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:

f4(0, 1, g2(x, y), z) -> f4(g2(x, y), g2(x, y), g2(x, y), h1(x))
g2(0, 1) -> 0
g2(0, 1) -> 1
h1(g2(x, y)) -> h1(x)

Q is empty.


QTRS
  ↳ DependencyPairsProof

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

f4(0, 1, g2(x, y), z) -> f4(g2(x, y), g2(x, y), g2(x, y), h1(x))
g2(0, 1) -> 0
g2(0, 1) -> 1
h1(g2(x, y)) -> h1(x)

Q is empty.

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

F4(0, 1, g2(x, y), z) -> H1(x)
H1(g2(x, y)) -> H1(x)
F4(0, 1, g2(x, y), z) -> F4(g2(x, y), g2(x, y), g2(x, y), h1(x))

The TRS R consists of the following rules:

f4(0, 1, g2(x, y), z) -> f4(g2(x, y), g2(x, y), g2(x, y), h1(x))
g2(0, 1) -> 0
g2(0, 1) -> 1
h1(g2(x, y)) -> h1(x)

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:

F4(0, 1, g2(x, y), z) -> H1(x)
H1(g2(x, y)) -> H1(x)
F4(0, 1, g2(x, y), z) -> F4(g2(x, y), g2(x, y), g2(x, y), h1(x))

The TRS R consists of the following rules:

f4(0, 1, g2(x, y), z) -> f4(g2(x, y), g2(x, y), g2(x, y), h1(x))
g2(0, 1) -> 0
g2(0, 1) -> 1
h1(g2(x, y)) -> h1(x)

Q is empty.
We have to consider all minimal (P,Q,R)-chains.
The approximation of the Dependency Graph contains 2 SCCs with 1 less node.

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

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

H1(g2(x, y)) -> H1(x)

The TRS R consists of the following rules:

f4(0, 1, g2(x, y), z) -> f4(g2(x, y), g2(x, y), g2(x, y), h1(x))
g2(0, 1) -> 0
g2(0, 1) -> 1
h1(g2(x, y)) -> h1(x)

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.

H1(g2(x, y)) -> H1(x)
Used argument filtering: H1(x1)  =  x1
g2(x1, x2)  =  g1(x1)
Used ordering: Quasi Precedence: trivial


↳ QTRS
  ↳ DependencyPairsProof
    ↳ QDP
      ↳ DependencyGraphProof
        ↳ AND
          ↳ QDP
            ↳ QDPAfsSolverProof
QDP
                ↳ PisEmptyProof
          ↳ QDP

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

f4(0, 1, g2(x, y), z) -> f4(g2(x, y), g2(x, y), g2(x, y), h1(x))
g2(0, 1) -> 0
g2(0, 1) -> 1
h1(g2(x, y)) -> h1(x)

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

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

F4(0, 1, g2(x, y), z) -> F4(g2(x, y), g2(x, y), g2(x, y), h1(x))

The TRS R consists of the following rules:

f4(0, 1, g2(x, y), z) -> f4(g2(x, y), g2(x, y), g2(x, y), h1(x))
g2(0, 1) -> 0
g2(0, 1) -> 1
h1(g2(x, y)) -> h1(x)

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