Termination w.r.t. Q proof of TRS/AG01#3.49.trs

Termination w.r.t. Q of the following Term Rewriting System could be proven:

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

f₁(c₂(s₁(x), y)) -> f₁(c₂(x, s₁(y)))
f₁(c₂(s₁(x), s₁(y))) -> g₁(c₂(x, y))
g₁(c₂(x, s₁(y))) -> g₁(c₂(s₁(x), y))
g₁(c₂(s₁(x), s₁(y))) -> f₁(c₂(x, y))

Q is empty.

↳ QTRS

  ↳ DependencyPairsProof

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

f₁(c₂(s₁(x), y)) -> f₁(c₂(x, s₁(y)))
f₁(c₂(s₁(x), s₁(y))) -> g₁(c₂(x, y))
g₁(c₂(x, s₁(y))) -> g₁(c₂(s₁(x), y))
g₁(c₂(s₁(x), s₁(y))) -> f₁(c₂(x, y))

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:

F₁(c₂(s₁(x), s₁(y))) -> G₁(c₂(x, y))
F₁(c₂(s₁(x), y)) -> F₁(c₂(x, s₁(y)))
G₁(c₂(x, s₁(y))) -> G₁(c₂(s₁(x), y))
G₁(c₂(s₁(x), s₁(y))) -> F₁(c₂(x, y))

The TRS R consists of the following rules:

f₁(c₂(s₁(x), y)) -> f₁(c₂(x, s₁(y)))
f₁(c₂(s₁(x), s₁(y))) -> g₁(c₂(x, y))
g₁(c₂(x, s₁(y))) -> g₁(c₂(s₁(x), y))
g₁(c₂(s₁(x), s₁(y))) -> f₁(c₂(x, y))

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

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ QDPOrderProof

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

F₁(c₂(s₁(x), s₁(y))) -> G₁(c₂(x, y))
F₁(c₂(s₁(x), y)) -> F₁(c₂(x, s₁(y)))
G₁(c₂(x, s₁(y))) -> G₁(c₂(s₁(x), y))
G₁(c₂(s₁(x), s₁(y))) -> F₁(c₂(x, y))

The TRS R consists of the following rules:

f₁(c₂(s₁(x), y)) -> f₁(c₂(x, s₁(y)))
f₁(c₂(s₁(x), s₁(y))) -> g₁(c₂(x, y))
g₁(c₂(x, s₁(y))) -> g₁(c₂(s₁(x), y))
g₁(c₂(s₁(x), s₁(y))) -> f₁(c₂(x, y))

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.

F₁(c₂(s₁(x), s₁(y))) -> G₁(c₂(x, y))
G₁(c₂(s₁(x), s₁(y))) -> F₁(c₂(x, y))

The remaining pairs can at least be oriented weakly.

F₁(c₂(s₁(x), y)) -> F₁(c₂(x, s₁(y)))
G₁(c₂(x, s₁(y))) -> G₁(c₂(s₁(x), y))

Used ordering: Polynomial interpretation [21]:

POL(F₁(x₁)) = x₁
POL(G₁(x₁)) = x₁
POL(c₂(x₁, x₂)) = 1 + x₁ + x₂
POL(s₁(x₁)) = 1 + x₁

The following usable rules [14] were oriented: none

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ QDPOrderProof

        ↳ QDP

          ↳ DependencyGraphProof

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

F₁(c₂(s₁(x), y)) -> F₁(c₂(x, s₁(y)))
G₁(c₂(x, s₁(y))) -> G₁(c₂(s₁(x), y))

The TRS R consists of the following rules:

f₁(c₂(s₁(x), y)) -> f₁(c₂(x, s₁(y)))
f₁(c₂(s₁(x), s₁(y))) -> g₁(c₂(x, y))
g₁(c₂(x, s₁(y))) -> g₁(c₂(s₁(x), y))
g₁(c₂(s₁(x), s₁(y))) -> f₁(c₂(x, y))

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.

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ QDPOrderProof

        ↳ QDP

          ↳ DependencyGraphProof

            ↳ AND

              ↳ QDP

                ↳ QDPOrderProof

              ↳ QDP

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

G₁(c₂(x, s₁(y))) -> G₁(c₂(s₁(x), y))

The TRS R consists of the following rules:

f₁(c₂(s₁(x), y)) -> f₁(c₂(x, s₁(y)))
f₁(c₂(s₁(x), s₁(y))) -> g₁(c₂(x, y))
g₁(c₂(x, s₁(y))) -> g₁(c₂(s₁(x), y))
g₁(c₂(s₁(x), s₁(y))) -> f₁(c₂(x, y))

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(c₂(x, s₁(y))) -> G₁(c₂(s₁(x), y))

The remaining pairs can at least be oriented weakly.
none
Used ordering: Polynomial interpretation [21]:

POL(G₁(x₁)) = x₁
POL(c₂(x₁, x₂)) = x₂
POL(s₁(x₁)) = 1 + x₁

The following usable rules [14] were oriented: none

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ QDPOrderProof

        ↳ QDP

          ↳ DependencyGraphProof

            ↳ AND

              ↳ QDP

                ↳ QDPOrderProof

                  ↳ QDP

                    ↳ PisEmptyProof

              ↳ QDP

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

f₁(c₂(s₁(x), y)) -> f₁(c₂(x, s₁(y)))
f₁(c₂(s₁(x), s₁(y))) -> g₁(c₂(x, y))
g₁(c₂(x, s₁(y))) -> g₁(c₂(s₁(x), y))
g₁(c₂(s₁(x), s₁(y))) -> f₁(c₂(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

      ↳ QDPOrderProof

        ↳ QDP

          ↳ DependencyGraphProof

            ↳ AND

              ↳ QDP

              ↳ QDP

                ↳ QDPOrderProof

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

F₁(c₂(s₁(x), y)) -> F₁(c₂(x, s₁(y)))

The TRS R consists of the following rules:

f₁(c₂(s₁(x), y)) -> f₁(c₂(x, s₁(y)))
f₁(c₂(s₁(x), s₁(y))) -> g₁(c₂(x, y))
g₁(c₂(x, s₁(y))) -> g₁(c₂(s₁(x), y))
g₁(c₂(s₁(x), s₁(y))) -> f₁(c₂(x, y))

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.

F₁(c₂(s₁(x), y)) -> F₁(c₂(x, s₁(y)))

The remaining pairs can at least be oriented weakly.
none
Used ordering: Polynomial interpretation [21]:

POL(F₁(x₁)) = x₁
POL(c₂(x₁, x₂)) = x₁
POL(s₁(x₁)) = 1 + x₁

The following usable rules [14] were oriented: none

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ QDPOrderProof

        ↳ QDP

          ↳ DependencyGraphProof

            ↳ AND

              ↳ QDP

              ↳ QDP

                ↳ QDPOrderProof

                  ↳ QDP

                    ↳ PisEmptyProof

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

f₁(c₂(s₁(x), y)) -> f₁(c₂(x, s₁(y)))
f₁(c₂(s₁(x), s₁(y))) -> g₁(c₂(x, y))
g₁(c₂(x, s₁(y))) -> g₁(c₂(s₁(x), y))
g₁(c₂(s₁(x), s₁(y))) -> f₁(c₂(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.