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:

++(nil, y) → y
++(x, nil) → x
++(.(x, y), z) → .(x, ++(y, z))
++(++(x, y), z) → ++(x, ++(y, z))

Q is empty.


↳ QTRS
  ↳ DependencyPairsProof

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

++(nil, y) → y
++(x, nil) → x
++(.(x, y), z) → .(x, ++(y, z))
++(++(x, y), z) → ++(x, ++(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:

++1(++(x, y), z) → ++1(x, ++(y, z))
++1(++(x, y), z) → ++1(y, z)
++1(.(x, y), z) → ++1(y, z)

The TRS R consists of the following rules:

++(nil, y) → y
++(x, nil) → x
++(.(x, y), z) → .(x, ++(y, z))
++(++(x, y), z) → ++(x, ++(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:

++1(++(x, y), z) → ++1(x, ++(y, z))
++1(++(x, y), z) → ++1(y, z)
++1(.(x, y), z) → ++1(y, z)

The TRS R consists of the following rules:

++(nil, y) → y
++(x, nil) → x
++(.(x, y), z) → .(x, ++(y, z))
++(++(x, y), z) → ++(x, ++(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.


++1(++(x, y), z) → ++1(x, ++(y, z))
++1(++(x, y), z) → ++1(y, z)
The remaining pairs can at least be oriented weakly.

++1(.(x, y), z) → ++1(y, z)
Used ordering: Combined order from the following AFS and order.
++1(x1, x2)  =  x1
++(x1, x2)  =  ++(x1, x2)
.(x1, x2)  =  x2

Recursive Path Order [2].
Precedence:
trivial

The following usable rules [14] were oriented: none



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

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

++1(.(x, y), z) → ++1(y, z)

The TRS R consists of the following rules:

++(nil, y) → y
++(x, nil) → x
++(.(x, y), z) → .(x, ++(y, z))
++(++(x, y), z) → ++(x, ++(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.


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

Recursive Path Order [2].
Precedence:
trivial

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:

++(nil, y) → y
++(x, nil) → x
++(.(x, y), z) → .(x, ++(y, z))
++(++(x, y), z) → ++(x, ++(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.