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.

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 contains 3 SCCs with 3 less nodes.

↳ QTRS
  ↳ DependencyPairsProof
    ↳ QDP
      ↳ DependencyGraphProof
        ↳ AND
QDP
            ↳ QDPAfsSolverProof
          ↳ 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.
By using an argument filtering and a montonic ordering, at least one Dependency Pair of this SCC can be strictly oriented.

J2(g1(x), g1(y)) -> J2(x, y)
Used argument filtering: J2(x1, x2)  =  x2
g1(x1)  =  g1(x1)
Used ordering: Quasi Precedence: trivial


↳ QTRS
  ↳ DependencyPairsProof
    ↳ QDP
      ↳ DependencyGraphProof
        ↳ AND
          ↳ QDP
            ↳ QDPAfsSolverProof
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
            ↳ QDPAfsSolverProof
          ↳ 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.
By using an argument filtering and a montonic ordering, at least one Dependency Pair of this SCC can be strictly oriented.

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)
Used argument filtering: I2(x1, x2)  =  x2
j2(x1, x2)  =  j2(x1, x2)
0  =  0
g1(x1)  =  g
Used ordering: Quasi Precedence: trivial


↳ QTRS
  ↳ DependencyPairsProof
    ↳ QDP
      ↳ DependencyGraphProof
        ↳ AND
          ↳ QDP
          ↳ QDP
            ↳ QDPAfsSolverProof
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.