Termination w.r.t. Q proof of /home/nowonder/forschung/aprove/satrung/aprove/lpar4/AProVESLASHLiveness

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:

active₁(f₁(x)) -> mark₁(f₁(f₁(x)))
chk₁(no₁(f₁(x))) -> f₁(chk₁(mat₂(f₁(f₁(f₁(f₁(f₁(f₁(f₁(f₁(f₁(f₁(X)))))))))), x)))
mat₂(f₁(x), f₁(y)) -> f₁(mat₂(x, y))
chk₁(no₁(c)) -> active₁(c)
mat₂(f₁(x), c) -> no₁(c)
f₁(active₁(x)) -> active₁(f₁(x))
f₁(no₁(x)) -> no₁(f₁(x))
f₁(mark₁(x)) -> mark₁(f₁(x))
tp₁(mark₁(x)) -> tp₁(chk₁(mat₂(f₁(f₁(f₁(f₁(f₁(f₁(f₁(f₁(f₁(f₁(X)))))))))), x)))

Q is empty.

↳ QTRS

  ↳ DependencyPairsProof

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

active₁(f₁(x)) -> mark₁(f₁(f₁(x)))
chk₁(no₁(f₁(x))) -> f₁(chk₁(mat₂(f₁(f₁(f₁(f₁(f₁(f₁(f₁(f₁(f₁(f₁(X)))))))))), x)))
mat₂(f₁(x), f₁(y)) -> f₁(mat₂(x, y))
chk₁(no₁(c)) -> active₁(c)
mat₂(f₁(x), c) -> no₁(c)
f₁(active₁(x)) -> active₁(f₁(x))
f₁(no₁(x)) -> no₁(f₁(x))
f₁(mark₁(x)) -> mark₁(f₁(x))
tp₁(mark₁(x)) -> tp₁(chk₁(mat₂(f₁(f₁(f₁(f₁(f₁(f₁(f₁(f₁(f₁(f₁(X)))))))))), x)))

Q is empty.

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

TP₁(mark₁(x)) -> F₁(f₁(f₁(f₁(f₁(f₁(f₁(f₁(f₁(X)))))))))
CHK₁(no₁(f₁(x))) -> F₁(f₁(f₁(f₁(f₁(f₁(f₁(f₁(X))))))))
CHK₁(no₁(f₁(x))) -> F₁(f₁(f₁(f₁(X))))
CHK₁(no₁(f₁(x))) -> F₁(chk₁(mat₂(f₁(f₁(f₁(f₁(f₁(f₁(f₁(f₁(f₁(f₁(X)))))))))), x)))
TP₁(mark₁(x)) -> F₁(X)
CHK₁(no₁(f₁(x))) -> F₁(f₁(f₁(X)))
CHK₁(no₁(f₁(x))) -> F₁(f₁(f₁(f₁(f₁(f₁(X))))))
CHK₁(no₁(c)) -> ACTIVE₁(c)
TP₁(mark₁(x)) -> F₁(f₁(f₁(f₁(f₁(X)))))
CHK₁(no₁(f₁(x))) -> F₁(f₁(f₁(f₁(f₁(f₁(f₁(f₁(f₁(f₁(X))))))))))
F₁(active₁(x)) -> F₁(x)
TP₁(mark₁(x)) -> CHK₁(mat₂(f₁(f₁(f₁(f₁(f₁(f₁(f₁(f₁(f₁(f₁(X)))))))))), x))
CHK₁(no₁(f₁(x))) -> CHK₁(mat₂(f₁(f₁(f₁(f₁(f₁(f₁(f₁(f₁(f₁(f₁(X)))))))))), x))
CHK₁(no₁(f₁(x))) -> F₁(f₁(X))
MAT₂(f₁(x), f₁(y)) -> MAT₂(x, y)
TP₁(mark₁(x)) -> F₁(f₁(f₁(f₁(f₁(f₁(f₁(f₁(X))))))))
CHK₁(no₁(f₁(x))) -> F₁(f₁(f₁(f₁(f₁(X)))))
TP₁(mark₁(x)) -> MAT₂(f₁(f₁(f₁(f₁(f₁(f₁(f₁(f₁(f₁(f₁(X)))))))))), x)
TP₁(mark₁(x)) -> F₁(f₁(f₁(X)))
F₁(no₁(x)) -> F₁(x)
F₁(active₁(x)) -> ACTIVE₁(f₁(x))
ACTIVE₁(f₁(x)) -> F₁(f₁(x))
TP₁(mark₁(x)) -> F₁(f₁(f₁(f₁(X))))
MAT₂(f₁(x), f₁(y)) -> F₁(mat₂(x, y))
F₁(mark₁(x)) -> F₁(x)
CHK₁(no₁(f₁(x))) -> F₁(f₁(f₁(f₁(f₁(f₁(f₁(X)))))))
CHK₁(no₁(f₁(x))) -> MAT₂(f₁(f₁(f₁(f₁(f₁(f₁(f₁(f₁(f₁(f₁(X)))))))))), x)
TP₁(mark₁(x)) -> F₁(f₁(f₁(f₁(f₁(f₁(X))))))
CHK₁(no₁(f₁(x))) -> F₁(X)
TP₁(mark₁(x)) -> TP₁(chk₁(mat₂(f₁(f₁(f₁(f₁(f₁(f₁(f₁(f₁(f₁(f₁(X)))))))))), x)))
TP₁(mark₁(x)) -> F₁(f₁(f₁(f₁(f₁(f₁(f₁(X)))))))
CHK₁(no₁(f₁(x))) -> F₁(f₁(f₁(f₁(f₁(f₁(f₁(f₁(f₁(X)))))))))
TP₁(mark₁(x)) -> F₁(f₁(f₁(f₁(f₁(f₁(f₁(f₁(f₁(f₁(X))))))))))
TP₁(mark₁(x)) -> F₁(f₁(X))

The TRS R consists of the following rules:

active₁(f₁(x)) -> mark₁(f₁(f₁(x)))
chk₁(no₁(f₁(x))) -> f₁(chk₁(mat₂(f₁(f₁(f₁(f₁(f₁(f₁(f₁(f₁(f₁(f₁(X)))))))))), x)))
mat₂(f₁(x), f₁(y)) -> f₁(mat₂(x, y))
chk₁(no₁(c)) -> active₁(c)
mat₂(f₁(x), c) -> no₁(c)
f₁(active₁(x)) -> active₁(f₁(x))
f₁(no₁(x)) -> no₁(f₁(x))
f₁(mark₁(x)) -> mark₁(f₁(x))
tp₁(mark₁(x)) -> tp₁(chk₁(mat₂(f₁(f₁(f₁(f₁(f₁(f₁(f₁(f₁(f₁(f₁(X)))))))))), 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:

TP₁(mark₁(x)) -> F₁(f₁(f₁(f₁(f₁(f₁(f₁(f₁(f₁(X)))))))))
CHK₁(no₁(f₁(x))) -> F₁(f₁(f₁(f₁(f₁(f₁(f₁(f₁(X))))))))
CHK₁(no₁(f₁(x))) -> F₁(f₁(f₁(f₁(X))))
CHK₁(no₁(f₁(x))) -> F₁(chk₁(mat₂(f₁(f₁(f₁(f₁(f₁(f₁(f₁(f₁(f₁(f₁(X)))))))))), x)))
TP₁(mark₁(x)) -> F₁(X)
CHK₁(no₁(f₁(x))) -> F₁(f₁(f₁(X)))
CHK₁(no₁(f₁(x))) -> F₁(f₁(f₁(f₁(f₁(f₁(X))))))
CHK₁(no₁(c)) -> ACTIVE₁(c)
TP₁(mark₁(x)) -> F₁(f₁(f₁(f₁(f₁(X)))))
CHK₁(no₁(f₁(x))) -> F₁(f₁(f₁(f₁(f₁(f₁(f₁(f₁(f₁(f₁(X))))))))))
F₁(active₁(x)) -> F₁(x)
TP₁(mark₁(x)) -> CHK₁(mat₂(f₁(f₁(f₁(f₁(f₁(f₁(f₁(f₁(f₁(f₁(X)))))))))), x))
CHK₁(no₁(f₁(x))) -> CHK₁(mat₂(f₁(f₁(f₁(f₁(f₁(f₁(f₁(f₁(f₁(f₁(X)))))))))), x))
CHK₁(no₁(f₁(x))) -> F₁(f₁(X))
MAT₂(f₁(x), f₁(y)) -> MAT₂(x, y)
TP₁(mark₁(x)) -> F₁(f₁(f₁(f₁(f₁(f₁(f₁(f₁(X))))))))
CHK₁(no₁(f₁(x))) -> F₁(f₁(f₁(f₁(f₁(X)))))
TP₁(mark₁(x)) -> MAT₂(f₁(f₁(f₁(f₁(f₁(f₁(f₁(f₁(f₁(f₁(X)))))))))), x)
TP₁(mark₁(x)) -> F₁(f₁(f₁(X)))
F₁(no₁(x)) -> F₁(x)
F₁(active₁(x)) -> ACTIVE₁(f₁(x))
ACTIVE₁(f₁(x)) -> F₁(f₁(x))
TP₁(mark₁(x)) -> F₁(f₁(f₁(f₁(X))))
MAT₂(f₁(x), f₁(y)) -> F₁(mat₂(x, y))
F₁(mark₁(x)) -> F₁(x)
CHK₁(no₁(f₁(x))) -> F₁(f₁(f₁(f₁(f₁(f₁(f₁(X)))))))
CHK₁(no₁(f₁(x))) -> MAT₂(f₁(f₁(f₁(f₁(f₁(f₁(f₁(f₁(f₁(f₁(X)))))))))), x)
TP₁(mark₁(x)) -> F₁(f₁(f₁(f₁(f₁(f₁(X))))))
CHK₁(no₁(f₁(x))) -> F₁(X)
TP₁(mark₁(x)) -> TP₁(chk₁(mat₂(f₁(f₁(f₁(f₁(f₁(f₁(f₁(f₁(f₁(f₁(X)))))))))), x)))
TP₁(mark₁(x)) -> F₁(f₁(f₁(f₁(f₁(f₁(f₁(X)))))))
CHK₁(no₁(f₁(x))) -> F₁(f₁(f₁(f₁(f₁(f₁(f₁(f₁(f₁(X)))))))))
TP₁(mark₁(x)) -> F₁(f₁(f₁(f₁(f₁(f₁(f₁(f₁(f₁(f₁(X))))))))))
TP₁(mark₁(x)) -> F₁(f₁(X))

The TRS R consists of the following rules:

active₁(f₁(x)) -> mark₁(f₁(f₁(x)))
chk₁(no₁(f₁(x))) -> f₁(chk₁(mat₂(f₁(f₁(f₁(f₁(f₁(f₁(f₁(f₁(f₁(f₁(X)))))))))), x)))
mat₂(f₁(x), f₁(y)) -> f₁(mat₂(x, y))
chk₁(no₁(c)) -> active₁(c)
mat₂(f₁(x), c) -> no₁(c)
f₁(active₁(x)) -> active₁(f₁(x))
f₁(no₁(x)) -> no₁(f₁(x))
f₁(mark₁(x)) -> mark₁(f₁(x))
tp₁(mark₁(x)) -> tp₁(chk₁(mat₂(f₁(f₁(f₁(f₁(f₁(f₁(f₁(f₁(f₁(f₁(X)))))))))), x)))

Q is empty.
We have to consider all minimal (P,Q,R)-chains.
The approximation of the Dependency Graph contains 3 SCCs with 27 less nodes.

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ DependencyGraphProof

        ↳ AND

          ↳ QDP

            ↳ QDPAfsSolverProof

          ↳ QDP

          ↳ QDP

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

F₁(mark₁(x)) -> F₁(x)
F₁(active₁(x)) -> F₁(x)
F₁(no₁(x)) -> F₁(x)
F₁(active₁(x)) -> ACTIVE₁(f₁(x))
ACTIVE₁(f₁(x)) -> F₁(f₁(x))

The TRS R consists of the following rules:

active₁(f₁(x)) -> mark₁(f₁(f₁(x)))
chk₁(no₁(f₁(x))) -> f₁(chk₁(mat₂(f₁(f₁(f₁(f₁(f₁(f₁(f₁(f₁(f₁(f₁(X)))))))))), x)))
mat₂(f₁(x), f₁(y)) -> f₁(mat₂(x, y))
chk₁(no₁(c)) -> active₁(c)
mat₂(f₁(x), c) -> no₁(c)
f₁(active₁(x)) -> active₁(f₁(x))
f₁(no₁(x)) -> no₁(f₁(x))
f₁(mark₁(x)) -> mark₁(f₁(x))
tp₁(mark₁(x)) -> tp₁(chk₁(mat₂(f₁(f₁(f₁(f₁(f₁(f₁(f₁(f₁(f₁(f₁(X)))))))))), x)))

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.

F₁(no₁(x)) -> F₁(x)

Used argument filtering: F₁(x1) = x1
mark₁(x1) = x1
active₁(x1) = x1
no₁(x1) = no₁(x1)
ACTIVE₁(x1) = x1
f₁(x1) = x1
Used ordering: Quasi Precedence: trivial

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ DependencyGraphProof

        ↳ AND

          ↳ QDP

            ↳ QDPAfsSolverProof

              ↳ QDP

                ↳ QDPAfsSolverProof

          ↳ QDP

          ↳ QDP

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

F₁(mark₁(x)) -> F₁(x)
F₁(active₁(x)) -> F₁(x)
F₁(active₁(x)) -> ACTIVE₁(f₁(x))
ACTIVE₁(f₁(x)) -> F₁(f₁(x))

The TRS R consists of the following rules:

active₁(f₁(x)) -> mark₁(f₁(f₁(x)))
chk₁(no₁(f₁(x))) -> f₁(chk₁(mat₂(f₁(f₁(f₁(f₁(f₁(f₁(f₁(f₁(f₁(f₁(X)))))))))), x)))
mat₂(f₁(x), f₁(y)) -> f₁(mat₂(x, y))
chk₁(no₁(c)) -> active₁(c)
mat₂(f₁(x), c) -> no₁(c)
f₁(active₁(x)) -> active₁(f₁(x))
f₁(no₁(x)) -> no₁(f₁(x))
f₁(mark₁(x)) -> mark₁(f₁(x))
tp₁(mark₁(x)) -> tp₁(chk₁(mat₂(f₁(f₁(f₁(f₁(f₁(f₁(f₁(f₁(f₁(f₁(X)))))))))), x)))

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.

F₁(active₁(x)) -> F₁(x)
F₁(active₁(x)) -> ACTIVE₁(f₁(x))

Used argument filtering: F₁(x1) = x1
mark₁(x1) = x1
active₁(x1) = active₁(x1)
ACTIVE₁(x1) = x1
f₁(x1) = x1
no₁(x1) = no
Used ordering: Quasi Precedence: trivial

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ DependencyGraphProof

        ↳ AND

          ↳ QDP

            ↳ QDPAfsSolverProof

              ↳ QDP

                ↳ QDPAfsSolverProof

                  ↳ QDP

                    ↳ DependencyGraphProof

          ↳ QDP

          ↳ QDP

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

F₁(mark₁(x)) -> F₁(x)
ACTIVE₁(f₁(x)) -> F₁(f₁(x))

The TRS R consists of the following rules:

active₁(f₁(x)) -> mark₁(f₁(f₁(x)))
chk₁(no₁(f₁(x))) -> f₁(chk₁(mat₂(f₁(f₁(f₁(f₁(f₁(f₁(f₁(f₁(f₁(f₁(X)))))))))), x)))
mat₂(f₁(x), f₁(y)) -> f₁(mat₂(x, y))
chk₁(no₁(c)) -> active₁(c)
mat₂(f₁(x), c) -> no₁(c)
f₁(active₁(x)) -> active₁(f₁(x))
f₁(no₁(x)) -> no₁(f₁(x))
f₁(mark₁(x)) -> mark₁(f₁(x))
tp₁(mark₁(x)) -> tp₁(chk₁(mat₂(f₁(f₁(f₁(f₁(f₁(f₁(f₁(f₁(f₁(f₁(X)))))))))), x)))

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

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ DependencyGraphProof

        ↳ AND

          ↳ QDP

            ↳ QDPAfsSolverProof

              ↳ QDP

                ↳ QDPAfsSolverProof

                  ↳ QDP

                    ↳ DependencyGraphProof

                      ↳ QDP

                        ↳ QDPAfsSolverProof

          ↳ QDP

          ↳ QDP

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

F₁(mark₁(x)) -> F₁(x)

The TRS R consists of the following rules:

active₁(f₁(x)) -> mark₁(f₁(f₁(x)))
chk₁(no₁(f₁(x))) -> f₁(chk₁(mat₂(f₁(f₁(f₁(f₁(f₁(f₁(f₁(f₁(f₁(f₁(X)))))))))), x)))
mat₂(f₁(x), f₁(y)) -> f₁(mat₂(x, y))
chk₁(no₁(c)) -> active₁(c)
mat₂(f₁(x), c) -> no₁(c)
f₁(active₁(x)) -> active₁(f₁(x))
f₁(no₁(x)) -> no₁(f₁(x))
f₁(mark₁(x)) -> mark₁(f₁(x))
tp₁(mark₁(x)) -> tp₁(chk₁(mat₂(f₁(f₁(f₁(f₁(f₁(f₁(f₁(f₁(f₁(f₁(X)))))))))), x)))

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.

F₁(mark₁(x)) -> F₁(x)

Used argument filtering: F₁(x1) = x1
mark₁(x1) = mark₁(x1)
Used ordering: Quasi Precedence: trivial

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ DependencyGraphProof

        ↳ AND

          ↳ QDP

            ↳ QDPAfsSolverProof

              ↳ QDP

                ↳ QDPAfsSolverProof

                  ↳ QDP

                    ↳ DependencyGraphProof

                      ↳ QDP

                        ↳ QDPAfsSolverProof

                          ↳ QDP

                            ↳ PisEmptyProof

          ↳ QDP

          ↳ QDP

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

active₁(f₁(x)) -> mark₁(f₁(f₁(x)))
chk₁(no₁(f₁(x))) -> f₁(chk₁(mat₂(f₁(f₁(f₁(f₁(f₁(f₁(f₁(f₁(f₁(f₁(X)))))))))), x)))
mat₂(f₁(x), f₁(y)) -> f₁(mat₂(x, y))
chk₁(no₁(c)) -> active₁(c)
mat₂(f₁(x), c) -> no₁(c)
f₁(active₁(x)) -> active₁(f₁(x))
f₁(no₁(x)) -> no₁(f₁(x))
f₁(mark₁(x)) -> mark₁(f₁(x))
tp₁(mark₁(x)) -> tp₁(chk₁(mat₂(f₁(f₁(f₁(f₁(f₁(f₁(f₁(f₁(f₁(f₁(X)))))))))), x)))

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

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

CHK₁(no₁(f₁(x))) -> CHK₁(mat₂(f₁(f₁(f₁(f₁(f₁(f₁(f₁(f₁(f₁(f₁(X)))))))))), x))

The TRS R consists of the following rules:

active₁(f₁(x)) -> mark₁(f₁(f₁(x)))
chk₁(no₁(f₁(x))) -> f₁(chk₁(mat₂(f₁(f₁(f₁(f₁(f₁(f₁(f₁(f₁(f₁(f₁(X)))))))))), x)))
mat₂(f₁(x), f₁(y)) -> f₁(mat₂(x, y))
chk₁(no₁(c)) -> active₁(c)
mat₂(f₁(x), c) -> no₁(c)
f₁(active₁(x)) -> active₁(f₁(x))
f₁(no₁(x)) -> no₁(f₁(x))
f₁(mark₁(x)) -> mark₁(f₁(x))
tp₁(mark₁(x)) -> tp₁(chk₁(mat₂(f₁(f₁(f₁(f₁(f₁(f₁(f₁(f₁(f₁(f₁(X)))))))))), x)))

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.

CHK₁(no₁(f₁(x))) -> CHK₁(mat₂(f₁(f₁(f₁(f₁(f₁(f₁(f₁(f₁(f₁(f₁(X)))))))))), x))

Used argument filtering: CHK₁(x1) = x1
no₁(x1) = no₁(x1)
f₁(x1) = f
mat₂(x1, x2) = mat
c = c
Used ordering: Quasi Precedence: [f, mat] > no_1 [f, mat] > c

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ DependencyGraphProof

        ↳ AND

          ↳ QDP

          ↳ QDP

            ↳ QDPAfsSolverProof

              ↳ QDP

                ↳ PisEmptyProof

          ↳ QDP

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

active₁(f₁(x)) -> mark₁(f₁(f₁(x)))
chk₁(no₁(f₁(x))) -> f₁(chk₁(mat₂(f₁(f₁(f₁(f₁(f₁(f₁(f₁(f₁(f₁(f₁(X)))))))))), x)))
mat₂(f₁(x), f₁(y)) -> f₁(mat₂(x, y))
chk₁(no₁(c)) -> active₁(c)
mat₂(f₁(x), c) -> no₁(c)
f₁(active₁(x)) -> active₁(f₁(x))
f₁(no₁(x)) -> no₁(f₁(x))
f₁(mark₁(x)) -> mark₁(f₁(x))
tp₁(mark₁(x)) -> tp₁(chk₁(mat₂(f₁(f₁(f₁(f₁(f₁(f₁(f₁(f₁(f₁(f₁(X)))))))))), x)))

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

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

TP₁(mark₁(x)) -> TP₁(chk₁(mat₂(f₁(f₁(f₁(f₁(f₁(f₁(f₁(f₁(f₁(f₁(X)))))))))), x)))

The TRS R consists of the following rules:

active₁(f₁(x)) -> mark₁(f₁(f₁(x)))
chk₁(no₁(f₁(x))) -> f₁(chk₁(mat₂(f₁(f₁(f₁(f₁(f₁(f₁(f₁(f₁(f₁(f₁(X)))))))))), x)))
mat₂(f₁(x), f₁(y)) -> f₁(mat₂(x, y))
chk₁(no₁(c)) -> active₁(c)
mat₂(f₁(x), c) -> no₁(c)
f₁(active₁(x)) -> active₁(f₁(x))
f₁(no₁(x)) -> no₁(f₁(x))
f₁(mark₁(x)) -> mark₁(f₁(x))
tp₁(mark₁(x)) -> tp₁(chk₁(mat₂(f₁(f₁(f₁(f₁(f₁(f₁(f₁(f₁(f₁(f₁(X)))))))))), x)))

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