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 oriented strictly and are deleted.


++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 remaining pairs can at least be oriented weakly.
none
Used ordering: Polynomial Order [17,21] with Interpretation:

POL( ++2(x1, x2) ) = 2x1 + 2x2 + 2


POL( .2(x1, x2) ) = 2x1 + 2x2 + 2


POL( nil ) = 2


POL( ++12(x1, x2) ) = max{0, 2x1 - 1}



The following usable rules [14] were oriented: none



↳ QTRS
  ↳ DependencyPairsProof
    ↳ 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.