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, z), *(y, z))
*(x, +(y, f(z))) → *(g(x, z), +(y, y))

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, z), *(y, z))
*(x, +(y, f(z))) → *(g(x, z), +(y, y))

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

The TRS R consists of the following rules:

*(*(x, y), z) → *(x, *(y, z))
*(+(x, y), z) → +(*(x, z), *(y, z))
*(x, +(y, f(z))) → *(g(x, z), +(y, y))

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

The TRS R consists of the following rules:

*(*(x, y), z) → *(x, *(y, z))
*(+(x, y), z) → +(*(x, z), *(y, z))
*(x, +(y, f(z))) → *(g(x, z), +(y, y))

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.

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

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

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

The TRS R consists of the following rules:

*(*(x, y), z) → *(x, *(y, z))
*(+(x, y), z) → +(*(x, z), *(y, z))
*(x, +(y, f(z))) → *(g(x, z), +(y, y))

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

*1(g(y_0, y_1), +(y_2, f(x2))) → *1(g(g(y_0, y_1), x2), +(y_2, y_2))



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

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

*1(g(y_0, y_1), +(y_2, f(x2))) → *1(g(g(y_0, y_1), x2), +(y_2, y_2))

The TRS R consists of the following rules:

*(*(x, y), z) → *(x, *(y, z))
*(+(x, y), z) → +(*(x, z), *(y, z))
*(x, +(y, f(z))) → *(g(x, z), +(y, y))

Q is empty.
We have to consider all minimal (P,Q,R)-chains.
By instantiating [15] the rule *1(g(y_0, y_1), +(y_2, f(x2))) → *1(g(g(y_0, y_1), x2), +(y_2, y_2)) we obtained the following new rules:

*1(g(g(y_0, y_1), y_2), +(y_3, f(x3))) → *1(g(g(g(y_0, y_1), y_2), x3), +(y_3, y_3))



↳ QTRS
  ↳ DependencyPairsProof
    ↳ QDP
      ↳ DependencyGraphProof
        ↳ AND
          ↳ QDP
            ↳ Instantiation
              ↳ QDP
                ↳ Instantiation
                  ↳ QDP
                    ↳ ForwardInstantiation
          ↳ QDP

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

*1(g(g(y_0, y_1), y_2), +(y_3, f(x3))) → *1(g(g(g(y_0, y_1), y_2), x3), +(y_3, y_3))

The TRS R consists of the following rules:

*(*(x, y), z) → *(x, *(y, z))
*(+(x, y), z) → +(*(x, z), *(y, z))
*(x, +(y, f(z))) → *(g(x, z), +(y, y))

Q is empty.
We have to consider all minimal (P,Q,R)-chains.
By forward instantiating [14] the rule *1(g(g(y_0, y_1), y_2), +(y_3, f(x3))) → *1(g(g(g(y_0, y_1), y_2), x3), +(y_3, y_3)) we obtained the following new rules:

*1(g(g(x0, x1), x2), +(f(y_4), f(x4))) → *1(g(g(g(x0, x1), x2), x4), +(f(y_4), f(y_4)))



↳ QTRS
  ↳ DependencyPairsProof
    ↳ QDP
      ↳ DependencyGraphProof
        ↳ AND
          ↳ QDP
            ↳ Instantiation
              ↳ QDP
                ↳ Instantiation
                  ↳ QDP
                    ↳ ForwardInstantiation
                      ↳ QDP
                        ↳ NonTerminationProof
          ↳ QDP

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

*1(g(g(x0, x1), x2), +(f(y_4), f(x4))) → *1(g(g(g(x0, x1), x2), x4), +(f(y_4), f(y_4)))

The TRS R consists of the following rules:

*(*(x, y), z) → *(x, *(y, z))
*(+(x, y), z) → +(*(x, z), *(y, z))
*(x, +(y, f(z))) → *(g(x, z), +(y, y))

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 semiunifying a rule from P directly.

The TRS P consists of the following rules:

*1(g(g(x0, x1), x2), +(f(y_4), f(x4))) → *1(g(g(g(x0, x1), x2), x4), +(f(y_4), f(y_4)))

The TRS R consists of the following rules:

*(*(x, y), z) → *(x, *(y, z))
*(+(x, y), z) → +(*(x, z), *(y, z))
*(x, +(y, f(z))) → *(g(x, z), +(y, y))


s = *1(g(g(x0, x1), x2), +(f(y_4), f(x4))) evaluates to t =*1(g(g(g(x0, x1), x2), x4), +(f(y_4), f(y_4)))

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



Rewriting sequence

The DP semiunifies directly so there is only one rewrite step from *^1(g(g(x0, x1), x2), +(f(y_4), f(x4))) to *^1(g(g(g(x0, x1), x2), x4), +(f(y_4), f(y_4))).





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

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

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

The TRS R consists of the following rules:

*(*(x, y), z) → *(x, *(y, z))
*(+(x, y), z) → +(*(x, z), *(y, z))
*(x, +(y, f(z))) → *(g(x, z), +(y, y))

Q is empty.
We have to consider all minimal (P,Q,R)-chains.
By using the subterm criterion [20] together with the size-change analysis [32] we have proven that there are no infinite chains for this DP problem.

From the DPs we obtained the following set of size-change graphs: