Termination w.r.t. Q of the following Term Rewriting System could not be shown:

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

*(x, +(y, z)) → +(*(x, y), *(x, z))
*(+(y, z), x) → +(*(x, y), *(x, z))
*(*(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:

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

The TRS R consists of the following rules:

*(x, +(y, z)) → +(*(x, y), *(x, z))
*(+(y, z), x) → +(*(x, y), *(x, z))
*(*(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
      ↳ DependencyGraphProof

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

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

The TRS R consists of the following rules:

*(x, +(y, z)) → +(*(x, y), *(x, z))
*(+(y, z), x) → +(*(x, y), *(x, z))
*(*(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 approximation of the Dependency Graph [15,17,22] contains 2 SCCs with 2 less nodes.

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

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))

The TRS R consists of the following rules:

*(x, +(y, z)) → +(*(x, y), *(x, z))
*(+(y, z), x) → +(*(x, y), *(x, z))
*(*(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.
none
Used ordering: Polynomial interpretation [25,35]:

POL(+1(x1, x2)) = (5/4)x_1   
POL(+(x1, x2)) = 1/4 + (3)x_1 + (4)x_2   
The value of delta used in the strict ordering is 5/16.
The following usable rules [17] were oriented: none



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

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

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

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

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

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

The TRS R consists of the following rules:

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