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 Order [17,21] with Interpretation:

POL( CONCAT2(x1, x2) ) = x1 + 3


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



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 Order [17,21] with Interpretation:

POL( concat2(x1, x2) ) = max{0, 2x1 + x2 - 3}


POL( leaf ) = 3


POL( LESS_LEAVES2(x1, x2) ) = max{0, x1 - 2}


POL( cons2(x1, x2) ) = 2x1 + x2 + 3



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 Order [17,21] with Interpretation:

POL( APP2(x1, x2) ) = x1 + 3


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



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 Order [17,21] with Interpretation:

POL( REVERSE1(x1) ) = 2x1 + 2


POL( add2(x1, x2) ) = 3x1 + x2 + 1



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 Order [17,21] with Interpretation:

POL( reverse1(x1) ) = x1


POL( SHUFFLE1(x1) ) = max{0, 2x1 - 3}


POL( nil ) = max{0, -3}


POL( app2(x1, x2) ) = x1 + x2


POL( add2(x1, x2) ) = 3x1 + x2 + 2



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 Order [17,21] with Interpretation:

POL( MINUS2(x1, x2) ) = 3x2 + 3


POL( s1(x1) ) = x1 + 3



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 Order [17,21] with Interpretation:

POL( minus2(x1, x2) ) = x1


POL( 0 ) = 1


POL( s1(x1) ) = 2x1 + 3


POL( QUOT2(x1, x2) ) = max{0, 2x1 + 2x2 - 1}



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.