Termination w.r.t. Q proof of /tmp/tmpgduOye/Liveness

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 [1,15] we result in the following initial DP problem:
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)))
TP(mark(x)) → F(X)
TP(mark(x)) → F(f(f(X)))
TP(mark(x)) → F(f(f(f(f(f(f(f(f(f(X))))))))))
CHK(no(f(x))) → F(f(f(X)))
CHK(no(c)) → ACTIVE(c)
CHK(no(f(x))) → F(f(f(f(X))))
TP(mark(x)) → F(f(X))
CHK(no(f(x))) → F(chk(mat(f(f(f(f(f(f(f(f(f(f(X)))))))))), x)))
F(no(x)) → 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(X))))))))
F(active(x)) → F(x)
F(mark(x)) → F(x)
MAT(f(x), f(y)) → F(mat(x, y))
CHK(no(f(x))) → 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(X))
ACTIVE(f(x)) → F(f(x))
F(active(x)) → ACTIVE(f(x))
TP(mark(x)) → F(f(f(f(X))))
CHK(no(f(x))) → CHK(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)) → F(f(f(f(f(X)))))
CHK(no(f(x))) → MAT(f(f(f(f(f(f(f(f(f(f(X)))))))))), x)
CHK(no(f(x))) → F(f(f(f(f(f(f(X)))))))
CHK(no(f(x))) → F(f(f(f(f(f(f(f(f(f(X))))))))))
TP(mark(x)) → 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))))))
TP(mark(x)) → MAT(f(f(f(f(f(f(f(f(f(f(X)))))))))), x)
CHK(no(f(x))) → F(X)
MAT(f(x), f(y)) → MAT(x, y)
TP(mark(x)) → F(f(f(f(f(f(f(X)))))))
CHK(no(f(x))) → F(f(f(f(f(f(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

      ↳ DependencyGraphProof

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)))
TP(mark(x)) → F(X)
TP(mark(x)) → F(f(f(X)))
TP(mark(x)) → F(f(f(f(f(f(f(f(f(f(X))))))))))
CHK(no(f(x))) → F(f(f(X)))
CHK(no(c)) → ACTIVE(c)
CHK(no(f(x))) → F(f(f(f(X))))
TP(mark(x)) → F(f(X))
CHK(no(f(x))) → F(chk(mat(f(f(f(f(f(f(f(f(f(f(X)))))))))), x)))
F(no(x)) → 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(X))))))))
F(active(x)) → F(x)
F(mark(x)) → F(x)
MAT(f(x), f(y)) → F(mat(x, y))
CHK(no(f(x))) → 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(X))
ACTIVE(f(x)) → F(f(x))
F(active(x)) → ACTIVE(f(x))
TP(mark(x)) → F(f(f(f(X))))
CHK(no(f(x))) → CHK(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)) → F(f(f(f(f(X)))))
CHK(no(f(x))) → MAT(f(f(f(f(f(f(f(f(f(f(X)))))))))), x)
CHK(no(f(x))) → F(f(f(f(f(f(f(X)))))))
CHK(no(f(x))) → F(f(f(f(f(f(f(f(f(f(X))))))))))
TP(mark(x)) → 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))))))
TP(mark(x)) → MAT(f(f(f(f(f(f(f(f(f(f(X)))))))))), x)
CHK(no(f(x))) → F(X)
MAT(f(x), f(y)) → MAT(x, y)
TP(mark(x)) → F(f(f(f(f(f(f(X)))))))
CHK(no(f(x))) → F(f(f(f(f(f(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 [15,17,22] 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:

F(mark(x)) → F(x)
F(no(x)) → F(x)
ACTIVE(f(x)) → F(f(x))
F(active(x)) → ACTIVE(f(x))
F(active(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 [15].

The following pairs can be oriented strictly and are deleted.

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

The remaining pairs can at least be oriented weakly.

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

Used ordering: Polynomial interpretation [25,35]:

POL(active(x₁)) = 1 + (4)x_1
POL(f(x₁)) = x_1
POL(no(x₁)) = 1 + (2)x_1
POL(mark(x₁)) = x_1
POL(ACTIVE(x₁)) = 4 + x_1
POL(F(x₁)) = 3 + x_1

The value of delta used in the strict ordering is 1.
The following usable rules [17] were oriented:

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

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ DependencyGraphProof

        ↳ AND

          ↳ QDP

            ↳ QDPOrderProof

              ↳ QDP

                ↳ DependencyGraphProof

          ↳ QDP

          ↳ QDP

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

F(mark(x)) → F(x)
F(active(x)) → ACTIVE(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 [15,17,22] contains 1 SCC with 1 less node.

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ DependencyGraphProof

        ↳ AND

          ↳ 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 [15].

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: Polynomial interpretation [25,35]:

POL(mark(x₁)) = 1 + (4)x_1
POL(F(x₁)) = (4)x_1

The value of delta used in the strict ordering is 4.
The following usable rules [17] were oriented: none

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ DependencyGraphProof

        ↳ AND

          ↳ 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

      ↳ 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 [15].

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: Polynomial interpretation [25,35]:

POL(c) = 0
POL(y) = 0
POL(f(x₁)) = 4 + (2)x_1
POL(X) = 3
POL(CHK(x₁)) = (2)x_1
POL(no(x₁)) = (4)x_1
POL(mat(x₁, x₂)) = (3)x_2

The value of delta used in the strict ordering is 32.
The following usable rules [17] were oriented:

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

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

      ↳ 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.