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:

active(c) → mark(f(g(c)))
active(f(g(X))) → mark(g(X))
proper(c) → ok(c)
proper(f(X)) → f(proper(X))
proper(g(X)) → g(proper(X))
f(ok(X)) → ok(f(X))
g(ok(X)) → ok(g(X))
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(c) → mark(f(g(c)))
active(f(g(X))) → mark(g(X))
proper(c) → ok(c)
proper(f(X)) → f(proper(X))
proper(g(X)) → g(proper(X))
f(ok(X)) → ok(f(X))
g(ok(X)) → ok(g(X))
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:

TOP(mark(X)) → PROPER(X)
PROPER(g(X)) → G(proper(X))
PROPER(g(X)) → PROPER(X)
PROPER(f(X)) → PROPER(X)
TOP(ok(X)) → TOP(active(X))
ACTIVE(c) → F(g(c))
F(ok(X)) → F(X)
TOP(mark(X)) → TOP(proper(X))
G(ok(X)) → G(X)
ACTIVE(c) → G(c)
TOP(ok(X)) → ACTIVE(X)
PROPER(f(X)) → F(proper(X))

The TRS R consists of the following rules:

active(c) → mark(f(g(c)))
active(f(g(X))) → mark(g(X))
proper(c) → ok(c)
proper(f(X)) → f(proper(X))
proper(g(X)) → g(proper(X))
f(ok(X)) → ok(f(X))
g(ok(X)) → ok(g(X))
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:

TOP(mark(X)) → PROPER(X)
PROPER(g(X)) → G(proper(X))
PROPER(g(X)) → PROPER(X)
PROPER(f(X)) → PROPER(X)
TOP(ok(X)) → TOP(active(X))
ACTIVE(c) → F(g(c))
F(ok(X)) → F(X)
TOP(mark(X)) → TOP(proper(X))
G(ok(X)) → G(X)
ACTIVE(c) → G(c)
TOP(ok(X)) → ACTIVE(X)
PROPER(f(X)) → F(proper(X))

The TRS R consists of the following rules:

active(c) → mark(f(g(c)))
active(f(g(X))) → mark(g(X))
proper(c) → ok(c)
proper(f(X)) → f(proper(X))
proper(g(X)) → g(proper(X))
f(ok(X)) → ok(f(X))
g(ok(X)) → ok(g(X))
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:

TOP(mark(X)) → PROPER(X)
PROPER(g(X)) → G(proper(X))
PROPER(g(X)) → PROPER(X)
ACTIVE(c) → F(g(c))
TOP(ok(X)) → ACTIVE(X)
PROPER(f(X)) → PROPER(X)
TOP(ok(X)) → TOP(active(X))
F(ok(X)) → F(X)
TOP(mark(X)) → TOP(proper(X))
G(ok(X)) → G(X)
ACTIVE(c) → G(c)
PROPER(f(X)) → F(proper(X))

The TRS R consists of the following rules:

active(c) → mark(f(g(c)))
active(f(g(X))) → mark(g(X))
proper(c) → ok(c)
proper(f(X)) → f(proper(X))
proper(g(X)) → g(proper(X))
f(ok(X)) → ok(f(X))
g(ok(X)) → ok(g(X))
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:

G(ok(X)) → G(X)

The TRS R consists of the following rules:

active(c) → mark(f(g(c)))
active(f(g(X))) → mark(g(X))
proper(c) → ok(c)
proper(f(X)) → f(proper(X))
proper(g(X)) → g(proper(X))
f(ok(X)) → ok(f(X))
g(ok(X)) → ok(g(X))
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.


G(ok(X)) → G(X)
The remaining pairs can at least be oriented weakly.
none
Used ordering: Combined order from the following AFS and order.
G(x1)  =  G(x1)
ok(x1)  =  ok(x1)

Lexicographic path order with status [19].
Precedence:
ok1 > G1

Status:
G1: [1]
ok1: multiset

The following usable rules [14] were oriented: none



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

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

active(c) → mark(f(g(c)))
active(f(g(X))) → mark(g(X))
proper(c) → ok(c)
proper(f(X)) → f(proper(X))
proper(g(X)) → g(proper(X))
f(ok(X)) → ok(f(X))
g(ok(X)) → ok(g(X))
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:

F(ok(X)) → F(X)

The TRS R consists of the following rules:

active(c) → mark(f(g(c)))
active(f(g(X))) → mark(g(X))
proper(c) → ok(c)
proper(f(X)) → f(proper(X))
proper(g(X)) → g(proper(X))
f(ok(X)) → ok(f(X))
g(ok(X)) → ok(g(X))
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(X)) → F(X)
The remaining pairs can at least be oriented weakly.
none
Used ordering: Combined order from the following AFS and order.
F(x1)  =  F(x1)
ok(x1)  =  ok(x1)

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

Status:
ok1: multiset
F1: [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(c) → mark(f(g(c)))
active(f(g(X))) → mark(g(X))
proper(c) → ok(c)
proper(f(X)) → f(proper(X))
proper(g(X)) → g(proper(X))
f(ok(X)) → ok(f(X))
g(ok(X)) → ok(g(X))
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:

PROPER(g(X)) → PROPER(X)
PROPER(f(X)) → PROPER(X)

The TRS R consists of the following rules:

active(c) → mark(f(g(c)))
active(f(g(X))) → mark(g(X))
proper(c) → ok(c)
proper(f(X)) → f(proper(X))
proper(g(X)) → g(proper(X))
f(ok(X)) → ok(f(X))
g(ok(X)) → ok(g(X))
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(g(X)) → PROPER(X)
The remaining pairs can at least be oriented weakly.

PROPER(f(X)) → PROPER(X)
Used ordering: Combined order from the following AFS and order.
PROPER(x1)  =  PROPER(x1)
g(x1)  =  g(x1)
f(x1)  =  x1

Lexicographic path order with status [19].
Precedence:
g1 > PROPER1

Status:
PROPER1: [1]
g1: multiset

The following usable rules [14] were oriented: none



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

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

PROPER(f(X)) → PROPER(X)

The TRS R consists of the following rules:

active(c) → mark(f(g(c)))
active(f(g(X))) → mark(g(X))
proper(c) → ok(c)
proper(f(X)) → f(proper(X))
proper(g(X)) → g(proper(X))
f(ok(X)) → ok(f(X))
g(ok(X)) → ok(g(X))
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(X)) → PROPER(X)
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)  =  f(x1)

Lexicographic path order with status [19].
Precedence:
f1 > PROPER1

Status:
PROPER1: [1]
f1: multiset

The following usable rules [14] were oriented: none



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

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

active(c) → mark(f(g(c)))
active(f(g(X))) → mark(g(X))
proper(c) → ok(c)
proper(f(X)) → f(proper(X))
proper(g(X)) → g(proper(X))
f(ok(X)) → ok(f(X))
g(ok(X)) → ok(g(X))
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
                ↳ QDPOrderProof

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(c) → mark(f(g(c)))
active(f(g(X))) → mark(g(X))
proper(c) → ok(c)
proper(f(X)) → f(proper(X))
proper(g(X)) → g(proper(X))
f(ok(X)) → ok(f(X))
g(ok(X)) → ok(g(X))
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.


TOP(mark(X)) → TOP(proper(X))
The remaining pairs can at least be oriented weakly.

TOP(ok(X)) → TOP(active(X))
Used ordering: Combined order from the following AFS and order.
TOP(x1)  =  TOP(x1)
ok(x1)  =  x1
active(x1)  =  x1
mark(x1)  =  mark(x1)
proper(x1)  =  proper(x1)
c  =  c
f(x1)  =  f
g(x1)  =  g

Lexicographic path order with status [19].
Precedence:
c > f > mark1 > TOP1
c > f > mark1 > proper1
c > f > g > proper1

Status:
mark1: multiset
c: multiset
f: []
TOP1: [1]
g: []
proper1: [1]

The following usable rules [14] were oriented:

active(c) → mark(f(g(c)))
proper(f(X)) → f(proper(X))
f(ok(X)) → ok(f(X))
proper(c) → ok(c)
proper(g(X)) → g(proper(X))
g(ok(X)) → ok(g(X))
active(f(g(X))) → mark(g(X))



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

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

TOP(ok(X)) → TOP(active(X))

The TRS R consists of the following rules:

active(c) → mark(f(g(c)))
active(f(g(X))) → mark(g(X))
proper(c) → ok(c)
proper(f(X)) → f(proper(X))
proper(g(X)) → g(proper(X))
f(ok(X)) → ok(f(X))
g(ok(X)) → ok(g(X))
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.


TOP(ok(X)) → TOP(active(X))
The remaining pairs can at least be oriented weakly.
none
Used ordering: Combined order from the following AFS and order.
TOP(x1)  =  TOP(x1)
ok(x1)  =  ok(x1)
active(x1)  =  x1
c  =  c
mark(x1)  =  mark
f(x1)  =  f
g(x1)  =  g

Lexicographic path order with status [19].
Precedence:
c > mark > TOP1
f > g > ok1 > TOP1
f > g > mark > TOP1

Status:
c: multiset
f: multiset
TOP1: [1]
ok1: [1]
g: []
mark: multiset

The following usable rules [14] were oriented:

active(c) → mark(f(g(c)))
active(f(g(X))) → mark(g(X))



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

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

active(c) → mark(f(g(c)))
active(f(g(X))) → mark(g(X))
proper(c) → ok(c)
proper(f(X)) → f(proper(X))
proper(g(X)) → g(proper(X))
f(ok(X)) → ok(f(X))
g(ok(X)) → ok(g(X))
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.