Termination w.r.t. Q proof of /home/nowonder/forschung/aprove/satrung/aprove/lpar1/SK90SLASH2.52.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.

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 contains 4 SCCs with 1 less node.

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ DependencyGraphProof

        ↳ AND

          ↳ QDP

            ↳ QDPAfsSolverProof

          ↳ 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.
By using an argument filtering and a montonic ordering, at least one Dependency Pair of this SCC can be strictly oriented.

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

Used argument filtering: F₃(x1, x2, x3) = x3
s₁(x1) = s₁(x1)
Used ordering: Precedence:

trivial

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ DependencyGraphProof

        ↳ AND

          ↳ QDP

            ↳ QDPAfsSolverProof

              ↳ 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

            ↳ QDPAfsSolverProof

          ↳ 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.
By using an argument filtering and a montonic ordering, at least one Dependency Pair of this SCC can be strictly oriented.

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

Used argument filtering: F₃(x1, x2, x3) = x2
s₁(x1) = s₁(x1)
Used ordering: Precedence:

trivial

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ DependencyGraphProof

        ↳ AND

          ↳ QDP

          ↳ QDP

            ↳ QDPAfsSolverProof

              ↳ 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

            ↳ QDPAfsSolverProof

          ↳ 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.
By using an argument filtering and a montonic ordering, at least one Dependency Pair of this SCC can be strictly oriented.

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

Used argument filtering: F₃(x1, x2, x3) = x2
s₁(x1) = s₁(x1)
f₃(x1, x2, x3) = f₃(x1, x2, x3)
0 = 0
Used ordering: Precedence:

f₃ > s₁

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ DependencyGraphProof

        ↳ AND

          ↳ QDP

          ↳ QDP

          ↳ QDP

            ↳ QDPAfsSolverProof

              ↳ QDP

                ↳ QDPAfsSolverProof

          ↳ 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.
By using an argument filtering and a montonic ordering, at least one Dependency Pair of this SCC can be strictly oriented.

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

Used argument filtering: F₃(x1, x2, x3) = x3
s₁(x1) = s₁(x1)
Used ordering: Precedence:

trivial

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ DependencyGraphProof

        ↳ AND

          ↳ QDP

          ↳ QDP

          ↳ QDP

            ↳ QDPAfsSolverProof

              ↳ QDP

                ↳ QDPAfsSolverProof

                  ↳ 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

            ↳ QDPAfsSolverProof

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.
By using an argument filtering and a montonic ordering, at least one Dependency Pair of this SCC can be strictly oriented.

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))

Used argument filtering: F₃(x1, x2, x3) = x1
s₁(x1) = s₁(x1)
f₃(x1, x2, x3) = f₃(x1, x2, x3)
0 = 0
Used ordering: Precedence:

f₃ > s₁

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ DependencyGraphProof

        ↳ AND

          ↳ QDP

          ↳ QDP

          ↳ QDP

          ↳ QDP

            ↳ QDPAfsSolverProof

              ↳ QDP

                ↳ QDPAfsSolverProof

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.
By using an argument filtering and a montonic ordering, at least one Dependency Pair of this SCC can be strictly oriented.

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

Used argument filtering: F₃(x1, x2, x3) = x3
s₁(x1) = s₁(x1)
Used ordering: Precedence:

trivial

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ DependencyGraphProof

        ↳ AND

          ↳ QDP

          ↳ QDP

          ↳ QDP

          ↳ QDP

            ↳ QDPAfsSolverProof

              ↳ QDP

                ↳ QDPAfsSolverProof

                  ↳ 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.