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.

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
      ↳ QDPAfsSolverProof

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.
By using an argument filtering and a montonic ordering, at least one Dependency Pair of this SCC can be strictly oriented.

++12(++2(x, y), z) -> ++12(x, ++2(y, z))
++12(++2(x, y), z) -> ++12(y, z)
Used argument filtering: ++12(x1, x2)  =  x1
++2(x1, x2)  =  ++2(x1, x2)
.2(x1, x2)  =  x2
nil  =  nil
Used ordering: Quasi Precedence: trivial 


↳ QTRS
  ↳ DependencyPairsProof
    ↳ QDP
      ↳ QDPAfsSolverProof
        ↳ QDP
          ↳ QDPAfsSolverProof

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.
By using an argument filtering and a montonic ordering, at least one Dependency Pair of this SCC can be strictly oriented.

++12(.2(x, y), z) -> ++12(y, z)
Used argument filtering: ++12(x1, x2)  =  x1
.2(x1, x2)  =  .1(x2)
Used ordering: Quasi Precedence: trivial 


↳ QTRS
  ↳ DependencyPairsProof
    ↳ QDP
      ↳ QDPAfsSolverProof
        ↳ QDP
          ↳ QDPAfsSolverProof
            ↳ 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.