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, 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 [1,13] 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), z) → *1(y, z)
*1(x, +(y, f(z))) → *1(g(x, z), +(y, y))
*1(*(x, y), z) → *1(x, *(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
      ↳ EdgeDeletionProof

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(y, z)
*1(x, +(y, f(z))) → *1(g(x, z), +(y, y))
*1(*(x, y), z) → *1(x, *(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.
We deleted some edges using various graph approximations

↳ QTRS
  ↳ DependencyPairsProof
    ↳ QDP
      ↳ EdgeDeletionProof
        ↳ 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), z) → *1(y, z)
*1(x, +(y, f(z))) → *1(g(x, z), +(y, y))
*1(*(x, y), z) → *1(x, *(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 [13,14,18] contains 2 SCCs.

↳ QTRS
  ↳ DependencyPairsProof
    ↳ QDP
      ↳ EdgeDeletionProof
        ↳ QDP
          ↳ DependencyGraphProof
            ↳ AND
              ↳ QDP
              ↳ 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.

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

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(y, z)
*1(*(x, y), z) → *1(x, *(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.
We use the reduction pair processor [13].


The following pairs can be oriented strictly and are deleted.


*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))
The remaining pairs can at least be oriented weakly.
none
Used ordering: Combined order from the following AFS and order.
*1(x1, x2)  =  x1
+(x1, x2)  =  +(x1, x2)
*(x1, x2)  =  *(x1, x2)
f(x1)  =  f
g(x1, x2)  =  g(x1, x2)

Recursive path order with status [2].
Precedence:
*2 > +2 > g2
f > g2

Status:
g2: multiset
+2: multiset
f: multiset
*2: multiset

The following usable rules [14] were oriented: none



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

Q DP problem:
P is empty.
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 TRS P is empty. Hence, there is no (P,Q,R) chain.