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:

f2(x, g1(y)) -> f2(h1(x), i2(x, y))
i2(x, j2(0, 0)) -> g1(0)
i2(x, j2(y, z)) -> j2(g1(y), i2(x, z))
i2(h1(x), j2(j2(y, z), 0)) -> j2(i2(h1(x), j2(y, z)), i2(x, j2(y, z)))
j2(g1(x), g1(y)) -> g1(j2(x, y))

Q is empty.


QTRS
  ↳ DependencyPairsProof

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

f2(x, g1(y)) -> f2(h1(x), i2(x, y))
i2(x, j2(0, 0)) -> g1(0)
i2(x, j2(y, z)) -> j2(g1(y), i2(x, z))
i2(h1(x), j2(j2(y, z), 0)) -> j2(i2(h1(x), j2(y, z)), i2(x, j2(y, z)))
j2(g1(x), g1(y)) -> g1(j2(x, 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:

I2(h1(x), j2(j2(y, z), 0)) -> I2(h1(x), j2(y, z))
F2(x, g1(y)) -> I2(x, y)
I2(h1(x), j2(j2(y, z), 0)) -> J2(i2(h1(x), j2(y, z)), i2(x, j2(y, z)))
I2(x, j2(y, z)) -> J2(g1(y), i2(x, z))
J2(g1(x), g1(y)) -> J2(x, y)
I2(h1(x), j2(j2(y, z), 0)) -> I2(x, j2(y, z))
I2(x, j2(y, z)) -> I2(x, z)
F2(x, g1(y)) -> F2(h1(x), i2(x, y))

The TRS R consists of the following rules:

f2(x, g1(y)) -> f2(h1(x), i2(x, y))
i2(x, j2(0, 0)) -> g1(0)
i2(x, j2(y, z)) -> j2(g1(y), i2(x, z))
i2(h1(x), j2(j2(y, z), 0)) -> j2(i2(h1(x), j2(y, z)), i2(x, j2(y, z)))
j2(g1(x), g1(y)) -> g1(j2(x, 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:

I2(h1(x), j2(j2(y, z), 0)) -> I2(h1(x), j2(y, z))
F2(x, g1(y)) -> I2(x, y)
I2(h1(x), j2(j2(y, z), 0)) -> J2(i2(h1(x), j2(y, z)), i2(x, j2(y, z)))
I2(x, j2(y, z)) -> J2(g1(y), i2(x, z))
J2(g1(x), g1(y)) -> J2(x, y)
I2(h1(x), j2(j2(y, z), 0)) -> I2(x, j2(y, z))
I2(x, j2(y, z)) -> I2(x, z)
F2(x, g1(y)) -> F2(h1(x), i2(x, y))

The TRS R consists of the following rules:

f2(x, g1(y)) -> f2(h1(x), i2(x, y))
i2(x, j2(0, 0)) -> g1(0)
i2(x, j2(y, z)) -> j2(g1(y), i2(x, z))
i2(h1(x), j2(j2(y, z), 0)) -> j2(i2(h1(x), j2(y, z)), i2(x, j2(y, z)))
j2(g1(x), g1(y)) -> g1(j2(x, y))

Q is empty.
We have to consider all minimal (P,Q,R)-chains.
The approximation of the Dependency Graph [13,14,18] contains 3 SCCs with 3 less nodes.

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

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

J2(g1(x), g1(y)) -> J2(x, y)

The TRS R consists of the following rules:

f2(x, g1(y)) -> f2(h1(x), i2(x, y))
i2(x, j2(0, 0)) -> g1(0)
i2(x, j2(y, z)) -> j2(g1(y), i2(x, z))
i2(h1(x), j2(j2(y, z), 0)) -> j2(i2(h1(x), j2(y, z)), i2(x, j2(y, z)))
j2(g1(x), g1(y)) -> g1(j2(x, 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 strictly oriented and are deleted.


J2(g1(x), g1(y)) -> J2(x, y)
The remaining pairs can at least by weakly be oriented.
none
Used ordering: Combined order from the following AFS and order.
J2(x1, x2)  =  J1(x1)
g1(x1)  =  g1(x1)

Lexicographic Path Order [19].
Precedence:
trivial


The following usable rules [14] were oriented: none



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

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

f2(x, g1(y)) -> f2(h1(x), i2(x, y))
i2(x, j2(0, 0)) -> g1(0)
i2(x, j2(y, z)) -> j2(g1(y), i2(x, z))
i2(h1(x), j2(j2(y, z), 0)) -> j2(i2(h1(x), j2(y, z)), i2(x, j2(y, z)))
j2(g1(x), g1(y)) -> g1(j2(x, 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.

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

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

I2(h1(x), j2(j2(y, z), 0)) -> I2(h1(x), j2(y, z))
I2(h1(x), j2(j2(y, z), 0)) -> I2(x, j2(y, z))
I2(x, j2(y, z)) -> I2(x, z)

The TRS R consists of the following rules:

f2(x, g1(y)) -> f2(h1(x), i2(x, y))
i2(x, j2(0, 0)) -> g1(0)
i2(x, j2(y, z)) -> j2(g1(y), i2(x, z))
i2(h1(x), j2(j2(y, z), 0)) -> j2(i2(h1(x), j2(y, z)), i2(x, j2(y, z)))
j2(g1(x), g1(y)) -> g1(j2(x, 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 strictly oriented and are deleted.


I2(h1(x), j2(j2(y, z), 0)) -> I2(h1(x), j2(y, z))
I2(h1(x), j2(j2(y, z), 0)) -> I2(x, j2(y, z))
I2(x, j2(y, z)) -> I2(x, z)
The remaining pairs can at least by weakly be oriented.
none
Used ordering: Combined order from the following AFS and order.
I2(x1, x2)  =  I1(x2)
h1(x1)  =  h
j2(x1, x2)  =  j2(x1, x2)
0  =  0
g1(x1)  =  g

Lexicographic Path Order [19].
Precedence:
[h, 0] > [I1, j2] > g


The following usable rules [14] were oriented:

j2(g1(x), g1(y)) -> g1(j2(x, y))



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

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

f2(x, g1(y)) -> f2(h1(x), i2(x, y))
i2(x, j2(0, 0)) -> g1(0)
i2(x, j2(y, z)) -> j2(g1(y), i2(x, z))
i2(h1(x), j2(j2(y, z), 0)) -> j2(i2(h1(x), j2(y, z)), i2(x, j2(y, z)))
j2(g1(x), g1(y)) -> g1(j2(x, 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.

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

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

F2(x, g1(y)) -> F2(h1(x), i2(x, y))

The TRS R consists of the following rules:

f2(x, g1(y)) -> f2(h1(x), i2(x, y))
i2(x, j2(0, 0)) -> g1(0)
i2(x, j2(y, z)) -> j2(g1(y), i2(x, z))
i2(h1(x), j2(j2(y, z), 0)) -> j2(i2(h1(x), j2(y, z)), i2(x, j2(y, z)))
j2(g1(x), g1(y)) -> g1(j2(x, y))

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