Termination w.r.t. Q proof of TRS/nontermin/caribooex3.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:

f₁(g₁(a)) -> f₁(s₁(g₁(b)))
f₁(f₁(x)) -> b
g₁(x) -> f₁(g₁(x))

Q is empty.

↳ QTRS

  ↳ DependencyPairsProof

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

f₁(g₁(a)) -> f₁(s₁(g₁(b)))
f₁(f₁(x)) -> b
g₁(x) -> f₁(g₁(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:

G₁(x) -> F₁(g₁(x))
F₁(g₁(a)) -> G₁(b)
G₁(x) -> G₁(x)
F₁(g₁(a)) -> F₁(s₁(g₁(b)))

The TRS R consists of the following rules:

f₁(g₁(a)) -> f₁(s₁(g₁(b)))
f₁(f₁(x)) -> b
g₁(x) -> f₁(g₁(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:

G₁(x) -> F₁(g₁(x))
F₁(g₁(a)) -> G₁(b)
G₁(x) -> G₁(x)
F₁(g₁(a)) -> F₁(s₁(g₁(b)))

The TRS R consists of the following rules:

f₁(g₁(a)) -> f₁(s₁(g₁(b)))
f₁(f₁(x)) -> b
g₁(x) -> f₁(g₁(x))

Q is empty.
We have to consider all minimal (P,Q,R)-chains.
The approximation of the Dependency Graph [13,14,18] contains 1 SCC with 1 less node.

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ DependencyGraphProof

        ↳ QDP

          ↳ QDPOrderProof

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

G₁(x) -> F₁(g₁(x))
F₁(g₁(a)) -> G₁(b)
G₁(x) -> G₁(x)

The TRS R consists of the following rules:

f₁(g₁(a)) -> f₁(s₁(g₁(b)))
f₁(f₁(x)) -> b
g₁(x) -> f₁(g₁(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.

G₁(x) -> F₁(g₁(x))
F₁(g₁(a)) -> G₁(b)

The remaining pairs can at least be oriented weakly.

G₁(x) -> G₁(x)

Used ordering: Polynomial Order [17,21] with Interpretation:

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

POL( G₁(x₁) ) = 2x₁ + 2

POL( a ) = 2

POL( f₁(x₁) ) = max{0, -2}

POL( s₁(x₁) ) = 2x₁ + 2

POL( g₁(x₁) ) = x₁

POL( F₁(x₁) ) = 2x₁ + 1

The following usable rules [14] were oriented:

f₁(f₁(x)) -> b
g₁(x) -> f₁(g₁(x))
f₁(g₁(a)) -> f₁(s₁(g₁(b)))

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ DependencyGraphProof

        ↳ QDP

          ↳ QDPOrderProof

            ↳ QDP

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

G₁(x) -> G₁(x)

The TRS R consists of the following rules:

f₁(g₁(a)) -> f₁(s₁(g₁(b)))
f₁(f₁(x)) -> b
g₁(x) -> f₁(g₁(x))

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