Termination w.r.t. Q of the following Term Rewriting System could be disproven:

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

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

POL(*(x1, x2)) = x2   
POL(*1(x1, x2)) = 1 + x2   
POL(+(x1, x2)) = 1 + x1 + x2   
POL(+1(x1, x2)) = 1 + x1 + x2   

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)), *(y, u)) → *1(y, +(z, u))
+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.
The approximation of the Dependency Graph [15,17,22] contains 1 SCC with 1 less node.

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

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.
We used the non-termination processor [17] to show that the DP problem is infinite.
Found a loop by narrowing to the left:

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)))


s = +1(+(x'', *(y, z)), *(y, u)) evaluates to t =+1(+(x'', *(y, z)), *(y, u))

Thus s starts an infinite chain as s semiunifies with t with the following substitutions:



Rewriting sequence

+1(+(x'', *(y, z)), *(y, u)) → +1(x'', *(y, +(z, u)))
with rule +1(+(x''', *(y', z')), *(y', u')) → +1(x''', *(y', +(z', u'))) at position [] and matcher [x''' / x'', u' / u, z' / z, y' / y]

+1(x'', *(y, +(z, u))) → +1(x'', +(*(y, z), *(y, u)))
with rule *(x', +(y', z')) → +(*(x', y'), *(x', z')) at position [1] and matcher [z' / u, y' / z, x' / y]

+1(x'', +(*(y, z), *(y, u))) → +1(+(x'', *(y, z)), *(y, u))
with rule +1(x, +(y, z)) → +1(+(x, y), z)

Now applying the matcher to the start term leads to a term which is equal to the last term in the rewriting sequence


All these steps are and every following step will be a correct step w.r.t to Q.