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 [1,15] we result in the following initial DP problem:
Q DP problem:
The TRS P consists of the following rules:

++1(++(x, y), z) → ++1(y, z)
++1(++(x, y), z) → ++1(x, ++(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(y, z)
++1(++(x, y), z) → ++1(x, ++(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 [15].


The following pairs can be oriented strictly and are deleted.


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

++1(.(x, y), z) → ++1(y, z)
Used ordering: Polynomial interpretation [25,35]:

POL(++1(x1, x2)) = (1/4)x_1   
POL(++(x1, x2)) = 13/4 + x_1 + (11/4)x_2   
POL(.(x1, x2)) = (15/4)x_2   
POL(nil) = 0   
The value of delta used in the strict ordering is 13/16.
The following usable rules [17] 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 [15].


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: Polynomial interpretation [25,35]:

POL(++1(x1, x2)) = (4)x_1   
POL(.(x1, x2)) = 1 + (4)x_2   
The value of delta used in the strict ordering is 4.
The following usable rules [17] 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.