Termination w.r.t. Q of the following Term Rewriting System could be proven:

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

active1(f1(x)) -> mark1(x)
top1(active1(c)) -> top1(mark1(c))
top1(mark1(x)) -> top1(check1(x))
check1(f1(x)) -> f1(check1(x))
check1(x) -> start1(match2(f1(X), x))
match2(f1(x), f1(y)) -> f1(match2(x, y))
match2(X, x) -> proper1(x)
proper1(c) -> ok1(c)
proper1(f1(x)) -> f1(proper1(x))
f1(ok1(x)) -> ok1(f1(x))
start1(ok1(x)) -> found1(x)
f1(found1(x)) -> found1(f1(x))
top1(found1(x)) -> top1(active1(x))
active1(f1(x)) -> f1(active1(x))
f1(mark1(x)) -> mark1(f1(x))

Q is empty.


QTRS
  ↳ DependencyPairsProof

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

active1(f1(x)) -> mark1(x)
top1(active1(c)) -> top1(mark1(c))
top1(mark1(x)) -> top1(check1(x))
check1(f1(x)) -> f1(check1(x))
check1(x) -> start1(match2(f1(X), x))
match2(f1(x), f1(y)) -> f1(match2(x, y))
match2(X, x) -> proper1(x)
proper1(c) -> ok1(c)
proper1(f1(x)) -> f1(proper1(x))
f1(ok1(x)) -> ok1(f1(x))
start1(ok1(x)) -> found1(x)
f1(found1(x)) -> found1(f1(x))
top1(found1(x)) -> top1(active1(x))
active1(f1(x)) -> f1(active1(x))
f1(mark1(x)) -> mark1(f1(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:

TOP1(found1(x)) -> TOP1(active1(x))
CHECK1(f1(x)) -> CHECK1(x)
MATCH2(X, x) -> PROPER1(x)
ACTIVE1(f1(x)) -> ACTIVE1(x)
TOP1(mark1(x)) -> CHECK1(x)
ACTIVE1(f1(x)) -> F1(active1(x))
CHECK1(x) -> MATCH2(f1(X), x)
TOP1(mark1(x)) -> TOP1(check1(x))
CHECK1(f1(x)) -> F1(check1(x))
F1(mark1(x)) -> F1(x)
CHECK1(x) -> F1(X)
MATCH2(f1(x), f1(y)) -> F1(match2(x, y))
PROPER1(f1(x)) -> PROPER1(x)
TOP1(found1(x)) -> ACTIVE1(x)
F1(ok1(x)) -> F1(x)
CHECK1(x) -> START1(match2(f1(X), x))
TOP1(active1(c)) -> TOP1(mark1(c))
MATCH2(f1(x), f1(y)) -> MATCH2(x, y)
F1(found1(x)) -> F1(x)
PROPER1(f1(x)) -> F1(proper1(x))

The TRS R consists of the following rules:

active1(f1(x)) -> mark1(x)
top1(active1(c)) -> top1(mark1(c))
top1(mark1(x)) -> top1(check1(x))
check1(f1(x)) -> f1(check1(x))
check1(x) -> start1(match2(f1(X), x))
match2(f1(x), f1(y)) -> f1(match2(x, y))
match2(X, x) -> proper1(x)
proper1(c) -> ok1(c)
proper1(f1(x)) -> f1(proper1(x))
f1(ok1(x)) -> ok1(f1(x))
start1(ok1(x)) -> found1(x)
f1(found1(x)) -> found1(f1(x))
top1(found1(x)) -> top1(active1(x))
active1(f1(x)) -> f1(active1(x))
f1(mark1(x)) -> mark1(f1(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:

TOP1(found1(x)) -> TOP1(active1(x))
CHECK1(f1(x)) -> CHECK1(x)
MATCH2(X, x) -> PROPER1(x)
ACTIVE1(f1(x)) -> ACTIVE1(x)
TOP1(mark1(x)) -> CHECK1(x)
ACTIVE1(f1(x)) -> F1(active1(x))
CHECK1(x) -> MATCH2(f1(X), x)
TOP1(mark1(x)) -> TOP1(check1(x))
CHECK1(f1(x)) -> F1(check1(x))
F1(mark1(x)) -> F1(x)
CHECK1(x) -> F1(X)
MATCH2(f1(x), f1(y)) -> F1(match2(x, y))
PROPER1(f1(x)) -> PROPER1(x)
TOP1(found1(x)) -> ACTIVE1(x)
F1(ok1(x)) -> F1(x)
CHECK1(x) -> START1(match2(f1(X), x))
TOP1(active1(c)) -> TOP1(mark1(c))
MATCH2(f1(x), f1(y)) -> MATCH2(x, y)
F1(found1(x)) -> F1(x)
PROPER1(f1(x)) -> F1(proper1(x))

The TRS R consists of the following rules:

active1(f1(x)) -> mark1(x)
top1(active1(c)) -> top1(mark1(c))
top1(mark1(x)) -> top1(check1(x))
check1(f1(x)) -> f1(check1(x))
check1(x) -> start1(match2(f1(X), x))
match2(f1(x), f1(y)) -> f1(match2(x, y))
match2(X, x) -> proper1(x)
proper1(c) -> ok1(c)
proper1(f1(x)) -> f1(proper1(x))
f1(ok1(x)) -> ok1(f1(x))
start1(ok1(x)) -> found1(x)
f1(found1(x)) -> found1(f1(x))
top1(found1(x)) -> top1(active1(x))
active1(f1(x)) -> f1(active1(x))
f1(mark1(x)) -> mark1(f1(x))

Q is empty.
We have to consider all minimal (P,Q,R)-chains.
The approximation of the Dependency Graph [13,14,18] contains 6 SCCs with 10 less nodes.

↳ QTRS
  ↳ DependencyPairsProof
    ↳ QDP
      ↳ DependencyGraphProof
        ↳ AND
QDP
            ↳ QDPOrderProof
          ↳ QDP
          ↳ QDP
          ↳ QDP
          ↳ QDP
          ↳ QDP

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

F1(mark1(x)) -> F1(x)
F1(ok1(x)) -> F1(x)
F1(found1(x)) -> F1(x)

The TRS R consists of the following rules:

active1(f1(x)) -> mark1(x)
top1(active1(c)) -> top1(mark1(c))
top1(mark1(x)) -> top1(check1(x))
check1(f1(x)) -> f1(check1(x))
check1(x) -> start1(match2(f1(X), x))
match2(f1(x), f1(y)) -> f1(match2(x, y))
match2(X, x) -> proper1(x)
proper1(c) -> ok1(c)
proper1(f1(x)) -> f1(proper1(x))
f1(ok1(x)) -> ok1(f1(x))
start1(ok1(x)) -> found1(x)
f1(found1(x)) -> found1(f1(x))
top1(found1(x)) -> top1(active1(x))
active1(f1(x)) -> f1(active1(x))
f1(mark1(x)) -> mark1(f1(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 strictly oriented and are deleted.


F1(ok1(x)) -> F1(x)
The remaining pairs can at least by weakly be oriented.

F1(mark1(x)) -> F1(x)
F1(found1(x)) -> F1(x)
Used ordering: Combined order from the following AFS and order.
F1(x1)  =  F1(x1)
mark1(x1)  =  x1
ok1(x1)  =  ok1(x1)
found1(x1)  =  x1

Lexicographic Path Order [19].
Precedence:
[F1, ok1]


The following usable rules [14] were oriented: none



↳ QTRS
  ↳ DependencyPairsProof
    ↳ QDP
      ↳ DependencyGraphProof
        ↳ AND
          ↳ QDP
            ↳ QDPOrderProof
QDP
                ↳ QDPOrderProof
          ↳ QDP
          ↳ QDP
          ↳ QDP
          ↳ QDP
          ↳ QDP

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

F1(mark1(x)) -> F1(x)
F1(found1(x)) -> F1(x)

The TRS R consists of the following rules:

active1(f1(x)) -> mark1(x)
top1(active1(c)) -> top1(mark1(c))
top1(mark1(x)) -> top1(check1(x))
check1(f1(x)) -> f1(check1(x))
check1(x) -> start1(match2(f1(X), x))
match2(f1(x), f1(y)) -> f1(match2(x, y))
match2(X, x) -> proper1(x)
proper1(c) -> ok1(c)
proper1(f1(x)) -> f1(proper1(x))
f1(ok1(x)) -> ok1(f1(x))
start1(ok1(x)) -> found1(x)
f1(found1(x)) -> found1(f1(x))
top1(found1(x)) -> top1(active1(x))
active1(f1(x)) -> f1(active1(x))
f1(mark1(x)) -> mark1(f1(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 strictly oriented and are deleted.


F1(found1(x)) -> F1(x)
The remaining pairs can at least by weakly be oriented.

F1(mark1(x)) -> F1(x)
Used ordering: Combined order from the following AFS and order.
F1(x1)  =  F1(x1)
mark1(x1)  =  x1
found1(x1)  =  found1(x1)

Lexicographic Path Order [19].
Precedence:
trivial


The following usable rules [14] were oriented: none



↳ QTRS
  ↳ DependencyPairsProof
    ↳ QDP
      ↳ DependencyGraphProof
        ↳ AND
          ↳ QDP
            ↳ QDPOrderProof
              ↳ QDP
                ↳ QDPOrderProof
QDP
                    ↳ QDPOrderProof
          ↳ QDP
          ↳ QDP
          ↳ QDP
          ↳ 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(x)
top1(active1(c)) -> top1(mark1(c))
top1(mark1(x)) -> top1(check1(x))
check1(f1(x)) -> f1(check1(x))
check1(x) -> start1(match2(f1(X), x))
match2(f1(x), f1(y)) -> f1(match2(x, y))
match2(X, x) -> proper1(x)
proper1(c) -> ok1(c)
proper1(f1(x)) -> f1(proper1(x))
f1(ok1(x)) -> ok1(f1(x))
start1(ok1(x)) -> found1(x)
f1(found1(x)) -> found1(f1(x))
top1(found1(x)) -> top1(active1(x))
active1(f1(x)) -> f1(active1(x))
f1(mark1(x)) -> mark1(f1(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 strictly oriented and are deleted.


F1(mark1(x)) -> F1(x)
The remaining pairs can at least by weakly be oriented.
none
Used ordering: Combined order from the following AFS and order.
F1(x1)  =  F1(x1)
mark1(x1)  =  mark1(x1)

Lexicographic Path Order [19].
Precedence:
mark1 > F1


The following usable rules [14] were oriented: none



↳ QTRS
  ↳ DependencyPairsProof
    ↳ QDP
      ↳ DependencyGraphProof
        ↳ AND
          ↳ QDP
            ↳ QDPOrderProof
              ↳ QDP
                ↳ QDPOrderProof
                  ↳ QDP
                    ↳ QDPOrderProof
QDP
                        ↳ PisEmptyProof
          ↳ QDP
          ↳ QDP
          ↳ QDP
          ↳ QDP
          ↳ QDP

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

active1(f1(x)) -> mark1(x)
top1(active1(c)) -> top1(mark1(c))
top1(mark1(x)) -> top1(check1(x))
check1(f1(x)) -> f1(check1(x))
check1(x) -> start1(match2(f1(X), x))
match2(f1(x), f1(y)) -> f1(match2(x, y))
match2(X, x) -> proper1(x)
proper1(c) -> ok1(c)
proper1(f1(x)) -> f1(proper1(x))
f1(ok1(x)) -> ok1(f1(x))
start1(ok1(x)) -> found1(x)
f1(found1(x)) -> found1(f1(x))
top1(found1(x)) -> top1(active1(x))
active1(f1(x)) -> f1(active1(x))
f1(mark1(x)) -> mark1(f1(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
          ↳ QDP
          ↳ QDP
          ↳ QDP

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

ACTIVE1(f1(x)) -> ACTIVE1(x)

The TRS R consists of the following rules:

active1(f1(x)) -> mark1(x)
top1(active1(c)) -> top1(mark1(c))
top1(mark1(x)) -> top1(check1(x))
check1(f1(x)) -> f1(check1(x))
check1(x) -> start1(match2(f1(X), x))
match2(f1(x), f1(y)) -> f1(match2(x, y))
match2(X, x) -> proper1(x)
proper1(c) -> ok1(c)
proper1(f1(x)) -> f1(proper1(x))
f1(ok1(x)) -> ok1(f1(x))
start1(ok1(x)) -> found1(x)
f1(found1(x)) -> found1(f1(x))
top1(found1(x)) -> top1(active1(x))
active1(f1(x)) -> f1(active1(x))
f1(mark1(x)) -> mark1(f1(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 strictly oriented and are deleted.


ACTIVE1(f1(x)) -> ACTIVE1(x)
The remaining pairs can at least by weakly be oriented.
none
Used ordering: Combined order from the following AFS and order.
ACTIVE1(x1)  =  ACTIVE1(x1)
f1(x1)  =  f1(x1)

Lexicographic Path Order [19].
Precedence:
f1 > ACTIVE1


The following usable rules [14] were oriented: none



↳ QTRS
  ↳ DependencyPairsProof
    ↳ QDP
      ↳ DependencyGraphProof
        ↳ AND
          ↳ QDP
          ↳ QDP
            ↳ QDPOrderProof
QDP
                ↳ PisEmptyProof
          ↳ QDP
          ↳ QDP
          ↳ QDP
          ↳ QDP

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

active1(f1(x)) -> mark1(x)
top1(active1(c)) -> top1(mark1(c))
top1(mark1(x)) -> top1(check1(x))
check1(f1(x)) -> f1(check1(x))
check1(x) -> start1(match2(f1(X), x))
match2(f1(x), f1(y)) -> f1(match2(x, y))
match2(X, x) -> proper1(x)
proper1(c) -> ok1(c)
proper1(f1(x)) -> f1(proper1(x))
f1(ok1(x)) -> ok1(f1(x))
start1(ok1(x)) -> found1(x)
f1(found1(x)) -> found1(f1(x))
top1(found1(x)) -> top1(active1(x))
active1(f1(x)) -> f1(active1(x))
f1(mark1(x)) -> mark1(f1(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
            ↳ QDPOrderProof
          ↳ QDP
          ↳ QDP
          ↳ QDP

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

PROPER1(f1(x)) -> PROPER1(x)

The TRS R consists of the following rules:

active1(f1(x)) -> mark1(x)
top1(active1(c)) -> top1(mark1(c))
top1(mark1(x)) -> top1(check1(x))
check1(f1(x)) -> f1(check1(x))
check1(x) -> start1(match2(f1(X), x))
match2(f1(x), f1(y)) -> f1(match2(x, y))
match2(X, x) -> proper1(x)
proper1(c) -> ok1(c)
proper1(f1(x)) -> f1(proper1(x))
f1(ok1(x)) -> ok1(f1(x))
start1(ok1(x)) -> found1(x)
f1(found1(x)) -> found1(f1(x))
top1(found1(x)) -> top1(active1(x))
active1(f1(x)) -> f1(active1(x))
f1(mark1(x)) -> mark1(f1(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 strictly oriented and are deleted.


PROPER1(f1(x)) -> PROPER1(x)
The remaining pairs can at least by weakly be oriented.
none
Used ordering: Combined order from the following AFS and order.
PROPER1(x1)  =  PROPER1(x1)
f1(x1)  =  f1(x1)

Lexicographic Path Order [19].
Precedence:
f1 > PROPER1


The following usable rules [14] were oriented: none



↳ QTRS
  ↳ DependencyPairsProof
    ↳ QDP
      ↳ DependencyGraphProof
        ↳ AND
          ↳ QDP
          ↳ QDP
          ↳ QDP
            ↳ QDPOrderProof
QDP
                ↳ PisEmptyProof
          ↳ QDP
          ↳ QDP
          ↳ QDP

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

active1(f1(x)) -> mark1(x)
top1(active1(c)) -> top1(mark1(c))
top1(mark1(x)) -> top1(check1(x))
check1(f1(x)) -> f1(check1(x))
check1(x) -> start1(match2(f1(X), x))
match2(f1(x), f1(y)) -> f1(match2(x, y))
match2(X, x) -> proper1(x)
proper1(c) -> ok1(c)
proper1(f1(x)) -> f1(proper1(x))
f1(ok1(x)) -> ok1(f1(x))
start1(ok1(x)) -> found1(x)
f1(found1(x)) -> found1(f1(x))
top1(found1(x)) -> top1(active1(x))
active1(f1(x)) -> f1(active1(x))
f1(mark1(x)) -> mark1(f1(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
QDP
            ↳ QDPOrderProof
          ↳ QDP
          ↳ QDP

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

MATCH2(f1(x), f1(y)) -> MATCH2(x, y)

The TRS R consists of the following rules:

active1(f1(x)) -> mark1(x)
top1(active1(c)) -> top1(mark1(c))
top1(mark1(x)) -> top1(check1(x))
check1(f1(x)) -> f1(check1(x))
check1(x) -> start1(match2(f1(X), x))
match2(f1(x), f1(y)) -> f1(match2(x, y))
match2(X, x) -> proper1(x)
proper1(c) -> ok1(c)
proper1(f1(x)) -> f1(proper1(x))
f1(ok1(x)) -> ok1(f1(x))
start1(ok1(x)) -> found1(x)
f1(found1(x)) -> found1(f1(x))
top1(found1(x)) -> top1(active1(x))
active1(f1(x)) -> f1(active1(x))
f1(mark1(x)) -> mark1(f1(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 strictly oriented and are deleted.


MATCH2(f1(x), f1(y)) -> MATCH2(x, y)
The remaining pairs can at least by weakly be oriented.
none
Used ordering: Combined order from the following AFS and order.
MATCH2(x1, x2)  =  MATCH1(x1)
f1(x1)  =  f1(x1)

Lexicographic Path Order [19].
Precedence:
trivial


The following usable rules [14] were oriented: none



↳ QTRS
  ↳ DependencyPairsProof
    ↳ QDP
      ↳ DependencyGraphProof
        ↳ AND
          ↳ QDP
          ↳ QDP
          ↳ QDP
          ↳ QDP
            ↳ QDPOrderProof
QDP
                ↳ PisEmptyProof
          ↳ QDP
          ↳ QDP

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

active1(f1(x)) -> mark1(x)
top1(active1(c)) -> top1(mark1(c))
top1(mark1(x)) -> top1(check1(x))
check1(f1(x)) -> f1(check1(x))
check1(x) -> start1(match2(f1(X), x))
match2(f1(x), f1(y)) -> f1(match2(x, y))
match2(X, x) -> proper1(x)
proper1(c) -> ok1(c)
proper1(f1(x)) -> f1(proper1(x))
f1(ok1(x)) -> ok1(f1(x))
start1(ok1(x)) -> found1(x)
f1(found1(x)) -> found1(f1(x))
top1(found1(x)) -> top1(active1(x))
active1(f1(x)) -> f1(active1(x))
f1(mark1(x)) -> mark1(f1(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
          ↳ QDP
QDP
            ↳ QDPOrderProof
          ↳ QDP

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

CHECK1(f1(x)) -> CHECK1(x)

The TRS R consists of the following rules:

active1(f1(x)) -> mark1(x)
top1(active1(c)) -> top1(mark1(c))
top1(mark1(x)) -> top1(check1(x))
check1(f1(x)) -> f1(check1(x))
check1(x) -> start1(match2(f1(X), x))
match2(f1(x), f1(y)) -> f1(match2(x, y))
match2(X, x) -> proper1(x)
proper1(c) -> ok1(c)
proper1(f1(x)) -> f1(proper1(x))
f1(ok1(x)) -> ok1(f1(x))
start1(ok1(x)) -> found1(x)
f1(found1(x)) -> found1(f1(x))
top1(found1(x)) -> top1(active1(x))
active1(f1(x)) -> f1(active1(x))
f1(mark1(x)) -> mark1(f1(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 strictly oriented and are deleted.


CHECK1(f1(x)) -> CHECK1(x)
The remaining pairs can at least by weakly be oriented.
none
Used ordering: Combined order from the following AFS and order.
CHECK1(x1)  =  CHECK1(x1)
f1(x1)  =  f1(x1)

Lexicographic Path Order [19].
Precedence:
f1 > CHECK1


The following usable rules [14] were oriented: none



↳ QTRS
  ↳ DependencyPairsProof
    ↳ QDP
      ↳ DependencyGraphProof
        ↳ AND
          ↳ QDP
          ↳ QDP
          ↳ QDP
          ↳ QDP
          ↳ QDP
            ↳ QDPOrderProof
QDP
                ↳ PisEmptyProof
          ↳ QDP

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

active1(f1(x)) -> mark1(x)
top1(active1(c)) -> top1(mark1(c))
top1(mark1(x)) -> top1(check1(x))
check1(f1(x)) -> f1(check1(x))
check1(x) -> start1(match2(f1(X), x))
match2(f1(x), f1(y)) -> f1(match2(x, y))
match2(X, x) -> proper1(x)
proper1(c) -> ok1(c)
proper1(f1(x)) -> f1(proper1(x))
f1(ok1(x)) -> ok1(f1(x))
start1(ok1(x)) -> found1(x)
f1(found1(x)) -> found1(f1(x))
top1(found1(x)) -> top1(active1(x))
active1(f1(x)) -> f1(active1(x))
f1(mark1(x)) -> mark1(f1(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
          ↳ QDP
          ↳ QDP
QDP
            ↳ QDPOrderProof

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

TOP1(found1(x)) -> TOP1(active1(x))
TOP1(active1(c)) -> TOP1(mark1(c))
TOP1(mark1(x)) -> TOP1(check1(x))

The TRS R consists of the following rules:

active1(f1(x)) -> mark1(x)
top1(active1(c)) -> top1(mark1(c))
top1(mark1(x)) -> top1(check1(x))
check1(f1(x)) -> f1(check1(x))
check1(x) -> start1(match2(f1(X), x))
match2(f1(x), f1(y)) -> f1(match2(x, y))
match2(X, x) -> proper1(x)
proper1(c) -> ok1(c)
proper1(f1(x)) -> f1(proper1(x))
f1(ok1(x)) -> ok1(f1(x))
start1(ok1(x)) -> found1(x)
f1(found1(x)) -> found1(f1(x))
top1(found1(x)) -> top1(active1(x))
active1(f1(x)) -> f1(active1(x))
f1(mark1(x)) -> mark1(f1(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 strictly oriented and are deleted.


TOP1(active1(c)) -> TOP1(mark1(c))
The remaining pairs can at least by weakly be oriented.

TOP1(found1(x)) -> TOP1(active1(x))
TOP1(mark1(x)) -> TOP1(check1(x))
Used ordering: Combined order from the following AFS and order.
TOP1(x1)  =  x1
found1(x1)  =  x1
active1(x1)  =  x1
c  =  c
mark1(x1)  =  mark
check1(x1)  =  check
match2(x1, x2)  =  x1
f1(x1)  =  f
proper1(x1)  =  proper1(x1)
start1(x1)  =  x1
ok1(x1)  =  x1
X  =  X

Lexicographic Path Order [19].
Precedence:
c > [mark, check, f] > proper1
X > proper1


The following usable rules [14] were oriented:

active1(f1(x)) -> mark1(x)
active1(f1(x)) -> f1(active1(x))
f1(ok1(x)) -> ok1(f1(x))
f1(found1(x)) -> found1(f1(x))
f1(mark1(x)) -> mark1(f1(x))
check1(f1(x)) -> f1(check1(x))
check1(x) -> start1(match2(f1(X), x))
match2(f1(x), f1(y)) -> f1(match2(x, y))
start1(ok1(x)) -> found1(x)



↳ QTRS
  ↳ DependencyPairsProof
    ↳ QDP
      ↳ DependencyGraphProof
        ↳ AND
          ↳ QDP
          ↳ QDP
          ↳ QDP
          ↳ QDP
          ↳ QDP
          ↳ QDP
            ↳ QDPOrderProof
QDP
                ↳ QDPOrderProof

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

TOP1(found1(x)) -> TOP1(active1(x))
TOP1(mark1(x)) -> TOP1(check1(x))

The TRS R consists of the following rules:

active1(f1(x)) -> mark1(x)
top1(active1(c)) -> top1(mark1(c))
top1(mark1(x)) -> top1(check1(x))
check1(f1(x)) -> f1(check1(x))
check1(x) -> start1(match2(f1(X), x))
match2(f1(x), f1(y)) -> f1(match2(x, y))
match2(X, x) -> proper1(x)
proper1(c) -> ok1(c)
proper1(f1(x)) -> f1(proper1(x))
f1(ok1(x)) -> ok1(f1(x))
start1(ok1(x)) -> found1(x)
f1(found1(x)) -> found1(f1(x))
top1(found1(x)) -> top1(active1(x))
active1(f1(x)) -> f1(active1(x))
f1(mark1(x)) -> mark1(f1(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 strictly oriented and are deleted.


TOP1(found1(x)) -> TOP1(active1(x))
The remaining pairs can at least by weakly be oriented.

TOP1(mark1(x)) -> TOP1(check1(x))
Used ordering: Combined order from the following AFS and order.
TOP1(x1)  =  x1
found1(x1)  =  found1(x1)
active1(x1)  =  x1
mark1(x1)  =  mark1(x1)
check1(x1)  =  check1(x1)
match2(x1, x2)  =  match1(x2)
f1(x1)  =  f1(x1)
proper1(x1)  =  proper1(x1)
start1(x1)  =  x1
ok1(x1)  =  ok1(x1)
X  =  X
c  =  c

Lexicographic Path Order [19].
Precedence:
c > [found1, mark1, check1, match1, f1, proper1, ok1, X]


The following usable rules [14] were oriented:

active1(f1(x)) -> mark1(x)
active1(f1(x)) -> f1(active1(x))
f1(ok1(x)) -> ok1(f1(x))
f1(found1(x)) -> found1(f1(x))
f1(mark1(x)) -> mark1(f1(x))
check1(f1(x)) -> f1(check1(x))
check1(x) -> start1(match2(f1(X), x))
match2(f1(x), f1(y)) -> f1(match2(x, y))
start1(ok1(x)) -> found1(x)
match2(X, x) -> proper1(x)
proper1(c) -> ok1(c)
proper1(f1(x)) -> f1(proper1(x))



↳ QTRS
  ↳ DependencyPairsProof
    ↳ QDP
      ↳ DependencyGraphProof
        ↳ AND
          ↳ QDP
          ↳ QDP
          ↳ QDP
          ↳ QDP
          ↳ QDP
          ↳ QDP
            ↳ QDPOrderProof
              ↳ QDP
                ↳ QDPOrderProof
QDP
                    ↳ QDPOrderProof

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

TOP1(mark1(x)) -> TOP1(check1(x))

The TRS R consists of the following rules:

active1(f1(x)) -> mark1(x)
top1(active1(c)) -> top1(mark1(c))
top1(mark1(x)) -> top1(check1(x))
check1(f1(x)) -> f1(check1(x))
check1(x) -> start1(match2(f1(X), x))
match2(f1(x), f1(y)) -> f1(match2(x, y))
match2(X, x) -> proper1(x)
proper1(c) -> ok1(c)
proper1(f1(x)) -> f1(proper1(x))
f1(ok1(x)) -> ok1(f1(x))
start1(ok1(x)) -> found1(x)
f1(found1(x)) -> found1(f1(x))
top1(found1(x)) -> top1(active1(x))
active1(f1(x)) -> f1(active1(x))
f1(mark1(x)) -> mark1(f1(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 strictly oriented and are deleted.


TOP1(mark1(x)) -> TOP1(check1(x))
The remaining pairs can at least by weakly be oriented.
none
Used ordering: Combined order from the following AFS and order.
TOP1(x1)  =  TOP1(x1)
mark1(x1)  =  mark1(x1)
check1(x1)  =  check
match2(x1, x2)  =  x1
f1(x1)  =  x1
proper1(x1)  =  proper
start1(x1)  =  x1
ok1(x1)  =  ok
found1(x1)  =  found
X  =  X
c  =  c

Lexicographic Path Order [19].
Precedence:
[TOP1, mark1] > [check, proper, ok, found, X, c]


The following usable rules [14] were oriented:

check1(f1(x)) -> f1(check1(x))
check1(x) -> start1(match2(f1(X), x))
match2(f1(x), f1(y)) -> f1(match2(x, y))
start1(ok1(x)) -> found1(x)
match2(X, x) -> proper1(x)
f1(ok1(x)) -> ok1(f1(x))
f1(found1(x)) -> found1(f1(x))
f1(mark1(x)) -> mark1(f1(x))
proper1(c) -> ok1(c)
proper1(f1(x)) -> f1(proper1(x))



↳ QTRS
  ↳ DependencyPairsProof
    ↳ QDP
      ↳ DependencyGraphProof
        ↳ AND
          ↳ QDP
          ↳ QDP
          ↳ QDP
          ↳ QDP
          ↳ QDP
          ↳ QDP
            ↳ QDPOrderProof
              ↳ QDP
                ↳ QDPOrderProof
                  ↳ QDP
                    ↳ QDPOrderProof
QDP
                        ↳ PisEmptyProof

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

active1(f1(x)) -> mark1(x)
top1(active1(c)) -> top1(mark1(c))
top1(mark1(x)) -> top1(check1(x))
check1(f1(x)) -> f1(check1(x))
check1(x) -> start1(match2(f1(X), x))
match2(f1(x), f1(y)) -> f1(match2(x, y))
match2(X, x) -> proper1(x)
proper1(c) -> ok1(c)
proper1(f1(x)) -> f1(proper1(x))
f1(ok1(x)) -> ok1(f1(x))
start1(ok1(x)) -> found1(x)
f1(found1(x)) -> found1(f1(x))
top1(found1(x)) -> top1(active1(x))
active1(f1(x)) -> f1(active1(x))
f1(mark1(x)) -> mark1(f1(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.