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(g1(h1(f1(X))))
active1(f1(X)) -> f1(active1(X))
active1(h1(X)) -> h1(active1(X))
f1(mark1(X)) -> mark1(f1(X))
h1(mark1(X)) -> mark1(h1(X))
proper1(f1(X)) -> f1(proper1(X))
proper1(g1(X)) -> g1(proper1(X))
proper1(h1(X)) -> h1(proper1(X))
f1(ok1(X)) -> ok1(f1(X))
g1(ok1(X)) -> ok1(g1(X))
h1(ok1(X)) -> ok1(h1(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(f1(X)) -> mark1(g1(h1(f1(X))))
active1(f1(X)) -> f1(active1(X))
active1(h1(X)) -> h1(active1(X))
f1(mark1(X)) -> mark1(f1(X))
h1(mark1(X)) -> mark1(h1(X))
proper1(f1(X)) -> f1(proper1(X))
proper1(g1(X)) -> g1(proper1(X))
proper1(h1(X)) -> h1(proper1(X))
f1(ok1(X)) -> ok1(f1(X))
g1(ok1(X)) -> ok1(g1(X))
h1(ok1(X)) -> ok1(h1(X))
top1(mark1(X)) -> top1(proper1(X))
top1(ok1(X)) -> top1(active1(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(mark1(X)) -> PROPER1(X)
ACTIVE1(f1(X)) -> ACTIVE1(X)
PROPER1(g1(X)) -> G1(proper1(X))
PROPER1(g1(X)) -> PROPER1(X)
ACTIVE1(h1(X)) -> ACTIVE1(X)
ACTIVE1(f1(X)) -> F1(active1(X))
PROPER1(h1(X)) -> H1(proper1(X))
TOP1(ok1(X)) -> ACTIVE1(X)
H1(mark1(X)) -> H1(X)
H1(ok1(X)) -> H1(X)
ACTIVE1(f1(X)) -> H1(f1(X))
F1(mark1(X)) -> F1(X)
PROPER1(h1(X)) -> PROPER1(X)
PROPER1(f1(X)) -> PROPER1(X)
TOP1(ok1(X)) -> TOP1(active1(X))
F1(ok1(X)) -> F1(X)
TOP1(mark1(X)) -> TOP1(proper1(X))
G1(ok1(X)) -> G1(X)
ACTIVE1(h1(X)) -> H1(active1(X))
ACTIVE1(f1(X)) -> G1(h1(f1(X)))
PROPER1(f1(X)) -> F1(proper1(X))

The TRS R consists of the following rules:

active1(f1(X)) -> mark1(g1(h1(f1(X))))
active1(f1(X)) -> f1(active1(X))
active1(h1(X)) -> h1(active1(X))
f1(mark1(X)) -> mark1(f1(X))
h1(mark1(X)) -> mark1(h1(X))
proper1(f1(X)) -> f1(proper1(X))
proper1(g1(X)) -> g1(proper1(X))
proper1(h1(X)) -> h1(proper1(X))
f1(ok1(X)) -> ok1(f1(X))
g1(ok1(X)) -> ok1(g1(X))
h1(ok1(X)) -> ok1(h1(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)
ACTIVE1(f1(X)) -> ACTIVE1(X)
PROPER1(g1(X)) -> G1(proper1(X))
PROPER1(g1(X)) -> PROPER1(X)
ACTIVE1(h1(X)) -> ACTIVE1(X)
ACTIVE1(f1(X)) -> F1(active1(X))
PROPER1(h1(X)) -> H1(proper1(X))
TOP1(ok1(X)) -> ACTIVE1(X)
H1(mark1(X)) -> H1(X)
H1(ok1(X)) -> H1(X)
ACTIVE1(f1(X)) -> H1(f1(X))
F1(mark1(X)) -> F1(X)
PROPER1(h1(X)) -> PROPER1(X)
PROPER1(f1(X)) -> PROPER1(X)
TOP1(ok1(X)) -> TOP1(active1(X))
F1(ok1(X)) -> F1(X)
TOP1(mark1(X)) -> TOP1(proper1(X))
G1(ok1(X)) -> G1(X)
ACTIVE1(h1(X)) -> H1(active1(X))
ACTIVE1(f1(X)) -> G1(h1(f1(X)))
PROPER1(f1(X)) -> F1(proper1(X))

The TRS R consists of the following rules:

active1(f1(X)) -> mark1(g1(h1(f1(X))))
active1(f1(X)) -> f1(active1(X))
active1(h1(X)) -> h1(active1(X))
f1(mark1(X)) -> mark1(f1(X))
h1(mark1(X)) -> mark1(h1(X))
proper1(f1(X)) -> f1(proper1(X))
proper1(g1(X)) -> g1(proper1(X))
proper1(h1(X)) -> h1(proper1(X))
f1(ok1(X)) -> ok1(f1(X))
g1(ok1(X)) -> ok1(g1(X))
h1(ok1(X)) -> ok1(h1(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 [13,14,18] contains 6 SCCs with 9 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:

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

The TRS R consists of the following rules:

active1(f1(X)) -> mark1(g1(h1(f1(X))))
active1(f1(X)) -> f1(active1(X))
active1(h1(X)) -> h1(active1(X))
f1(mark1(X)) -> mark1(f1(X))
h1(mark1(X)) -> mark1(h1(X))
proper1(f1(X)) -> f1(proper1(X))
proper1(g1(X)) -> g1(proper1(X))
proper1(h1(X)) -> h1(proper1(X))
f1(ok1(X)) -> ok1(f1(X))
g1(ok1(X)) -> ok1(g1(X))
h1(ok1(X)) -> ok1(h1(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.
We use the reduction pair processor [13].


The following pairs can be strictly oriented and are deleted.


G1(ok1(X)) -> G1(X)
The remaining pairs can at least by weakly be oriented.
none
Used ordering: Combined order from the following AFS and order.
G1(x1)  =  G1(x1)
ok1(x1)  =  ok1(x1)

Lexicographic Path Order [19].
Precedence:
ok1 > G1


The following usable rules [14] were oriented: none



↳ QTRS
  ↳ DependencyPairsProof
    ↳ QDP
      ↳ DependencyGraphProof
        ↳ AND
          ↳ 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(g1(h1(f1(X))))
active1(f1(X)) -> f1(active1(X))
active1(h1(X)) -> h1(active1(X))
f1(mark1(X)) -> mark1(f1(X))
h1(mark1(X)) -> mark1(h1(X))
proper1(f1(X)) -> f1(proper1(X))
proper1(g1(X)) -> g1(proper1(X))
proper1(h1(X)) -> h1(proper1(X))
f1(ok1(X)) -> ok1(f1(X))
g1(ok1(X)) -> ok1(g1(X))
h1(ok1(X)) -> ok1(h1(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
            ↳ QDPOrderProof
          ↳ QDP
          ↳ QDP
          ↳ QDP
          ↳ QDP

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

H1(mark1(X)) -> H1(X)
H1(ok1(X)) -> H1(X)

The TRS R consists of the following rules:

active1(f1(X)) -> mark1(g1(h1(f1(X))))
active1(f1(X)) -> f1(active1(X))
active1(h1(X)) -> h1(active1(X))
f1(mark1(X)) -> mark1(f1(X))
h1(mark1(X)) -> mark1(h1(X))
proper1(f1(X)) -> f1(proper1(X))
proper1(g1(X)) -> g1(proper1(X))
proper1(h1(X)) -> h1(proper1(X))
f1(ok1(X)) -> ok1(f1(X))
g1(ok1(X)) -> ok1(g1(X))
h1(ok1(X)) -> ok1(h1(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.
We use the reduction pair processor [13].


The following pairs can be strictly oriented and are deleted.


H1(ok1(X)) -> H1(X)
The remaining pairs can at least by weakly be oriented.

H1(mark1(X)) -> H1(X)
Used ordering: Combined order from the following AFS and order.
H1(x1)  =  H1(x1)
mark1(x1)  =  x1
ok1(x1)  =  ok1(x1)

Lexicographic Path Order [19].
Precedence:
trivial


The following usable rules [14] were oriented: none



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

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

H1(mark1(X)) -> H1(X)

The TRS R consists of the following rules:

active1(f1(X)) -> mark1(g1(h1(f1(X))))
active1(f1(X)) -> f1(active1(X))
active1(h1(X)) -> h1(active1(X))
f1(mark1(X)) -> mark1(f1(X))
h1(mark1(X)) -> mark1(h1(X))
proper1(f1(X)) -> f1(proper1(X))
proper1(g1(X)) -> g1(proper1(X))
proper1(h1(X)) -> h1(proper1(X))
f1(ok1(X)) -> ok1(f1(X))
g1(ok1(X)) -> ok1(g1(X))
h1(ok1(X)) -> ok1(h1(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.
We use the reduction pair processor [13].


The following pairs can be strictly oriented and are deleted.


H1(mark1(X)) -> H1(X)
The remaining pairs can at least by weakly be oriented.
none
Used ordering: Combined order from the following AFS and order.
H1(x1)  =  H1(x1)
mark1(x1)  =  mark1(x1)

Lexicographic Path Order [19].
Precedence:
mark1 > H1


The following usable rules [14] were oriented: none



↳ QTRS
  ↳ DependencyPairsProof
    ↳ QDP
      ↳ DependencyGraphProof
        ↳ AND
          ↳ QDP
          ↳ QDP
            ↳ QDPOrderProof
              ↳ 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(g1(h1(f1(X))))
active1(f1(X)) -> f1(active1(X))
active1(h1(X)) -> h1(active1(X))
f1(mark1(X)) -> mark1(f1(X))
h1(mark1(X)) -> mark1(h1(X))
proper1(f1(X)) -> f1(proper1(X))
proper1(g1(X)) -> g1(proper1(X))
proper1(h1(X)) -> h1(proper1(X))
f1(ok1(X)) -> ok1(f1(X))
g1(ok1(X)) -> ok1(g1(X))
h1(ok1(X)) -> ok1(h1(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
            ↳ QDPOrderProof
          ↳ 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(g1(h1(f1(X))))
active1(f1(X)) -> f1(active1(X))
active1(h1(X)) -> h1(active1(X))
f1(mark1(X)) -> mark1(f1(X))
h1(mark1(X)) -> mark1(h1(X))
proper1(f1(X)) -> f1(proper1(X))
proper1(g1(X)) -> g1(proper1(X))
proper1(h1(X)) -> h1(proper1(X))
f1(ok1(X)) -> ok1(f1(X))
g1(ok1(X)) -> ok1(g1(X))
h1(ok1(X)) -> ok1(h1(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.
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)
Used ordering: Combined order from the following AFS and order.
F1(x1)  =  F1(x1)
mark1(x1)  =  x1
ok1(x1)  =  ok1(x1)

Lexicographic Path Order [19].
Precedence:
trivial


The following usable rules [14] were oriented: none



↳ QTRS
  ↳ DependencyPairsProof
    ↳ QDP
      ↳ DependencyGraphProof
        ↳ AND
          ↳ QDP
          ↳ QDP
          ↳ QDP
            ↳ QDPOrderProof
QDP
                ↳ QDPOrderProof
          ↳ 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(g1(h1(f1(X))))
active1(f1(X)) -> f1(active1(X))
active1(h1(X)) -> h1(active1(X))
f1(mark1(X)) -> mark1(f1(X))
h1(mark1(X)) -> mark1(h1(X))
proper1(f1(X)) -> f1(proper1(X))
proper1(g1(X)) -> g1(proper1(X))
proper1(h1(X)) -> h1(proper1(X))
f1(ok1(X)) -> ok1(f1(X))
g1(ok1(X)) -> ok1(g1(X))
h1(ok1(X)) -> ok1(h1(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.
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
          ↳ QDP
          ↳ QDP
            ↳ QDPOrderProof
              ↳ 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(g1(h1(f1(X))))
active1(f1(X)) -> f1(active1(X))
active1(h1(X)) -> h1(active1(X))
f1(mark1(X)) -> mark1(f1(X))
h1(mark1(X)) -> mark1(h1(X))
proper1(f1(X)) -> f1(proper1(X))
proper1(g1(X)) -> g1(proper1(X))
proper1(h1(X)) -> h1(proper1(X))
f1(ok1(X)) -> ok1(f1(X))
g1(ok1(X)) -> ok1(g1(X))
h1(ok1(X)) -> ok1(h1(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
            ↳ QDPOrderProof
          ↳ QDP
          ↳ QDP

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

PROPER1(g1(X)) -> PROPER1(X)
PROPER1(h1(X)) -> PROPER1(X)
PROPER1(f1(X)) -> PROPER1(X)

The TRS R consists of the following rules:

active1(f1(X)) -> mark1(g1(h1(f1(X))))
active1(f1(X)) -> f1(active1(X))
active1(h1(X)) -> h1(active1(X))
f1(mark1(X)) -> mark1(f1(X))
h1(mark1(X)) -> mark1(h1(X))
proper1(f1(X)) -> f1(proper1(X))
proper1(g1(X)) -> g1(proper1(X))
proper1(h1(X)) -> h1(proper1(X))
f1(ok1(X)) -> ok1(f1(X))
g1(ok1(X)) -> ok1(g1(X))
h1(ok1(X)) -> ok1(h1(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.
We use the reduction pair processor [13].


The following pairs can be strictly oriented and are deleted.


PROPER1(h1(X)) -> PROPER1(X)
The remaining pairs can at least by weakly be oriented.

PROPER1(g1(X)) -> PROPER1(X)
PROPER1(f1(X)) -> PROPER1(X)
Used ordering: Combined order from the following AFS and order.
PROPER1(x1)  =  PROPER1(x1)
g1(x1)  =  x1
h1(x1)  =  h1(x1)
f1(x1)  =  x1

Lexicographic Path Order [19].
Precedence:
[PROPER1, h1]


The following usable rules [14] were oriented: none



↳ QTRS
  ↳ DependencyPairsProof
    ↳ QDP
      ↳ DependencyGraphProof
        ↳ AND
          ↳ QDP
          ↳ QDP
          ↳ QDP
          ↳ QDP
            ↳ QDPOrderProof
QDP
                ↳ QDPOrderProof
          ↳ QDP
          ↳ 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(f1(X)) -> mark1(g1(h1(f1(X))))
active1(f1(X)) -> f1(active1(X))
active1(h1(X)) -> h1(active1(X))
f1(mark1(X)) -> mark1(f1(X))
h1(mark1(X)) -> mark1(h1(X))
proper1(f1(X)) -> f1(proper1(X))
proper1(g1(X)) -> g1(proper1(X))
proper1(h1(X)) -> h1(proper1(X))
f1(ok1(X)) -> ok1(f1(X))
g1(ok1(X)) -> ok1(g1(X))
h1(ok1(X)) -> ok1(h1(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.
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.

PROPER1(g1(X)) -> PROPER1(X)
Used ordering: Combined order from the following AFS and order.
PROPER1(x1)  =  PROPER1(x1)
g1(x1)  =  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
                ↳ QDPOrderProof
QDP
                    ↳ QDPOrderProof
          ↳ 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(f1(X)) -> mark1(g1(h1(f1(X))))
active1(f1(X)) -> f1(active1(X))
active1(h1(X)) -> h1(active1(X))
f1(mark1(X)) -> mark1(f1(X))
h1(mark1(X)) -> mark1(h1(X))
proper1(f1(X)) -> f1(proper1(X))
proper1(g1(X)) -> g1(proper1(X))
proper1(h1(X)) -> h1(proper1(X))
f1(ok1(X)) -> ok1(f1(X))
g1(ok1(X)) -> ok1(g1(X))
h1(ok1(X)) -> ok1(h1(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.
We use the reduction pair processor [13].


The following pairs can be strictly oriented and are deleted.


PROPER1(g1(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)
g1(x1)  =  g1(x1)

Lexicographic Path Order [19].
Precedence:
g1 > PROPER1


The following usable rules [14] were oriented: none



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

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

active1(f1(X)) -> mark1(g1(h1(f1(X))))
active1(f1(X)) -> f1(active1(X))
active1(h1(X)) -> h1(active1(X))
f1(mark1(X)) -> mark1(f1(X))
h1(mark1(X)) -> mark1(h1(X))
proper1(f1(X)) -> f1(proper1(X))
proper1(g1(X)) -> g1(proper1(X))
proper1(h1(X)) -> h1(proper1(X))
f1(ok1(X)) -> ok1(f1(X))
g1(ok1(X)) -> ok1(g1(X))
h1(ok1(X)) -> ok1(h1(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
            ↳ QDPOrderProof
          ↳ QDP

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

ACTIVE1(f1(X)) -> ACTIVE1(X)
ACTIVE1(h1(X)) -> ACTIVE1(X)

The TRS R consists of the following rules:

active1(f1(X)) -> mark1(g1(h1(f1(X))))
active1(f1(X)) -> f1(active1(X))
active1(h1(X)) -> h1(active1(X))
f1(mark1(X)) -> mark1(f1(X))
h1(mark1(X)) -> mark1(h1(X))
proper1(f1(X)) -> f1(proper1(X))
proper1(g1(X)) -> g1(proper1(X))
proper1(h1(X)) -> h1(proper1(X))
f1(ok1(X)) -> ok1(f1(X))
g1(ok1(X)) -> ok1(g1(X))
h1(ok1(X)) -> ok1(h1(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.
We use the reduction pair processor [13].


The following pairs can be strictly oriented and are deleted.


ACTIVE1(h1(X)) -> ACTIVE1(X)
The remaining pairs can at least by weakly be oriented.

ACTIVE1(f1(X)) -> ACTIVE1(X)
Used ordering: Combined order from the following AFS and order.
ACTIVE1(x1)  =  ACTIVE1(x1)
f1(x1)  =  x1
h1(x1)  =  h1(x1)

Lexicographic Path Order [19].
Precedence:
trivial


The following usable rules [14] were oriented: none



↳ QTRS
  ↳ DependencyPairsProof
    ↳ QDP
      ↳ DependencyGraphProof
        ↳ AND
          ↳ QDP
          ↳ QDP
          ↳ QDP
          ↳ QDP
          ↳ QDP
            ↳ QDPOrderProof
QDP
                ↳ QDPOrderProof
          ↳ 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(g1(h1(f1(X))))
active1(f1(X)) -> f1(active1(X))
active1(h1(X)) -> h1(active1(X))
f1(mark1(X)) -> mark1(f1(X))
h1(mark1(X)) -> mark1(h1(X))
proper1(f1(X)) -> f1(proper1(X))
proper1(g1(X)) -> g1(proper1(X))
proper1(h1(X)) -> h1(proper1(X))
f1(ok1(X)) -> ok1(f1(X))
g1(ok1(X)) -> ok1(g1(X))
h1(ok1(X)) -> ok1(h1(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.
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
          ↳ QDP
          ↳ QDP
          ↳ QDP
            ↳ QDPOrderProof
              ↳ QDP
                ↳ QDPOrderProof
QDP
                    ↳ PisEmptyProof
          ↳ QDP

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

active1(f1(X)) -> mark1(g1(h1(f1(X))))
active1(f1(X)) -> f1(active1(X))
active1(h1(X)) -> h1(active1(X))
f1(mark1(X)) -> mark1(f1(X))
h1(mark1(X)) -> mark1(h1(X))
proper1(f1(X)) -> f1(proper1(X))
proper1(g1(X)) -> g1(proper1(X))
proper1(h1(X)) -> h1(proper1(X))
f1(ok1(X)) -> ok1(f1(X))
g1(ok1(X)) -> ok1(g1(X))
h1(ok1(X)) -> ok1(h1(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
QDP
            ↳ QDPOrderProof

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(f1(X)) -> mark1(g1(h1(f1(X))))
active1(f1(X)) -> f1(active1(X))
active1(h1(X)) -> h1(active1(X))
f1(mark1(X)) -> mark1(f1(X))
h1(mark1(X)) -> mark1(h1(X))
proper1(f1(X)) -> f1(proper1(X))
proper1(g1(X)) -> g1(proper1(X))
proper1(h1(X)) -> h1(proper1(X))
f1(ok1(X)) -> ok1(f1(X))
g1(ok1(X)) -> ok1(g1(X))
h1(ok1(X)) -> ok1(h1(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.
We use the reduction pair processor [13].


The following pairs can be strictly oriented and are deleted.


TOP1(mark1(X)) -> TOP1(proper1(X))
The remaining pairs can at least by weakly be oriented.

TOP1(ok1(X)) -> TOP1(active1(X))
Used ordering: Combined order from the following AFS and order.
TOP1(x1)  =  x1
ok1(x1)  =  x1
active1(x1)  =  x1
mark1(x1)  =  mark1(x1)
proper1(x1)  =  x1
f1(x1)  =  f1(x1)
g1(x1)  =  g
h1(x1)  =  h1(x1)

Lexicographic Path Order [19].
Precedence:
[f1, h1] > mark1
[f1, h1] > g


The following usable rules [14] were oriented:

proper1(f1(X)) -> f1(proper1(X))
proper1(g1(X)) -> g1(proper1(X))
proper1(h1(X)) -> h1(proper1(X))
h1(mark1(X)) -> mark1(h1(X))
h1(ok1(X)) -> ok1(h1(X))
g1(ok1(X)) -> ok1(g1(X))
f1(mark1(X)) -> mark1(f1(X))
f1(ok1(X)) -> ok1(f1(X))
active1(f1(X)) -> mark1(g1(h1(f1(X))))
active1(f1(X)) -> f1(active1(X))
active1(h1(X)) -> h1(active1(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(ok1(X)) -> TOP1(active1(X))

The TRS R consists of the following rules:

active1(f1(X)) -> mark1(g1(h1(f1(X))))
active1(f1(X)) -> f1(active1(X))
active1(h1(X)) -> h1(active1(X))
f1(mark1(X)) -> mark1(f1(X))
h1(mark1(X)) -> mark1(h1(X))
proper1(f1(X)) -> f1(proper1(X))
proper1(g1(X)) -> g1(proper1(X))
proper1(h1(X)) -> h1(proper1(X))
f1(ok1(X)) -> ok1(f1(X))
g1(ok1(X)) -> ok1(g1(X))
h1(ok1(X)) -> ok1(h1(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.
We use the reduction pair processor [13].


The following pairs can be strictly oriented and are deleted.


TOP1(ok1(X)) -> TOP1(active1(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)
ok1(x1)  =  ok
active1(x1)  =  active
f1(x1)  =  x1
mark1(x1)  =  mark
g1(x1)  =  g1(x1)
h1(x1)  =  x1

Lexicographic Path Order [19].
Precedence:
[TOP1, ok] > [active, mark, g1]


The following usable rules [14] were oriented:

active1(f1(X)) -> mark1(g1(h1(f1(X))))
active1(f1(X)) -> f1(active1(X))
active1(h1(X)) -> h1(active1(X))
h1(mark1(X)) -> mark1(h1(X))
h1(ok1(X)) -> ok1(h1(X))
f1(mark1(X)) -> mark1(f1(X))
f1(ok1(X)) -> ok1(f1(X))
g1(ok1(X)) -> ok1(g1(X))



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

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

active1(f1(X)) -> mark1(g1(h1(f1(X))))
active1(f1(X)) -> f1(active1(X))
active1(h1(X)) -> h1(active1(X))
f1(mark1(X)) -> mark1(f1(X))
h1(mark1(X)) -> mark1(h1(X))
proper1(f1(X)) -> f1(proper1(X))
proper1(g1(X)) -> g1(proper1(X))
proper1(h1(X)) -> h1(proper1(X))
f1(ok1(X)) -> ok1(f1(X))
g1(ok1(X)) -> ok1(g1(X))
h1(ok1(X)) -> ok1(h1(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.