Termination w.r.t. Q proof of TRS/SK902.52.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:

f₃(x, 0, 0) -> s₁(x)
f₃(0, y, 0) -> s₁(y)
f₃(0, 0, z) -> s₁(z)
f₃(s₁(0), y, z) -> f₃(0, s₁(y), s₁(z))
f₃(s₁(x), s₁(y), 0) -> f₃(x, y, s₁(0))
f₃(s₁(x), 0, s₁(z)) -> f₃(x, s₁(0), z)
f₃(0, s₁(0), s₁(0)) -> s₁(s₁(0))
f₃(s₁(x), s₁(y), s₁(z)) -> f₃(x, y, f₃(s₁(x), s₁(y), z))
f₃(0, s₁(s₁(y)), s₁(0)) -> f₃(0, y, s₁(0))
f₃(0, s₁(0), s₁(s₁(z))) -> f₃(0, s₁(0), z)
f₃(0, s₁(s₁(y)), s₁(s₁(z))) -> f₃(0, y, f₃(0, s₁(s₁(y)), s₁(z)))

Q is empty.

↳ QTRS

  ↳ DependencyPairsProof

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

f₃(x, 0, 0) -> s₁(x)
f₃(0, y, 0) -> s₁(y)
f₃(0, 0, z) -> s₁(z)
f₃(s₁(0), y, z) -> f₃(0, s₁(y), s₁(z))
f₃(s₁(x), s₁(y), 0) -> f₃(x, y, s₁(0))
f₃(s₁(x), 0, s₁(z)) -> f₃(x, s₁(0), z)
f₃(0, s₁(0), s₁(0)) -> s₁(s₁(0))
f₃(s₁(x), s₁(y), s₁(z)) -> f₃(x, y, f₃(s₁(x), s₁(y), z))
f₃(0, s₁(s₁(y)), s₁(0)) -> f₃(0, y, s₁(0))
f₃(0, s₁(0), s₁(s₁(z))) -> f₃(0, s₁(0), z)
f₃(0, s₁(s₁(y)), s₁(s₁(z))) -> f₃(0, y, f₃(0, s₁(s₁(y)), s₁(z)))

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:

F₃(s₁(0), y, z) -> F₃(0, s₁(y), s₁(z))
F₃(s₁(x), 0, s₁(z)) -> F₃(x, s₁(0), z)
F₃(s₁(x), s₁(y), 0) -> F₃(x, y, s₁(0))
F₃(0, s₁(s₁(y)), s₁(s₁(z))) -> F₃(0, s₁(s₁(y)), s₁(z))
F₃(s₁(x), s₁(y), s₁(z)) -> F₃(s₁(x), s₁(y), z)
F₃(0, s₁(s₁(y)), s₁(s₁(z))) -> F₃(0, y, f₃(0, s₁(s₁(y)), s₁(z)))
F₃(0, s₁(0), s₁(s₁(z))) -> F₃(0, s₁(0), z)
F₃(0, s₁(s₁(y)), s₁(0)) -> F₃(0, y, s₁(0))
F₃(s₁(x), s₁(y), s₁(z)) -> F₃(x, y, f₃(s₁(x), s₁(y), z))

The TRS R consists of the following rules:

f₃(x, 0, 0) -> s₁(x)
f₃(0, y, 0) -> s₁(y)
f₃(0, 0, z) -> s₁(z)
f₃(s₁(0), y, z) -> f₃(0, s₁(y), s₁(z))
f₃(s₁(x), s₁(y), 0) -> f₃(x, y, s₁(0))
f₃(s₁(x), 0, s₁(z)) -> f₃(x, s₁(0), z)
f₃(0, s₁(0), s₁(0)) -> s₁(s₁(0))
f₃(s₁(x), s₁(y), s₁(z)) -> f₃(x, y, f₃(s₁(x), s₁(y), z))
f₃(0, s₁(s₁(y)), s₁(0)) -> f₃(0, y, s₁(0))
f₃(0, s₁(0), s₁(s₁(z))) -> f₃(0, s₁(0), z)
f₃(0, s₁(s₁(y)), s₁(s₁(z))) -> f₃(0, y, f₃(0, s₁(s₁(y)), s₁(z)))

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:

F₃(s₁(0), y, z) -> F₃(0, s₁(y), s₁(z))
F₃(s₁(x), 0, s₁(z)) -> F₃(x, s₁(0), z)
F₃(s₁(x), s₁(y), 0) -> F₃(x, y, s₁(0))
F₃(0, s₁(s₁(y)), s₁(s₁(z))) -> F₃(0, s₁(s₁(y)), s₁(z))
F₃(s₁(x), s₁(y), s₁(z)) -> F₃(s₁(x), s₁(y), z)
F₃(0, s₁(s₁(y)), s₁(s₁(z))) -> F₃(0, y, f₃(0, s₁(s₁(y)), s₁(z)))
F₃(0, s₁(0), s₁(s₁(z))) -> F₃(0, s₁(0), z)
F₃(0, s₁(s₁(y)), s₁(0)) -> F₃(0, y, s₁(0))
F₃(s₁(x), s₁(y), s₁(z)) -> F₃(x, y, f₃(s₁(x), s₁(y), z))

The TRS R consists of the following rules:

f₃(x, 0, 0) -> s₁(x)
f₃(0, y, 0) -> s₁(y)
f₃(0, 0, z) -> s₁(z)
f₃(s₁(0), y, z) -> f₃(0, s₁(y), s₁(z))
f₃(s₁(x), s₁(y), 0) -> f₃(x, y, s₁(0))
f₃(s₁(x), 0, s₁(z)) -> f₃(x, s₁(0), z)
f₃(0, s₁(0), s₁(0)) -> s₁(s₁(0))
f₃(s₁(x), s₁(y), s₁(z)) -> f₃(x, y, f₃(s₁(x), s₁(y), z))
f₃(0, s₁(s₁(y)), s₁(0)) -> f₃(0, y, s₁(0))
f₃(0, s₁(0), s₁(s₁(z))) -> f₃(0, s₁(0), z)
f₃(0, s₁(s₁(y)), s₁(s₁(z))) -> f₃(0, y, f₃(0, s₁(s₁(y)), s₁(z)))

Q is empty.
We have to consider all minimal (P,Q,R)-chains.
The approximation of the Dependency Graph [13,14,18] contains 4 SCCs with 1 less node.

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ DependencyGraphProof

        ↳ AND

          ↳ QDP

            ↳ QDPOrderProof

          ↳ QDP

          ↳ QDP

          ↳ QDP

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

F₃(0, s₁(0), s₁(s₁(z))) -> F₃(0, s₁(0), z)

The TRS R consists of the following rules:

f₃(x, 0, 0) -> s₁(x)
f₃(0, y, 0) -> s₁(y)
f₃(0, 0, z) -> s₁(z)
f₃(s₁(0), y, z) -> f₃(0, s₁(y), s₁(z))
f₃(s₁(x), s₁(y), 0) -> f₃(x, y, s₁(0))
f₃(s₁(x), 0, s₁(z)) -> f₃(x, s₁(0), z)
f₃(0, s₁(0), s₁(0)) -> s₁(s₁(0))
f₃(s₁(x), s₁(y), s₁(z)) -> f₃(x, y, f₃(s₁(x), s₁(y), z))
f₃(0, s₁(s₁(y)), s₁(0)) -> f₃(0, y, s₁(0))
f₃(0, s₁(0), s₁(s₁(z))) -> f₃(0, s₁(0), z)
f₃(0, s₁(s₁(y)), s₁(s₁(z))) -> f₃(0, y, f₃(0, s₁(s₁(y)), s₁(z)))

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.

F₃(0, s₁(0), s₁(s₁(z))) -> F₃(0, s₁(0), z)

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

POL(0) = 0
POL(F₃(x₁, x₂, x₃)) = x₃
POL(s₁(x₁)) = 1 + x₁

The following usable rules [14] were oriented: none

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ DependencyGraphProof

        ↳ AND

          ↳ QDP

            ↳ QDPOrderProof

              ↳ QDP

                ↳ PisEmptyProof

          ↳ QDP

          ↳ QDP

          ↳ QDP

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

f₃(x, 0, 0) -> s₁(x)
f₃(0, y, 0) -> s₁(y)
f₃(0, 0, z) -> s₁(z)
f₃(s₁(0), y, z) -> f₃(0, s₁(y), s₁(z))
f₃(s₁(x), s₁(y), 0) -> f₃(x, y, s₁(0))
f₃(s₁(x), 0, s₁(z)) -> f₃(x, s₁(0), z)
f₃(0, s₁(0), s₁(0)) -> s₁(s₁(0))
f₃(s₁(x), s₁(y), s₁(z)) -> f₃(x, y, f₃(s₁(x), s₁(y), z))
f₃(0, s₁(s₁(y)), s₁(0)) -> f₃(0, y, s₁(0))
f₃(0, s₁(0), s₁(s₁(z))) -> f₃(0, s₁(0), z)
f₃(0, s₁(s₁(y)), s₁(s₁(z))) -> f₃(0, y, f₃(0, s₁(s₁(y)), s₁(z)))

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

            ↳ QDPOrderProof

          ↳ QDP

          ↳ QDP

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

F₃(0, s₁(s₁(y)), s₁(0)) -> F₃(0, y, s₁(0))

The TRS R consists of the following rules:

f₃(x, 0, 0) -> s₁(x)
f₃(0, y, 0) -> s₁(y)
f₃(0, 0, z) -> s₁(z)
f₃(s₁(0), y, z) -> f₃(0, s₁(y), s₁(z))
f₃(s₁(x), s₁(y), 0) -> f₃(x, y, s₁(0))
f₃(s₁(x), 0, s₁(z)) -> f₃(x, s₁(0), z)
f₃(0, s₁(0), s₁(0)) -> s₁(s₁(0))
f₃(s₁(x), s₁(y), s₁(z)) -> f₃(x, y, f₃(s₁(x), s₁(y), z))
f₃(0, s₁(s₁(y)), s₁(0)) -> f₃(0, y, s₁(0))
f₃(0, s₁(0), s₁(s₁(z))) -> f₃(0, s₁(0), z)
f₃(0, s₁(s₁(y)), s₁(s₁(z))) -> f₃(0, y, f₃(0, s₁(s₁(y)), s₁(z)))

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.

F₃(0, s₁(s₁(y)), s₁(0)) -> F₃(0, y, s₁(0))

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

POL(0) = 0
POL(F₃(x₁, x₂, x₃)) = x₂
POL(s₁(x₁)) = 1 + x₁

The following usable rules [14] were oriented: none

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ DependencyGraphProof

        ↳ AND

          ↳ QDP

          ↳ QDP

            ↳ QDPOrderProof

              ↳ QDP

                ↳ PisEmptyProof

          ↳ QDP

          ↳ QDP

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

f₃(x, 0, 0) -> s₁(x)
f₃(0, y, 0) -> s₁(y)
f₃(0, 0, z) -> s₁(z)
f₃(s₁(0), y, z) -> f₃(0, s₁(y), s₁(z))
f₃(s₁(x), s₁(y), 0) -> f₃(x, y, s₁(0))
f₃(s₁(x), 0, s₁(z)) -> f₃(x, s₁(0), z)
f₃(0, s₁(0), s₁(0)) -> s₁(s₁(0))
f₃(s₁(x), s₁(y), s₁(z)) -> f₃(x, y, f₃(s₁(x), s₁(y), z))
f₃(0, s₁(s₁(y)), s₁(0)) -> f₃(0, y, s₁(0))
f₃(0, s₁(0), s₁(s₁(z))) -> f₃(0, s₁(0), z)
f₃(0, s₁(s₁(y)), s₁(s₁(z))) -> f₃(0, y, f₃(0, s₁(s₁(y)), s₁(z)))

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

          ↳ QDP

            ↳ QDPOrderProof

          ↳ QDP

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

F₃(0, s₁(s₁(y)), s₁(s₁(z))) -> F₃(0, s₁(s₁(y)), s₁(z))
F₃(0, s₁(s₁(y)), s₁(s₁(z))) -> F₃(0, y, f₃(0, s₁(s₁(y)), s₁(z)))

The TRS R consists of the following rules:

f₃(x, 0, 0) -> s₁(x)
f₃(0, y, 0) -> s₁(y)
f₃(0, 0, z) -> s₁(z)
f₃(s₁(0), y, z) -> f₃(0, s₁(y), s₁(z))
f₃(s₁(x), s₁(y), 0) -> f₃(x, y, s₁(0))
f₃(s₁(x), 0, s₁(z)) -> f₃(x, s₁(0), z)
f₃(0, s₁(0), s₁(0)) -> s₁(s₁(0))
f₃(s₁(x), s₁(y), s₁(z)) -> f₃(x, y, f₃(s₁(x), s₁(y), z))
f₃(0, s₁(s₁(y)), s₁(0)) -> f₃(0, y, s₁(0))
f₃(0, s₁(0), s₁(s₁(z))) -> f₃(0, s₁(0), z)
f₃(0, s₁(s₁(y)), s₁(s₁(z))) -> f₃(0, y, f₃(0, s₁(s₁(y)), s₁(z)))

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.

F₃(0, s₁(s₁(y)), s₁(s₁(z))) -> F₃(0, y, f₃(0, s₁(s₁(y)), s₁(z)))

The remaining pairs can at least be oriented weakly.

F₃(0, s₁(s₁(y)), s₁(s₁(z))) -> F₃(0, s₁(s₁(y)), s₁(z))

Used ordering: Polynomial interpretation [21]:

POL(0) = 0
POL(F₃(x₁, x₂, x₃)) = x₂
POL(f₃(x₁, x₂, x₃)) = 0
POL(s₁(x₁)) = 1 + x₁

The following usable rules [14] were oriented: none

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ DependencyGraphProof

        ↳ AND

          ↳ QDP

          ↳ QDP

          ↳ QDP

            ↳ QDPOrderProof

              ↳ QDP

                ↳ QDPOrderProof

          ↳ QDP

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

F₃(0, s₁(s₁(y)), s₁(s₁(z))) -> F₃(0, s₁(s₁(y)), s₁(z))

The TRS R consists of the following rules:

f₃(x, 0, 0) -> s₁(x)
f₃(0, y, 0) -> s₁(y)
f₃(0, 0, z) -> s₁(z)
f₃(s₁(0), y, z) -> f₃(0, s₁(y), s₁(z))
f₃(s₁(x), s₁(y), 0) -> f₃(x, y, s₁(0))
f₃(s₁(x), 0, s₁(z)) -> f₃(x, s₁(0), z)
f₃(0, s₁(0), s₁(0)) -> s₁(s₁(0))
f₃(s₁(x), s₁(y), s₁(z)) -> f₃(x, y, f₃(s₁(x), s₁(y), z))
f₃(0, s₁(s₁(y)), s₁(0)) -> f₃(0, y, s₁(0))
f₃(0, s₁(0), s₁(s₁(z))) -> f₃(0, s₁(0), z)
f₃(0, s₁(s₁(y)), s₁(s₁(z))) -> f₃(0, y, f₃(0, s₁(s₁(y)), s₁(z)))

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.

F₃(0, s₁(s₁(y)), s₁(s₁(z))) -> F₃(0, s₁(s₁(y)), s₁(z))

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

POL(0) = 0
POL(F₃(x₁, x₂, x₃)) = x₃
POL(s₁(x₁)) = 1 + x₁

The following usable rules [14] were oriented: none

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ DependencyGraphProof

        ↳ AND

          ↳ QDP

          ↳ QDP

          ↳ QDP

            ↳ QDPOrderProof

              ↳ QDP

                ↳ QDPOrderProof

                  ↳ QDP

                    ↳ PisEmptyProof

          ↳ QDP

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

f₃(x, 0, 0) -> s₁(x)
f₃(0, y, 0) -> s₁(y)
f₃(0, 0, z) -> s₁(z)
f₃(s₁(0), y, z) -> f₃(0, s₁(y), s₁(z))
f₃(s₁(x), s₁(y), 0) -> f₃(x, y, s₁(0))
f₃(s₁(x), 0, s₁(z)) -> f₃(x, s₁(0), z)
f₃(0, s₁(0), s₁(0)) -> s₁(s₁(0))
f₃(s₁(x), s₁(y), s₁(z)) -> f₃(x, y, f₃(s₁(x), s₁(y), z))
f₃(0, s₁(s₁(y)), s₁(0)) -> f₃(0, y, s₁(0))
f₃(0, s₁(0), s₁(s₁(z))) -> f₃(0, s₁(0), z)
f₃(0, s₁(s₁(y)), s₁(s₁(z))) -> f₃(0, y, f₃(0, s₁(s₁(y)), s₁(z)))

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

          ↳ QDP

          ↳ QDP

            ↳ QDPOrderProof

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

F₃(s₁(x), s₁(y), 0) -> F₃(x, y, s₁(0))
F₃(s₁(x), 0, s₁(z)) -> F₃(x, s₁(0), z)
F₃(s₁(x), s₁(y), s₁(z)) -> F₃(s₁(x), s₁(y), z)
F₃(s₁(x), s₁(y), s₁(z)) -> F₃(x, y, f₃(s₁(x), s₁(y), z))

The TRS R consists of the following rules:

f₃(x, 0, 0) -> s₁(x)
f₃(0, y, 0) -> s₁(y)
f₃(0, 0, z) -> s₁(z)
f₃(s₁(0), y, z) -> f₃(0, s₁(y), s₁(z))
f₃(s₁(x), s₁(y), 0) -> f₃(x, y, s₁(0))
f₃(s₁(x), 0, s₁(z)) -> f₃(x, s₁(0), z)
f₃(0, s₁(0), s₁(0)) -> s₁(s₁(0))
f₃(s₁(x), s₁(y), s₁(z)) -> f₃(x, y, f₃(s₁(x), s₁(y), z))
f₃(0, s₁(s₁(y)), s₁(0)) -> f₃(0, y, s₁(0))
f₃(0, s₁(0), s₁(s₁(z))) -> f₃(0, s₁(0), z)
f₃(0, s₁(s₁(y)), s₁(s₁(z))) -> f₃(0, y, f₃(0, s₁(s₁(y)), s₁(z)))

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.

F₃(s₁(x), s₁(y), 0) -> F₃(x, y, s₁(0))
F₃(s₁(x), s₁(y), s₁(z)) -> F₃(x, y, f₃(s₁(x), s₁(y), z))

The remaining pairs can at least be oriented weakly.

F₃(s₁(x), 0, s₁(z)) -> F₃(x, s₁(0), z)
F₃(s₁(x), s₁(y), s₁(z)) -> F₃(s₁(x), s₁(y), z)

Used ordering: Polynomial interpretation [21]:

POL(0) = 1
POL(F₃(x₁, x₂, x₃)) = x₁ + x₂
POL(f₃(x₁, x₂, x₃)) = 0
POL(s₁(x₁)) = 1 + x₁

The following usable rules [14] were oriented: none

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ DependencyGraphProof

        ↳ AND

          ↳ QDP

          ↳ QDP

          ↳ QDP

          ↳ QDP

            ↳ QDPOrderProof

              ↳ QDP

                ↳ DependencyGraphProof

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

F₃(s₁(x), 0, s₁(z)) -> F₃(x, s₁(0), z)
F₃(s₁(x), s₁(y), s₁(z)) -> F₃(s₁(x), s₁(y), z)

The TRS R consists of the following rules:

f₃(x, 0, 0) -> s₁(x)
f₃(0, y, 0) -> s₁(y)
f₃(0, 0, z) -> s₁(z)
f₃(s₁(0), y, z) -> f₃(0, s₁(y), s₁(z))
f₃(s₁(x), s₁(y), 0) -> f₃(x, y, s₁(0))
f₃(s₁(x), 0, s₁(z)) -> f₃(x, s₁(0), z)
f₃(0, s₁(0), s₁(0)) -> s₁(s₁(0))
f₃(s₁(x), s₁(y), s₁(z)) -> f₃(x, y, f₃(s₁(x), s₁(y), z))
f₃(0, s₁(s₁(y)), s₁(0)) -> f₃(0, y, s₁(0))
f₃(0, s₁(0), s₁(s₁(z))) -> f₃(0, s₁(0), z)
f₃(0, s₁(s₁(y)), s₁(s₁(z))) -> f₃(0, y, f₃(0, s₁(s₁(y)), s₁(z)))

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

        ↳ AND

          ↳ QDP

          ↳ QDP

          ↳ QDP

          ↳ QDP

            ↳ QDPOrderProof

              ↳ QDP

                ↳ DependencyGraphProof

                  ↳ QDP

                    ↳ QDPOrderProof

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

F₃(s₁(x), s₁(y), s₁(z)) -> F₃(s₁(x), s₁(y), z)

The TRS R consists of the following rules:

f₃(x, 0, 0) -> s₁(x)
f₃(0, y, 0) -> s₁(y)
f₃(0, 0, z) -> s₁(z)
f₃(s₁(0), y, z) -> f₃(0, s₁(y), s₁(z))
f₃(s₁(x), s₁(y), 0) -> f₃(x, y, s₁(0))
f₃(s₁(x), 0, s₁(z)) -> f₃(x, s₁(0), z)
f₃(0, s₁(0), s₁(0)) -> s₁(s₁(0))
f₃(s₁(x), s₁(y), s₁(z)) -> f₃(x, y, f₃(s₁(x), s₁(y), z))
f₃(0, s₁(s₁(y)), s₁(0)) -> f₃(0, y, s₁(0))
f₃(0, s₁(0), s₁(s₁(z))) -> f₃(0, s₁(0), z)
f₃(0, s₁(s₁(y)), s₁(s₁(z))) -> f₃(0, y, f₃(0, s₁(s₁(y)), s₁(z)))

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.

F₃(s₁(x), s₁(y), s₁(z)) -> F₃(s₁(x), s₁(y), z)

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

POL(F₃(x₁, x₂, x₃)) = x₃
POL(s₁(x₁)) = 1 + x₁

The following usable rules [14] were oriented: none

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ DependencyGraphProof

        ↳ AND

          ↳ QDP

          ↳ QDP

          ↳ QDP

          ↳ QDP

            ↳ QDPOrderProof

              ↳ QDP

                ↳ DependencyGraphProof

                  ↳ QDP

                    ↳ QDPOrderProof

                      ↳ QDP

                        ↳ PisEmptyProof

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

f₃(x, 0, 0) -> s₁(x)
f₃(0, y, 0) -> s₁(y)
f₃(0, 0, z) -> s₁(z)
f₃(s₁(0), y, z) -> f₃(0, s₁(y), s₁(z))
f₃(s₁(x), s₁(y), 0) -> f₃(x, y, s₁(0))
f₃(s₁(x), 0, s₁(z)) -> f₃(x, s₁(0), z)
f₃(0, s₁(0), s₁(0)) -> s₁(s₁(0))
f₃(s₁(x), s₁(y), s₁(z)) -> f₃(x, y, f₃(s₁(x), s₁(y), z))
f₃(0, s₁(s₁(y)), s₁(0)) -> f₃(0, y, s₁(0))
f₃(0, s₁(0), s₁(s₁(z))) -> f₃(0, s₁(0), z)
f₃(0, s₁(s₁(y)), s₁(s₁(z))) -> f₃(0, y, f₃(0, s₁(s₁(y)), s₁(z)))

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.