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

The TRS R consists of the following rules:

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

The TRS R consists of the following rules:

active1(f3(a, b, X)) -> mark1(f3(X, X, X))
active1(c) -> mark1(a)
active1(c) -> mark1(b)
active1(f3(X1, X2, X3)) -> f3(active1(X1), X2, X3)
active1(f3(X1, X2, X3)) -> f3(X1, X2, active1(X3))
f3(mark1(X1), X2, X3) -> mark1(f3(X1, X2, X3))
f3(X1, X2, mark1(X3)) -> mark1(f3(X1, X2, X3))
proper1(f3(X1, X2, X3)) -> f3(proper1(X1), proper1(X2), proper1(X3))
proper1(a) -> ok1(a)
proper1(b) -> ok1(b)
proper1(c) -> ok1(c)
f3(ok1(X1), ok1(X2), ok1(X3)) -> ok1(f3(X1, X2, X3))
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 4 SCCs with 6 less nodes.

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

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

F3(mark1(X1), X2, X3) -> F3(X1, X2, X3)
F3(ok1(X1), ok1(X2), ok1(X3)) -> F3(X1, X2, X3)
F3(X1, X2, mark1(X3)) -> F3(X1, X2, X3)

The TRS R consists of the following rules:

active1(f3(a, b, X)) -> mark1(f3(X, X, X))
active1(c) -> mark1(a)
active1(c) -> mark1(b)
active1(f3(X1, X2, X3)) -> f3(active1(X1), X2, X3)
active1(f3(X1, X2, X3)) -> f3(X1, X2, active1(X3))
f3(mark1(X1), X2, X3) -> mark1(f3(X1, X2, X3))
f3(X1, X2, mark1(X3)) -> mark1(f3(X1, X2, X3))
proper1(f3(X1, X2, X3)) -> f3(proper1(X1), proper1(X2), proper1(X3))
proper1(a) -> ok1(a)
proper1(b) -> ok1(b)
proper1(c) -> ok1(c)
f3(ok1(X1), ok1(X2), ok1(X3)) -> ok1(f3(X1, X2, X3))
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(mark1(X1), X2, X3) -> F3(X1, X2, X3)
The remaining pairs can at least be oriented weakly.

F3(ok1(X1), ok1(X2), ok1(X3)) -> F3(X1, X2, X3)
F3(X1, X2, mark1(X3)) -> F3(X1, X2, X3)
Used ordering: Polynomial Order [17,21] with Interpretation:

POL( F3(x1, ..., x3) ) = x1


POL( mark1(x1) ) = x1 + 1


POL( ok1(x1) ) = x1



The following usable rules [14] were oriented: none



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

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

F3(X1, X2, mark1(X3)) -> F3(X1, X2, X3)
F3(ok1(X1), ok1(X2), ok1(X3)) -> F3(X1, X2, X3)

The TRS R consists of the following rules:

active1(f3(a, b, X)) -> mark1(f3(X, X, X))
active1(c) -> mark1(a)
active1(c) -> mark1(b)
active1(f3(X1, X2, X3)) -> f3(active1(X1), X2, X3)
active1(f3(X1, X2, X3)) -> f3(X1, X2, active1(X3))
f3(mark1(X1), X2, X3) -> mark1(f3(X1, X2, X3))
f3(X1, X2, mark1(X3)) -> mark1(f3(X1, X2, X3))
proper1(f3(X1, X2, X3)) -> f3(proper1(X1), proper1(X2), proper1(X3))
proper1(a) -> ok1(a)
proper1(b) -> ok1(b)
proper1(c) -> ok1(c)
f3(ok1(X1), ok1(X2), ok1(X3)) -> ok1(f3(X1, X2, X3))
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(X1, X2, mark1(X3)) -> F3(X1, X2, X3)
The remaining pairs can at least be oriented weakly.

F3(ok1(X1), ok1(X2), ok1(X3)) -> F3(X1, X2, X3)
Used ordering: Polynomial Order [17,21] with Interpretation:

POL( F3(x1, ..., x3) ) = x3


POL( mark1(x1) ) = x1 + 1


POL( ok1(x1) ) = x1



The following usable rules [14] were oriented: none



↳ QTRS
  ↳ DependencyPairsProof
    ↳ QDP
      ↳ DependencyGraphProof
        ↳ AND
          ↳ QDP
            ↳ QDPOrderProof
              ↳ QDP
                ↳ QDPOrderProof
                  ↳ QDP
                    ↳ QDPOrderProof
          ↳ 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(a, b, X)) -> mark1(f3(X, X, X))
active1(c) -> mark1(a)
active1(c) -> mark1(b)
active1(f3(X1, X2, X3)) -> f3(active1(X1), X2, X3)
active1(f3(X1, X2, X3)) -> f3(X1, X2, active1(X3))
f3(mark1(X1), X2, X3) -> mark1(f3(X1, X2, X3))
f3(X1, X2, mark1(X3)) -> mark1(f3(X1, X2, X3))
proper1(f3(X1, X2, X3)) -> f3(proper1(X1), proper1(X2), proper1(X3))
proper1(a) -> ok1(a)
proper1(b) -> ok1(b)
proper1(c) -> ok1(c)
f3(ok1(X1), ok1(X2), ok1(X3)) -> ok1(f3(X1, X2, X3))
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 Order [17,21] with Interpretation:

POL( F3(x1, ..., x3) ) = x3


POL( ok1(x1) ) = x1 + 1



The following usable rules [14] were oriented: none



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

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

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

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

PROPER1(f3(X1, X2, X3)) -> PROPER1(X1)
PROPER1(f3(X1, X2, X3)) -> PROPER1(X3)
PROPER1(f3(X1, X2, X3)) -> PROPER1(X2)

The TRS R consists of the following rules:

active1(f3(a, b, X)) -> mark1(f3(X, X, X))
active1(c) -> mark1(a)
active1(c) -> mark1(b)
active1(f3(X1, X2, X3)) -> f3(active1(X1), X2, X3)
active1(f3(X1, X2, X3)) -> f3(X1, X2, active1(X3))
f3(mark1(X1), X2, X3) -> mark1(f3(X1, X2, X3))
f3(X1, X2, mark1(X3)) -> mark1(f3(X1, X2, X3))
proper1(f3(X1, X2, X3)) -> f3(proper1(X1), proper1(X2), proper1(X3))
proper1(a) -> ok1(a)
proper1(b) -> ok1(b)
proper1(c) -> ok1(c)
f3(ok1(X1), ok1(X2), ok1(X3)) -> ok1(f3(X1, X2, X3))
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(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 Order [17,21] with Interpretation:

POL( PROPER1(x1) ) = x1


POL( f3(x1, ..., x3) ) = x1 + x2 + x3 + 1



The following usable rules [14] were oriented: none



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

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

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

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

ACTIVE1(f3(X1, X2, X3)) -> ACTIVE1(X1)
ACTIVE1(f3(X1, X2, X3)) -> ACTIVE1(X3)

The TRS R consists of the following rules:

active1(f3(a, b, X)) -> mark1(f3(X, X, X))
active1(c) -> mark1(a)
active1(c) -> mark1(b)
active1(f3(X1, X2, X3)) -> f3(active1(X1), X2, X3)
active1(f3(X1, X2, X3)) -> f3(X1, X2, active1(X3))
f3(mark1(X1), X2, X3) -> mark1(f3(X1, X2, X3))
f3(X1, X2, mark1(X3)) -> mark1(f3(X1, X2, X3))
proper1(f3(X1, X2, X3)) -> f3(proper1(X1), proper1(X2), proper1(X3))
proper1(a) -> ok1(a)
proper1(b) -> ok1(b)
proper1(c) -> ok1(c)
f3(ok1(X1), ok1(X2), ok1(X3)) -> ok1(f3(X1, X2, X3))
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(f3(X1, X2, X3)) -> ACTIVE1(X1)
ACTIVE1(f3(X1, X2, X3)) -> ACTIVE1(X3)
The remaining pairs can at least be oriented weakly.
none
Used ordering: Polynomial Order [17,21] with Interpretation:

POL( ACTIVE1(x1) ) = x1


POL( f3(x1, ..., x3) ) = x1 + x3 + 1



The following usable rules [14] were oriented: none



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

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

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

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