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

The following usable rules [14] were oriented:

f1(mark1(x)) -> mark1(f1(x))
f1(no1(x)) -> no1(f1(x))
active1(f1(x)) -> mark1(f1(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)
ACTIVE1(f1(x)) -> F1(f1(x))
The remaining pairs can at least be oriented weakly.

F1(mark1(x)) -> F1(x)
F1(active1(x)) -> ACTIVE1(f1(x))
Used ordering: Polynomial interpretation [21]:

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

The following usable rules [14] were oriented:

f1(mark1(x)) -> mark1(f1(x))
f1(no1(x)) -> no1(f1(x))
active1(f1(x)) -> mark1(f1(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)
F1(active1(x)) -> ACTIVE1(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)) = x1   
POL(mark1(x1)) = 1 + 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)) = x1   
POL(X) = 0   
POL(c) = 0   
POL(f1(x1)) = 1   
POL(mat2(x1, x2)) = x1   
POL(no1(x1)) = 1 + 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.