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

Q is empty.


↳ QTRS
  ↳ DependencyPairsProof

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

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

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

The TRS R consists of the following rules:

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

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

The TRS R consists of the following rules:

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

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

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

POL(*1(x1, x2)) = 4 + (2)x_2   
POL(*(x1, x2)) = 9/4 + (4)x_2   
POL(+1(x1, x2)) = 13/4 + (1/2)x_1 + (1/2)x_2   
POL(+(x1, x2)) = 3/4 + x_1 + x_2   
The value of delta used in the strict ordering is 3/8.
The following usable rules [17] were oriented:

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



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

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

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

The TRS R consists of the following rules:

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

Q is empty.
We have to consider all minimal (P,Q,R)-chains.
The approximation of the Dependency Graph [15,17,22] contains 1 SCC with 1 less node.

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

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

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

The TRS R consists of the following rules:

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

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