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))


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 contains 7 SCCs with 5 less nodes.

↳ QTRS
  ↳ Non-Overlap Check
    ↳ QTRS
      ↳ DependencyPairsProof
        ↳ QDP
          ↳ DependencyGraphProof
            ↳ AND
              ↳ QDP
                ↳ QDPAfsSolverProof
              ↳ 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.
By using an argument filtering and a montonic ordering, at least one Dependency Pair of this SCC can be strictly oriented.

CONCAT2(cons2(u, v), y) -> CONCAT2(v, y)
Used argument filtering: CONCAT2(x1, x2)  =  x1
cons2(x1, x2)  =  cons1(x2)
Used ordering: Quasi Precedence: trivial 


↳ QTRS
  ↳ Non-Overlap Check
    ↳ QTRS
      ↳ DependencyPairsProof
        ↳ QDP
          ↳ DependencyGraphProof
            ↳ AND
              ↳ QDP
                ↳ QDPAfsSolverProof
                  ↳ 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
                ↳ QDPAfsSolverProof
              ↳ 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.
By using an argument filtering and a montonic ordering, at least one Dependency Pair of this SCC can be strictly oriented.

APP2(add2(n, x), y) -> APP2(x, y)
Used argument filtering: APP2(x1, x2)  =  x1
add2(x1, x2)  =  add1(x2)
Used ordering: Quasi Precedence: trivial 


↳ QTRS
  ↳ Non-Overlap Check
    ↳ QTRS
      ↳ DependencyPairsProof
        ↳ QDP
          ↳ DependencyGraphProof
            ↳ AND
              ↳ QDP
              ↳ QDP
              ↳ QDP
                ↳ QDPAfsSolverProof
                  ↳ 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
                ↳ QDPAfsSolverProof
              ↳ 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.
By using an argument filtering and a montonic ordering, at least one Dependency Pair of this SCC can be strictly oriented.

REVERSE1(add2(n, x)) -> REVERSE1(x)
Used argument filtering: REVERSE1(x1)  =  x1
add2(x1, x2)  =  add1(x2)
Used ordering: Quasi Precedence: trivial 


↳ QTRS
  ↳ Non-Overlap Check
    ↳ QTRS
      ↳ DependencyPairsProof
        ↳ QDP
          ↳ DependencyGraphProof
            ↳ AND
              ↳ QDP
              ↳ QDP
              ↳ QDP
              ↳ QDP
                ↳ QDPAfsSolverProof
                  ↳ 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
                ↳ QDPAfsSolverProof
              ↳ 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.
By using an argument filtering and a montonic ordering, at least one Dependency Pair of this SCC can be strictly oriented.

MINUS2(s1(x), s1(y)) -> MINUS2(x, y)
Used argument filtering: MINUS2(x1, x2)  =  x2
s1(x1)  =  s1(x1)
Used ordering: Quasi Precedence: trivial 


↳ QTRS
  ↳ Non-Overlap Check
    ↳ QTRS
      ↳ DependencyPairsProof
        ↳ QDP
          ↳ DependencyGraphProof
            ↳ AND
              ↳ QDP
              ↳ QDP
              ↳ QDP
              ↳ QDP
              ↳ QDP
              ↳ QDP
                ↳ QDPAfsSolverProof
                  ↳ 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
                ↳ QDPAfsSolverProof

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.
By using an argument filtering and a montonic ordering, at least one Dependency Pair of this SCC can be strictly oriented.

QUOT2(s1(x), s1(y)) -> QUOT2(minus2(x, y), s1(y))
Used argument filtering: QUOT2(x1, x2)  =  x1
s1(x1)  =  s1(x1)
minus2(x1, x2)  =  x1
Used ordering: Quasi Precedence: trivial 


↳ QTRS
  ↳ Non-Overlap Check
    ↳ QTRS
      ↳ DependencyPairsProof
        ↳ QDP
          ↳ DependencyGraphProof
            ↳ AND
              ↳ QDP
              ↳ QDP
              ↳ QDP
              ↳ QDP
              ↳ QDP
              ↳ QDP
              ↳ QDP
                ↳ QDPAfsSolverProof
                  ↳ 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.