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 Order [17,21] with Interpretation:

POL( 0 ) = 1

POL( s₁(x₁) ) = x₁ + 1

POL( F₃(x₁, ..., x₃) ) = 3x₃ + 1

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 Order [17,21] with Interpretation:

POL( 0 ) = max{0, -3}

POL( s₁(x₁) ) = 3x₁ + 2

POL( F₃(x₁, ..., x₃) ) = max{0, 3x₂ - 3}

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 Order [17,21] with Interpretation:

POL( s₁(x₁) ) = x₁ + 2

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

POL( F₃(x₁, ..., x₃) ) = max{0, x₂ - 3}

POL( f₃(x₁, ..., x₃) ) = 0

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 Order [17,21] with Interpretation:

POL( 0 ) = 1

POL( s₁(x₁) ) = 3x₁ + 1

POL( F₃(x₁, ..., x₃) ) = max{0, 3x₃ - 2}

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

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

POL( s₁(x₁) ) = 2x₁ + 1

POL( 0 ) = 2

POL( F₃(x₁, ..., x₃) ) = x₁

POL( f₃(x₁, ..., x₃) ) = max{0, -2}

The following usable rules [14] were oriented: none

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ DependencyGraphProof

        ↳ AND

          ↳ QDP

          ↳ QDP

          ↳ QDP

          ↳ QDP

            ↳ QDPOrderProof

              ↳ 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 Order [17,21] with Interpretation:

POL( s₁(x₁) ) = 3x₁ + 1

POL( F₃(x₁, ..., x₃) ) = max{0, x₁ + 3x₃ - 2}

The following usable rules [14] were oriented: none

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ DependencyGraphProof

        ↳ AND

          ↳ QDP

          ↳ QDP

          ↳ QDP

          ↳ QDP

            ↳ QDPOrderProof

              ↳ 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.