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), *2(x, z)) -> *2(x, +2(y, z))
+2(+2(x, y), z) -> +2(x, +2(y, z))
+2(*2(x, y), +2(*2(x, z), u)) -> +2(*2(x, +2(y, z)), u)

Q is empty.


↳ QTRS
  ↳ DependencyPairsProof

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

+2(*2(x, y), *2(x, z)) -> *2(x, +2(y, z))
+2(+2(x, y), z) -> +2(x, +2(y, z))
+2(*2(x, y), +2(*2(x, z), u)) -> +2(*2(x, +2(y, z)), u)

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

The TRS R consists of the following rules:

+2(*2(x, y), *2(x, z)) -> *2(x, +2(y, z))
+2(+2(x, y), z) -> +2(x, +2(y, z))
+2(*2(x, y), +2(*2(x, z), u)) -> +2(*2(x, +2(y, z)), u)

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

The TRS R consists of the following rules:

+2(*2(x, y), *2(x, z)) -> *2(x, +2(y, z))
+2(+2(x, y), z) -> +2(x, +2(y, z))
+2(*2(x, y), +2(*2(x, z), u)) -> +2(*2(x, +2(y, z)), u)

Q is empty.
We have to consider all minimal (P,Q,R)-chains.
The approximation of the Dependency Graph [13,14,18] contains 1 SCC with 1 less node.

↳ QTRS
  ↳ DependencyPairsProof
    ↳ QDP
      ↳ DependencyGraphProof
        ↳ 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), +2(*2(x, z), u)) -> +12(y, z)
+12(*2(x, y), *2(x, 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), *2(x, z)) -> *2(x, +2(y, z))
+2(+2(x, y), z) -> +2(x, +2(y, z))
+2(*2(x, y), +2(*2(x, z), u)) -> +2(*2(x, +2(y, z)), u)

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), +2(*2(x, z), u)) -> +12(y, z)
+12(*2(x, y), *2(x, 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)) = 1 + x2   
POL(+2(x1, x2)) = 3 + 3·x1 + x1·x2 + 2·x2   
POL(+12(x1, x2)) = 3·x1 + x1·x2 + 2·x2   
POL(u) = 1   

The following usable rules [14] were oriented:

+2(*2(x, y), +2(*2(x, z), u)) -> +2(*2(x, +2(y, z)), u)
+2(*2(x, y), *2(x, z)) -> *2(x, +2(y, z))
+2(+2(x, y), z) -> +2(x, +2(y, z))



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

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

+2(*2(x, y), *2(x, z)) -> *2(x, +2(y, z))
+2(+2(x, y), z) -> +2(x, +2(y, z))
+2(*2(x, y), +2(*2(x, z), u)) -> +2(*2(x, +2(y, z)), u)

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.