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:

flatten1(nil) -> nil
flatten1(unit1(x)) -> flatten1(x)
flatten1(++2(x, y)) -> ++2(flatten1(x), flatten1(y))
flatten1(++2(unit1(x), y)) -> ++2(flatten1(x), flatten1(y))
flatten1(flatten1(x)) -> flatten1(x)
rev1(nil) -> nil
rev1(unit1(x)) -> unit1(x)
rev1(++2(x, y)) -> ++2(rev1(y), rev1(x))
rev1(rev1(x)) -> x
++2(x, nil) -> x
++2(nil, y) -> y
++2(++2(x, y), z) -> ++2(x, ++2(y, z))

Q is empty.


↳ QTRS
  ↳ DependencyPairsProof

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

flatten1(nil) -> nil
flatten1(unit1(x)) -> flatten1(x)
flatten1(++2(x, y)) -> ++2(flatten1(x), flatten1(y))
flatten1(++2(unit1(x), y)) -> ++2(flatten1(x), flatten1(y))
flatten1(flatten1(x)) -> flatten1(x)
rev1(nil) -> nil
rev1(unit1(x)) -> unit1(x)
rev1(++2(x, y)) -> ++2(rev1(y), rev1(x))
rev1(rev1(x)) -> x
++2(x, nil) -> x
++2(nil, y) -> y
++2(++2(x, y), z) -> ++2(x, ++2(y, z))

Q is empty.

Q DP problem:
The TRS P consists of the following rules:

FLATTEN1(++2(x, y)) -> FLATTEN1(x)
REV1(++2(x, y)) -> REV1(x)
REV1(++2(x, y)) -> REV1(y)
FLATTEN1(++2(x, y)) -> FLATTEN1(y)
FLATTEN1(++2(unit1(x), y)) -> FLATTEN1(y)
FLATTEN1(++2(unit1(x), y)) -> FLATTEN1(x)
++12(++2(x, y), z) -> ++12(x, ++2(y, z))
FLATTEN1(++2(x, y)) -> ++12(flatten1(x), flatten1(y))
REV1(++2(x, y)) -> ++12(rev1(y), rev1(x))
FLATTEN1(++2(unit1(x), y)) -> ++12(flatten1(x), flatten1(y))
FLATTEN1(unit1(x)) -> FLATTEN1(x)
++12(++2(x, y), z) -> ++12(y, z)

The TRS R consists of the following rules:

flatten1(nil) -> nil
flatten1(unit1(x)) -> flatten1(x)
flatten1(++2(x, y)) -> ++2(flatten1(x), flatten1(y))
flatten1(++2(unit1(x), y)) -> ++2(flatten1(x), flatten1(y))
flatten1(flatten1(x)) -> flatten1(x)
rev1(nil) -> nil
rev1(unit1(x)) -> unit1(x)
rev1(++2(x, y)) -> ++2(rev1(y), rev1(x))
rev1(rev1(x)) -> x
++2(x, nil) -> x
++2(nil, y) -> y
++2(++2(x, y), z) -> ++2(x, ++2(y, 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:

FLATTEN1(++2(x, y)) -> FLATTEN1(x)
REV1(++2(x, y)) -> REV1(x)
REV1(++2(x, y)) -> REV1(y)
FLATTEN1(++2(x, y)) -> FLATTEN1(y)
FLATTEN1(++2(unit1(x), y)) -> FLATTEN1(y)
FLATTEN1(++2(unit1(x), y)) -> FLATTEN1(x)
++12(++2(x, y), z) -> ++12(x, ++2(y, z))
FLATTEN1(++2(x, y)) -> ++12(flatten1(x), flatten1(y))
REV1(++2(x, y)) -> ++12(rev1(y), rev1(x))
FLATTEN1(++2(unit1(x), y)) -> ++12(flatten1(x), flatten1(y))
FLATTEN1(unit1(x)) -> FLATTEN1(x)
++12(++2(x, y), z) -> ++12(y, z)

The TRS R consists of the following rules:

flatten1(nil) -> nil
flatten1(unit1(x)) -> flatten1(x)
flatten1(++2(x, y)) -> ++2(flatten1(x), flatten1(y))
flatten1(++2(unit1(x), y)) -> ++2(flatten1(x), flatten1(y))
flatten1(flatten1(x)) -> flatten1(x)
rev1(nil) -> nil
rev1(unit1(x)) -> unit1(x)
rev1(++2(x, y)) -> ++2(rev1(y), rev1(x))
rev1(rev1(x)) -> x
++2(x, nil) -> x
++2(nil, y) -> y
++2(++2(x, y), z) -> ++2(x, ++2(y, z))

Q is empty.
We have to consider all minimal (P,Q,R)-chains.
The approximation of the Dependency Graph contains 3 SCCs with 3 less nodes.

↳ QTRS
  ↳ DependencyPairsProof
    ↳ QDP
      ↳ DependencyGraphProof
        ↳ AND
          ↳ QDP
            ↳ QDPAfsSolverProof
          ↳ QDP
          ↳ QDP

Q DP problem:
The TRS P consists of the following rules:

++12(++2(x, y), z) -> ++12(x, ++2(y, z))
++12(++2(x, y), z) -> ++12(y, z)

The TRS R consists of the following rules:

flatten1(nil) -> nil
flatten1(unit1(x)) -> flatten1(x)
flatten1(++2(x, y)) -> ++2(flatten1(x), flatten1(y))
flatten1(++2(unit1(x), y)) -> ++2(flatten1(x), flatten1(y))
flatten1(flatten1(x)) -> flatten1(x)
rev1(nil) -> nil
rev1(unit1(x)) -> unit1(x)
rev1(++2(x, y)) -> ++2(rev1(y), rev1(x))
rev1(rev1(x)) -> x
++2(x, nil) -> x
++2(nil, y) -> y
++2(++2(x, y), z) -> ++2(x, ++2(y, z))

Q is empty.
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.

++12(++2(x, y), z) -> ++12(x, ++2(y, z))
++12(++2(x, y), z) -> ++12(y, z)
Used argument filtering: ++12(x1, x2)  =  x1
++2(x1, x2)  =  ++2(x1, x2)
nil  =  nil
Used ordering: Precedence:
trivial



↳ QTRS
  ↳ DependencyPairsProof
    ↳ QDP
      ↳ DependencyGraphProof
        ↳ AND
          ↳ QDP
            ↳ QDPAfsSolverProof
              ↳ QDP
                ↳ PisEmptyProof
          ↳ QDP
          ↳ QDP

Q DP problem:
P is empty.
The TRS R consists of the following rules:

flatten1(nil) -> nil
flatten1(unit1(x)) -> flatten1(x)
flatten1(++2(x, y)) -> ++2(flatten1(x), flatten1(y))
flatten1(++2(unit1(x), y)) -> ++2(flatten1(x), flatten1(y))
flatten1(flatten1(x)) -> flatten1(x)
rev1(nil) -> nil
rev1(unit1(x)) -> unit1(x)
rev1(++2(x, y)) -> ++2(rev1(y), rev1(x))
rev1(rev1(x)) -> x
++2(x, nil) -> x
++2(nil, y) -> y
++2(++2(x, y), z) -> ++2(x, ++2(y, 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
            ↳ QDPAfsSolverProof
          ↳ QDP

Q DP problem:
The TRS P consists of the following rules:

REV1(++2(x, y)) -> REV1(y)
REV1(++2(x, y)) -> REV1(x)

The TRS R consists of the following rules:

flatten1(nil) -> nil
flatten1(unit1(x)) -> flatten1(x)
flatten1(++2(x, y)) -> ++2(flatten1(x), flatten1(y))
flatten1(++2(unit1(x), y)) -> ++2(flatten1(x), flatten1(y))
flatten1(flatten1(x)) -> flatten1(x)
rev1(nil) -> nil
rev1(unit1(x)) -> unit1(x)
rev1(++2(x, y)) -> ++2(rev1(y), rev1(x))
rev1(rev1(x)) -> x
++2(x, nil) -> x
++2(nil, y) -> y
++2(++2(x, y), z) -> ++2(x, ++2(y, z))

Q is empty.
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.

REV1(++2(x, y)) -> REV1(y)
REV1(++2(x, y)) -> REV1(x)
Used argument filtering: REV1(x1)  =  x1
++2(x1, x2)  =  ++2(x1, x2)
Used ordering: Precedence:
trivial



↳ QTRS
  ↳ DependencyPairsProof
    ↳ QDP
      ↳ DependencyGraphProof
        ↳ AND
          ↳ QDP
          ↳ QDP
            ↳ QDPAfsSolverProof
              ↳ QDP
                ↳ PisEmptyProof
          ↳ QDP

Q DP problem:
P is empty.
The TRS R consists of the following rules:

flatten1(nil) -> nil
flatten1(unit1(x)) -> flatten1(x)
flatten1(++2(x, y)) -> ++2(flatten1(x), flatten1(y))
flatten1(++2(unit1(x), y)) -> ++2(flatten1(x), flatten1(y))
flatten1(flatten1(x)) -> flatten1(x)
rev1(nil) -> nil
rev1(unit1(x)) -> unit1(x)
rev1(++2(x, y)) -> ++2(rev1(y), rev1(x))
rev1(rev1(x)) -> x
++2(x, nil) -> x
++2(nil, y) -> y
++2(++2(x, y), z) -> ++2(x, ++2(y, 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
            ↳ QDPAfsSolverProof

Q DP problem:
The TRS P consists of the following rules:

FLATTEN1(++2(x, y)) -> FLATTEN1(x)
FLATTEN1(++2(x, y)) -> FLATTEN1(y)
FLATTEN1(++2(unit1(x), y)) -> FLATTEN1(y)
FLATTEN1(++2(unit1(x), y)) -> FLATTEN1(x)
FLATTEN1(unit1(x)) -> FLATTEN1(x)

The TRS R consists of the following rules:

flatten1(nil) -> nil
flatten1(unit1(x)) -> flatten1(x)
flatten1(++2(x, y)) -> ++2(flatten1(x), flatten1(y))
flatten1(++2(unit1(x), y)) -> ++2(flatten1(x), flatten1(y))
flatten1(flatten1(x)) -> flatten1(x)
rev1(nil) -> nil
rev1(unit1(x)) -> unit1(x)
rev1(++2(x, y)) -> ++2(rev1(y), rev1(x))
rev1(rev1(x)) -> x
++2(x, nil) -> x
++2(nil, y) -> y
++2(++2(x, y), z) -> ++2(x, ++2(y, z))

Q is empty.
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.

FLATTEN1(++2(x, y)) -> FLATTEN1(x)
FLATTEN1(++2(x, y)) -> FLATTEN1(y)
FLATTEN1(++2(unit1(x), y)) -> FLATTEN1(y)
FLATTEN1(++2(unit1(x), y)) -> FLATTEN1(x)
Used argument filtering: FLATTEN1(x1)  =  x1
++2(x1, x2)  =  ++2(x1, x2)
unit1(x1)  =  x1
Used ordering: Precedence:
trivial



↳ QTRS
  ↳ DependencyPairsProof
    ↳ QDP
      ↳ DependencyGraphProof
        ↳ AND
          ↳ QDP
          ↳ QDP
          ↳ QDP
            ↳ QDPAfsSolverProof
              ↳ QDP
                ↳ QDPAfsSolverProof

Q DP problem:
The TRS P consists of the following rules:

FLATTEN1(unit1(x)) -> FLATTEN1(x)

The TRS R consists of the following rules:

flatten1(nil) -> nil
flatten1(unit1(x)) -> flatten1(x)
flatten1(++2(x, y)) -> ++2(flatten1(x), flatten1(y))
flatten1(++2(unit1(x), y)) -> ++2(flatten1(x), flatten1(y))
flatten1(flatten1(x)) -> flatten1(x)
rev1(nil) -> nil
rev1(unit1(x)) -> unit1(x)
rev1(++2(x, y)) -> ++2(rev1(y), rev1(x))
rev1(rev1(x)) -> x
++2(x, nil) -> x
++2(nil, y) -> y
++2(++2(x, y), z) -> ++2(x, ++2(y, z))

Q is empty.
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.

FLATTEN1(unit1(x)) -> FLATTEN1(x)
Used argument filtering: FLATTEN1(x1)  =  x1
unit1(x1)  =  unit1(x1)
Used ordering: Precedence:
trivial



↳ QTRS
  ↳ DependencyPairsProof
    ↳ QDP
      ↳ DependencyGraphProof
        ↳ AND
          ↳ QDP
          ↳ QDP
          ↳ QDP
            ↳ QDPAfsSolverProof
              ↳ QDP
                ↳ QDPAfsSolverProof
                  ↳ QDP
                    ↳ PisEmptyProof

Q DP problem:
P is empty.
The TRS R consists of the following rules:

flatten1(nil) -> nil
flatten1(unit1(x)) -> flatten1(x)
flatten1(++2(x, y)) -> ++2(flatten1(x), flatten1(y))
flatten1(++2(unit1(x), y)) -> ++2(flatten1(x), flatten1(y))
flatten1(flatten1(x)) -> flatten1(x)
rev1(nil) -> nil
rev1(unit1(x)) -> unit1(x)
rev1(++2(x, y)) -> ++2(rev1(y), rev1(x))
rev1(rev1(x)) -> x
++2(x, nil) -> x
++2(nil, y) -> y
++2(++2(x, y), z) -> ++2(x, ++2(y, 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.