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(c) -> mark1(f1(g1(c)))
active1(f1(g1(X))) -> mark1(g1(X))
proper1(c) -> ok1(c)
proper1(f1(X)) -> f1(proper1(X))
proper1(g1(X)) -> g1(proper1(X))
f1(ok1(X)) -> ok1(f1(X))
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(c) -> mark1(f1(g1(c)))
active1(f1(g1(X))) -> mark1(g1(X))
proper1(c) -> ok1(c)
proper1(f1(X)) -> f1(proper1(X))
proper1(g1(X)) -> g1(proper1(X))
f1(ok1(X)) -> ok1(f1(X))
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:

TOP1(mark1(X)) -> PROPER1(X)
PROPER1(g1(X)) -> G1(proper1(X))
PROPER1(g1(X)) -> PROPER1(X)
PROPER1(f1(X)) -> PROPER1(X)
TOP1(ok1(X)) -> TOP1(active1(X))
ACTIVE1(c) -> F1(g1(c))
F1(ok1(X)) -> F1(X)
TOP1(mark1(X)) -> TOP1(proper1(X))
G1(ok1(X)) -> G1(X)
ACTIVE1(c) -> G1(c)
TOP1(ok1(X)) -> ACTIVE1(X)
PROPER1(f1(X)) -> F1(proper1(X))

The TRS R consists of the following rules:

active1(c) -> mark1(f1(g1(c)))
active1(f1(g1(X))) -> mark1(g1(X))
proper1(c) -> ok1(c)
proper1(f1(X)) -> f1(proper1(X))
proper1(g1(X)) -> g1(proper1(X))
f1(ok1(X)) -> ok1(f1(X))
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:

TOP1(mark1(X)) -> PROPER1(X)
PROPER1(g1(X)) -> G1(proper1(X))
PROPER1(g1(X)) -> PROPER1(X)
PROPER1(f1(X)) -> PROPER1(X)
TOP1(ok1(X)) -> TOP1(active1(X))
ACTIVE1(c) -> F1(g1(c))
F1(ok1(X)) -> F1(X)
TOP1(mark1(X)) -> TOP1(proper1(X))
G1(ok1(X)) -> G1(X)
ACTIVE1(c) -> G1(c)
TOP1(ok1(X)) -> ACTIVE1(X)
PROPER1(f1(X)) -> F1(proper1(X))

The TRS R consists of the following rules:

active1(c) -> mark1(f1(g1(c)))
active1(f1(g1(X))) -> mark1(g1(X))
proper1(c) -> ok1(c)
proper1(f1(X)) -> f1(proper1(X))
proper1(g1(X)) -> g1(proper1(X))
f1(ok1(X)) -> ok1(f1(X))
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 4 SCCs with 6 less nodes.

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

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

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

The TRS R consists of the following rules:

active1(c) -> mark1(f1(g1(c)))
active1(f1(g1(X))) -> mark1(g1(X))
proper1(c) -> ok1(c)
proper1(f1(X)) -> f1(proper1(X))
proper1(g1(X)) -> g1(proper1(X))
f1(ok1(X)) -> ok1(f1(X))
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
ok1(x1)  =  ok1(x1)
Used ordering: Quasi Precedence: trivial 


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

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

active1(c) -> mark1(f1(g1(c)))
active1(f1(g1(X))) -> mark1(g1(X))
proper1(c) -> ok1(c)
proper1(f1(X)) -> f1(proper1(X))
proper1(g1(X)) -> g1(proper1(X))
f1(ok1(X)) -> ok1(f1(X))
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

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

F1(ok1(X)) -> F1(X)

The TRS R consists of the following rules:

active1(c) -> mark1(f1(g1(c)))
active1(f1(g1(X))) -> mark1(g1(X))
proper1(c) -> ok1(c)
proper1(f1(X)) -> f1(proper1(X))
proper1(g1(X)) -> g1(proper1(X))
f1(ok1(X)) -> ok1(f1(X))
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.

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


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

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

active1(c) -> mark1(f1(g1(c)))
active1(f1(g1(X))) -> mark1(g1(X))
proper1(c) -> ok1(c)
proper1(f1(X)) -> f1(proper1(X))
proper1(g1(X)) -> g1(proper1(X))
f1(ok1(X)) -> ok1(f1(X))
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

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

PROPER1(g1(X)) -> PROPER1(X)
PROPER1(f1(X)) -> PROPER1(X)

The TRS R consists of the following rules:

active1(c) -> mark1(f1(g1(c)))
active1(f1(g1(X))) -> mark1(g1(X))
proper1(c) -> ok1(c)
proper1(f1(X)) -> f1(proper1(X))
proper1(g1(X)) -> g1(proper1(X))
f1(ok1(X)) -> ok1(f1(X))
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(f1(X)) -> PROPER1(X)
Used argument filtering: PROPER1(x1)  =  x1
g1(x1)  =  x1
f1(x1)  =  f1(x1)
Used ordering: Quasi Precedence: trivial 


↳ QTRS
  ↳ DependencyPairsProof
    ↳ QDP
      ↳ DependencyGraphProof
        ↳ AND
          ↳ QDP
          ↳ QDP
          ↳ QDP
            ↳ QDPAfsSolverProof
              ↳ QDP
                ↳ QDPAfsSolverProof
          ↳ 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(c) -> mark1(f1(g1(c)))
active1(f1(g1(X))) -> mark1(g1(X))
proper1(c) -> ok1(c)
proper1(f1(X)) -> f1(proper1(X))
proper1(g1(X)) -> g1(proper1(X))
f1(ok1(X)) -> ok1(f1(X))
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

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

active1(c) -> mark1(f1(g1(c)))
active1(f1(g1(X))) -> mark1(g1(X))
proper1(c) -> ok1(c)
proper1(f1(X)) -> f1(proper1(X))
proper1(g1(X)) -> g1(proper1(X))
f1(ok1(X)) -> ok1(f1(X))
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

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(c) -> mark1(f1(g1(c)))
active1(f1(g1(X))) -> mark1(g1(X))
proper1(c) -> ok1(c)
proper1(f1(X)) -> f1(proper1(X))
proper1(g1(X)) -> g1(proper1(X))
f1(ok1(X)) -> ok1(f1(X))
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
ok1(x1)  =  x1
active1(x1)  =  x1
mark1(x1)  =  mark1(x1)
proper1(x1)  =  x1
c  =  c
f1(x1)  =  f
g1(x1)  =  g
Used ordering: Quasi Precedence: c > f > mark_1 c > f > g 


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

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

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

The TRS R consists of the following rules:

active1(c) -> mark1(f1(g1(c)))
active1(f1(g1(X))) -> mark1(g1(X))
proper1(c) -> ok1(c)
proper1(f1(X)) -> f1(proper1(X))
proper1(g1(X)) -> g1(proper1(X))
f1(ok1(X)) -> ok1(f1(X))
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)  =  ok
active1(x1)  =  active
mark1(x1)  =  mark
g1(x1)  =  g
Used ordering: Quasi Precedence: [ok, g] > [active, mark] 


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

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

active1(c) -> mark1(f1(g1(c)))
active1(f1(g1(X))) -> mark1(g1(X))
proper1(c) -> ok1(c)
proper1(f1(X)) -> f1(proper1(X))
proper1(g1(X)) -> g1(proper1(X))
f1(ok1(X)) -> ok1(f1(X))
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.