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.

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 contains 6 SCCs with 10 less nodes.

↳ QTRS
  ↳ DependencyPairsProof
    ↳ QDP
      ↳ DependencyGraphProof
        ↳ AND
QDP
            ↳ QDPAfsSolverProof
          ↳ 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.
By using an argument filtering and a montonic ordering, at least one Dependency Pair of this SCC can be strictly oriented.

F1(found1(x)) -> F1(x)
Used argument filtering: F1(x1)  =  x1
mark1(x1)  =  x1
ok1(x1)  =  x1
found1(x1)  =  found1(x1)
Used ordering: Quasi Precedence: trivial


↳ QTRS
  ↳ DependencyPairsProof
    ↳ QDP
      ↳ DependencyGraphProof
        ↳ AND
          ↳ QDP
            ↳ QDPAfsSolverProof
QDP
                ↳ QDPAfsSolverProof
          ↳ 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)

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.
By using an argument filtering and a montonic ordering, at least one Dependency Pair of this SCC can be strictly oriented.

F1(ok1(x)) -> F1(x)
Used argument filtering: F1(x1)  =  x1
mark1(x1)  =  x1
ok1(x1)  =  ok1(x1)
Used ordering: Quasi Precedence: trivial


↳ QTRS
  ↳ DependencyPairsProof
    ↳ QDP
      ↳ DependencyGraphProof
        ↳ AND
          ↳ QDP
            ↳ QDPAfsSolverProof
              ↳ QDP
                ↳ QDPAfsSolverProof
QDP
                    ↳ QDPAfsSolverProof
          ↳ 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.
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
                    ↳ QDPAfsSolverProof
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
            ↳ QDPAfsSolverProof
          ↳ 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.
By using an argument filtering and a montonic ordering, at least one Dependency Pair of this SCC can be strictly oriented.

ACTIVE1(f1(x)) -> ACTIVE1(x)
Used argument filtering: ACTIVE1(x1)  =  x1
f1(x1)  =  f1(x1)
Used ordering: Quasi Precedence: trivial


↳ QTRS
  ↳ DependencyPairsProof
    ↳ QDP
      ↳ DependencyGraphProof
        ↳ AND
          ↳ QDP
          ↳ QDP
            ↳ QDPAfsSolverProof
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
            ↳ QDPAfsSolverProof
          ↳ 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.
By using an argument filtering and a montonic ordering, at least one Dependency Pair of this SCC can be strictly oriented.

PROPER1(f1(x)) -> PROPER1(x)
Used argument filtering: PROPER1(x1)  =  x1
f1(x1)  =  f1(x1)
Used ordering: Quasi Precedence: trivial


↳ QTRS
  ↳ DependencyPairsProof
    ↳ QDP
      ↳ DependencyGraphProof
        ↳ AND
          ↳ QDP
          ↳ QDP
          ↳ QDP
            ↳ QDPAfsSolverProof
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
            ↳ QDPAfsSolverProof
          ↳ 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.
By using an argument filtering and a montonic ordering, at least one Dependency Pair of this SCC can be strictly oriented.

MATCH2(f1(x), f1(y)) -> MATCH2(x, y)
Used argument filtering: MATCH2(x1, x2)  =  x2
f1(x1)  =  f1(x1)
Used ordering: Quasi Precedence: trivial


↳ QTRS
  ↳ DependencyPairsProof
    ↳ QDP
      ↳ DependencyGraphProof
        ↳ AND
          ↳ QDP
          ↳ QDP
          ↳ QDP
          ↳ QDP
            ↳ QDPAfsSolverProof
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
            ↳ QDPAfsSolverProof
          ↳ 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.
By using an argument filtering and a montonic ordering, at least one Dependency Pair of this SCC can be strictly oriented.

CHECK1(f1(x)) -> CHECK1(x)
Used argument filtering: CHECK1(x1)  =  x1
f1(x1)  =  f1(x1)
Used ordering: Quasi Precedence: trivial


↳ QTRS
  ↳ DependencyPairsProof
    ↳ QDP
      ↳ DependencyGraphProof
        ↳ AND
          ↳ QDP
          ↳ QDP
          ↳ QDP
          ↳ QDP
          ↳ QDP
            ↳ QDPAfsSolverProof
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
            ↳ QDPAfsSolverProof

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.
By using an argument filtering and a montonic ordering, at least one Dependency Pair of this SCC can be strictly oriented.

TOP1(active1(c)) -> TOP1(mark1(c))
Used argument filtering: TOP1(x1)  =  x1
found1(x1)  =  x1
active1(x1)  =  x1
c  =  c
mark1(x1)  =  mark
check1(x1)  =  check
f1(x1)  =  f
ok1(x1)  =  x1
start1(x1)  =  x1
match2(x1, x2)  =  match
Used ordering: Quasi Precedence: c > [mark, check, f, match]


↳ QTRS
  ↳ DependencyPairsProof
    ↳ QDP
      ↳ DependencyGraphProof
        ↳ AND
          ↳ QDP
          ↳ QDP
          ↳ QDP
          ↳ QDP
          ↳ QDP
          ↳ QDP
            ↳ QDPAfsSolverProof
QDP
                ↳ QDPAfsSolverProof

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.
By using an argument filtering and a montonic ordering, at least one Dependency Pair of this SCC can be strictly oriented.

TOP1(mark1(x)) -> TOP1(check1(x))
Used argument filtering: TOP1(x1)  =  x1
found1(x1)  =  x1
active1(x1)  =  x1
mark1(x1)  =  mark1(x1)
check1(x1)  =  x1
f1(x1)  =  f1(x1)
start1(x1)  =  x1
match2(x1, x2)  =  x2
ok1(x1)  =  x1
proper1(x1)  =  x1
c  =  c
Used ordering: Quasi Precedence: f_1 > mark_1


↳ QTRS
  ↳ DependencyPairsProof
    ↳ QDP
      ↳ DependencyGraphProof
        ↳ AND
          ↳ QDP
          ↳ QDP
          ↳ QDP
          ↳ QDP
          ↳ QDP
          ↳ QDP
            ↳ QDPAfsSolverProof
              ↳ QDP
                ↳ QDPAfsSolverProof
QDP
                    ↳ QDPAfsSolverProof

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

TOP1(found1(x)) -> TOP1(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.
By using an argument filtering and a montonic ordering, at least one Dependency Pair of this SCC can be strictly oriented.

TOP1(found1(x)) -> TOP1(active1(x))
Used argument filtering: TOP1(x1)  =  x1
found1(x1)  =  found
active1(x1)  =  active
mark1(x1)  =  mark
f1(x1)  =  x1
ok1(x1)  =  ok
Used ordering: Quasi Precedence: found > active > mark


↳ QTRS
  ↳ DependencyPairsProof
    ↳ QDP
      ↳ DependencyGraphProof
        ↳ AND
          ↳ QDP
          ↳ QDP
          ↳ QDP
          ↳ QDP
          ↳ QDP
          ↳ QDP
            ↳ QDPAfsSolverProof
              ↳ QDP
                ↳ QDPAfsSolverProof
                  ↳ QDP
                    ↳ QDPAfsSolverProof
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.