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.