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)))

The set Q consists of the following terms:

*(x0, +(x1, x2))
+(+(x0, *(x1, x2)), *(x1, x3))
+(x0, +(x1, x2))



↳ 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)))

The set Q consists of the following terms:

*(x0, +(x1, x2))
+(+(x0, *(x1, x2)), *(x1, x3))
+(x0, +(x1, x2))


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)))

The set Q consists of the following terms:

*(x0, +(x1, x2))
+(+(x0, *(x1, x2)), *(x1, x3))
+(x0, +(x1, x2))

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(+(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)))

The set Q consists of the following terms:

*(x0, +(x1, x2))
+(+(x0, *(x1, x2)), *(x1, x3))
+(x0, +(x1, x2))

We have to consider all minimal (P,Q,R)-chains.
The approximation of the Dependency Graph [15,17,22] contains 1 SCC with 2 less nodes.

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

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

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

The set Q consists of the following terms:

*(x0, +(x1, x2))
+(+(x0, *(x1, x2)), *(x1, x3))
+(x0, +(x1, x2))

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)) → +1(x, y)
*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)) = x2   
POL(+(x1, x2)) = 1 + x1 + x2   
POL(+1(x1, x2)) = 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
      ↳ DependencyGraphProof
        ↳ QDP
          ↳ QDPOrderProof
            ↳ QDP
              ↳ DependencyGraphProof
          ↳ NonTerminationProof

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)))

The set Q consists of the following terms:

*(x0, +(x1, x2))
+(+(x0, *(x1, x2)), *(x1, x3))
+(x0, +(x1, x2))

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
      ↳ DependencyGraphProof
        ↳ QDP
          ↳ QDPOrderProof
            ↳ QDP
              ↳ DependencyGraphProof
                ↳ QDP
                  ↳ Narrowing
          ↳ 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)))

The set Q consists of the following terms:

*(x0, +(x1, x2))
+(+(x0, *(x1, x2)), *(x1, x3))
+(x0, +(x1, x2))

We have to consider all minimal (P,Q,R)-chains.
By narrowing [15] the rule +1(+(x, *(y, z)), *(y, u)) → +1(x, *(y, +(z, u))) at position [1] we obtained the following new rules:

+1(+(y0, *(x0, x1)), *(x0, x2)) → +1(y0, +(*(x0, x1), *(x0, x2)))



↳ QTRS
  ↳ DependencyPairsProof
    ↳ QDP
      ↳ DependencyGraphProof
        ↳ QDP
          ↳ QDPOrderProof
            ↳ QDP
              ↳ DependencyGraphProof
                ↳ QDP
                  ↳ Narrowing
                    ↳ QDP
                      ↳ Instantiation
          ↳ NonTerminationProof

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

+1(x, +(y, z)) → +1(+(x, y), z)
+1(+(y0, *(x0, x1)), *(x0, x2)) → +1(y0, +(*(x0, x1), *(x0, x2)))

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)))

The set Q consists of the following terms:

*(x0, +(x1, x2))
+(+(x0, *(x1, x2)), *(x1, x3))
+(x0, +(x1, x2))

We have to consider all minimal (P,Q,R)-chains.
By instantiating [15] the rule +1(x, +(y, z)) → +1(+(x, y), z) we obtained the following new rules:

+1(z0, +(*(z1, z2), *(z1, z3))) → +1(+(z0, *(z1, z2)), *(z1, z3))



↳ QTRS
  ↳ DependencyPairsProof
    ↳ QDP
      ↳ DependencyGraphProof
        ↳ QDP
          ↳ QDPOrderProof
            ↳ QDP
              ↳ DependencyGraphProof
                ↳ QDP
                  ↳ Narrowing
                    ↳ QDP
                      ↳ Instantiation
                        ↳ QDP
                          ↳ NonTerminationProof
          ↳ NonTerminationProof

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

+1(z0, +(*(z1, z2), *(z1, z3))) → +1(+(z0, *(z1, z2)), *(z1, z3))
+1(+(y0, *(x0, x1)), *(x0, x2)) → +1(y0, +(*(x0, x1), *(x0, x2)))

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)))

The set Q consists of the following terms:

*(x0, +(x1, x2))
+(+(x0, *(x1, x2)), *(x1, x3))
+(x0, +(x1, x2))

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(z0, +(*(z1, z2), *(z1, z3))) → +1(+(z0, *(z1, z2)), *(z1, z3))
+1(+(y0, *(x0, x1)), *(x0, x2)) → +1(y0, +(*(x0, x1), *(x0, x2)))

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(+(y0, *(x0, x1)), *(x0, x2)) evaluates to t =+1(+(y0, *(x0, x1)), *(x0, x2))

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



Rewriting sequence

+1(+(y0, *(x0, x1)), *(x0, x2)) → +1(y0, +(*(x0, x1), *(x0, x2)))
with rule +1(+(y0', *(x0', x1')), *(x0', x2')) → +1(y0', +(*(x0', x1'), *(x0', x2'))) at position [] and matcher [y0' / y0, x0' / x0, x2' / x2, x1' / x1]

+1(y0, +(*(x0, x1), *(x0, x2))) → +1(+(y0, *(x0, x1)), *(x0, x2))
with rule +1(z0, +(*(z1, z2), *(z1, z3))) → +1(+(z0, *(z1, z2)), *(z1, z3))

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.




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)
+1(x, +(y, z)) → +1(x, y)
+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)))
*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)))


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

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



Rewriting sequence

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

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

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

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.