Termination w.r.t. Q of the given QTRS could not be shown:

(0) Obligation:

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.

(1) DependencyPairsProof (EQUIVALENT transformation)

Using Dependency Pairs [AG00,LPAR04] we result in the following initial DP problem.

(2) Obligation:

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

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

(3) DependencyGraphProof (EQUIVALENT transformation)

The approximation of the Dependency Graph [LPAR04,FROCOS05,EDGSTAR] contains 3 SCCs with 27 less nodes.

(4) Complex Obligation (AND)

(5) Obligation:

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

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

(6) QDPOrderProof (EQUIVALENT transformation)

We use the reduction pair processor [LPAR04].

The following pairs can be oriented strictly and are deleted.

F(no(x)) → F(x)

The remaining pairs can at least be oriented weakly.
Used ordering: Combined order from the following AFS and order.
F(x1)  =  x1
active(x1)  =  x1
ACTIVE(x1)  =  x1
f(x1)  =  x1
no(x1)  =  no(x1)
mark(x1)  =  x1
chk(x1)  =  chk
mat(x1, x2)  =  mat
X  =  X
y  =  y
c  =  c
tp(x1)  =  tp

Lexicographic path order with status [LPO].
Precedence:

chk > c
mat > no₁
mat > c

Status:

trivial

The following usable rules [FROCOS05] were oriented:

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)))

(7) Obligation:

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

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

(8) QDPOrderProof (EQUIVALENT transformation)

We use the reduction pair processor [LPAR04].

The following pairs can be oriented strictly and are deleted.

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

The remaining pairs can at least be oriented weakly.
Used ordering: Combined order from the following AFS and order.
F(x1)  =  x1
active(x1)  =  active(x1)
ACTIVE(x1)  =  x1
f(x1)  =  x1
mark(x1)  =  x1
chk(x1)  =  chk
no(x1)  =  no
mat(x1, x2)  =  mat
X  =  X
y  =  y
c  =  c
tp(x1)  =  tp

Lexicographic path order with status [LPO].
Precedence:

chk > active₁
chk > c
mat > no

Status:

trivial

The following usable rules [FROCOS05] were oriented:

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)))

(9) Obligation:

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

ACTIVE(f(x)) → F(f(x))
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.

(10) DependencyGraphProof (EQUIVALENT transformation)

The approximation of the Dependency Graph [LPAR04,FROCOS05,EDGSTAR] contains 1 SCC with 1 less node.

(11) Obligation:

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.

(12) QDPOrderProof (EQUIVALENT transformation)

We use the reduction pair processor [LPAR04].

The following pairs can be oriented strictly and are deleted.

F(mark(x)) → F(x)

The remaining pairs can at least be oriented weakly.
Used ordering: Combined order from the following AFS and order.
F(x1)  =  x1
mark(x1)  =  mark(x1)
active(x1)  =  active(x1)
f(x1)  =  x1
chk(x1)  =  chk
no(x1)  =  no
mat(x1, x2)  =  mat
X  =  X
y  =  y
c  =  c
tp(x1)  =  tp

Lexicographic path order with status [LPO].
Precedence:

chk > active₁ > mark₁
chk > c
mat > no

Status:

trivial

The following usable rules [FROCOS05] were oriented:

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)))

(13) Obligation:

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.

(14) PisEmptyProof (EQUIVALENT transformation)

The TRS P is empty. Hence, there is no (P,Q,R) chain.

(15) TRUE

(16) Obligation:

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.

(17) QDPOrderProof (EQUIVALENT transformation)

We use the reduction pair processor [LPAR04].

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.
Used ordering: Combined order from the following AFS and order.
CHK(x1)  =  x1
no(x1)  =  x1
f(x1)  =  f(x1)
mat(x1, x2)  =  x2
active(x1)  =  active
mark(x1)  =  mark
chk(x1)  =  x1
y  =  y
c  =  c
tp(x1)  =  tp

Lexicographic path order with status [LPO].
Precedence:

c > active > mark

Status:

trivial

The following usable rules [FROCOS05] were oriented:

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)))

(18) Obligation:

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.

(19) PisEmptyProof (EQUIVALENT transformation)

The TRS P is empty. Hence, there is no (P,Q,R) chain.

(20) TRUE

(21) Obligation:

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.