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(f2(g1(X), Y)) -> mark1(f2(X, f2(g1(X), Y)))
active1(f2(X1, X2)) -> f2(active1(X1), X2)
active1(g1(X)) -> g1(active1(X))
f2(mark1(X1), X2) -> mark1(f2(X1, X2))
g1(mark1(X)) -> mark1(g1(X))
proper1(f2(X1, X2)) -> f2(proper1(X1), proper1(X2))
proper1(g1(X)) -> g1(proper1(X))
f2(ok1(X1), ok1(X2)) -> ok1(f2(X1, X2))
g1(ok1(X)) -> ok1(g1(X))
top1(mark1(X)) -> top1(proper1(X))
top1(ok1(X)) -> top1(active1(X))

Q is empty.


QTRS
  ↳ DependencyPairsProof

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

active1(f2(g1(X), Y)) -> mark1(f2(X, f2(g1(X), Y)))
active1(f2(X1, X2)) -> f2(active1(X1), X2)
active1(g1(X)) -> g1(active1(X))
f2(mark1(X1), X2) -> mark1(f2(X1, X2))
g1(mark1(X)) -> mark1(g1(X))
proper1(f2(X1, X2)) -> f2(proper1(X1), proper1(X2))
proper1(g1(X)) -> g1(proper1(X))
f2(ok1(X1), ok1(X2)) -> ok1(f2(X1, X2))
g1(ok1(X)) -> ok1(g1(X))
top1(mark1(X)) -> top1(proper1(X))
top1(ok1(X)) -> top1(active1(X))

Q is empty.

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

ACTIVE1(f2(g1(X), Y)) -> F2(X, f2(g1(X), Y))
ACTIVE1(g1(X)) -> ACTIVE1(X)
TOP1(mark1(X)) -> PROPER1(X)
PROPER1(g1(X)) -> G1(proper1(X))
PROPER1(g1(X)) -> PROPER1(X)
TOP1(ok1(X)) -> ACTIVE1(X)
F2(ok1(X1), ok1(X2)) -> F2(X1, X2)
G1(mark1(X)) -> G1(X)
ACTIVE1(g1(X)) -> G1(active1(X))
PROPER1(f2(X1, X2)) -> PROPER1(X2)
TOP1(ok1(X)) -> TOP1(active1(X))
PROPER1(f2(X1, X2)) -> F2(proper1(X1), proper1(X2))
TOP1(mark1(X)) -> TOP1(proper1(X))
ACTIVE1(f2(X1, X2)) -> F2(active1(X1), X2)
G1(ok1(X)) -> G1(X)
F2(mark1(X1), X2) -> F2(X1, X2)
ACTIVE1(f2(X1, X2)) -> ACTIVE1(X1)
PROPER1(f2(X1, X2)) -> PROPER1(X1)

The TRS R consists of the following rules:

active1(f2(g1(X), Y)) -> mark1(f2(X, f2(g1(X), Y)))
active1(f2(X1, X2)) -> f2(active1(X1), X2)
active1(g1(X)) -> g1(active1(X))
f2(mark1(X1), X2) -> mark1(f2(X1, X2))
g1(mark1(X)) -> mark1(g1(X))
proper1(f2(X1, X2)) -> f2(proper1(X1), proper1(X2))
proper1(g1(X)) -> g1(proper1(X))
f2(ok1(X1), ok1(X2)) -> ok1(f2(X1, X2))
g1(ok1(X)) -> ok1(g1(X))
top1(mark1(X)) -> top1(proper1(X))
top1(ok1(X)) -> top1(active1(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:

ACTIVE1(f2(g1(X), Y)) -> F2(X, f2(g1(X), Y))
ACTIVE1(g1(X)) -> ACTIVE1(X)
TOP1(mark1(X)) -> PROPER1(X)
PROPER1(g1(X)) -> G1(proper1(X))
PROPER1(g1(X)) -> PROPER1(X)
TOP1(ok1(X)) -> ACTIVE1(X)
F2(ok1(X1), ok1(X2)) -> F2(X1, X2)
G1(mark1(X)) -> G1(X)
ACTIVE1(g1(X)) -> G1(active1(X))
PROPER1(f2(X1, X2)) -> PROPER1(X2)
TOP1(ok1(X)) -> TOP1(active1(X))
PROPER1(f2(X1, X2)) -> F2(proper1(X1), proper1(X2))
TOP1(mark1(X)) -> TOP1(proper1(X))
ACTIVE1(f2(X1, X2)) -> F2(active1(X1), X2)
G1(ok1(X)) -> G1(X)
F2(mark1(X1), X2) -> F2(X1, X2)
ACTIVE1(f2(X1, X2)) -> ACTIVE1(X1)
PROPER1(f2(X1, X2)) -> PROPER1(X1)

The TRS R consists of the following rules:

active1(f2(g1(X), Y)) -> mark1(f2(X, f2(g1(X), Y)))
active1(f2(X1, X2)) -> f2(active1(X1), X2)
active1(g1(X)) -> g1(active1(X))
f2(mark1(X1), X2) -> mark1(f2(X1, X2))
g1(mark1(X)) -> mark1(g1(X))
proper1(f2(X1, X2)) -> f2(proper1(X1), proper1(X2))
proper1(g1(X)) -> g1(proper1(X))
f2(ok1(X1), ok1(X2)) -> ok1(f2(X1, X2))
g1(ok1(X)) -> ok1(g1(X))
top1(mark1(X)) -> top1(proper1(X))
top1(ok1(X)) -> top1(active1(X))

Q is empty.
We have to consider all minimal (P,Q,R)-chains.
The approximation of the Dependency Graph contains 5 SCCs with 7 less nodes.

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

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

G1(mark1(X)) -> G1(X)
G1(ok1(X)) -> G1(X)

The TRS R consists of the following rules:

active1(f2(g1(X), Y)) -> mark1(f2(X, f2(g1(X), Y)))
active1(f2(X1, X2)) -> f2(active1(X1), X2)
active1(g1(X)) -> g1(active1(X))
f2(mark1(X1), X2) -> mark1(f2(X1, X2))
g1(mark1(X)) -> mark1(g1(X))
proper1(f2(X1, X2)) -> f2(proper1(X1), proper1(X2))
proper1(g1(X)) -> g1(proper1(X))
f2(ok1(X1), ok1(X2)) -> ok1(f2(X1, X2))
g1(ok1(X)) -> ok1(g1(X))
top1(mark1(X)) -> top1(proper1(X))
top1(ok1(X)) -> top1(active1(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.

G1(ok1(X)) -> G1(X)
Used argument filtering: G1(x1)  =  x1
mark1(x1)  =  x1
ok1(x1)  =  ok1(x1)
Used ordering: Quasi Precedence: trivial


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

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

G1(mark1(X)) -> G1(X)

The TRS R consists of the following rules:

active1(f2(g1(X), Y)) -> mark1(f2(X, f2(g1(X), Y)))
active1(f2(X1, X2)) -> f2(active1(X1), X2)
active1(g1(X)) -> g1(active1(X))
f2(mark1(X1), X2) -> mark1(f2(X1, X2))
g1(mark1(X)) -> mark1(g1(X))
proper1(f2(X1, X2)) -> f2(proper1(X1), proper1(X2))
proper1(g1(X)) -> g1(proper1(X))
f2(ok1(X1), ok1(X2)) -> ok1(f2(X1, X2))
g1(ok1(X)) -> ok1(g1(X))
top1(mark1(X)) -> top1(proper1(X))
top1(ok1(X)) -> top1(active1(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.

G1(mark1(X)) -> G1(X)
Used argument filtering: G1(x1)  =  x1
mark1(x1)  =  mark1(x1)
Used ordering: Quasi Precedence: trivial


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

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

active1(f2(g1(X), Y)) -> mark1(f2(X, f2(g1(X), Y)))
active1(f2(X1, X2)) -> f2(active1(X1), X2)
active1(g1(X)) -> g1(active1(X))
f2(mark1(X1), X2) -> mark1(f2(X1, X2))
g1(mark1(X)) -> mark1(g1(X))
proper1(f2(X1, X2)) -> f2(proper1(X1), proper1(X2))
proper1(g1(X)) -> g1(proper1(X))
f2(ok1(X1), ok1(X2)) -> ok1(f2(X1, X2))
g1(ok1(X)) -> ok1(g1(X))
top1(mark1(X)) -> top1(proper1(X))
top1(ok1(X)) -> top1(active1(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

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

F2(mark1(X1), X2) -> F2(X1, X2)
F2(ok1(X1), ok1(X2)) -> F2(X1, X2)

The TRS R consists of the following rules:

active1(f2(g1(X), Y)) -> mark1(f2(X, f2(g1(X), Y)))
active1(f2(X1, X2)) -> f2(active1(X1), X2)
active1(g1(X)) -> g1(active1(X))
f2(mark1(X1), X2) -> mark1(f2(X1, X2))
g1(mark1(X)) -> mark1(g1(X))
proper1(f2(X1, X2)) -> f2(proper1(X1), proper1(X2))
proper1(g1(X)) -> g1(proper1(X))
f2(ok1(X1), ok1(X2)) -> ok1(f2(X1, X2))
g1(ok1(X)) -> ok1(g1(X))
top1(mark1(X)) -> top1(proper1(X))
top1(ok1(X)) -> top1(active1(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.

F2(ok1(X1), ok1(X2)) -> F2(X1, X2)
Used argument filtering: F2(x1, x2)  =  x2
ok1(x1)  =  ok1(x1)
Used ordering: Quasi Precedence: trivial


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

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

F2(mark1(X1), X2) -> F2(X1, X2)

The TRS R consists of the following rules:

active1(f2(g1(X), Y)) -> mark1(f2(X, f2(g1(X), Y)))
active1(f2(X1, X2)) -> f2(active1(X1), X2)
active1(g1(X)) -> g1(active1(X))
f2(mark1(X1), X2) -> mark1(f2(X1, X2))
g1(mark1(X)) -> mark1(g1(X))
proper1(f2(X1, X2)) -> f2(proper1(X1), proper1(X2))
proper1(g1(X)) -> g1(proper1(X))
f2(ok1(X1), ok1(X2)) -> ok1(f2(X1, X2))
g1(ok1(X)) -> ok1(g1(X))
top1(mark1(X)) -> top1(proper1(X))
top1(ok1(X)) -> top1(active1(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.

F2(mark1(X1), X2) -> F2(X1, X2)
Used argument filtering: F2(x1, x2)  =  x1
mark1(x1)  =  mark1(x1)
Used ordering: Quasi Precedence: trivial


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

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

active1(f2(g1(X), Y)) -> mark1(f2(X, f2(g1(X), Y)))
active1(f2(X1, X2)) -> f2(active1(X1), X2)
active1(g1(X)) -> g1(active1(X))
f2(mark1(X1), X2) -> mark1(f2(X1, X2))
g1(mark1(X)) -> mark1(g1(X))
proper1(f2(X1, X2)) -> f2(proper1(X1), proper1(X2))
proper1(g1(X)) -> g1(proper1(X))
f2(ok1(X1), ok1(X2)) -> ok1(f2(X1, X2))
g1(ok1(X)) -> ok1(g1(X))
top1(mark1(X)) -> top1(proper1(X))
top1(ok1(X)) -> top1(active1(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

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

PROPER1(g1(X)) -> PROPER1(X)
PROPER1(f2(X1, X2)) -> PROPER1(X2)
PROPER1(f2(X1, X2)) -> PROPER1(X1)

The TRS R consists of the following rules:

active1(f2(g1(X), Y)) -> mark1(f2(X, f2(g1(X), Y)))
active1(f2(X1, X2)) -> f2(active1(X1), X2)
active1(g1(X)) -> g1(active1(X))
f2(mark1(X1), X2) -> mark1(f2(X1, X2))
g1(mark1(X)) -> mark1(g1(X))
proper1(f2(X1, X2)) -> f2(proper1(X1), proper1(X2))
proper1(g1(X)) -> g1(proper1(X))
f2(ok1(X1), ok1(X2)) -> ok1(f2(X1, X2))
g1(ok1(X)) -> ok1(g1(X))
top1(mark1(X)) -> top1(proper1(X))
top1(ok1(X)) -> top1(active1(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(f2(X1, X2)) -> PROPER1(X2)
PROPER1(f2(X1, X2)) -> PROPER1(X1)
Used argument filtering: PROPER1(x1)  =  x1
g1(x1)  =  x1
f2(x1, x2)  =  f2(x1, x2)
Used ordering: Quasi Precedence: trivial


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

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

PROPER1(g1(X)) -> PROPER1(X)

The TRS R consists of the following rules:

active1(f2(g1(X), Y)) -> mark1(f2(X, f2(g1(X), Y)))
active1(f2(X1, X2)) -> f2(active1(X1), X2)
active1(g1(X)) -> g1(active1(X))
f2(mark1(X1), X2) -> mark1(f2(X1, X2))
g1(mark1(X)) -> mark1(g1(X))
proper1(f2(X1, X2)) -> f2(proper1(X1), proper1(X2))
proper1(g1(X)) -> g1(proper1(X))
f2(ok1(X1), ok1(X2)) -> ok1(f2(X1, X2))
g1(ok1(X)) -> ok1(g1(X))
top1(mark1(X)) -> top1(proper1(X))
top1(ok1(X)) -> top1(active1(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(g1(X)) -> PROPER1(X)
Used argument filtering: PROPER1(x1)  =  x1
g1(x1)  =  g1(x1)
Used ordering: Quasi Precedence: trivial


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

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

active1(f2(g1(X), Y)) -> mark1(f2(X, f2(g1(X), Y)))
active1(f2(X1, X2)) -> f2(active1(X1), X2)
active1(g1(X)) -> g1(active1(X))
f2(mark1(X1), X2) -> mark1(f2(X1, X2))
g1(mark1(X)) -> mark1(g1(X))
proper1(f2(X1, X2)) -> f2(proper1(X1), proper1(X2))
proper1(g1(X)) -> g1(proper1(X))
f2(ok1(X1), ok1(X2)) -> ok1(f2(X1, X2))
g1(ok1(X)) -> ok1(g1(X))
top1(mark1(X)) -> top1(proper1(X))
top1(ok1(X)) -> top1(active1(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

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

ACTIVE1(g1(X)) -> ACTIVE1(X)
ACTIVE1(f2(X1, X2)) -> ACTIVE1(X1)

The TRS R consists of the following rules:

active1(f2(g1(X), Y)) -> mark1(f2(X, f2(g1(X), Y)))
active1(f2(X1, X2)) -> f2(active1(X1), X2)
active1(g1(X)) -> g1(active1(X))
f2(mark1(X1), X2) -> mark1(f2(X1, X2))
g1(mark1(X)) -> mark1(g1(X))
proper1(f2(X1, X2)) -> f2(proper1(X1), proper1(X2))
proper1(g1(X)) -> g1(proper1(X))
f2(ok1(X1), ok1(X2)) -> ok1(f2(X1, X2))
g1(ok1(X)) -> ok1(g1(X))
top1(mark1(X)) -> top1(proper1(X))
top1(ok1(X)) -> top1(active1(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(f2(X1, X2)) -> ACTIVE1(X1)
Used argument filtering: ACTIVE1(x1)  =  x1
g1(x1)  =  x1
f2(x1, x2)  =  f1(x1)
Used ordering: Quasi Precedence: trivial


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

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

ACTIVE1(g1(X)) -> ACTIVE1(X)

The TRS R consists of the following rules:

active1(f2(g1(X), Y)) -> mark1(f2(X, f2(g1(X), Y)))
active1(f2(X1, X2)) -> f2(active1(X1), X2)
active1(g1(X)) -> g1(active1(X))
f2(mark1(X1), X2) -> mark1(f2(X1, X2))
g1(mark1(X)) -> mark1(g1(X))
proper1(f2(X1, X2)) -> f2(proper1(X1), proper1(X2))
proper1(g1(X)) -> g1(proper1(X))
f2(ok1(X1), ok1(X2)) -> ok1(f2(X1, X2))
g1(ok1(X)) -> ok1(g1(X))
top1(mark1(X)) -> top1(proper1(X))
top1(ok1(X)) -> top1(active1(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(g1(X)) -> ACTIVE1(X)
Used argument filtering: ACTIVE1(x1)  =  x1
g1(x1)  =  g1(x1)
Used ordering: Quasi Precedence: trivial


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

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

active1(f2(g1(X), Y)) -> mark1(f2(X, f2(g1(X), Y)))
active1(f2(X1, X2)) -> f2(active1(X1), X2)
active1(g1(X)) -> g1(active1(X))
f2(mark1(X1), X2) -> mark1(f2(X1, X2))
g1(mark1(X)) -> mark1(g1(X))
proper1(f2(X1, X2)) -> f2(proper1(X1), proper1(X2))
proper1(g1(X)) -> g1(proper1(X))
f2(ok1(X1), ok1(X2)) -> ok1(f2(X1, X2))
g1(ok1(X)) -> ok1(g1(X))
top1(mark1(X)) -> top1(proper1(X))
top1(ok1(X)) -> top1(active1(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

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

TOP1(ok1(X)) -> TOP1(active1(X))
TOP1(mark1(X)) -> TOP1(proper1(X))

The TRS R consists of the following rules:

active1(f2(g1(X), Y)) -> mark1(f2(X, f2(g1(X), Y)))
active1(f2(X1, X2)) -> f2(active1(X1), X2)
active1(g1(X)) -> g1(active1(X))
f2(mark1(X1), X2) -> mark1(f2(X1, X2))
g1(mark1(X)) -> mark1(g1(X))
proper1(f2(X1, X2)) -> f2(proper1(X1), proper1(X2))
proper1(g1(X)) -> g1(proper1(X))
f2(ok1(X1), ok1(X2)) -> ok1(f2(X1, X2))
g1(ok1(X)) -> ok1(g1(X))
top1(mark1(X)) -> top1(proper1(X))
top1(ok1(X)) -> top1(active1(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(ok1(X)) -> TOP1(active1(X))
Used argument filtering: TOP1(x1)  =  x1
ok1(x1)  =  ok1(x1)
active1(x1)  =  x1
mark1(x1)  =  mark
proper1(x1)  =  proper
f2(x1, x2)  =  x2
g1(x1)  =  x1
Used ordering: Quasi Precedence: ok_1 > [mark, proper]


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

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

TOP1(mark1(X)) -> TOP1(proper1(X))

The TRS R consists of the following rules:

active1(f2(g1(X), Y)) -> mark1(f2(X, f2(g1(X), Y)))
active1(f2(X1, X2)) -> f2(active1(X1), X2)
active1(g1(X)) -> g1(active1(X))
f2(mark1(X1), X2) -> mark1(f2(X1, X2))
g1(mark1(X)) -> mark1(g1(X))
proper1(f2(X1, X2)) -> f2(proper1(X1), proper1(X2))
proper1(g1(X)) -> g1(proper1(X))
f2(ok1(X1), ok1(X2)) -> ok1(f2(X1, X2))
g1(ok1(X)) -> ok1(g1(X))
top1(mark1(X)) -> top1(proper1(X))
top1(ok1(X)) -> top1(active1(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(proper1(X))
Used argument filtering: TOP1(x1)  =  x1
mark1(x1)  =  mark
proper1(x1)  =  proper
f2(x1, x2)  =  x1
g1(x1)  =  x1
ok1(x1)  =  ok
Used ordering: Quasi Precedence: mark > proper


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

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

active1(f2(g1(X), Y)) -> mark1(f2(X, f2(g1(X), Y)))
active1(f2(X1, X2)) -> f2(active1(X1), X2)
active1(g1(X)) -> g1(active1(X))
f2(mark1(X1), X2) -> mark1(f2(X1, X2))
g1(mark1(X)) -> mark1(g1(X))
proper1(f2(X1, X2)) -> f2(proper1(X1), proper1(X2))
proper1(g1(X)) -> g1(proper1(X))
f2(ok1(X1), ok1(X2)) -> ok1(f2(X1, X2))
g1(ok1(X)) -> ok1(g1(X))
top1(mark1(X)) -> top1(proper1(X))
top1(ok1(X)) -> top1(active1(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.