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 strictly oriented and are deleted.


++12(x, ++2(y, z)) -> ++12(++2(x, y), z)
++12(x, ++2(y, z)) -> ++12(x, y)
The remaining pairs can at least by weakly be oriented.

++12(.2(x, y), z) -> ++12(y, z)
Used ordering: Combined order from the following AFS and order.
++12(x1, x2)  =  ++11(x2)
++2(x1, x2)  =  ++2(x1, x2)
.2(x1, x2)  =  .1(x2)
nil  =  nil

Lexicographic Path Order [19].
Precedence:
++^11 > [++2, .1]
nil > [++2, .1]


The following usable rules [14] were oriented: none



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

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

++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 strictly oriented and are deleted.


++12(.2(x, y), z) -> ++12(y, z)
The remaining pairs can at least by weakly be oriented.
none
Used ordering: Combined order from the following AFS and order.
++12(x1, x2)  =  ++11(x1)
.2(x1, x2)  =  .1(x2)

Lexicographic Path Order [19].
Precedence:
[++^11, .1]


The following usable rules [14] were oriented: none



↳ QTRS
  ↳ DependencyPairsProof
    ↳ QDP
      ↳ DependencyGraphProof
        ↳ AND
          ↳ QDP
            ↳ QDPOrderProof
              ↳ 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 strictly oriented and are deleted.


REV1(++2(x, y)) -> REV1(y)
REV1(++2(x, y)) -> REV1(x)
The remaining pairs can at least by weakly be oriented.
none
Used ordering: Combined order from the following AFS and order.
REV1(x1)  =  REV1(x1)
++2(x1, x2)  =  ++2(x1, x2)

Lexicographic Path Order [19].
Precedence:
++2 > REV1


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.