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)) = 2·x1 + x2   
POL(cons2(x1, x2)) = 2 + 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:

reverse1(add2(n, x)) -> app2(reverse1(x), add2(n, nil))
app2(nil, y) -> y
reverse1(nil) -> 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)) = 2·x2   
POL(s1(x1)) = 2 + 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)) = 2 + 2·x1   

The following usable rules [14] were oriented:

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

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.