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:

rev1(nil) -> nil
rev1(rev1(x)) -> x
rev1(++2(x, y)) -> ++2(rev1(y), rev1(x))
++2(nil, y) -> y
++2(x, nil) -> x
++2(.2(x, y), z) -> .2(x, ++2(y, z))
++2(x, ++2(y, z)) -> ++2(++2(x, y), z)
make1(x) -> .2(x, nil)

Q is empty.


↳ QTRS
  ↳ DependencyPairsProof

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

rev1(nil) -> nil
rev1(rev1(x)) -> x
rev1(++2(x, y)) -> ++2(rev1(y), rev1(x))
++2(nil, y) -> y
++2(x, nil) -> x
++2(.2(x, y), z) -> .2(x, ++2(y, z))
++2(x, ++2(y, z)) -> ++2(++2(x, y), z)
make1(x) -> .2(x, nil)

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:

REV1(++2(x, y)) -> REV1(x)
REV1(++2(x, y)) -> REV1(y)
++12(x, ++2(y, z)) -> ++12(++2(x, y), z)
REV1(++2(x, y)) -> ++12(rev1(y), rev1(x))
++12(x, ++2(y, z)) -> ++12(x, y)
++12(.2(x, y), z) -> ++12(y, z)

The TRS R consists of the following rules:

rev1(nil) -> nil
rev1(rev1(x)) -> x
rev1(++2(x, y)) -> ++2(rev1(y), rev1(x))
++2(nil, y) -> y
++2(x, nil) -> x
++2(.2(x, y), z) -> .2(x, ++2(y, z))
++2(x, ++2(y, z)) -> ++2(++2(x, y), z)
make1(x) -> .2(x, nil)

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:

REV1(++2(x, y)) -> REV1(x)
REV1(++2(x, y)) -> REV1(y)
++12(x, ++2(y, z)) -> ++12(++2(x, y), z)
REV1(++2(x, y)) -> ++12(rev1(y), rev1(x))
++12(x, ++2(y, z)) -> ++12(x, y)
++12(.2(x, y), z) -> ++12(y, z)

The TRS R consists of the following rules:

rev1(nil) -> nil
rev1(rev1(x)) -> x
rev1(++2(x, y)) -> ++2(rev1(y), rev1(x))
++2(nil, y) -> y
++2(x, nil) -> x
++2(.2(x, y), z) -> .2(x, ++2(y, z))
++2(x, ++2(y, z)) -> ++2(++2(x, y), z)
make1(x) -> .2(x, nil)

Q is empty.
We have to consider all minimal (P,Q,R)-chains.
The approximation of the Dependency Graph [13,14,18] contains 2 SCCs with 1 less node.

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

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

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

The TRS R consists of the following rules:

rev1(nil) -> nil
rev1(rev1(x)) -> x
rev1(++2(x, y)) -> ++2(rev1(y), rev1(x))
++2(nil, y) -> y
++2(x, nil) -> x
++2(.2(x, y), z) -> .2(x, ++2(y, z))
++2(x, ++2(y, z)) -> ++2(++2(x, y), z)
make1(x) -> .2(x, nil)

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.


++12(x, ++2(y, z)) -> ++12(++2(x, y), z)
++12(x, ++2(y, z)) -> ++12(x, y)
++12(.2(x, y), z) -> ++12(y, z)
The remaining pairs can at least be oriented weakly.
none
Used ordering: Polynomial interpretation [21]:

POL(++2(x1, x2)) = 2 + x1 + 3·x2   
POL(++12(x1, x2)) = x1 + 3·x2   
POL(.2(x1, x2)) = 2 + 3·x1 + x2   
POL(nil) = 0   

The following usable rules [14] were oriented:

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



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

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

rev1(nil) -> nil
rev1(rev1(x)) -> x
rev1(++2(x, y)) -> ++2(rev1(y), rev1(x))
++2(nil, y) -> y
++2(x, nil) -> x
++2(.2(x, y), z) -> .2(x, ++2(y, z))
++2(x, ++2(y, z)) -> ++2(++2(x, y), z)
make1(x) -> .2(x, nil)

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

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:

rev1(nil) -> nil
rev1(rev1(x)) -> x
rev1(++2(x, y)) -> ++2(rev1(y), rev1(x))
++2(nil, y) -> y
++2(x, nil) -> x
++2(.2(x, y), z) -> .2(x, ++2(y, z))
++2(x, ++2(y, z)) -> ++2(++2(x, y), z)
make1(x) -> .2(x, nil)

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.


REV1(++2(x, y)) -> REV1(y)
REV1(++2(x, y)) -> REV1(x)
The remaining pairs can at least be oriented weakly.
none
Used ordering: Polynomial interpretation [21]:

POL(++2(x1, x2)) = 1 + 2·x1 + x2   
POL(REV1(x1)) = 3·x1   

The following usable rules [14] were oriented: none



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

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

rev1(nil) -> nil
rev1(rev1(x)) -> x
rev1(++2(x, y)) -> ++2(rev1(y), rev1(x))
++2(nil, y) -> y
++2(x, nil) -> x
++2(.2(x, y), z) -> .2(x, ++2(y, z))
++2(x, ++2(y, z)) -> ++2(++2(x, y), z)
make1(x) -> .2(x, nil)

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.