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.

Using Dependency Pairs [1,13] we result in the following initial DP problem:
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 [13,14,18] contains 5 SCCs with 7 less nodes.

↳ QTRS
  ↳ DependencyPairsProof
    ↳ QDP
      ↳ DependencyGraphProof
        ↳ AND
QDP
            ↳ QDPOrderProof
          ↳ 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.
We use the reduction pair processor [13].


The following pairs can be oriented strictly and are deleted.


G1(mark1(X)) -> G1(X)
The remaining pairs can at least be oriented weakly.

G1(ok1(X)) -> G1(X)
Used ordering: Polynomial interpretation [21]:

POL(G1(x1)) = x12   
POL(mark1(x1)) = 1 + x12   
POL(ok1(x1)) = x1   

The following usable rules [14] were oriented: none



↳ QTRS
  ↳ DependencyPairsProof
    ↳ QDP
      ↳ DependencyGraphProof
        ↳ AND
          ↳ QDP
            ↳ QDPOrderProof
QDP
                ↳ QDPOrderProof
          ↳ 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(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.
We use the reduction pair processor [13].


The following pairs can be oriented strictly and are deleted.


G1(ok1(X)) -> G1(X)
The remaining pairs can at least be oriented weakly.
none
Used ordering: Polynomial interpretation [21]:

POL(G1(x1)) = x12   
POL(ok1(x1)) = 1 + x12   

The following usable rules [14] were oriented: none



↳ QTRS
  ↳ DependencyPairsProof
    ↳ QDP
      ↳ DependencyGraphProof
        ↳ AND
          ↳ 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(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
            ↳ QDPOrderProof
          ↳ 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.
We use the reduction pair processor [13].


The following pairs can be oriented strictly and are deleted.


F2(ok1(X1), ok1(X2)) -> F2(X1, X2)
The remaining pairs can at least be oriented weakly.

F2(mark1(X1), X2) -> F2(X1, X2)
Used ordering: Polynomial interpretation [21]:

POL(F2(x1, x2)) = x2   
POL(mark1(x1)) = 0   
POL(ok1(x1)) = 1 + x1   

The following usable rules [14] were oriented: none



↳ QTRS
  ↳ DependencyPairsProof
    ↳ QDP
      ↳ DependencyGraphProof
        ↳ AND
          ↳ QDP
          ↳ QDP
            ↳ QDPOrderProof
QDP
                ↳ QDPOrderProof
          ↳ 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.
We use the reduction pair processor [13].


The following pairs can be oriented strictly and are deleted.


F2(mark1(X1), X2) -> F2(X1, X2)
The remaining pairs can at least be oriented weakly.
none
Used ordering: Polynomial interpretation [21]:

POL(F2(x1, x2)) = x1   
POL(mark1(x1)) = 1 + x1   

The following usable rules [14] were oriented: none



↳ QTRS
  ↳ DependencyPairsProof
    ↳ QDP
      ↳ DependencyGraphProof
        ↳ AND
          ↳ 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(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
            ↳ QDPOrderProof
          ↳ 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.
We use the reduction pair processor [13].


The following pairs can be oriented strictly and are deleted.


PROPER1(g1(X)) -> PROPER1(X)
The remaining pairs can at least be oriented weakly.

PROPER1(f2(X1, X2)) -> PROPER1(X2)
PROPER1(f2(X1, X2)) -> PROPER1(X1)
Used ordering: Polynomial interpretation [21]:

POL(PROPER1(x1)) = x12   
POL(f2(x1, x2)) = x1 + x2   
POL(g1(x1)) = 1 + x12   

The following usable rules [14] were oriented: none



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

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

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.
We use the reduction pair processor [13].


The following pairs can be oriented strictly and are deleted.


PROPER1(f2(X1, X2)) -> PROPER1(X2)
PROPER1(f2(X1, X2)) -> PROPER1(X1)
The remaining pairs can at least be oriented weakly.
none
Used ordering: Polynomial interpretation [21]:

POL(PROPER1(x1)) = x12   
POL(f2(x1, x2)) = 1 + x1 + x2   

The following usable rules [14] were oriented: none



↳ QTRS
  ↳ DependencyPairsProof
    ↳ QDP
      ↳ DependencyGraphProof
        ↳ AND
          ↳ QDP
          ↳ QDP
          ↳ QDP
            ↳ QDPOrderProof
              ↳ QDP
                ↳ QDPOrderProof
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
            ↳ QDPOrderProof
          ↳ 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.
We use the reduction pair processor [13].


The following pairs can be oriented strictly and are deleted.


ACTIVE1(g1(X)) -> ACTIVE1(X)
The remaining pairs can at least be oriented weakly.

ACTIVE1(f2(X1, X2)) -> ACTIVE1(X1)
Used ordering: Polynomial interpretation [21]:

POL(ACTIVE1(x1)) = x12   
POL(f2(x1, x2)) = x1   
POL(g1(x1)) = 1 + x12   

The following usable rules [14] were oriented: none



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

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

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.
We use the reduction pair processor [13].


The following pairs can be oriented strictly and are deleted.


ACTIVE1(f2(X1, X2)) -> ACTIVE1(X1)
The remaining pairs can at least be oriented weakly.
none
Used ordering: Polynomial interpretation [21]:

POL(ACTIVE1(x1)) = x12   
POL(f2(x1, x2)) = 1 + x1   

The following usable rules [14] were oriented: none



↳ QTRS
  ↳ DependencyPairsProof
    ↳ QDP
      ↳ DependencyGraphProof
        ↳ AND
          ↳ 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(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
            ↳ 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(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.
We use the reduction pair processor [13].


The following pairs can be oriented strictly and are deleted.


TOP1(ok1(X)) -> TOP1(active1(X))
The remaining pairs can at least be oriented weakly.

TOP1(mark1(X)) -> TOP1(proper1(X))
Used ordering: Polynomial interpretation [21]:

POL(TOP1(x1)) = x12   
POL(active1(x1)) = 0   
POL(f2(x1, x2)) = x1   
POL(g1(x1)) = x12   
POL(mark1(x1)) = 0   
POL(ok1(x1)) = 1   
POL(proper1(x1)) = 0   

The following usable rules [14] were oriented:

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



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

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.
We use the reduction pair processor [13].


The following pairs can be oriented strictly and are deleted.


TOP1(mark1(X)) -> TOP1(proper1(X))
The remaining pairs can at least be oriented weakly.
none
Used ordering: Polynomial interpretation [21]:

POL(TOP1(x1)) = x12   
POL(f2(x1, x2)) = x1   
POL(g1(x1)) = x1   
POL(mark1(x1)) = 1   
POL(ok1(x1)) = 0   
POL(proper1(x1)) = 0   

The following usable rules [14] were oriented:

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



↳ QTRS
  ↳ DependencyPairsProof
    ↳ QDP
      ↳ DependencyGraphProof
        ↳ AND
          ↳ 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(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.