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(f(a, X, X)) → mark(f(X, b, b))
active(b) → mark(a)
active(f(X1, X2, X3)) → f(X1, active(X2), X3)
f(X1, mark(X2), 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)
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, X, X)) → mark(f(X, b, b))
active(b) → mark(a)
active(f(X1, X2, X3)) → f(X1, active(X2), X3)
f(X1, mark(X2), 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)
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 [1,15] we result in the following initial DP problem:
Q DP problem:
The TRS P consists of the following rules:

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

The TRS R consists of the following rules:

active(f(a, X, X)) → mark(f(X, b, b))
active(b) → mark(a)
active(f(X1, X2, X3)) → f(X1, active(X2), X3)
f(X1, mark(X2), 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)
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
      ↳ DependencyGraphProof

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

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

The TRS R consists of the following rules:

active(f(a, X, X)) → mark(f(X, b, b))
active(b) → mark(a)
active(f(X1, X2, X3)) → f(X1, active(X2), X3)
f(X1, mark(X2), 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)
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 [15,17,22] contains 4 SCCs with 5 less nodes.

↳ QTRS
  ↳ DependencyPairsProof
    ↳ QDP
      ↳ DependencyGraphProof
        ↳ AND
          ↳ QDP
            ↳ UsableRulesProof
          ↳ QDP
          ↳ QDP
          ↳ QDP

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

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

The TRS R consists of the following rules:

active(f(a, X, X)) → mark(f(X, b, b))
active(b) → mark(a)
active(f(X1, X2, X3)) → f(X1, active(X2), X3)
f(X1, mark(X2), 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)
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 can use the usable rules and reduction pair processor [15] with the Ce-compatible extension of the polynomial order that maps every function symbol to the sum of its argument. Then, we can delete all non-usable rules [17] from R.

↳ QTRS
  ↳ DependencyPairsProof
    ↳ QDP
      ↳ DependencyGraphProof
        ↳ AND
          ↳ QDP
            ↳ UsableRulesProof
              ↳ QDP
                ↳ QDPSizeChangeProof
          ↳ QDP
          ↳ QDP
          ↳ QDP

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

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

R is empty.
Q is empty.
We have to consider all minimal (P,Q,R)-chains.
By using the subterm criterion [20] together with the size-change analysis [32] we have proven that there are no infinite chains for this DP problem.

From the DPs we obtained the following set of size-change graphs: 



↳ QTRS
  ↳ DependencyPairsProof
    ↳ QDP
      ↳ DependencyGraphProof
        ↳ AND
          ↳ QDP
          ↳ QDP
            ↳ UsableRulesProof
          ↳ 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(X2)
PROPER(f(X1, X2, X3)) → PROPER(X3)

The TRS R consists of the following rules:

active(f(a, X, X)) → mark(f(X, b, b))
active(b) → mark(a)
active(f(X1, X2, X3)) → f(X1, active(X2), X3)
f(X1, mark(X2), 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)
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 can use the usable rules and reduction pair processor [15] with the Ce-compatible extension of the polynomial order that maps every function symbol to the sum of its argument. Then, we can delete all non-usable rules [17] from R.

↳ QTRS
  ↳ DependencyPairsProof
    ↳ QDP
      ↳ DependencyGraphProof
        ↳ AND
          ↳ QDP
          ↳ QDP
            ↳ UsableRulesProof
              ↳ QDP
                ↳ QDPSizeChangeProof
          ↳ 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(X2)
PROPER(f(X1, X2, X3)) → PROPER(X3)

R is empty.
Q is empty.
We have to consider all minimal (P,Q,R)-chains.
By using the subterm criterion [20] together with the size-change analysis [32] we have proven that there are no infinite chains for this DP problem.

From the DPs we obtained the following set of size-change graphs: 



↳ QTRS
  ↳ DependencyPairsProof
    ↳ QDP
      ↳ DependencyGraphProof
        ↳ AND
          ↳ QDP
          ↳ QDP
          ↳ QDP
            ↳ UsableRulesProof
          ↳ QDP

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

ACTIVE(f(X1, X2, X3)) → ACTIVE(X2)

The TRS R consists of the following rules:

active(f(a, X, X)) → mark(f(X, b, b))
active(b) → mark(a)
active(f(X1, X2, X3)) → f(X1, active(X2), X3)
f(X1, mark(X2), 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)
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 can use the usable rules and reduction pair processor [15] with the Ce-compatible extension of the polynomial order that maps every function symbol to the sum of its argument. Then, we can delete all non-usable rules [17] from R.

↳ QTRS
  ↳ DependencyPairsProof
    ↳ QDP
      ↳ DependencyGraphProof
        ↳ AND
          ↳ QDP
          ↳ QDP
          ↳ QDP
            ↳ UsableRulesProof
              ↳ QDP
                ↳ QDPSizeChangeProof
          ↳ QDP

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

ACTIVE(f(X1, X2, X3)) → ACTIVE(X2)

R is empty.
Q is empty.
We have to consider all minimal (P,Q,R)-chains.
By using the subterm criterion [20] together with the size-change analysis [32] we have proven that there are no infinite chains for this DP problem.

From the DPs we obtained the following set of size-change graphs: 



↳ QTRS
  ↳ DependencyPairsProof
    ↳ QDP
      ↳ DependencyGraphProof
        ↳ AND
          ↳ QDP
          ↳ QDP
          ↳ QDP
          ↳ QDP
            ↳ UsableRulesProof

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

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

The TRS R consists of the following rules:

active(f(a, X, X)) → mark(f(X, b, b))
active(b) → mark(a)
active(f(X1, X2, X3)) → f(X1, active(X2), X3)
f(X1, mark(X2), 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)
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 can use the usable rules and reduction pair processor [15] with the Ce-compatible extension of the polynomial order that maps every function symbol to the sum of its argument. Then, we can delete all non-usable rules [17] from R.

↳ QTRS
  ↳ DependencyPairsProof
    ↳ QDP
      ↳ DependencyGraphProof
        ↳ AND
          ↳ QDP
          ↳ QDP
          ↳ QDP
          ↳ QDP
            ↳ UsableRulesProof
              ↳ QDP
                ↳ Narrowing

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

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

The TRS R consists of the following rules:

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

Q is empty.
We have to consider all minimal (P,Q,R)-chains.
By narrowing [15] the rule TOP(mark(X)) → TOP(proper(X)) at position [0] we obtained the following new rules:

TOP(mark(b)) → TOP(ok(b))
TOP(mark(f(x0, x1, x2))) → TOP(f(proper(x0), proper(x1), proper(x2)))
TOP(mark(a)) → TOP(ok(a))



↳ QTRS
  ↳ DependencyPairsProof
    ↳ QDP
      ↳ DependencyGraphProof
        ↳ AND
          ↳ QDP
          ↳ QDP
          ↳ QDP
          ↳ QDP
            ↳ UsableRulesProof
              ↳ QDP
                ↳ Narrowing
                  ↳ QDP
                    ↳ DependencyGraphProof

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

TOP(mark(f(x0, x1, x2))) → TOP(f(proper(x0), proper(x1), proper(x2)))
TOP(mark(b)) → TOP(ok(b))
TOP(mark(a)) → TOP(ok(a))
TOP(ok(X)) → TOP(active(X))

The TRS R consists of the following rules:

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

Q is empty.
We have to consider all minimal (P,Q,R)-chains.
The approximation of the Dependency Graph [15,17,22] contains 1 SCC with 1 less node.

↳ QTRS
  ↳ DependencyPairsProof
    ↳ QDP
      ↳ DependencyGraphProof
        ↳ AND
          ↳ QDP
          ↳ QDP
          ↳ QDP
          ↳ QDP
            ↳ UsableRulesProof
              ↳ QDP
                ↳ Narrowing
                  ↳ QDP
                    ↳ DependencyGraphProof
                      ↳ QDP
                        ↳ Narrowing

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

TOP(mark(f(x0, x1, x2))) → TOP(f(proper(x0), proper(x1), proper(x2)))
TOP(mark(a)) → TOP(ok(a))
TOP(ok(X)) → TOP(active(X))

The TRS R consists of the following rules:

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

Q is empty.
We have to consider all minimal (P,Q,R)-chains.
By narrowing [15] the rule TOP(ok(X)) → TOP(active(X)) at position [0] we obtained the following new rules:

TOP(ok(b)) → TOP(mark(a))
TOP(ok(f(x0, x1, x2))) → TOP(f(x0, active(x1), x2))
TOP(ok(f(a, x0, x0))) → TOP(mark(f(x0, b, b)))



↳ QTRS
  ↳ DependencyPairsProof
    ↳ QDP
      ↳ DependencyGraphProof
        ↳ AND
          ↳ QDP
          ↳ QDP
          ↳ QDP
          ↳ QDP
            ↳ UsableRulesProof
              ↳ QDP
                ↳ Narrowing
                  ↳ QDP
                    ↳ DependencyGraphProof
                      ↳ QDP
                        ↳ Narrowing
                          ↳ QDP
                            ↳ DependencyGraphProof

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

TOP(ok(f(a, x0, x0))) → TOP(mark(f(x0, b, b)))
TOP(ok(f(x0, x1, x2))) → TOP(f(x0, active(x1), x2))
TOP(mark(f(x0, x1, x2))) → TOP(f(proper(x0), proper(x1), proper(x2)))
TOP(ok(b)) → TOP(mark(a))
TOP(mark(a)) → TOP(ok(a))

The TRS R consists of the following rules:

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

Q is empty.
We have to consider all minimal (P,Q,R)-chains.
The approximation of the Dependency Graph [15,17,22] contains 1 SCC with 2 less nodes.

↳ QTRS
  ↳ DependencyPairsProof
    ↳ QDP
      ↳ DependencyGraphProof
        ↳ AND
          ↳ QDP
          ↳ QDP
          ↳ QDP
          ↳ QDP
            ↳ UsableRulesProof
              ↳ QDP
                ↳ Narrowing
                  ↳ QDP
                    ↳ DependencyGraphProof
                      ↳ QDP
                        ↳ Narrowing
                          ↳ QDP
                            ↳ DependencyGraphProof
                              ↳ QDP
                                ↳ QDPOrderProof

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

TOP(ok(f(a, x0, x0))) → TOP(mark(f(x0, b, b)))
TOP(ok(f(x0, x1, x2))) → TOP(f(x0, active(x1), x2))
TOP(mark(f(x0, x1, x2))) → TOP(f(proper(x0), proper(x1), proper(x2)))

The TRS R consists of the following rules:

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

Q is empty.
We have to consider all minimal (P,Q,R)-chains.
We use the reduction pair processor [15].


The following pairs can be oriented strictly and are deleted.


TOP(ok(f(a, x0, x0))) → TOP(mark(f(x0, b, b)))
The remaining pairs can at least be oriented weakly.

TOP(ok(f(x0, x1, x2))) → TOP(f(x0, active(x1), x2))
TOP(mark(f(x0, x1, x2))) → TOP(f(proper(x0), proper(x1), proper(x2)))
Used ordering: Polynomial interpretation [25]:

POL(TOP(x1)) = x1   
POL(a) = 1   
POL(active(x1)) = 0   
POL(b) = 0   
POL(f(x1, x2, x3)) = x1 + x3   
POL(mark(x1)) = x1   
POL(ok(x1)) = x1   
POL(proper(x1)) = x1   

The following usable rules [17] were oriented:

proper(b) → ok(b)
proper(a) → ok(a)
proper(f(X1, X2, X3)) → f(proper(X1), proper(X2), proper(X3))
f(ok(X1), ok(X2), ok(X3)) → ok(f(X1, X2, X3))
f(X1, mark(X2), X3) → mark(f(X1, X2, X3))



↳ QTRS
  ↳ DependencyPairsProof
    ↳ QDP
      ↳ DependencyGraphProof
        ↳ AND
          ↳ QDP
          ↳ QDP
          ↳ QDP
          ↳ QDP
            ↳ UsableRulesProof
              ↳ QDP
                ↳ Narrowing
                  ↳ QDP
                    ↳ DependencyGraphProof
                      ↳ QDP
                        ↳ Narrowing
                          ↳ QDP
                            ↳ DependencyGraphProof
                              ↳ QDP
                                ↳ QDPOrderProof
                                  ↳ QDP
                                    ↳ QDPOrderProof

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

TOP(ok(f(x0, x1, x2))) → TOP(f(x0, active(x1), x2))
TOP(mark(f(x0, x1, x2))) → TOP(f(proper(x0), proper(x1), proper(x2)))

The TRS R consists of the following rules:

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

Q is empty.
We have to consider all minimal (P,Q,R)-chains.
We use the reduction pair processor [15].


The following pairs can be oriented strictly and are deleted.


TOP(mark(f(x0, x1, x2))) → TOP(f(proper(x0), proper(x1), proper(x2)))
The remaining pairs can at least be oriented weakly.

TOP(ok(f(x0, x1, x2))) → TOP(f(x0, active(x1), x2))
Used ordering: Matrix interpretation [3]:
Non-tuple symbols:
M( active(x1) ) =
/0\
\0/
+
/00\
\01/
·x1

M( a ) =
/1\
\0/

M( f(x1, ..., x3) ) =
/0\
\0/
+
/00\
\20/
·x1+
/00\
\01/
·x2+
/00\
\20/
·x3

M( ok(x1) ) =
/0\
\0/
+
/10\
\01/
·x1

M( mark(x1) ) =
/0\
\1/
+
/00\
\01/
·x1

M( b ) =
/0\
\1/

M( proper(x1) ) =
/0\
\0/
+
/10\
\01/
·x1

Tuple symbols:
M( TOP(x1) ) = 0+
[0,2]
·x1


Matrix type:
We used a basic matrix type which is not further parametrizeable.


As matrix orders are CE-compatible, we used usable rules w.r.t. argument filtering in the order.
The following usable rules [17] were oriented:

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



↳ QTRS
  ↳ DependencyPairsProof
    ↳ QDP
      ↳ DependencyGraphProof
        ↳ AND
          ↳ QDP
          ↳ QDP
          ↳ QDP
          ↳ QDP
            ↳ UsableRulesProof
              ↳ QDP
                ↳ Narrowing
                  ↳ QDP
                    ↳ DependencyGraphProof
                      ↳ QDP
                        ↳ Narrowing
                          ↳ QDP
                            ↳ DependencyGraphProof
                              ↳ QDP
                                ↳ QDPOrderProof
                                  ↳ QDP
                                    ↳ QDPOrderProof
                                      ↳ QDP
                                        ↳ UsableRulesProof

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

TOP(ok(f(x0, x1, x2))) → TOP(f(x0, active(x1), x2))

The TRS R consists of the following rules:

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

Q is empty.
We have to consider all minimal (P,Q,R)-chains.
We can use the usable rules and reduction pair processor [15] with the Ce-compatible extension of the polynomial order that maps every function symbol to the sum of its argument. Then, we can delete all non-usable rules [17] from R.

↳ QTRS
  ↳ DependencyPairsProof
    ↳ QDP
      ↳ DependencyGraphProof
        ↳ AND
          ↳ QDP
          ↳ QDP
          ↳ QDP
          ↳ QDP
            ↳ UsableRulesProof
              ↳ QDP
                ↳ Narrowing
                  ↳ QDP
                    ↳ DependencyGraphProof
                      ↳ QDP
                        ↳ Narrowing
                          ↳ QDP
                            ↳ DependencyGraphProof
                              ↳ QDP
                                ↳ QDPOrderProof
                                  ↳ QDP
                                    ↳ QDPOrderProof
                                      ↳ QDP
                                        ↳ UsableRulesProof
                                          ↳ QDP
                                            ↳ RuleRemovalProof

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

TOP(ok(f(x0, x1, x2))) → TOP(f(x0, active(x1), x2))

The TRS R consists of the following rules:

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

Q is empty.
We have to consider all minimal (P,Q,R)-chains.
By using the rule removal processor [15] with the following polynomial ordering [25], at least one Dependency Pair or term rewrite system rule of this QDP problem can be strictly oriented.

Strictly oriented rules of the TRS R:

active(b) → mark(a)
f(ok(X1), ok(X2), ok(X3)) → ok(f(X1, X2, X3))

Used ordering: POLO with Polynomial interpretation [25]:

POL(TOP(x1)) = 2·x1   
POL(a) = 1   
POL(active(x1)) = 1 + x1   
POL(b) = 1   
POL(f(x1, x2, x3)) = x1 + x2 + x3   
POL(mark(x1)) = x1   
POL(ok(x1)) = 1 + x1   



↳ QTRS
  ↳ DependencyPairsProof
    ↳ QDP
      ↳ DependencyGraphProof
        ↳ AND
          ↳ QDP
          ↳ QDP
          ↳ QDP
          ↳ QDP
            ↳ UsableRulesProof
              ↳ QDP
                ↳ Narrowing
                  ↳ QDP
                    ↳ DependencyGraphProof
                      ↳ QDP
                        ↳ Narrowing
                          ↳ QDP
                            ↳ DependencyGraphProof
                              ↳ QDP
                                ↳ QDPOrderProof
                                  ↳ QDP
                                    ↳ QDPOrderProof
                                      ↳ QDP
                                        ↳ UsableRulesProof
                                          ↳ QDP
                                            ↳ RuleRemovalProof
                                              ↳ QDP
                                                ↳ DependencyGraphProof

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

TOP(ok(f(x0, x1, x2))) → TOP(f(x0, active(x1), x2))

The TRS R consists of the following rules:

active(f(a, X, X)) → mark(f(X, b, b))
active(f(X1, X2, X3)) → f(X1, active(X2), X3)
f(X1, mark(X2), X3) → mark(f(X1, X2, X3))

Q is empty.
We have to consider all minimal (P,Q,R)-chains.
The approximation of the Dependency Graph [15,17,22] contains 0 SCCs with 1 less node.