Termination w.r.t. Q of the following Term Rewriting System could not be shown:

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

active1(f3(X, g1(X), Y)) -> mark1(f3(Y, Y, Y))
active1(g1(b)) -> mark1(c)
active1(b) -> mark1(c)
active1(g1(X)) -> g1(active1(X))
g1(mark1(X)) -> mark1(g1(X))
proper1(f3(X1, X2, X3)) -> f3(proper1(X1), proper1(X2), proper1(X3))
proper1(g1(X)) -> g1(proper1(X))
proper1(b) -> ok1(b)
proper1(c) -> ok1(c)
f3(ok1(X1), ok1(X2), ok1(X3)) -> ok1(f3(X1, X2, X3))
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(f3(X, g1(X), Y)) -> mark1(f3(Y, Y, Y))
active1(g1(b)) -> mark1(c)
active1(b) -> mark1(c)
active1(g1(X)) -> g1(active1(X))
g1(mark1(X)) -> mark1(g1(X))
proper1(f3(X1, X2, X3)) -> f3(proper1(X1), proper1(X2), proper1(X3))
proper1(g1(X)) -> g1(proper1(X))
proper1(b) -> ok1(b)
proper1(c) -> ok1(c)
f3(ok1(X1), ok1(X2), ok1(X3)) -> ok1(f3(X1, X2, X3))
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(g1(X)) -> ACTIVE1(X)
TOP1(mark1(X)) -> PROPER1(X)
PROPER1(f3(X1, X2, X3)) -> PROPER1(X1)
PROPER1(g1(X)) -> G1(proper1(X))
ACTIVE1(f3(X, g1(X), Y)) -> F3(Y, Y, Y)
PROPER1(g1(X)) -> PROPER1(X)
PROPER1(f3(X1, X2, X3)) -> PROPER1(X3)
TOP1(ok1(X)) -> ACTIVE1(X)
G1(mark1(X)) -> G1(X)
ACTIVE1(g1(X)) -> G1(active1(X))
TOP1(ok1(X)) -> TOP1(active1(X))
TOP1(mark1(X)) -> TOP1(proper1(X))
PROPER1(f3(X1, X2, X3)) -> PROPER1(X2)
G1(ok1(X)) -> G1(X)
PROPER1(f3(X1, X2, X3)) -> F3(proper1(X1), proper1(X2), proper1(X3))
F3(ok1(X1), ok1(X2), ok1(X3)) -> F3(X1, X2, X3)

The TRS R consists of the following rules:

active1(f3(X, g1(X), Y)) -> mark1(f3(Y, Y, Y))
active1(g1(b)) -> mark1(c)
active1(b) -> mark1(c)
active1(g1(X)) -> g1(active1(X))
g1(mark1(X)) -> mark1(g1(X))
proper1(f3(X1, X2, X3)) -> f3(proper1(X1), proper1(X2), proper1(X3))
proper1(g1(X)) -> g1(proper1(X))
proper1(b) -> ok1(b)
proper1(c) -> ok1(c)
f3(ok1(X1), ok1(X2), ok1(X3)) -> ok1(f3(X1, X2, X3))
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(g1(X)) -> ACTIVE1(X)
TOP1(mark1(X)) -> PROPER1(X)
PROPER1(f3(X1, X2, X3)) -> PROPER1(X1)
PROPER1(g1(X)) -> G1(proper1(X))
ACTIVE1(f3(X, g1(X), Y)) -> F3(Y, Y, Y)
PROPER1(g1(X)) -> PROPER1(X)
PROPER1(f3(X1, X2, X3)) -> PROPER1(X3)
TOP1(ok1(X)) -> ACTIVE1(X)
G1(mark1(X)) -> G1(X)
ACTIVE1(g1(X)) -> G1(active1(X))
TOP1(ok1(X)) -> TOP1(active1(X))
TOP1(mark1(X)) -> TOP1(proper1(X))
PROPER1(f3(X1, X2, X3)) -> PROPER1(X2)
G1(ok1(X)) -> G1(X)
PROPER1(f3(X1, X2, X3)) -> F3(proper1(X1), proper1(X2), proper1(X3))
F3(ok1(X1), ok1(X2), ok1(X3)) -> F3(X1, X2, X3)

The TRS R consists of the following rules:

active1(f3(X, g1(X), Y)) -> mark1(f3(Y, Y, Y))
active1(g1(b)) -> mark1(c)
active1(b) -> mark1(c)
active1(g1(X)) -> g1(active1(X))
g1(mark1(X)) -> mark1(g1(X))
proper1(f3(X1, X2, X3)) -> f3(proper1(X1), proper1(X2), proper1(X3))
proper1(g1(X)) -> g1(proper1(X))
proper1(b) -> ok1(b)
proper1(c) -> ok1(c)
f3(ok1(X1), ok1(X2), ok1(X3)) -> ok1(f3(X1, X2, X3))
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 6 less nodes.

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

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

F3(ok1(X1), ok1(X2), ok1(X3)) -> F3(X1, X2, X3)

The TRS R consists of the following rules:

active1(f3(X, g1(X), Y)) -> mark1(f3(Y, Y, Y))
active1(g1(b)) -> mark1(c)
active1(b) -> mark1(c)
active1(g1(X)) -> g1(active1(X))
g1(mark1(X)) -> mark1(g1(X))
proper1(f3(X1, X2, X3)) -> f3(proper1(X1), proper1(X2), proper1(X3))
proper1(g1(X)) -> g1(proper1(X))
proper1(b) -> ok1(b)
proper1(c) -> ok1(c)
f3(ok1(X1), ok1(X2), ok1(X3)) -> ok1(f3(X1, X2, X3))
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.


F3(ok1(X1), ok1(X2), ok1(X3)) -> F3(X1, X2, X3)
The remaining pairs can at least be oriented weakly.
none
Used ordering: Polynomial interpretation [21]:

POL(F3(x1, x2, x3)) = x3   
POL(ok1(x1)) = 1 + 2·x1   

The following usable rules [14] were oriented: none



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

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

active1(f3(X, g1(X), Y)) -> mark1(f3(Y, Y, Y))
active1(g1(b)) -> mark1(c)
active1(b) -> mark1(c)
active1(g1(X)) -> g1(active1(X))
g1(mark1(X)) -> mark1(g1(X))
proper1(f3(X1, X2, X3)) -> f3(proper1(X1), proper1(X2), proper1(X3))
proper1(g1(X)) -> g1(proper1(X))
proper1(b) -> ok1(b)
proper1(c) -> ok1(c)
f3(ok1(X1), ok1(X2), ok1(X3)) -> ok1(f3(X1, X2, X3))
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:

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

The TRS R consists of the following rules:

active1(f3(X, g1(X), Y)) -> mark1(f3(Y, Y, Y))
active1(g1(b)) -> mark1(c)
active1(b) -> mark1(c)
active1(g1(X)) -> g1(active1(X))
g1(mark1(X)) -> mark1(g1(X))
proper1(f3(X1, X2, X3)) -> f3(proper1(X1), proper1(X2), proper1(X3))
proper1(g1(X)) -> g1(proper1(X))
proper1(b) -> ok1(b)
proper1(c) -> ok1(c)
f3(ok1(X1), ok1(X2), ok1(X3)) -> ok1(f3(X1, X2, X3))
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)
G1(ok1(X)) -> G1(X)
The remaining pairs can at least be oriented weakly.
none
Used ordering: Polynomial interpretation [21]:

POL(G1(x1)) = 2·x1   
POL(mark1(x1)) = 2 + 2·x1   
POL(ok1(x1)) = 2 + 2·x1   

The following usable rules [14] were oriented: none



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

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

active1(f3(X, g1(X), Y)) -> mark1(f3(Y, Y, Y))
active1(g1(b)) -> mark1(c)
active1(b) -> mark1(c)
active1(g1(X)) -> g1(active1(X))
g1(mark1(X)) -> mark1(g1(X))
proper1(f3(X1, X2, X3)) -> f3(proper1(X1), proper1(X2), proper1(X3))
proper1(g1(X)) -> g1(proper1(X))
proper1(b) -> ok1(b)
proper1(c) -> ok1(c)
f3(ok1(X1), ok1(X2), ok1(X3)) -> ok1(f3(X1, X2, X3))
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(f3(X1, X2, X3)) -> PROPER1(X1)
PROPER1(g1(X)) -> PROPER1(X)
PROPER1(f3(X1, X2, X3)) -> PROPER1(X3)
PROPER1(f3(X1, X2, X3)) -> PROPER1(X2)

The TRS R consists of the following rules:

active1(f3(X, g1(X), Y)) -> mark1(f3(Y, Y, Y))
active1(g1(b)) -> mark1(c)
active1(b) -> mark1(c)
active1(g1(X)) -> g1(active1(X))
g1(mark1(X)) -> mark1(g1(X))
proper1(f3(X1, X2, X3)) -> f3(proper1(X1), proper1(X2), proper1(X3))
proper1(g1(X)) -> g1(proper1(X))
proper1(b) -> ok1(b)
proper1(c) -> ok1(c)
f3(ok1(X1), ok1(X2), ok1(X3)) -> ok1(f3(X1, X2, X3))
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(f3(X1, X2, X3)) -> PROPER1(X1)
PROPER1(g1(X)) -> PROPER1(X)
PROPER1(f3(X1, X2, X3)) -> PROPER1(X3)
PROPER1(f3(X1, X2, X3)) -> PROPER1(X2)
The remaining pairs can at least be oriented weakly.
none
Used ordering: Polynomial interpretation [21]:

POL(PROPER1(x1)) = 2·x1   
POL(f3(x1, x2, x3)) = 2 + 2·x1 + 2·x2 + 2·x3   
POL(g1(x1)) = 2 + 2·x1   

The following usable rules [14] were oriented: none



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

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

active1(f3(X, g1(X), Y)) -> mark1(f3(Y, Y, Y))
active1(g1(b)) -> mark1(c)
active1(b) -> mark1(c)
active1(g1(X)) -> g1(active1(X))
g1(mark1(X)) -> mark1(g1(X))
proper1(f3(X1, X2, X3)) -> f3(proper1(X1), proper1(X2), proper1(X3))
proper1(g1(X)) -> g1(proper1(X))
proper1(b) -> ok1(b)
proper1(c) -> ok1(c)
f3(ok1(X1), ok1(X2), ok1(X3)) -> ok1(f3(X1, X2, X3))
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)

The TRS R consists of the following rules:

active1(f3(X, g1(X), Y)) -> mark1(f3(Y, Y, Y))
active1(g1(b)) -> mark1(c)
active1(b) -> mark1(c)
active1(g1(X)) -> g1(active1(X))
g1(mark1(X)) -> mark1(g1(X))
proper1(f3(X1, X2, X3)) -> f3(proper1(X1), proper1(X2), proper1(X3))
proper1(g1(X)) -> g1(proper1(X))
proper1(b) -> ok1(b)
proper1(c) -> ok1(c)
f3(ok1(X1), ok1(X2), ok1(X3)) -> ok1(f3(X1, X2, X3))
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.
none
Used ordering: Polynomial interpretation [21]:

POL(ACTIVE1(x1)) = 2·x1   
POL(g1(x1)) = 1 + 2·x1   

The following usable rules [14] were oriented: none



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

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

active1(f3(X, g1(X), Y)) -> mark1(f3(Y, Y, Y))
active1(g1(b)) -> mark1(c)
active1(b) -> mark1(c)
active1(g1(X)) -> g1(active1(X))
g1(mark1(X)) -> mark1(g1(X))
proper1(f3(X1, X2, X3)) -> f3(proper1(X1), proper1(X2), proper1(X3))
proper1(g1(X)) -> g1(proper1(X))
proper1(b) -> ok1(b)
proper1(c) -> ok1(c)
f3(ok1(X1), ok1(X2), ok1(X3)) -> ok1(f3(X1, X2, X3))
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

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(f3(X, g1(X), Y)) -> mark1(f3(Y, Y, Y))
active1(g1(b)) -> mark1(c)
active1(b) -> mark1(c)
active1(g1(X)) -> g1(active1(X))
g1(mark1(X)) -> mark1(g1(X))
proper1(f3(X1, X2, X3)) -> f3(proper1(X1), proper1(X2), proper1(X3))
proper1(g1(X)) -> g1(proper1(X))
proper1(b) -> ok1(b)
proper1(c) -> ok1(c)
f3(ok1(X1), ok1(X2), ok1(X3)) -> ok1(f3(X1, X2, X3))
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.