Termination w.r.t. Q proof of TRS/AProVELiveness

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.

Using Dependency Pairs [1,13] we result in the following initial DP problem:
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 [13,14,18] contains 3 SCCs with 27 less nodes.

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ DependencyGraphProof

        ↳ AND

          ↳ QDP

            ↳ QDPOrderProof

          ↳ 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.
We use the reduction pair processor [13].

The following pairs can be oriented strictly and are deleted.

F₁(no₁(x)) -> F₁(x)

The remaining pairs can at least be oriented weakly.

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

Used ordering: Polynomial interpretation [21]:

POL(ACTIVE₁(x₁)) = 2·x₁
POL(F₁(x₁)) = 2·x₁
POL(active₁(x₁)) = 2·x₁
POL(f₁(x₁)) = x₁
POL(mark₁(x₁)) = x₁
POL(no₁(x₁)) = 1 + x₁

The following usable rules [14] were oriented:

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

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ DependencyGraphProof

        ↳ AND

          ↳ QDP

            ↳ QDPOrderProof

              ↳ QDP

                ↳ QDPOrderProof

          ↳ 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.
We use the reduction pair processor [13].

The following pairs can be oriented strictly and are deleted.

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

The remaining pairs can at least be oriented weakly.

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

Used ordering: Polynomial interpretation [21]:

POL(ACTIVE₁(x₁)) = 2·x₁
POL(F₁(x₁)) = 2·x₁
POL(active₁(x₁)) = 1 + x₁
POL(f₁(x₁)) = x₁
POL(mark₁(x₁)) = x₁
POL(no₁(x₁)) = 0

The following usable rules [14] were oriented:

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

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ DependencyGraphProof

        ↳ AND

          ↳ QDP

            ↳ QDPOrderProof

              ↳ QDP

                ↳ QDPOrderProof

                  ↳ 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 [13,14,18] contains 1 SCC with 1 less node.

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ DependencyGraphProof

        ↳ AND

          ↳ QDP

            ↳ QDPOrderProof

              ↳ QDP

                ↳ QDPOrderProof

                  ↳ QDP

                    ↳ DependencyGraphProof

                      ↳ QDP

                        ↳ QDPOrderProof

          ↳ 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.
We use the reduction pair processor [13].

The following pairs can be oriented strictly and are deleted.

F₁(mark₁(x)) -> F₁(x)

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

POL(F₁(x₁)) = 2·x₁
POL(mark₁(x₁)) = 1 + 2·x₁

The following usable rules [14] were oriented: none

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ DependencyGraphProof

        ↳ AND

          ↳ QDP

            ↳ QDPOrderProof

              ↳ QDP

                ↳ QDPOrderProof

                  ↳ QDP

                    ↳ DependencyGraphProof

                      ↳ QDP

                        ↳ QDPOrderProof

                          ↳ 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

            ↳ QDPOrderProof

          ↳ 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.
We use the reduction pair processor [13].

The following pairs can be oriented strictly and are deleted.

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

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

POL(CHK₁(x₁)) = 2·x₁
POL(X) = 0
POL(c) = 2
POL(f₁(x₁)) = 2·x₁
POL(mat₂(x₁, x₂)) = 2·x₂
POL(no₁(x₁)) = 2 + x₁
POL(y) = 0

The following usable rules [14] were oriented:

mat₂(f₁(x), c) -> no₁(c)
mat₂(f₁(x), f₁(y)) -> f₁(mat₂(x, y))

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ DependencyGraphProof

        ↳ AND

          ↳ QDP

          ↳ QDP

            ↳ QDPOrderProof

              ↳ 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.