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.

Using Dependency Pairs [1,13] we result in the following initial DP problem:
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 [13,14,18] contains 2 SCCs with 1 less node.

↳ QTRS
  ↳ DependencyPairsProof
    ↳ QDP
      ↳ DependencyGraphProof
        ↳ AND
          ↳ QDP
            ↳ QDPOrderProof
          ↳ 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.
We use the reduction pair processor [13].


The following pairs can be oriented strictly and are deleted.


H1(g2(x, y)) -> H1(x)
The remaining pairs can at least be oriented weakly.
none
Used ordering: Polynomial Order [17,21] with Interpretation:

POL( g2(x1, x2) ) = 3x1 + x2 + 1


POL( H1(x1) ) = 2x1 + 2



The following usable rules [14] were oriented: none



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