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:

active(f(a, b, X)) → mark(f(X, X, X))
active(c) → mark(a)
active(c) → mark(b)
active(f(X1, X2, X3)) → f(active(X1), X2, X3)
active(f(X1, X2, X3)) → f(X1, X2, active(X3))
f(mark(X1), X2, X3) → mark(f(X1, X2, X3))
f(X1, X2, mark(X3)) → mark(f(X1, X2, X3))
proper(f(X1, X2, X3)) → f(proper(X1), proper(X2), proper(X3))
proper(a) → ok(a)
proper(b) → ok(b)
proper(c) → ok(c)
f(ok(X1), ok(X2), ok(X3)) → ok(f(X1, X2, X3))
top(mark(X)) → top(proper(X))
top(ok(X)) → top(active(X))

Q is empty.


QTRS
  ↳ DependencyPairsProof

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

active(f(a, b, X)) → mark(f(X, X, X))
active(c) → mark(a)
active(c) → mark(b)
active(f(X1, X2, X3)) → f(active(X1), X2, X3)
active(f(X1, X2, X3)) → f(X1, X2, active(X3))
f(mark(X1), X2, X3) → mark(f(X1, X2, X3))
f(X1, X2, mark(X3)) → mark(f(X1, X2, X3))
proper(f(X1, X2, X3)) → f(proper(X1), proper(X2), proper(X3))
proper(a) → ok(a)
proper(b) → ok(b)
proper(c) → ok(c)
f(ok(X1), ok(X2), ok(X3)) → ok(f(X1, X2, X3))
top(mark(X)) → top(proper(X))
top(ok(X)) → top(active(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:

ACTIVE(f(X1, X2, X3)) → ACTIVE(X1)
TOP(mark(X)) → PROPER(X)
ACTIVE(f(X1, X2, X3)) → ACTIVE(X3)
PROPER(f(X1, X2, X3)) → PROPER(X1)
ACTIVE(f(X1, X2, X3)) → F(active(X1), X2, X3)
PROPER(f(X1, X2, X3)) → PROPER(X3)
ACTIVE(f(a, b, X)) → F(X, X, X)
F(mark(X1), X2, X3) → F(X1, X2, X3)
TOP(ok(X)) → ACTIVE(X)
ACTIVE(f(X1, X2, X3)) → F(X1, X2, active(X3))
TOP(ok(X)) → TOP(active(X))
TOP(mark(X)) → TOP(proper(X))
PROPER(f(X1, X2, X3)) → PROPER(X2)
PROPER(f(X1, X2, X3)) → F(proper(X1), proper(X2), proper(X3))
F(X1, X2, mark(X3)) → F(X1, X2, X3)
F(ok(X1), ok(X2), ok(X3)) → F(X1, X2, X3)

The TRS R consists of the following rules:

active(f(a, b, X)) → mark(f(X, X, X))
active(c) → mark(a)
active(c) → mark(b)
active(f(X1, X2, X3)) → f(active(X1), X2, X3)
active(f(X1, X2, X3)) → f(X1, X2, active(X3))
f(mark(X1), X2, X3) → mark(f(X1, X2, X3))
f(X1, X2, mark(X3)) → mark(f(X1, X2, X3))
proper(f(X1, X2, X3)) → f(proper(X1), proper(X2), proper(X3))
proper(a) → ok(a)
proper(b) → ok(b)
proper(c) → ok(c)
f(ok(X1), ok(X2), ok(X3)) → ok(f(X1, X2, X3))
top(mark(X)) → top(proper(X))
top(ok(X)) → top(active(X))

Q is empty.
We have to consider all minimal (P,Q,R)-chains.

↳ QTRS
  ↳ DependencyPairsProof
QDP
      ↳ EdgeDeletionProof

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

ACTIVE(f(X1, X2, X3)) → ACTIVE(X1)
TOP(mark(X)) → PROPER(X)
ACTIVE(f(X1, X2, X3)) → ACTIVE(X3)
PROPER(f(X1, X2, X3)) → PROPER(X1)
ACTIVE(f(X1, X2, X3)) → F(active(X1), X2, X3)
PROPER(f(X1, X2, X3)) → PROPER(X3)
ACTIVE(f(a, b, X)) → F(X, X, X)
F(mark(X1), X2, X3) → F(X1, X2, X3)
TOP(ok(X)) → ACTIVE(X)
ACTIVE(f(X1, X2, X3)) → F(X1, X2, active(X3))
TOP(ok(X)) → TOP(active(X))
TOP(mark(X)) → TOP(proper(X))
PROPER(f(X1, X2, X3)) → PROPER(X2)
PROPER(f(X1, X2, X3)) → F(proper(X1), proper(X2), proper(X3))
F(X1, X2, mark(X3)) → F(X1, X2, X3)
F(ok(X1), ok(X2), ok(X3)) → F(X1, X2, X3)

The TRS R consists of the following rules:

active(f(a, b, X)) → mark(f(X, X, X))
active(c) → mark(a)
active(c) → mark(b)
active(f(X1, X2, X3)) → f(active(X1), X2, X3)
active(f(X1, X2, X3)) → f(X1, X2, active(X3))
f(mark(X1), X2, X3) → mark(f(X1, X2, X3))
f(X1, X2, mark(X3)) → mark(f(X1, X2, X3))
proper(f(X1, X2, X3)) → f(proper(X1), proper(X2), proper(X3))
proper(a) → ok(a)
proper(b) → ok(b)
proper(c) → ok(c)
f(ok(X1), ok(X2), ok(X3)) → ok(f(X1, X2, X3))
top(mark(X)) → top(proper(X))
top(ok(X)) → top(active(X))

Q is empty.
We have to consider all minimal (P,Q,R)-chains.
We deleted some edges using various graph approximations

↳ QTRS
  ↳ DependencyPairsProof
    ↳ QDP
      ↳ EdgeDeletionProof
QDP
          ↳ DependencyGraphProof

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

ACTIVE(f(X1, X2, X3)) → ACTIVE(X1)
PROPER(f(X1, X2, X3)) → PROPER(X1)
ACTIVE(f(X1, X2, X3)) → ACTIVE(X3)
TOP(mark(X)) → PROPER(X)
ACTIVE(f(X1, X2, X3)) → F(active(X1), X2, X3)
PROPER(f(X1, X2, X3)) → PROPER(X3)
ACTIVE(f(a, b, X)) → F(X, X, X)
F(mark(X1), X2, X3) → F(X1, X2, X3)
TOP(ok(X)) → ACTIVE(X)
ACTIVE(f(X1, X2, X3)) → F(X1, X2, active(X3))
TOP(ok(X)) → TOP(active(X))
TOP(mark(X)) → TOP(proper(X))
PROPER(f(X1, X2, X3)) → PROPER(X2)
PROPER(f(X1, X2, X3)) → F(proper(X1), proper(X2), proper(X3))
F(ok(X1), ok(X2), ok(X3)) → F(X1, X2, X3)
F(X1, X2, mark(X3)) → F(X1, X2, X3)

The TRS R consists of the following rules:

active(f(a, b, X)) → mark(f(X, X, X))
active(c) → mark(a)
active(c) → mark(b)
active(f(X1, X2, X3)) → f(active(X1), X2, X3)
active(f(X1, X2, X3)) → f(X1, X2, active(X3))
f(mark(X1), X2, X3) → mark(f(X1, X2, X3))
f(X1, X2, mark(X3)) → mark(f(X1, X2, X3))
proper(f(X1, X2, X3)) → f(proper(X1), proper(X2), proper(X3))
proper(a) → ok(a)
proper(b) → ok(b)
proper(c) → ok(c)
f(ok(X1), ok(X2), ok(X3)) → ok(f(X1, X2, X3))
top(mark(X)) → top(proper(X))
top(ok(X)) → top(active(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
      ↳ EdgeDeletionProof
        ↳ QDP
          ↳ DependencyGraphProof
            ↳ AND
QDP
                ↳ QDPOrderProof
              ↳ QDP
              ↳ QDP
              ↳ QDP

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

F(mark(X1), X2, X3) → F(X1, X2, X3)
F(ok(X1), ok(X2), ok(X3)) → F(X1, X2, X3)
F(X1, X2, mark(X3)) → F(X1, X2, X3)

The TRS R consists of the following rules:

active(f(a, b, X)) → mark(f(X, X, X))
active(c) → mark(a)
active(c) → mark(b)
active(f(X1, X2, X3)) → f(active(X1), X2, X3)
active(f(X1, X2, X3)) → f(X1, X2, active(X3))
f(mark(X1), X2, X3) → mark(f(X1, X2, X3))
f(X1, X2, mark(X3)) → mark(f(X1, X2, X3))
proper(f(X1, X2, X3)) → f(proper(X1), proper(X2), proper(X3))
proper(a) → ok(a)
proper(b) → ok(b)
proper(c) → ok(c)
f(ok(X1), ok(X2), ok(X3)) → ok(f(X1, X2, X3))
top(mark(X)) → top(proper(X))
top(ok(X)) → top(active(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.


F(ok(X1), ok(X2), ok(X3)) → F(X1, X2, X3)
The remaining pairs can at least be oriented weakly.

F(mark(X1), X2, X3) → F(X1, X2, X3)
F(X1, X2, mark(X3)) → F(X1, X2, X3)
Used ordering: Combined order from the following AFS and order.
F(x1, x2, x3)  =  F(x3)
mark(x1)  =  x1
ok(x1)  =  ok(x1)

Lexicographic path order with status [19].
Quasi-Precedence:
ok1 > F1

Status:
ok1: [1]
F1: [1]


The following usable rules [14] were oriented: none



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

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

F(mark(X1), X2, X3) → F(X1, X2, X3)
F(X1, X2, mark(X3)) → F(X1, X2, X3)

The TRS R consists of the following rules:

active(f(a, b, X)) → mark(f(X, X, X))
active(c) → mark(a)
active(c) → mark(b)
active(f(X1, X2, X3)) → f(active(X1), X2, X3)
active(f(X1, X2, X3)) → f(X1, X2, active(X3))
f(mark(X1), X2, X3) → mark(f(X1, X2, X3))
f(X1, X2, mark(X3)) → mark(f(X1, X2, X3))
proper(f(X1, X2, X3)) → f(proper(X1), proper(X2), proper(X3))
proper(a) → ok(a)
proper(b) → ok(b)
proper(c) → ok(c)
f(ok(X1), ok(X2), ok(X3)) → ok(f(X1, X2, X3))
top(mark(X)) → top(proper(X))
top(ok(X)) → top(active(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.


F(mark(X1), X2, X3) → F(X1, X2, X3)
F(X1, X2, mark(X3)) → F(X1, X2, X3)
The remaining pairs can at least be oriented weakly.
none
Used ordering: Combined order from the following AFS and order.
F(x1, x2, x3)  =  F(x1, x2, x3)
mark(x1)  =  mark(x1)

Lexicographic path order with status [19].
Quasi-Precedence:
[F3, mark1]

Status:
mark1: [1]
F3: [3,2,1]


The following usable rules [14] were oriented: none



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

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

active(f(a, b, X)) → mark(f(X, X, X))
active(c) → mark(a)
active(c) → mark(b)
active(f(X1, X2, X3)) → f(active(X1), X2, X3)
active(f(X1, X2, X3)) → f(X1, X2, active(X3))
f(mark(X1), X2, X3) → mark(f(X1, X2, X3))
f(X1, X2, mark(X3)) → mark(f(X1, X2, X3))
proper(f(X1, X2, X3)) → f(proper(X1), proper(X2), proper(X3))
proper(a) → ok(a)
proper(b) → ok(b)
proper(c) → ok(c)
f(ok(X1), ok(X2), ok(X3)) → ok(f(X1, X2, X3))
top(mark(X)) → top(proper(X))
top(ok(X)) → top(active(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
      ↳ EdgeDeletionProof
        ↳ QDP
          ↳ DependencyGraphProof
            ↳ AND
              ↳ QDP
QDP
                ↳ QDPOrderProof
              ↳ QDP
              ↳ QDP

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

PROPER(f(X1, X2, X3)) → PROPER(X1)
PROPER(f(X1, X2, X3)) → PROPER(X3)
PROPER(f(X1, X2, X3)) → PROPER(X2)

The TRS R consists of the following rules:

active(f(a, b, X)) → mark(f(X, X, X))
active(c) → mark(a)
active(c) → mark(b)
active(f(X1, X2, X3)) → f(active(X1), X2, X3)
active(f(X1, X2, X3)) → f(X1, X2, active(X3))
f(mark(X1), X2, X3) → mark(f(X1, X2, X3))
f(X1, X2, mark(X3)) → mark(f(X1, X2, X3))
proper(f(X1, X2, X3)) → f(proper(X1), proper(X2), proper(X3))
proper(a) → ok(a)
proper(b) → ok(b)
proper(c) → ok(c)
f(ok(X1), ok(X2), ok(X3)) → ok(f(X1, X2, X3))
top(mark(X)) → top(proper(X))
top(ok(X)) → top(active(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.


PROPER(f(X1, X2, X3)) → PROPER(X1)
PROPER(f(X1, X2, X3)) → PROPER(X3)
PROPER(f(X1, X2, X3)) → PROPER(X2)
The remaining pairs can at least be oriented weakly.
none
Used ordering: Combined order from the following AFS and order.
PROPER(x1)  =  PROPER(x1)
f(x1, x2, x3)  =  f(x1, x2, x3)

Lexicographic path order with status [19].
Quasi-Precedence:
[PROPER1, f3]

Status:
f3: [3,2,1]
PROPER1: [1]


The following usable rules [14] were oriented: none



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

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

active(f(a, b, X)) → mark(f(X, X, X))
active(c) → mark(a)
active(c) → mark(b)
active(f(X1, X2, X3)) → f(active(X1), X2, X3)
active(f(X1, X2, X3)) → f(X1, X2, active(X3))
f(mark(X1), X2, X3) → mark(f(X1, X2, X3))
f(X1, X2, mark(X3)) → mark(f(X1, X2, X3))
proper(f(X1, X2, X3)) → f(proper(X1), proper(X2), proper(X3))
proper(a) → ok(a)
proper(b) → ok(b)
proper(c) → ok(c)
f(ok(X1), ok(X2), ok(X3)) → ok(f(X1, X2, X3))
top(mark(X)) → top(proper(X))
top(ok(X)) → top(active(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
      ↳ EdgeDeletionProof
        ↳ QDP
          ↳ DependencyGraphProof
            ↳ AND
              ↳ QDP
              ↳ QDP
QDP
                ↳ QDPOrderProof
              ↳ QDP

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

ACTIVE(f(X1, X2, X3)) → ACTIVE(X1)
ACTIVE(f(X1, X2, X3)) → ACTIVE(X3)

The TRS R consists of the following rules:

active(f(a, b, X)) → mark(f(X, X, X))
active(c) → mark(a)
active(c) → mark(b)
active(f(X1, X2, X3)) → f(active(X1), X2, X3)
active(f(X1, X2, X3)) → f(X1, X2, active(X3))
f(mark(X1), X2, X3) → mark(f(X1, X2, X3))
f(X1, X2, mark(X3)) → mark(f(X1, X2, X3))
proper(f(X1, X2, X3)) → f(proper(X1), proper(X2), proper(X3))
proper(a) → ok(a)
proper(b) → ok(b)
proper(c) → ok(c)
f(ok(X1), ok(X2), ok(X3)) → ok(f(X1, X2, X3))
top(mark(X)) → top(proper(X))
top(ok(X)) → top(active(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.


ACTIVE(f(X1, X2, X3)) → ACTIVE(X1)
ACTIVE(f(X1, X2, X3)) → ACTIVE(X3)
The remaining pairs can at least be oriented weakly.
none
Used ordering: Combined order from the following AFS and order.
ACTIVE(x1)  =  ACTIVE(x1)
f(x1, x2, x3)  =  f(x1, x2, x3)

Lexicographic path order with status [19].
Quasi-Precedence:
trivial

Status:
ACTIVE1: [1]
f3: [3,2,1]


The following usable rules [14] were oriented: none



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

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

active(f(a, b, X)) → mark(f(X, X, X))
active(c) → mark(a)
active(c) → mark(b)
active(f(X1, X2, X3)) → f(active(X1), X2, X3)
active(f(X1, X2, X3)) → f(X1, X2, active(X3))
f(mark(X1), X2, X3) → mark(f(X1, X2, X3))
f(X1, X2, mark(X3)) → mark(f(X1, X2, X3))
proper(f(X1, X2, X3)) → f(proper(X1), proper(X2), proper(X3))
proper(a) → ok(a)
proper(b) → ok(b)
proper(c) → ok(c)
f(ok(X1), ok(X2), ok(X3)) → ok(f(X1, X2, X3))
top(mark(X)) → top(proper(X))
top(ok(X)) → top(active(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
      ↳ EdgeDeletionProof
        ↳ QDP
          ↳ DependencyGraphProof
            ↳ AND
              ↳ QDP
              ↳ QDP
              ↳ QDP
QDP

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

TOP(ok(X)) → TOP(active(X))
TOP(mark(X)) → TOP(proper(X))

The TRS R consists of the following rules:

active(f(a, b, X)) → mark(f(X, X, X))
active(c) → mark(a)
active(c) → mark(b)
active(f(X1, X2, X3)) → f(active(X1), X2, X3)
active(f(X1, X2, X3)) → f(X1, X2, active(X3))
f(mark(X1), X2, X3) → mark(f(X1, X2, X3))
f(X1, X2, mark(X3)) → mark(f(X1, X2, X3))
proper(f(X1, X2, X3)) → f(proper(X1), proper(X2), proper(X3))
proper(a) → ok(a)
proper(b) → ok(b)
proper(c) → ok(c)
f(ok(X1), ok(X2), ok(X3)) → ok(f(X1, X2, X3))
top(mark(X)) → top(proper(X))
top(ok(X)) → top(active(X))

Q is empty.
We have to consider all minimal (P,Q,R)-chains.