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(+2(x, y), z) -> +2(x, +2(y, z))
+2(f1(x), f1(y)) -> f1(+2(x, y))
+2(f1(x), +2(f1(y), z)) -> +2(f1(+2(x, y)), z)

Q is empty.


↳ QTRS
  ↳ DependencyPairsProof

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

+2(+2(x, y), z) -> +2(x, +2(y, z))
+2(f1(x), f1(y)) -> f1(+2(x, y))
+2(f1(x), +2(f1(y), z)) -> +2(f1(+2(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:

+12(f1(x), f1(y)) -> +12(x, y)
+12(f1(x), +2(f1(y), z)) -> +12(x, y)
+12(+2(x, y), z) -> +12(y, z)
+12(f1(x), +2(f1(y), z)) -> +12(f1(+2(x, y)), z)
+12(+2(x, y), z) -> +12(x, +2(y, z))

The TRS R consists of the following rules:

+2(+2(x, y), z) -> +2(x, +2(y, z))
+2(f1(x), f1(y)) -> f1(+2(x, y))
+2(f1(x), +2(f1(y), z)) -> +2(f1(+2(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:

+12(f1(x), f1(y)) -> +12(x, y)
+12(f1(x), +2(f1(y), z)) -> +12(x, y)
+12(+2(x, y), z) -> +12(y, z)
+12(f1(x), +2(f1(y), z)) -> +12(f1(+2(x, y)), z)
+12(+2(x, y), z) -> +12(x, +2(y, z))

The TRS R consists of the following rules:

+2(+2(x, y), z) -> +2(x, +2(y, z))
+2(f1(x), f1(y)) -> f1(+2(x, y))
+2(f1(x), +2(f1(y), z)) -> +2(f1(+2(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.


+12(f1(x), f1(y)) -> +12(x, y)
+12(f1(x), +2(f1(y), z)) -> +12(x, y)
+12(f1(x), +2(f1(y), z)) -> +12(f1(+2(x, y)), z)
The remaining pairs can at least be oriented weakly.

+12(+2(x, y), z) -> +12(y, z)
+12(+2(x, y), z) -> +12(x, +2(y, z))
Used ordering: Polynomial interpretation [21]:

POL(+2(x1, x2)) = x1 + x1·x2 + x2   
POL(+12(x1, x2)) = x1 + x1·x2   
POL(f1(x1)) = 1 + x1   

The following usable rules [14] were oriented:

+2(f1(x), f1(y)) -> f1(+2(x, y))
+2(f1(x), +2(f1(y), z)) -> +2(f1(+2(x, y)), z)
+2(+2(x, y), z) -> +2(x, +2(y, z))



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

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(+2(x, y), z) -> +2(x, +2(y, z))
+2(f1(x), f1(y)) -> f1(+2(x, y))
+2(f1(x), +2(f1(y), z)) -> +2(f1(+2(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.


+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)) = 3 + 3·x1 + 2·x1·x2 + 3·x2   
POL(+12(x1, x2)) = 3·x1   
POL(f1(x1)) = 0   

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:

+2(+2(x, y), z) -> +2(x, +2(y, z))
+2(f1(x), f1(y)) -> f1(+2(x, y))
+2(f1(x), +2(f1(y), z)) -> +2(f1(+2(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.