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:

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

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

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:

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

The TRS R consists of the following rules:

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

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

The TRS R consists of the following rules:

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

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(x, ++2(y, z))
++12(++2(x, y), z) -> ++12(y, z)
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)  =  x1
++2(x1, x2)  =  ++2(x1, x2)
.2(x1, x2)  =  x2
nil  =  nil

Lexicographic Path Order [19].
Precedence:
trivial

The following usable rules [14] were oriented:

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



↳ QTRS
  ↳ DependencyPairsProof
    ↳ QDP
      ↳ QDPOrderProof
        ↳ QDP
          ↳ QDPOrderProof

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:

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

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:
.1 > ++^11

The following usable rules [14] were oriented: none



↳ QTRS
  ↳ DependencyPairsProof
    ↳ QDP
      ↳ QDPOrderProof
        ↳ QDP
          ↳ QDPOrderProof
            ↳ QDP
              ↳ PisEmptyProof

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

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