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:

active(f(x)) → mark(f(f(x)))
chk(no(f(x))) → f(chk(mat(f(f(f(f(f(f(f(f(f(f(X)))))))))), x)))
mat(f(x), f(y)) → f(mat(x, y))
chk(no(c)) → active(c)
mat(f(x), c) → no(c)
f(active(x)) → active(f(x))
f(no(x)) → no(f(x))
f(mark(x)) → mark(f(x))
tp(mark(x)) → tp(chk(mat(f(f(f(f(f(f(f(f(f(f(X)))))))))), x)))

Q is empty.

↳ QTRS

  ↳ DependencyPairsProof

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

active(f(x)) → mark(f(f(x)))
chk(no(f(x))) → f(chk(mat(f(f(f(f(f(f(f(f(f(f(X)))))))))), x)))
mat(f(x), f(y)) → f(mat(x, y))
chk(no(c)) → active(c)
mat(f(x), c) → no(c)
f(active(x)) → active(f(x))
f(no(x)) → no(f(x))
f(mark(x)) → mark(f(x))
tp(mark(x)) → tp(chk(mat(f(f(f(f(f(f(f(f(f(f(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:

TP(mark(x)) → F(f(f(f(f(f(f(f(f(X)))))))))
CHK(no(f(x))) → F(f(f(f(f(f(f(f(X))))))))
CHK(no(f(x))) → F(f(f(f(X))))
CHK(no(f(x))) → F(chk(mat(f(f(f(f(f(f(f(f(f(f(X)))))))))), x)))
TP(mark(x)) → F(X)
CHK(no(f(x))) → F(f(f(X)))
CHK(no(f(x))) → F(f(f(f(f(f(X))))))
CHK(no(c)) → ACTIVE(c)
TP(mark(x)) → F(f(f(f(f(X)))))
CHK(no(f(x))) → F(f(f(f(f(f(f(f(f(f(X))))))))))
F(active(x)) → F(x)
TP(mark(x)) → CHK(mat(f(f(f(f(f(f(f(f(f(f(X)))))))))), x))
CHK(no(f(x))) → CHK(mat(f(f(f(f(f(f(f(f(f(f(X)))))))))), x))
CHK(no(f(x))) → F(f(X))
MAT(f(x), f(y)) → MAT(x, y)
TP(mark(x)) → F(f(f(f(f(f(f(f(X))))))))
CHK(no(f(x))) → F(f(f(f(f(X)))))
TP(mark(x)) → MAT(f(f(f(f(f(f(f(f(f(f(X)))))))))), x)
TP(mark(x)) → F(f(f(X)))
F(no(x)) → F(x)
F(active(x)) → ACTIVE(f(x))
ACTIVE(f(x)) → F(f(x))
TP(mark(x)) → F(f(f(f(X))))
MAT(f(x), f(y)) → F(mat(x, y))
F(mark(x)) → F(x)
CHK(no(f(x))) → F(f(f(f(f(f(f(X)))))))
CHK(no(f(x))) → MAT(f(f(f(f(f(f(f(f(f(f(X)))))))))), x)
TP(mark(x)) → F(f(f(f(f(f(X))))))
CHK(no(f(x))) → F(X)
TP(mark(x)) → TP(chk(mat(f(f(f(f(f(f(f(f(f(f(X)))))))))), x)))
TP(mark(x)) → F(f(f(f(f(f(f(X)))))))
CHK(no(f(x))) → F(f(f(f(f(f(f(f(f(X)))))))))
TP(mark(x)) → F(f(f(f(f(f(f(f(f(f(X))))))))))
TP(mark(x)) → F(f(X))

The TRS R consists of the following rules:

active(f(x)) → mark(f(f(x)))
chk(no(f(x))) → f(chk(mat(f(f(f(f(f(f(f(f(f(f(X)))))))))), x)))
mat(f(x), f(y)) → f(mat(x, y))
chk(no(c)) → active(c)
mat(f(x), c) → no(c)
f(active(x)) → active(f(x))
f(no(x)) → no(f(x))
f(mark(x)) → mark(f(x))
tp(mark(x)) → tp(chk(mat(f(f(f(f(f(f(f(f(f(f(X)))))))))), x)))

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

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ EdgeDeletionProof

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

TP(mark(x)) → F(f(f(f(f(f(f(f(f(X)))))))))
CHK(no(f(x))) → F(f(f(f(f(f(f(f(X))))))))
CHK(no(f(x))) → F(f(f(f(X))))
CHK(no(f(x))) → F(chk(mat(f(f(f(f(f(f(f(f(f(f(X)))))))))), x)))
TP(mark(x)) → F(X)
CHK(no(f(x))) → F(f(f(X)))
CHK(no(f(x))) → F(f(f(f(f(f(X))))))
CHK(no(c)) → ACTIVE(c)
TP(mark(x)) → F(f(f(f(f(X)))))
CHK(no(f(x))) → F(f(f(f(f(f(f(f(f(f(X))))))))))
F(active(x)) → F(x)
TP(mark(x)) → CHK(mat(f(f(f(f(f(f(f(f(f(f(X)))))))))), x))
CHK(no(f(x))) → CHK(mat(f(f(f(f(f(f(f(f(f(f(X)))))))))), x))
CHK(no(f(x))) → F(f(X))
MAT(f(x), f(y)) → MAT(x, y)
TP(mark(x)) → F(f(f(f(f(f(f(f(X))))))))
CHK(no(f(x))) → F(f(f(f(f(X)))))
TP(mark(x)) → MAT(f(f(f(f(f(f(f(f(f(f(X)))))))))), x)
TP(mark(x)) → F(f(f(X)))
F(no(x)) → F(x)
F(active(x)) → ACTIVE(f(x))
ACTIVE(f(x)) → F(f(x))
TP(mark(x)) → F(f(f(f(X))))
MAT(f(x), f(y)) → F(mat(x, y))
F(mark(x)) → F(x)
CHK(no(f(x))) → F(f(f(f(f(f(f(X)))))))
CHK(no(f(x))) → MAT(f(f(f(f(f(f(f(f(f(f(X)))))))))), x)
TP(mark(x)) → F(f(f(f(f(f(X))))))
CHK(no(f(x))) → F(X)
TP(mark(x)) → TP(chk(mat(f(f(f(f(f(f(f(f(f(f(X)))))))))), x)))
TP(mark(x)) → F(f(f(f(f(f(f(X)))))))
CHK(no(f(x))) → F(f(f(f(f(f(f(f(f(X)))))))))
TP(mark(x)) → F(f(f(f(f(f(f(f(f(f(X))))))))))
TP(mark(x)) → F(f(X))

The TRS R consists of the following rules:

active(f(x)) → mark(f(f(x)))
chk(no(f(x))) → f(chk(mat(f(f(f(f(f(f(f(f(f(f(X)))))))))), x)))
mat(f(x), f(y)) → f(mat(x, y))
chk(no(c)) → active(c)
mat(f(x), c) → no(c)
f(active(x)) → active(f(x))
f(no(x)) → no(f(x))
f(mark(x)) → mark(f(x))
tp(mark(x)) → tp(chk(mat(f(f(f(f(f(f(f(f(f(f(X)))))))))), x)))

Q is empty.
We have to consider all minimal (P,Q,R)-chains.
We deleted some edges using various graph approximations

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ EdgeDeletionProof

        ↳ QDP

          ↳ DependencyGraphProof

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

CHK(no(f(x))) → F(f(f(f(f(f(f(f(X))))))))
TP(mark(x)) → F(f(f(f(f(f(f(f(f(X)))))))))
CHK(no(f(x))) → F(f(f(f(X))))
CHK(no(f(x))) → F(chk(mat(f(f(f(f(f(f(f(f(f(f(X)))))))))), x)))
CHK(no(f(x))) → F(f(f(f(f(f(X))))))
CHK(no(f(x))) → F(f(f(X)))
TP(mark(x)) → F(X)
CHK(no(c)) → ACTIVE(c)
CHK(no(f(x))) → F(f(f(f(f(f(f(f(f(f(X))))))))))
TP(mark(x)) → F(f(f(f(f(X)))))
F(active(x)) → F(x)
TP(mark(x)) → CHK(mat(f(f(f(f(f(f(f(f(f(f(X)))))))))), x))
CHK(no(f(x))) → CHK(mat(f(f(f(f(f(f(f(f(f(f(X)))))))))), x))
CHK(no(f(x))) → F(f(X))
MAT(f(x), f(y)) → MAT(x, y)
TP(mark(x)) → F(f(f(f(f(f(f(f(X))))))))
CHK(no(f(x))) → F(f(f(f(f(X)))))
TP(mark(x)) → MAT(f(f(f(f(f(f(f(f(f(f(X)))))))))), x)
TP(mark(x)) → F(f(f(X)))
F(no(x)) → F(x)
F(active(x)) → ACTIVE(f(x))
ACTIVE(f(x)) → F(f(x))
TP(mark(x)) → F(f(f(f(X))))
MAT(f(x), f(y)) → F(mat(x, y))
F(mark(x)) → F(x)
CHK(no(f(x))) → F(f(f(f(f(f(f(X)))))))
CHK(no(f(x))) → MAT(f(f(f(f(f(f(f(f(f(f(X)))))))))), x)
TP(mark(x)) → F(f(f(f(f(f(X))))))
TP(mark(x)) → TP(chk(mat(f(f(f(f(f(f(f(f(f(f(X)))))))))), x)))
CHK(no(f(x))) → F(X)
TP(mark(x)) → F(f(f(f(f(f(f(f(f(f(X))))))))))
CHK(no(f(x))) → F(f(f(f(f(f(f(f(f(X)))))))))
TP(mark(x)) → F(f(f(f(f(f(f(X)))))))
TP(mark(x)) → F(f(X))

The TRS R consists of the following rules:

active(f(x)) → mark(f(f(x)))
chk(no(f(x))) → f(chk(mat(f(f(f(f(f(f(f(f(f(f(X)))))))))), x)))
mat(f(x), f(y)) → f(mat(x, y))
chk(no(c)) → active(c)
mat(f(x), c) → no(c)
f(active(x)) → active(f(x))
f(no(x)) → no(f(x))
f(mark(x)) → mark(f(x))
tp(mark(x)) → tp(chk(mat(f(f(f(f(f(f(f(f(f(f(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

      ↳ EdgeDeletionProof

        ↳ QDP

          ↳ DependencyGraphProof

            ↳ AND

              ↳ QDP

                ↳ QDPOrderProof

              ↳ QDP

              ↳ QDP

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

F(mark(x)) → F(x)
F(active(x)) → F(x)
F(no(x)) → F(x)
F(active(x)) → ACTIVE(f(x))
ACTIVE(f(x)) → F(f(x))

The TRS R consists of the following rules:

active(f(x)) → mark(f(f(x)))
chk(no(f(x))) → f(chk(mat(f(f(f(f(f(f(f(f(f(f(X)))))))))), x)))
mat(f(x), f(y)) → f(mat(x, y))
chk(no(c)) → active(c)
mat(f(x), c) → no(c)
f(active(x)) → active(f(x))
f(no(x)) → no(f(x))
f(mark(x)) → mark(f(x))
tp(mark(x)) → tp(chk(mat(f(f(f(f(f(f(f(f(f(f(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.

F(no(x)) → F(x)

The remaining pairs can at least be oriented weakly.

F(mark(x)) → F(x)
F(active(x)) → F(x)
F(active(x)) → ACTIVE(f(x))
ACTIVE(f(x)) → F(f(x))

Used ordering: Combined order from the following AFS and order.
F(x1) = x1
mark(x1) = x1
active(x1) = x1
no(x1) = no(x1)
ACTIVE(x1) = x1
f(x1) = x1

Lexicographic Path Order [19].
Precedence:

trivial

The following usable rules [14] were oriented:

f(active(x)) → active(f(x))
active(f(x)) → mark(f(f(x)))
f(mark(x)) → mark(f(x))
f(no(x)) → no(f(x))

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ EdgeDeletionProof

        ↳ QDP

          ↳ DependencyGraphProof

            ↳ AND

              ↳ QDP

                ↳ QDPOrderProof

                  ↳ QDP

                    ↳ QDPOrderProof

              ↳ QDP

              ↳ QDP

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

F(mark(x)) → F(x)
F(active(x)) → F(x)
F(active(x)) → ACTIVE(f(x))
ACTIVE(f(x)) → F(f(x))

The TRS R consists of the following rules:

active(f(x)) → mark(f(f(x)))
chk(no(f(x))) → f(chk(mat(f(f(f(f(f(f(f(f(f(f(X)))))))))), x)))
mat(f(x), f(y)) → f(mat(x, y))
chk(no(c)) → active(c)
mat(f(x), c) → no(c)
f(active(x)) → active(f(x))
f(no(x)) → no(f(x))
f(mark(x)) → mark(f(x))
tp(mark(x)) → tp(chk(mat(f(f(f(f(f(f(f(f(f(f(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.

F(active(x)) → F(x)
F(active(x)) → ACTIVE(f(x))

The remaining pairs can at least be oriented weakly.

F(mark(x)) → F(x)
ACTIVE(f(x)) → F(f(x))

Used ordering: Combined order from the following AFS and order.
F(x1) = x1
mark(x1) = x1
active(x1) = active(x1)
ACTIVE(x1) = x1
f(x1) = x1
no(x1) = no

Lexicographic Path Order [19].
Precedence:

trivial

The following usable rules [14] were oriented:

f(active(x)) → active(f(x))
active(f(x)) → mark(f(f(x)))
f(mark(x)) → mark(f(x))
f(no(x)) → no(f(x))

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ EdgeDeletionProof

        ↳ QDP

          ↳ DependencyGraphProof

            ↳ AND

              ↳ QDP

                ↳ QDPOrderProof

                  ↳ QDP

                    ↳ QDPOrderProof

                      ↳ QDP

                        ↳ DependencyGraphProof

              ↳ QDP

              ↳ QDP

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

F(mark(x)) → F(x)
ACTIVE(f(x)) → F(f(x))

The TRS R consists of the following rules:

active(f(x)) → mark(f(f(x)))
chk(no(f(x))) → f(chk(mat(f(f(f(f(f(f(f(f(f(f(X)))))))))), x)))
mat(f(x), f(y)) → f(mat(x, y))
chk(no(c)) → active(c)
mat(f(x), c) → no(c)
f(active(x)) → active(f(x))
f(no(x)) → no(f(x))
f(mark(x)) → mark(f(x))
tp(mark(x)) → tp(chk(mat(f(f(f(f(f(f(f(f(f(f(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

      ↳ EdgeDeletionProof

        ↳ QDP

          ↳ DependencyGraphProof

            ↳ AND

              ↳ QDP

                ↳ QDPOrderProof

                  ↳ QDP

                    ↳ QDPOrderProof

                      ↳ QDP

                        ↳ DependencyGraphProof

                          ↳ QDP

                            ↳ QDPOrderProof

              ↳ QDP

              ↳ QDP

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

F(mark(x)) → F(x)

The TRS R consists of the following rules:

active(f(x)) → mark(f(f(x)))
chk(no(f(x))) → f(chk(mat(f(f(f(f(f(f(f(f(f(f(X)))))))))), x)))
mat(f(x), f(y)) → f(mat(x, y))
chk(no(c)) → active(c)
mat(f(x), c) → no(c)
f(active(x)) → active(f(x))
f(no(x)) → no(f(x))
f(mark(x)) → mark(f(x))
tp(mark(x)) → tp(chk(mat(f(f(f(f(f(f(f(f(f(f(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.

F(mark(x)) → F(x)

The remaining pairs can at least be oriented weakly.
none
Used ordering: Combined order from the following AFS and order.
F(x1) = x1
mark(x1) = mark(x1)

Lexicographic Path Order [19].
Precedence:

trivial

The following usable rules [14] were oriented: none

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ EdgeDeletionProof

        ↳ 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:

active(f(x)) → mark(f(f(x)))
chk(no(f(x))) → f(chk(mat(f(f(f(f(f(f(f(f(f(f(X)))))))))), x)))
mat(f(x), f(y)) → f(mat(x, y))
chk(no(c)) → active(c)
mat(f(x), c) → no(c)
f(active(x)) → active(f(x))
f(no(x)) → no(f(x))
f(mark(x)) → mark(f(x))
tp(mark(x)) → tp(chk(mat(f(f(f(f(f(f(f(f(f(f(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

      ↳ EdgeDeletionProof

        ↳ QDP

          ↳ DependencyGraphProof

            ↳ AND

              ↳ QDP

              ↳ QDP

                ↳ QDPOrderProof

              ↳ QDP

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

CHK(no(f(x))) → CHK(mat(f(f(f(f(f(f(f(f(f(f(X)))))))))), x))

The TRS R consists of the following rules:

active(f(x)) → mark(f(f(x)))
chk(no(f(x))) → f(chk(mat(f(f(f(f(f(f(f(f(f(f(X)))))))))), x)))
mat(f(x), f(y)) → f(mat(x, y))
chk(no(c)) → active(c)
mat(f(x), c) → no(c)
f(active(x)) → active(f(x))
f(no(x)) → no(f(x))
f(mark(x)) → mark(f(x))
tp(mark(x)) → tp(chk(mat(f(f(f(f(f(f(f(f(f(f(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.

CHK(no(f(x))) → CHK(mat(f(f(f(f(f(f(f(f(f(f(X)))))))))), x))

The remaining pairs can at least be oriented weakly.
none
Used ordering: Combined order from the following AFS and order.
CHK(x1) = x1
no(x1) = x1
f(x1) = f(x1)
mat(x1, x2) = x2
c = c
y = y

Lexicographic Path Order [19].
Precedence:

trivial

The following usable rules [14] were oriented:

mat(f(x), c) → no(c)
mat(f(x), f(y)) → f(mat(x, y))

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ EdgeDeletionProof

        ↳ QDP

          ↳ DependencyGraphProof

            ↳ AND

              ↳ QDP

              ↳ QDP

                ↳ QDPOrderProof

                  ↳ QDP

                    ↳ PisEmptyProof

              ↳ QDP

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

active(f(x)) → mark(f(f(x)))
chk(no(f(x))) → f(chk(mat(f(f(f(f(f(f(f(f(f(f(X)))))))))), x)))
mat(f(x), f(y)) → f(mat(x, y))
chk(no(c)) → active(c)
mat(f(x), c) → no(c)
f(active(x)) → active(f(x))
f(no(x)) → no(f(x))
f(mark(x)) → mark(f(x))
tp(mark(x)) → tp(chk(mat(f(f(f(f(f(f(f(f(f(f(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

      ↳ EdgeDeletionProof

        ↳ QDP

          ↳ DependencyGraphProof

            ↳ AND

              ↳ QDP

              ↳ QDP

              ↳ QDP

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

TP(mark(x)) → TP(chk(mat(f(f(f(f(f(f(f(f(f(f(X)))))))))), x)))

The TRS R consists of the following rules:

active(f(x)) → mark(f(f(x)))
chk(no(f(x))) → f(chk(mat(f(f(f(f(f(f(f(f(f(f(X)))))))))), x)))
mat(f(x), f(y)) → f(mat(x, y))
chk(no(c)) → active(c)
mat(f(x), c) → no(c)
f(active(x)) → active(f(x))
f(no(x)) → no(f(x))
f(mark(x)) → mark(f(x))
tp(mark(x)) → tp(chk(mat(f(f(f(f(f(f(f(f(f(f(X)))))))))), x)))

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