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:

flatten(nil) → nil
flatten(unit(x)) → flatten(x)
flatten(++(x, y)) → ++(flatten(x), flatten(y))
flatten(++(unit(x), y)) → ++(flatten(x), flatten(y))
flatten(flatten(x)) → flatten(x)
rev(nil) → nil
rev(unit(x)) → unit(x)
rev(++(x, y)) → ++(rev(y), rev(x))
rev(rev(x)) → x
++(x, nil) → x
++(nil, y) → y
++(++(x, y), z) → ++(x, ++(y, z))

Q is empty.


↳ QTRS
  ↳ DependencyPairsProof

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

flatten(nil) → nil
flatten(unit(x)) → flatten(x)
flatten(++(x, y)) → ++(flatten(x), flatten(y))
flatten(++(unit(x), y)) → ++(flatten(x), flatten(y))
flatten(flatten(x)) → flatten(x)
rev(nil) → nil
rev(unit(x)) → unit(x)
rev(++(x, y)) → ++(rev(y), rev(x))
rev(rev(x)) → x
++(x, nil) → x
++(nil, y) → y
++(++(x, y), z) → ++(x, ++(y, 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:

++1(++(x, y), z) → ++1(y, z)
FLATTEN(++(unit(x), y)) → FLATTEN(y)
++1(++(x, y), z) → ++1(x, ++(y, z))
REV(++(x, y)) → ++1(rev(y), rev(x))
FLATTEN(++(unit(x), y)) → FLATTEN(x)
FLATTEN(++(x, y)) → FLATTEN(y)
FLATTEN(++(x, y)) → FLATTEN(x)
FLATTEN(unit(x)) → FLATTEN(x)
REV(++(x, y)) → REV(x)
REV(++(x, y)) → REV(y)
FLATTEN(++(x, y)) → ++1(flatten(x), flatten(y))
FLATTEN(++(unit(x), y)) → ++1(flatten(x), flatten(y))

The TRS R consists of the following rules:

flatten(nil) → nil
flatten(unit(x)) → flatten(x)
flatten(++(x, y)) → ++(flatten(x), flatten(y))
flatten(++(unit(x), y)) → ++(flatten(x), flatten(y))
flatten(flatten(x)) → flatten(x)
rev(nil) → nil
rev(unit(x)) → unit(x)
rev(++(x, y)) → ++(rev(y), rev(x))
rev(rev(x)) → x
++(x, nil) → x
++(nil, y) → y
++(++(x, y), z) → ++(x, ++(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:

++1(++(x, y), z) → ++1(y, z)
FLATTEN(++(unit(x), y)) → FLATTEN(y)
++1(++(x, y), z) → ++1(x, ++(y, z))
REV(++(x, y)) → ++1(rev(y), rev(x))
FLATTEN(++(unit(x), y)) → FLATTEN(x)
FLATTEN(++(x, y)) → FLATTEN(y)
FLATTEN(++(x, y)) → FLATTEN(x)
FLATTEN(unit(x)) → FLATTEN(x)
REV(++(x, y)) → REV(x)
REV(++(x, y)) → REV(y)
FLATTEN(++(x, y)) → ++1(flatten(x), flatten(y))
FLATTEN(++(unit(x), y)) → ++1(flatten(x), flatten(y))

The TRS R consists of the following rules:

flatten(nil) → nil
flatten(unit(x)) → flatten(x)
flatten(++(x, y)) → ++(flatten(x), flatten(y))
flatten(++(unit(x), y)) → ++(flatten(x), flatten(y))
flatten(flatten(x)) → flatten(x)
rev(nil) → nil
rev(unit(x)) → unit(x)
rev(++(x, y)) → ++(rev(y), rev(x))
rev(rev(x)) → x
++(x, nil) → x
++(nil, y) → y
++(++(x, y), z) → ++(x, ++(y, z))

Q is empty.
We have to consider all minimal (P,Q,R)-chains.
The approximation of the Dependency Graph [15,17,22] contains 3 SCCs with 3 less nodes.

↳ QTRS
  ↳ DependencyPairsProof
    ↳ QDP
      ↳ DependencyGraphProof
        ↳ AND
          ↳ QDP
            ↳ QDPOrderProof
          ↳ QDP
          ↳ QDP

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

++1(++(x, y), z) → ++1(y, z)
++1(++(x, y), z) → ++1(x, ++(y, z))

The TRS R consists of the following rules:

flatten(nil) → nil
flatten(unit(x)) → flatten(x)
flatten(++(x, y)) → ++(flatten(x), flatten(y))
flatten(++(unit(x), y)) → ++(flatten(x), flatten(y))
flatten(flatten(x)) → flatten(x)
rev(nil) → nil
rev(unit(x)) → unit(x)
rev(++(x, y)) → ++(rev(y), rev(x))
rev(rev(x)) → x
++(x, nil) → x
++(nil, y) → y
++(++(x, y), z) → ++(x, ++(y, 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.


++1(++(x, y), z) → ++1(y, z)
++1(++(x, y), z) → ++1(x, ++(y, z))
The remaining pairs can at least be oriented weakly.
none
Used ordering: Polynomial interpretation [25,35]:

POL(++1(x1, x2)) = x_1 + (1/4)x_2   
POL(++(x1, x2)) = 1 + (2)x_1 + x_2   
POL(nil) = 0   
The value of delta used in the strict ordering is 3/4.
The following usable rules [17] were oriented:

++(nil, y) → y
++(x, nil) → x
++(++(x, y), z) → ++(x, ++(y, z))



↳ QTRS
  ↳ DependencyPairsProof
    ↳ QDP
      ↳ DependencyGraphProof
        ↳ AND
          ↳ QDP
            ↳ QDPOrderProof
              ↳ QDP
                ↳ PisEmptyProof
          ↳ QDP
          ↳ QDP

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

flatten(nil) → nil
flatten(unit(x)) → flatten(x)
flatten(++(x, y)) → ++(flatten(x), flatten(y))
flatten(++(unit(x), y)) → ++(flatten(x), flatten(y))
flatten(flatten(x)) → flatten(x)
rev(nil) → nil
rev(unit(x)) → unit(x)
rev(++(x, y)) → ++(rev(y), rev(x))
rev(rev(x)) → x
++(x, nil) → x
++(nil, y) → y
++(++(x, y), z) → ++(x, ++(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
            ↳ QDPOrderProof
          ↳ QDP

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

REV(++(x, y)) → REV(y)
REV(++(x, y)) → REV(x)

The TRS R consists of the following rules:

flatten(nil) → nil
flatten(unit(x)) → flatten(x)
flatten(++(x, y)) → ++(flatten(x), flatten(y))
flatten(++(unit(x), y)) → ++(flatten(x), flatten(y))
flatten(flatten(x)) → flatten(x)
rev(nil) → nil
rev(unit(x)) → unit(x)
rev(++(x, y)) → ++(rev(y), rev(x))
rev(rev(x)) → x
++(x, nil) → x
++(nil, y) → y
++(++(x, y), z) → ++(x, ++(y, 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.


REV(++(x, y)) → REV(y)
REV(++(x, y)) → REV(x)
The remaining pairs can at least be oriented weakly.
none
Used ordering: Polynomial interpretation [25,35]:

POL(++(x1, x2)) = 1/2 + (4)x_1 + (4)x_2   
POL(REV(x1)) = (1/4)x_1   
The value of delta used in the strict ordering is 1/8.
The following usable rules [17] were oriented: none



↳ QTRS
  ↳ DependencyPairsProof
    ↳ QDP
      ↳ DependencyGraphProof
        ↳ AND
          ↳ QDP
          ↳ QDP
            ↳ QDPOrderProof
              ↳ QDP
                ↳ PisEmptyProof
          ↳ QDP

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

flatten(nil) → nil
flatten(unit(x)) → flatten(x)
flatten(++(x, y)) → ++(flatten(x), flatten(y))
flatten(++(unit(x), y)) → ++(flatten(x), flatten(y))
flatten(flatten(x)) → flatten(x)
rev(nil) → nil
rev(unit(x)) → unit(x)
rev(++(x, y)) → ++(rev(y), rev(x))
rev(rev(x)) → x
++(x, nil) → x
++(nil, y) → y
++(++(x, y), z) → ++(x, ++(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
            ↳ QDPOrderProof

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

FLATTEN(++(unit(x), y)) → FLATTEN(y)
FLATTEN(++(unit(x), y)) → FLATTEN(x)
FLATTEN(++(x, y)) → FLATTEN(y)
FLATTEN(unit(x)) → FLATTEN(x)
FLATTEN(++(x, y)) → FLATTEN(x)

The TRS R consists of the following rules:

flatten(nil) → nil
flatten(unit(x)) → flatten(x)
flatten(++(x, y)) → ++(flatten(x), flatten(y))
flatten(++(unit(x), y)) → ++(flatten(x), flatten(y))
flatten(flatten(x)) → flatten(x)
rev(nil) → nil
rev(unit(x)) → unit(x)
rev(++(x, y)) → ++(rev(y), rev(x))
rev(rev(x)) → x
++(x, nil) → x
++(nil, y) → y
++(++(x, y), z) → ++(x, ++(y, 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.


FLATTEN(++(unit(x), y)) → FLATTEN(y)
FLATTEN(++(unit(x), y)) → FLATTEN(x)
FLATTEN(unit(x)) → FLATTEN(x)
The remaining pairs can at least be oriented weakly.

FLATTEN(++(x, y)) → FLATTEN(y)
FLATTEN(++(x, y)) → FLATTEN(x)
Used ordering: Polynomial interpretation [25,35]:

POL(++(x1, x2)) = (4)x_1 + (4)x_2   
POL(FLATTEN(x1)) = (1/4)x_1   
POL(unit(x1)) = 1/4 + (4)x_1   
The value of delta used in the strict ordering is 1/16.
The following usable rules [17] were oriented: none



↳ QTRS
  ↳ DependencyPairsProof
    ↳ QDP
      ↳ DependencyGraphProof
        ↳ AND
          ↳ QDP
          ↳ QDP
          ↳ QDP
            ↳ QDPOrderProof
              ↳ QDP
                ↳ QDPOrderProof

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

FLATTEN(++(x, y)) → FLATTEN(y)
FLATTEN(++(x, y)) → FLATTEN(x)

The TRS R consists of the following rules:

flatten(nil) → nil
flatten(unit(x)) → flatten(x)
flatten(++(x, y)) → ++(flatten(x), flatten(y))
flatten(++(unit(x), y)) → ++(flatten(x), flatten(y))
flatten(flatten(x)) → flatten(x)
rev(nil) → nil
rev(unit(x)) → unit(x)
rev(++(x, y)) → ++(rev(y), rev(x))
rev(rev(x)) → x
++(x, nil) → x
++(nil, y) → y
++(++(x, y), z) → ++(x, ++(y, 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.


FLATTEN(++(x, y)) → FLATTEN(y)
FLATTEN(++(x, y)) → FLATTEN(x)
The remaining pairs can at least be oriented weakly.
none
Used ordering: Polynomial interpretation [25,35]:

POL(++(x1, x2)) = 1/4 + (2)x_1 + (2)x_2   
POL(FLATTEN(x1)) = (1/4)x_1   
The value of delta used in the strict ordering is 1/16.
The following usable rules [17] were oriented: none



↳ QTRS
  ↳ DependencyPairsProof
    ↳ QDP
      ↳ DependencyGraphProof
        ↳ AND
          ↳ QDP
          ↳ QDP
          ↳ QDP
            ↳ QDPOrderProof
              ↳ QDP
                ↳ QDPOrderProof
                  ↳ QDP
                    ↳ PisEmptyProof

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

flatten(nil) → nil
flatten(unit(x)) → flatten(x)
flatten(++(x, y)) → ++(flatten(x), flatten(y))
flatten(++(unit(x), y)) → ++(flatten(x), flatten(y))
flatten(flatten(x)) → flatten(x)
rev(nil) → nil
rev(unit(x)) → unit(x)
rev(++(x, y)) → ++(rev(y), rev(x))
rev(rev(x)) → x
++(x, nil) → x
++(nil, y) → y
++(++(x, y), z) → ++(x, ++(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.