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, 1) → x
*(1, y) → y
*(i(x), x) → 1
*(x, i(x)) → 1
*(x, *(y, z)) → *(*(x, y), z)
i(1) → 1
*(*(x, y), i(y)) → x
*(*(x, i(y)), y) → x
i(i(x)) → x
i(*(x, y)) → *(i(y), i(x))
k(x, 1) → 1
k(x, x) → 1
*(k(x, y), k(y, x)) → 1
*(*(i(x), k(y, z)), x) → k(*(*(i(x), y), x), *(*(i(x), z), x))
k(*(x, i(y)), *(y, i(x))) → 1

Q is empty.


↳ QTRS
  ↳ DependencyPairsProof

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

*(x, 1) → x
*(1, y) → y
*(i(x), x) → 1
*(x, i(x)) → 1
*(x, *(y, z)) → *(*(x, y), z)
i(1) → 1
*(*(x, y), i(y)) → x
*(*(x, i(y)), y) → x
i(i(x)) → x
i(*(x, y)) → *(i(y), i(x))
k(x, 1) → 1
k(x, x) → 1
*(k(x, y), k(y, x)) → 1
*(*(i(x), k(y, z)), x) → k(*(*(i(x), y), x), *(*(i(x), z), x))
k(*(x, i(y)), *(y, i(x))) → 1

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(*(i(x), k(y, z)), x) → *1(*(i(x), z), x)
*1(*(i(x), k(y, z)), x) → K(*(*(i(x), y), x), *(*(i(x), z), x))
I(*(x, y)) → I(x)
*1(x, *(y, z)) → *1(x, y)
*1(*(i(x), k(y, z)), x) → *1(i(x), z)
I(*(x, y)) → *1(i(y), i(x))
*1(x, *(y, z)) → *1(*(x, y), z)
*1(*(i(x), k(y, z)), x) → *1(*(i(x), y), x)
I(*(x, y)) → I(y)
*1(*(i(x), k(y, z)), x) → *1(i(x), y)

The TRS R consists of the following rules:

*(x, 1) → x
*(1, y) → y
*(i(x), x) → 1
*(x, i(x)) → 1
*(x, *(y, z)) → *(*(x, y), z)
i(1) → 1
*(*(x, y), i(y)) → x
*(*(x, i(y)), y) → x
i(i(x)) → x
i(*(x, y)) → *(i(y), i(x))
k(x, 1) → 1
k(x, x) → 1
*(k(x, y), k(y, x)) → 1
*(*(i(x), k(y, z)), x) → k(*(*(i(x), y), x), *(*(i(x), z), x))
k(*(x, i(y)), *(y, i(x))) → 1

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(*(i(x), k(y, z)), x) → *1(*(i(x), z), x)
*1(*(i(x), k(y, z)), x) → K(*(*(i(x), y), x), *(*(i(x), z), x))
I(*(x, y)) → I(x)
*1(x, *(y, z)) → *1(x, y)
*1(*(i(x), k(y, z)), x) → *1(i(x), z)
I(*(x, y)) → *1(i(y), i(x))
*1(x, *(y, z)) → *1(*(x, y), z)
*1(*(i(x), k(y, z)), x) → *1(*(i(x), y), x)
I(*(x, y)) → I(y)
*1(*(i(x), k(y, z)), x) → *1(i(x), y)

The TRS R consists of the following rules:

*(x, 1) → x
*(1, y) → y
*(i(x), x) → 1
*(x, i(x)) → 1
*(x, *(y, z)) → *(*(x, y), z)
i(1) → 1
*(*(x, y), i(y)) → x
*(*(x, i(y)), y) → x
i(i(x)) → x
i(*(x, y)) → *(i(y), i(x))
k(x, 1) → 1
k(x, x) → 1
*(k(x, y), k(y, x)) → 1
*(*(i(x), k(y, z)), x) → k(*(*(i(x), y), x), *(*(i(x), z), x))
k(*(x, i(y)), *(y, i(x))) → 1

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 with 2 less nodes.

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

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

*1(*(i(x), k(y, z)), x) → *1(*(i(x), z), x)
*1(x, *(y, z)) → *1(x, y)
*1(*(i(x), k(y, z)), x) → *1(i(x), z)
*1(x, *(y, z)) → *1(*(x, y), z)
*1(*(i(x), k(y, z)), x) → *1(*(i(x), y), x)
*1(*(i(x), k(y, z)), x) → *1(i(x), y)

The TRS R consists of the following rules:

*(x, 1) → x
*(1, y) → y
*(i(x), x) → 1
*(x, i(x)) → 1
*(x, *(y, z)) → *(*(x, y), z)
i(1) → 1
*(*(x, y), i(y)) → x
*(*(x, i(y)), y) → x
i(i(x)) → x
i(*(x, y)) → *(i(y), i(x))
k(x, 1) → 1
k(x, x) → 1
*(k(x, y), k(y, x)) → 1
*(*(i(x), k(y, z)), x) → k(*(*(i(x), y), x), *(*(i(x), z), x))
k(*(x, i(y)), *(y, i(x))) → 1

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

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

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

I(*(x, y)) → I(x)
I(*(x, y)) → I(y)

The TRS R consists of the following rules:

*(x, 1) → x
*(1, y) → y
*(i(x), x) → 1
*(x, i(x)) → 1
*(x, *(y, z)) → *(*(x, y), z)
i(1) → 1
*(*(x, y), i(y)) → x
*(*(x, i(y)), y) → x
i(i(x)) → x
i(*(x, y)) → *(i(y), i(x))
k(x, 1) → 1
k(x, x) → 1
*(k(x, y), k(y, x)) → 1
*(*(i(x), k(y, z)), x) → k(*(*(i(x), y), x), *(*(i(x), z), x))
k(*(x, i(y)), *(y, i(x))) → 1

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.


I(*(x, y)) → I(x)
I(*(x, y)) → I(y)
The remaining pairs can at least be oriented weakly.
none
Used ordering: Polynomial interpretation [25,35]:

POL(*(x1, x2)) = 1/4 + (5/2)x_1 + (5/2)x_2   
POL(I(x1)) = (2)x_1   
The value of delta used in the strict ordering is 1/2.
The following usable rules [17] were oriented: none



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

Q DP problem:
P is empty.
The TRS R consists of the following rules:

*(x, 1) → x
*(1, y) → y
*(i(x), x) → 1
*(x, i(x)) → 1
*(x, *(y, z)) → *(*(x, y), z)
i(1) → 1
*(*(x, y), i(y)) → x
*(*(x, i(y)), y) → x
i(i(x)) → x
i(*(x, y)) → *(i(y), i(x))
k(x, 1) → 1
k(x, x) → 1
*(k(x, y), k(y, x)) → 1
*(*(i(x), k(y, z)), x) → k(*(*(i(x), y), x), *(*(i(x), z), x))
k(*(x, i(y)), *(y, i(x))) → 1

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.