Termination w.r.t. Q proof of TRS/Zantema06beans.trs

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, f₂(s₁(s₁(y)), f₂(z, w))) -> f₂(s₁(x), f₂(y, f₂(s₁(z), w)))
L₁(f₂(s₁(s₁(y)), f₂(z, w))) -> L₁(f₂(s₁(0), f₂(y, f₂(s₁(z), w))))
f₂(x, f₂(s₁(s₁(y)), nil)) -> f₂(s₁(x), f₂(y, f₂(s₁(0), nil)))

Q is empty.

↳ QTRS

  ↳ DependencyPairsProof

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

f₂(x, f₂(s₁(s₁(y)), f₂(z, w))) -> f₂(s₁(x), f₂(y, f₂(s₁(z), w)))
L₁(f₂(s₁(s₁(y)), f₂(z, w))) -> L₁(f₂(s₁(0), f₂(y, f₂(s₁(z), w))))
f₂(x, f₂(s₁(s₁(y)), nil)) -> f₂(s₁(x), f₂(y, f₂(s₁(0), nil)))

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:

L¹₁(f₂(s₁(s₁(y)), f₂(z, w))) -> F₂(s₁(z), w)
L¹₁(f₂(s₁(s₁(y)), f₂(z, w))) -> F₂(y, f₂(s₁(z), w))
F₂(x, f₂(s₁(s₁(y)), nil)) -> F₂(s₁(0), nil)
F₂(x, f₂(s₁(s₁(y)), f₂(z, w))) -> F₂(s₁(z), w)
F₂(x, f₂(s₁(s₁(y)), nil)) -> F₂(s₁(x), f₂(y, f₂(s₁(0), nil)))
L¹₁(f₂(s₁(s₁(y)), f₂(z, w))) -> L¹₁(f₂(s₁(0), f₂(y, f₂(s₁(z), w))))
F₂(x, f₂(s₁(s₁(y)), f₂(z, w))) -> F₂(s₁(x), f₂(y, f₂(s₁(z), w)))
F₂(x, f₂(s₁(s₁(y)), f₂(z, w))) -> F₂(y, f₂(s₁(z), w))
L¹₁(f₂(s₁(s₁(y)), f₂(z, w))) -> F₂(s₁(0), f₂(y, f₂(s₁(z), w)))
F₂(x, f₂(s₁(s₁(y)), nil)) -> F₂(y, f₂(s₁(0), nil))

The TRS R consists of the following rules:

f₂(x, f₂(s₁(s₁(y)), f₂(z, w))) -> f₂(s₁(x), f₂(y, f₂(s₁(z), w)))
L₁(f₂(s₁(s₁(y)), f₂(z, w))) -> L₁(f₂(s₁(0), f₂(y, f₂(s₁(z), w))))
f₂(x, f₂(s₁(s₁(y)), nil)) -> f₂(s₁(x), f₂(y, f₂(s₁(0), nil)))

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:

L¹₁(f₂(s₁(s₁(y)), f₂(z, w))) -> F₂(s₁(z), w)
L¹₁(f₂(s₁(s₁(y)), f₂(z, w))) -> F₂(y, f₂(s₁(z), w))
F₂(x, f₂(s₁(s₁(y)), nil)) -> F₂(s₁(0), nil)
F₂(x, f₂(s₁(s₁(y)), f₂(z, w))) -> F₂(s₁(z), w)
F₂(x, f₂(s₁(s₁(y)), nil)) -> F₂(s₁(x), f₂(y, f₂(s₁(0), nil)))
L¹₁(f₂(s₁(s₁(y)), f₂(z, w))) -> L¹₁(f₂(s₁(0), f₂(y, f₂(s₁(z), w))))
F₂(x, f₂(s₁(s₁(y)), f₂(z, w))) -> F₂(s₁(x), f₂(y, f₂(s₁(z), w)))
F₂(x, f₂(s₁(s₁(y)), f₂(z, w))) -> F₂(y, f₂(s₁(z), w))
L¹₁(f₂(s₁(s₁(y)), f₂(z, w))) -> F₂(s₁(0), f₂(y, f₂(s₁(z), w)))
F₂(x, f₂(s₁(s₁(y)), nil)) -> F₂(y, f₂(s₁(0), nil))

The TRS R consists of the following rules:

f₂(x, f₂(s₁(s₁(y)), f₂(z, w))) -> f₂(s₁(x), f₂(y, f₂(s₁(z), w)))
L₁(f₂(s₁(s₁(y)), f₂(z, w))) -> L₁(f₂(s₁(0), f₂(y, f₂(s₁(z), w))))
f₂(x, f₂(s₁(s₁(y)), nil)) -> f₂(s₁(x), f₂(y, f₂(s₁(0), nil)))

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 5 less nodes.

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ DependencyGraphProof

        ↳ AND

          ↳ QDP

            ↳ QDPOrderProof

          ↳ QDP

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

F₂(x, f₂(s₁(s₁(y)), f₂(z, w))) -> F₂(s₁(z), w)
F₂(x, f₂(s₁(s₁(y)), nil)) -> F₂(s₁(x), f₂(y, f₂(s₁(0), nil)))
F₂(x, f₂(s₁(s₁(y)), f₂(z, w))) -> F₂(s₁(x), f₂(y, f₂(s₁(z), w)))
F₂(x, f₂(s₁(s₁(y)), f₂(z, w))) -> F₂(y, f₂(s₁(z), w))

The TRS R consists of the following rules:

f₂(x, f₂(s₁(s₁(y)), f₂(z, w))) -> f₂(s₁(x), f₂(y, f₂(s₁(z), w)))
L₁(f₂(s₁(s₁(y)), f₂(z, w))) -> L₁(f₂(s₁(0), f₂(y, f₂(s₁(z), w))))
f₂(x, f₂(s₁(s₁(y)), nil)) -> f₂(s₁(x), f₂(y, f₂(s₁(0), nil)))

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.

F₂(x, f₂(s₁(s₁(y)), f₂(z, w))) -> F₂(s₁(z), w)
F₂(x, f₂(s₁(s₁(y)), nil)) -> F₂(s₁(x), f₂(y, f₂(s₁(0), nil)))
F₂(x, f₂(s₁(s₁(y)), f₂(z, w))) -> F₂(s₁(x), f₂(y, f₂(s₁(z), w)))
F₂(x, f₂(s₁(s₁(y)), f₂(z, w))) -> F₂(y, f₂(s₁(z), w))

The remaining pairs can at least be oriented weakly.
none
Used ordering: Polynomial Order [17,21] with Interpretation:

POL( s₁(x₁) ) = x₁ + 1

POL( 0 ) = max{0, -1}

POL( F₂(x₁, x₂) ) = x₂ + 1

POL( nil ) = max{0, -1}

POL( f₂(x₁, x₂) ) = x₁ + x₂

The following usable rules [14] were oriented:

f₂(x, f₂(s₁(s₁(y)), nil)) -> f₂(s₁(x), f₂(y, f₂(s₁(0), nil)))
f₂(x, f₂(s₁(s₁(y)), f₂(z, w))) -> f₂(s₁(x), f₂(y, f₂(s₁(z), w)))

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ DependencyGraphProof

        ↳ AND

          ↳ QDP

            ↳ QDPOrderProof

              ↳ QDP

                ↳ PisEmptyProof

          ↳ QDP

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

f₂(x, f₂(s₁(s₁(y)), f₂(z, w))) -> f₂(s₁(x), f₂(y, f₂(s₁(z), w)))
L₁(f₂(s₁(s₁(y)), f₂(z, w))) -> L₁(f₂(s₁(0), f₂(y, f₂(s₁(z), w))))
f₂(x, f₂(s₁(s₁(y)), nil)) -> f₂(s₁(x), f₂(y, f₂(s₁(0), nil)))

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:

L¹₁(f₂(s₁(s₁(y)), f₂(z, w))) -> L¹₁(f₂(s₁(0), f₂(y, f₂(s₁(z), w))))

The TRS R consists of the following rules:

f₂(x, f₂(s₁(s₁(y)), f₂(z, w))) -> f₂(s₁(x), f₂(y, f₂(s₁(z), w)))
L₁(f₂(s₁(s₁(y)), f₂(z, w))) -> L₁(f₂(s₁(0), f₂(y, f₂(s₁(z), w))))
f₂(x, f₂(s₁(s₁(y)), nil)) -> f₂(s₁(x), f₂(y, f₂(s₁(0), nil)))

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