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(a, b) -> +2(b, a)
+2(a, +2(b, z)) -> +2(b, +2(a, z))
+2(+2(x, y), z) -> +2(x, +2(y, z))
f2(a, y) -> a
f2(b, y) -> b
f2(+2(x, y), z) -> +2(f2(x, z), f2(y, z))

Q is empty.


↳ QTRS
  ↳ DependencyPairsProof

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

+2(a, b) -> +2(b, a)
+2(a, +2(b, z)) -> +2(b, +2(a, z))
+2(+2(x, y), z) -> +2(x, +2(y, z))
f2(a, y) -> a
f2(b, y) -> b
f2(+2(x, y), z) -> +2(f2(x, z), f2(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(a, b) -> +12(b, a)
F2(+2(x, y), z) -> F2(y, z)
+12(+2(x, y), z) -> +12(y, z)
+12(a, +2(b, z)) -> +12(a, z)
+12(a, +2(b, z)) -> +12(b, +2(a, z))
F2(+2(x, y), z) -> F2(x, z)
F2(+2(x, y), z) -> +12(f2(x, z), f2(y, z))
+12(+2(x, y), z) -> +12(x, +2(y, z))

The TRS R consists of the following rules:

+2(a, b) -> +2(b, a)
+2(a, +2(b, z)) -> +2(b, +2(a, z))
+2(+2(x, y), z) -> +2(x, +2(y, z))
f2(a, y) -> a
f2(b, y) -> b
f2(+2(x, y), z) -> +2(f2(x, z), f2(y, z))

Q is empty.
We have to consider all minimal (P,Q,R)-chains.

↳ QTRS
  ↳ DependencyPairsProof
    ↳ QDP
      ↳ DependencyGraphProof

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

+12(a, b) -> +12(b, a)
F2(+2(x, y), z) -> F2(y, z)
+12(+2(x, y), z) -> +12(y, z)
+12(a, +2(b, z)) -> +12(a, z)
+12(a, +2(b, z)) -> +12(b, +2(a, z))
F2(+2(x, y), z) -> F2(x, z)
F2(+2(x, y), z) -> +12(f2(x, z), f2(y, z))
+12(+2(x, y), z) -> +12(x, +2(y, z))

The TRS R consists of the following rules:

+2(a, b) -> +2(b, a)
+2(a, +2(b, z)) -> +2(b, +2(a, z))
+2(+2(x, y), z) -> +2(x, +2(y, z))
f2(a, y) -> a
f2(b, y) -> b
f2(+2(x, y), z) -> +2(f2(x, z), f2(y, z))

Q is empty.
We have to consider all minimal (P,Q,R)-chains.
The approximation of the Dependency Graph [13,14,18] contains 3 SCCs with 3 less nodes.

↳ QTRS
  ↳ DependencyPairsProof
    ↳ QDP
      ↳ DependencyGraphProof
        ↳ AND
          ↳ QDP
            ↳ QDPOrderProof
          ↳ QDP
          ↳ QDP

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

+12(a, +2(b, z)) -> +12(a, z)

The TRS R consists of the following rules:

+2(a, b) -> +2(b, a)
+2(a, +2(b, z)) -> +2(b, +2(a, z))
+2(+2(x, y), z) -> +2(x, +2(y, z))
f2(a, y) -> a
f2(b, y) -> b
f2(+2(x, y), z) -> +2(f2(x, z), f2(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(a, +2(b, z)) -> +12(a, z)
The remaining pairs can at least be oriented weakly.
none
Used ordering: Polynomial interpretation [21]:

POL(+2(x1, x2)) = 2·x1 + 2·x1·x2   
POL(+12(x1, x2)) = 2·x1·x2   
POL(a) = 2   
POL(b) = 2   

The following usable rules [14] were oriented: none



↳ QTRS
  ↳ DependencyPairsProof
    ↳ QDP
      ↳ DependencyGraphProof
        ↳ AND
          ↳ QDP
            ↳ QDPOrderProof
              ↳ QDP
                ↳ PisEmptyProof
          ↳ QDP
          ↳ QDP

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

+2(a, b) -> +2(b, a)
+2(a, +2(b, z)) -> +2(b, +2(a, z))
+2(+2(x, y), z) -> +2(x, +2(y, z))
f2(a, y) -> a
f2(b, y) -> b
f2(+2(x, y), z) -> +2(f2(x, z), f2(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.

↳ QTRS
  ↳ DependencyPairsProof
    ↳ QDP
      ↳ DependencyGraphProof
        ↳ AND
          ↳ QDP
          ↳ QDP
            ↳ QDPOrderProof
          ↳ QDP

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

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

The TRS R consists of the following rules:

+2(a, b) -> +2(b, a)
+2(a, +2(b, z)) -> +2(b, +2(a, z))
+2(+2(x, y), z) -> +2(x, +2(y, z))
f2(a, y) -> a
f2(b, y) -> b
f2(+2(x, y), z) -> +2(f2(x, z), f2(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(y, z)
+12(+2(x, y), z) -> +12(x, +2(y, z))
The remaining pairs can at least be oriented weakly.
none
Used ordering: Polynomial interpretation [21]:

POL(+2(x1, x2)) = 2 + 2·x1 + 2·x2   
POL(+12(x1, x2)) = 2·x1   
POL(a) = 0   
POL(b) = 0   

The following usable rules [14] were oriented: none



↳ QTRS
  ↳ DependencyPairsProof
    ↳ QDP
      ↳ DependencyGraphProof
        ↳ AND
          ↳ QDP
          ↳ QDP
            ↳ QDPOrderProof
              ↳ QDP
                ↳ PisEmptyProof
          ↳ QDP

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

+2(a, b) -> +2(b, a)
+2(a, +2(b, z)) -> +2(b, +2(a, z))
+2(+2(x, y), z) -> +2(x, +2(y, z))
f2(a, y) -> a
f2(b, y) -> b
f2(+2(x, y), z) -> +2(f2(x, z), f2(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.

↳ QTRS
  ↳ DependencyPairsProof
    ↳ QDP
      ↳ DependencyGraphProof
        ↳ AND
          ↳ QDP
          ↳ QDP
          ↳ QDP
            ↳ QDPOrderProof

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

F2(+2(x, y), z) -> F2(y, z)
F2(+2(x, y), z) -> F2(x, z)

The TRS R consists of the following rules:

+2(a, b) -> +2(b, a)
+2(a, +2(b, z)) -> +2(b, +2(a, z))
+2(+2(x, y), z) -> +2(x, +2(y, z))
f2(a, y) -> a
f2(b, y) -> b
f2(+2(x, y), z) -> +2(f2(x, z), f2(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.


F2(+2(x, y), z) -> F2(y, z)
F2(+2(x, y), z) -> F2(x, z)
The remaining pairs can at least be oriented weakly.
none
Used ordering: Polynomial interpretation [21]:

POL(+2(x1, x2)) = 1 + 2·x1 + 2·x2   
POL(F2(x1, x2)) = 2·x1   

The following usable rules [14] were oriented: none



↳ QTRS
  ↳ DependencyPairsProof
    ↳ QDP
      ↳ DependencyGraphProof
        ↳ AND
          ↳ QDP
          ↳ QDP
          ↳ QDP
            ↳ QDPOrderProof
              ↳ QDP
                ↳ PisEmptyProof

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

+2(a, b) -> +2(b, a)
+2(a, +2(b, z)) -> +2(b, +2(a, z))
+2(+2(x, y), z) -> +2(x, +2(y, z))
f2(a, y) -> a
f2(b, y) -> b
f2(+2(x, y), z) -> +2(f2(x, z), f2(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.