) → app(reverse(x), add(n, nil))
shuffle(nil) → nil
shuffle(add(n, x)) → add(n, shuffle(reverse(x)))
concat(leaf, y) → y
concat(cons(u, v), y) → cons(u, concat(v, y))
less_leaves(x, leaf) → false
less_leaves(leaf, cons(w, z)) → true
less_leaves(cons(u, v), cons(w, z)) → less_leaves(concat(u, v), concat(w, z))

Q is empty.

Using Dependency Pairs [1,15] we result in the following initial DP problem:
Q DP problem:
The TRS P consists of the following rules:

QUOT(s(x), s(y)) → QUOT(minus(x, y), s(y))
SHUFFLE(add(n, x)) → REVERSE(x)
REVERSE(add(n, x)) → REVERSE(x)
REVERSE(add(n, x)) → APP(reverse(x), add(n, nil))
APP(add(n, x), y) → APP(x, y)
MINUS(s(x), s(y)) → MINUS(x, y)
LESS_LEAVES(cons(u, v), cons(w, z)) → CONCAT(w, z)
CONCAT(cons(u, v), y) → CONCAT(v, y)
SHUFFLE(add(n, x)) → SHUFFLE(reverse(x))
LESS_LEAVES(cons(u, v), cons(w, z)) → CONCAT(u, v)
LESS_LEAVES(cons(u, v), cons(w, z)) → LESS_LEAVES(concat(u, v), concat(w, z))
QUOT(s(x), s(y)) → MINUS(x, y)

The TRS R consists of the following rules:

minus(x, 0) → x
minus(s(x), s(y)) → minus(x, y)
quot(0, s(y)) → 0
quot(s(x), s(y)) → s(quot(minus(x, y), s(y)))
app(nil, y) → y
app(add(n, x), y) → add(n, app(x, y))
reverse(nil) → nil
reverse(add(n, x)) → app(reverse(x), add(n, nil))
shuffle(nil) → nil
shuffle(add(n, x)) → add(n, shuffle(reverse(x)))
concat(leaf, y) → y
concat(cons(u, v), y) → cons(u, concat(v, y))
less_leaves(x, leaf) → false
less_leaves(leaf, cons(w, z)) → true
less_leaves(cons(u, v), cons(w, z)) → less_leaves(concat(u, v), concat(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:

QUOT(s(x), s(y)) → QUOT(minus(x, y), s(y))
SHUFFLE(add(n, x)) → REVERSE(x)
REVERSE(add(n, x)) → REVERSE(x)
REVERSE(add(n, x)) → APP(reverse(x), add(n, nil))
APP(add(n, x), y) → APP(x, y)
MINUS(s(x), s(y)) → MINUS(x, y)
LESS_LEAVES(cons(u, v), cons(w, z)) → CONCAT(w, z)
CONCAT(cons(u, v), y) → CONCAT(v, y)
SHUFFLE(add(n, x)) → SHUFFLE(reverse(x))
LESS_LEAVES(cons(u, v), cons(w, z)) → CONCAT(u, v)
LESS_LEAVES(cons(u, v), cons(w, z)) → LESS_LEAVES(concat(u, v), concat(w, z))
QUOT(s(x), s(y)) → MINUS(x, y)

The TRS R consists of the following rules:

minus(x, 0) → x
minus(s(x), s(y)) → minus(x, y)
quot(0, s(y)) → 0
quot(s(x), s(y)) → s(quot(minus(x, y), s(y)))
app(nil, y) → y
app(add(n, x), y) → add(n, app(x, y))
reverse(nil) → nil
reverse(add(n, x)) → app(reverse(x), add(n, nil))
shuffle(nil) → nil
shuffle(add(n, x)) → add(n, shuffle(reverse(x)))
concat(leaf, y) → y
concat(cons(u, v), y) → cons(u, concat(v, y))
less_leaves(x, leaf) → false
less_leaves(leaf, cons(w, z)) → true
less_leaves(cons(u, v), cons(w, z)) → less_leaves(concat(u, v), concat(w, z))

Q is empty.
We have to consider all minimal (P,Q,R)-chains.
The approximation of the Dependency Graph [15,17,22] 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:

CONCAT(cons(u, v), y) → CONCAT(v, y)

The TRS R consists of the following rules:

minus(x, 0) → x
minus(s(x), s(y)) → minus(x, y)
quot(0, s(y)) → 0
quot(s(x), s(y)) → s(quot(minus(x, y), s(y)))
app(nil, y) → y
app(add(n, x), y) → add(n, app(x, y))
reverse(nil) → nil
reverse(add(n, x)) → app(reverse(x), add(n, nil))
shuffle(nil) → nil
shuffle(add(n, x)) → add(n, shuffle(reverse(x)))
concat(leaf, y) → y
concat(cons(u, v), y) → cons(u, concat(v, y))
less_leaves(x, leaf) → false
less_leaves(leaf, cons(w, z)) → true
less_leaves(cons(u, v), cons(w, z)) → less_leaves(concat(u, v), concat(w, z))

Q is empty.
We have to consider all minimal (P,Q,R)-chains.
We use the reduction pair processor [15].

The following pairs can be oriented strictly and are deleted.

CONCAT(cons(u, v), y) → CONCAT(v, y)

The remaining pairs can at least be oriented weakly.
none
Used ordering: Polynomial interpretation [25,35]:

POL(cons(x₁, x₂)) = 1 + (4)x_2
POL(CONCAT(x₁, x₂)) = (4)x_1

The value of delta used in the strict ordering is 4.
The following usable rules [17] 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:

minus(x, 0) → x
minus(s(x), s(y)) → minus(x, y)
quot(0, s(y)) → 0
quot(s(x), s(y)) → s(quot(minus(x, y), s(y)))
app(nil, y) → y
app(add(n, x), y) → add(n, app(x, y))
reverse(nil) → nil
reverse(add(n, x)) → app(reverse(x), add(n, nil))
shuffle(nil) → nil
shuffle(add(n, x)) → add(n, shuffle(reverse(x)))
concat(leaf, y) → y
concat(cons(u, v), y) → cons(u, concat(v, y))
less_leaves(x, leaf) → false
less_leaves(leaf, cons(w, z)) → true
less_leaves(cons(u, v), cons(w, z)) → less_leaves(concat(u, v), concat(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_LEAVES(cons(u, v), cons(w, z)) → LESS_LEAVES(concat(u, v), concat(w, z))

The TRS R consists of the following rules:

minus(x, 0) → x
minus(s(x), s(y)) → minus(x, y)
quot(0, s(y)) → 0
quot(s(x), s(y)) → s(quot(minus(x, y), s(y)))
app(nil, y) → y
app(add(n, x), y) → add(n, app(x, y))
reverse(nil) → nil
reverse(add(n, x)) → app(reverse(x), add(n, nil))
shuffle(nil) → nil
shuffle(add(n, x)) → add(n, shuffle(reverse(x)))
concat(leaf, y) → y
concat(cons(u, v), y) → cons(u, concat(v, y))
less_leaves(x, leaf) → false
less_leaves(leaf, cons(w, z)) → true
less_leaves(cons(u, v), cons(w, z)) → less_leaves(concat(u, v), concat(w, z))

Q is empty.
We have to consider all minimal (P,Q,R)-chains.
We use the reduction pair processor [15].

The following pairs can be oriented strictly and are deleted.

LESS_LEAVES(cons(u, v), cons(w, z)) → LESS_LEAVES(concat(u, v), concat(w, z))

The remaining pairs can at least be oriented weakly.
none
Used ordering: Polynomial interpretation [25,35]:

POL(LESS_LEAVES(x₁, x₂)) = (2)x_1
POL(cons(x₁, x₂)) = 1 + x_1 + x_2
POL(concat(x₁, x₂)) = x_1 + x_2
POL(leaf) = 0

The value of delta used in the strict ordering is 2.
The following usable rules [17] were oriented:

concat(leaf, y) → y
concat(cons(u, v), y) → cons(u, concat(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:

minus(x, 0) → x
minus(s(x), s(y)) → minus(x, y)
quot(0, s(y)) → 0
quot(s(x), s(y)) → s(quot(minus(x, y), s(y)))
app(nil, y) → y
app(add(n, x), y) → add(n, app(x, y))
reverse(nil) → nil
reverse(add(n, x)) → app(reverse(x), add(n, nil))
shuffle(nil) → nil
shuffle(add(n, x)) → add(n, shuffle(reverse(x)))
concat(leaf, y) → y
concat(cons(u, v), y) → cons(u, concat(v, y))
less_leaves(x, leaf) → false
less_leaves(leaf, cons(w, z)) → true
less_leaves(cons(u, v), cons(w, z)) → less_leaves(concat(u, v), concat(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:

APP(add(n, x), y) → APP(x, y)

The TRS R consists of the following rules:

minus(x, 0) → x
minus(s(x), s(y)) → minus(x, y)
quot(0, s(y)) → 0
quot(s(x), s(y)) → s(quot(minus(x, y), s(y)))
app(nil, y) → y
app(add(n, x), y) → add(n, app(x, y))
reverse(nil) → nil
reverse(add(n, x)) → app(reverse(x), add(n, nil))
shuffle(nil) → nil
shuffle(add(n, x)) → add(n, shuffle(reverse(x)))
concat(leaf, y) → y
concat(cons(u, v), y) → cons(u, concat(v, y))
less_leaves(x, leaf) → false
less_leaves(leaf, cons(w, z)) → true
less_leaves(cons(u, v), cons(w, z)) → less_leaves(concat(u, v), concat(w, z))

Q is empty.
We have to consider all minimal (P,Q,R)-chains.
We use the reduction pair processor [15].

The following pairs can be oriented strictly and are deleted.

APP(add(n, x), y) → APP(x, y)

The remaining pairs can at least be oriented weakly.
none
Used ordering: Polynomial interpretation [25,35]:

POL(APP(x₁, x₂)) = (4)x_1
POL(add(x₁, x₂)) = 1 + (4)x_2

The value of delta used in the strict ordering is 4.
The following usable rules [17] 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:

minus(x, 0) → x
minus(s(x), s(y)) → minus(x, y)
quot(0, s(y)) → 0
quot(s(x), s(y)) → s(quot(minus(x, y), s(y)))
app(nil, y) → y
app(add(n, x), y) → add(n, app(x, y))
reverse(nil) → nil
reverse(add(n, x)) → app(reverse(x), add(n, nil))
shuffle(nil) → nil
shuffle(add(n, x)) → add(n, shuffle(reverse(x)))
concat(leaf, y) → y
concat(cons(u, v), y) → cons(u, concat(v, y))
less_leaves(x, leaf) → false
less_leaves(leaf, cons(w, z)) → true
less_leaves(cons(u, v), cons(w, z)) → less_leaves(concat(u, v), concat(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:

REVERSE(add(n, x)) → REVERSE(x)

The TRS R consists of the following rules:

minus(x, 0) → x
minus(s(x), s(y)) → minus(x, y)
quot(0, s(y)) → 0
quot(s(x), s(y)) → s(quot(minus(x, y), s(y)))
app(nil, y) → y
app(add(n, x), y) → add(n, app(x, y))
reverse(nil) → nil
reverse(add(n, x)) → app(reverse(x), add(n, nil))
shuffle(nil) → nil
shuffle(add(n, x)) → add(n, shuffle(reverse(x)))
concat(leaf, y) → y
concat(cons(u, v), y) → cons(u, concat(v, y))
less_leaves(x, leaf) → false
less_leaves(leaf, cons(w, z)) → true
less_leaves(cons(u, v), cons(w, z)) → less_leaves(concat(u, v), concat(w, z))

Q is empty.
We have to consider all minimal (P,Q,R)-chains.
We use the reduction pair processor [15].

The following pairs can be oriented strictly and are deleted.

REVERSE(add(n, x)) → REVERSE(x)

The remaining pairs can at least be oriented weakly.
none
Used ordering: Polynomial interpretation [25,35]:

POL(add(x₁, x₂)) = 1 + (4)x_2
POL(REVERSE(x₁)) = (4)x_1

The value of delta used in the strict ordering is 4.
The following usable rules [17] 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:

minus(x, 0) → x
minus(s(x), s(y)) → minus(x, y)
quot(0, s(y)) → 0
quot(s(x), s(y)) → s(quot(minus(x, y), s(y)))
app(nil, y) → y
app(add(n, x), y) → add(n, app(x, y))
reverse(nil) → nil
reverse(add(n, x)) → app(reverse(x), add(n, nil))
shuffle(nil) → nil
shuffle(add(n, x)) → add(n, shuffle(reverse(x)))
concat(leaf, y) → y
concat(cons(u, v), y) → cons(u, concat(v, y))
less_leaves(x, leaf) → false
less_leaves(leaf, cons(w, z)) → true
less_leaves(cons(u, v), cons(w, z)) → less_leaves(concat(u, v), concat(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:

SHUFFLE(add(n, x)) → SHUFFLE(reverse(x))

The TRS R consists of the following rules:

minus(x, 0) → x
minus(s(x), s(y)) → minus(x, y)
quot(0, s(y)) → 0
quot(s(x), s(y)) → s(quot(minus(x, y), s(y)))
app(nil, y) → y
app(add(n, x), y) → add(n, app(x, y))
reverse(nil) → nil
reverse(add(n, x)) → app(reverse(x), add(n, nil))
shuffle(nil) → nil
shuffle(add(n, x)) → add(n, shuffle(reverse(x)))
concat(leaf, y) → y
concat(cons(u, v), y) → cons(u, concat(v, y))
less_leaves(x, leaf) → false
less_leaves(leaf, cons(w, z)) → true
less_leaves(cons(u, v), cons(w, z)) → less_leaves(concat(u, v), concat(w, z))

Q is empty.
We have to consider all minimal (P,Q,R)-chains.
We use the reduction pair processor [15].

The following pairs can be oriented strictly and are deleted.

SHUFFLE(add(n, x)) → SHUFFLE(reverse(x))

The remaining pairs can at least be oriented weakly.
none
Used ordering: Polynomial interpretation [25,35]:

POL(SHUFFLE(x₁)) = (2)x_1
POL(add(x₁, x₂)) = 4 + (3)x_1 + x_2
POL(reverse(x₁)) = x_1
POL(app(x₁, x₂)) = x_1 + x_2
POL(nil) = 0

The value of delta used in the strict ordering is 8.
The following usable rules [17] were oriented:

app(nil, y) → y
reverse(nil) → nil
app(add(n, x), y) → add(n, app(x, y))
reverse(add(n, x)) → app(reverse(x), add(n, nil))

↳ 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:

minus(x, 0) → x
minus(s(x), s(y)) → minus(x, y)
quot(0, s(y)) → 0
quot(s(x), s(y)) → s(quot(minus(x, y), s(y)))
app(nil, y) → y
app(add(n, x), y) → add(n, app(x, y))
reverse(nil) → nil
reverse(add(n, x)) → app(reverse(x), add(n, nil))
shuffle(nil) → nil
shuffle(add(n, x)) → add(n, shuffle(reverse(x)))
concat(leaf, y) → y
concat(cons(u, v), y) → cons(u, concat(v, y))
less_leaves(x, leaf) → false
less_leaves(leaf, cons(w, z)) → true
less_leaves(cons(u, v), cons(w, z)) → less_leaves(concat(u, v), concat(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:

MINUS(s(x), s(y)) → MINUS(x, y)

The TRS R consists of the following rules:

minus(x, 0) → x
minus(s(x), s(y)) → minus(x, y)
quot(0, s(y)) → 0
quot(s(x), s(y)) → s(quot(minus(x, y), s(y)))
app(nil, y) → y
app(add(n, x), y) → add(n, app(x, y))
reverse(nil) → nil
reverse(add(n, x)) → app(reverse(x), add(n, nil))
shuffle(nil) → nil
shuffle(add(n, x)) → add(n, shuffle(reverse(x)))
concat(leaf, y) → y
concat(cons(u, v), y) → cons(u, concat(v, y))
less_leaves(x, leaf) → false
less_leaves(leaf, cons(w, z)) → true
less_leaves(cons(u, v), cons(w, z)) → less_leaves(concat(u, v), concat(w, z))

Q is empty.
We have to consider all minimal (P,Q,R)-chains.
We use the reduction pair processor [15].

The following pairs can be oriented strictly and are deleted.

MINUS(s(x), s(y)) → MINUS(x, y)

The remaining pairs can at least be oriented weakly.
none
Used ordering: Polynomial interpretation [25,35]:

POL(MINUS(x₁, x₂)) = (3)x_2
POL(s(x₁)) = 4 + (2)x_1

The value of delta used in the strict ordering is 12.
The following usable rules [17] 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:

minus(x, 0) → x
minus(s(x), s(y)) → minus(x, y)
quot(0, s(y)) → 0
quot(s(x), s(y)) → s(quot(minus(x, y), s(y)))
app(nil, y) → y
app(add(n, x), y) → add(n, app(x, y))
reverse(nil) → nil
reverse(add(n, x)) → app(reverse(x), add(n, nil))
shuffle(nil) → nil
shuffle(add(n, x)) → add(n, shuffle(reverse(x)))
concat(leaf, y) → y
concat(cons(u, v), y) → cons(u, concat(v, y))
less_leaves(x, leaf) → false
less_leaves(leaf, cons(w, z)) → true
less_leaves(cons(u, v), cons(w, z)) → less_leaves(concat(u, v), concat(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:

QUOT(s(x), s(y)) → QUOT(minus(x, y), s(y))

The TRS R consists of the following rules:

minus(x, 0) → x
minus(s(x), s(y)) → minus(x, y)
quot(0, s(y)) → 0
quot(s(x), s(y)) → s(quot(minus(x, y), s(y)))
app(nil, y) → y
app(add(n, x), y) → add(n, app(x, y))
reverse(nil) → nil
reverse(add(n, x)) → app(reverse(x), add(n, nil))
shuffle(nil) → nil
shuffle(add(n, x)) → add(n, shuffle(reverse(x)))
concat(leaf, y) → y
concat(cons(u, v), y) → cons(u, concat(v, y))
less_leaves(x, leaf) → false
less_leaves(leaf, cons(w, z)) → true
less_leaves(cons(u, v), cons(w, z)) → less_leaves(concat(u, v), concat(w, z))

Q is empty.
We have to consider all minimal (P,Q,R)-chains.
We use the reduction pair processor [15].

The following pairs can be oriented strictly and are deleted.

QUOT(s(x), s(y)) → QUOT(minus(x, y), s(y))

The remaining pairs can at least be oriented weakly.
none
Used ordering: Polynomial interpretation [25,35]:

POL(minus(x₁, x₂)) = 3 + x_1
POL(QUOT(x₁, x₂)) = (2)x_1
POL(s(x₁)) = 4 + (4)x_1
POL(0) = 0

The value of delta used in the strict ordering is 2.
The following usable rules [17] were oriented:

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

↳ 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:

minus(x, 0) → x
minus(s(x), s(y)) → minus(x, y)
quot(0, s(y)) → 0
quot(s(x), s(y)) → s(quot(minus(x, y), s(y)))
app(nil, y) → y
app(add(n, x), y) → add(n, app(x, y))
reverse(nil) → nil
reverse(add(n, x)) → app(reverse(x), add(n, nil))
shuffle(nil) → nil
shuffle(add(n, x)) → add(n, shuffle(reverse(x)))
concat(leaf, y) → y
concat(cons(u, v), y) → cons(u, concat(v, y))
less_leaves(x, leaf) → false
less_leaves(leaf, cons(w, z)) → true
less_leaves(cons(u, v), cons(w, z)) → less_leaves(concat(u, v), concat(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.