Termination w.r.t. Q proof of TRS/TRCSREx5_Zan97

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(X)) → mark(if(X, c, f(true)))
active(if(true, X, Y)) → mark(X)
active(if(false, X, Y)) → mark(Y)
active(f(X)) → f(active(X))
active(if(X1, X2, X3)) → if(active(X1), X2, X3)
active(if(X1, X2, X3)) → if(X1, active(X2), X3)
f(mark(X)) → mark(f(X))
if(mark(X1), X2, X3) → mark(if(X1, X2, X3))
if(X1, mark(X2), X3) → mark(if(X1, X2, X3))
proper(f(X)) → f(proper(X))
proper(if(X1, X2, X3)) → if(proper(X1), proper(X2), proper(X3))
proper(c) → ok(c)
proper(true) → ok(true)
proper(false) → ok(false)
f(ok(X)) → ok(f(X))
if(ok(X1), ok(X2), ok(X3)) → ok(if(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(X)) → mark(if(X, c, f(true)))
active(if(true, X, Y)) → mark(X)
active(if(false, X, Y)) → mark(Y)
active(f(X)) → f(active(X))
active(if(X1, X2, X3)) → if(active(X1), X2, X3)
active(if(X1, X2, X3)) → if(X1, active(X2), X3)
f(mark(X)) → mark(f(X))
if(mark(X1), X2, X3) → mark(if(X1, X2, X3))
if(X1, mark(X2), X3) → mark(if(X1, X2, X3))
proper(f(X)) → f(proper(X))
proper(if(X1, X2, X3)) → if(proper(X1), proper(X2), proper(X3))
proper(c) → ok(c)
proper(true) → ok(true)
proper(false) → ok(false)
f(ok(X)) → ok(f(X))
if(ok(X1), ok(X2), ok(X3)) → ok(if(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:

TOP(mark(X)) → PROPER(X)
ACTIVE(f(X)) → ACTIVE(X)
IF(X1, mark(X2), X3) → IF(X1, X2, X3)
ACTIVE(f(X)) → F(active(X))
PROPER(if(X1, X2, X3)) → PROPER(X1)
ACTIVE(if(X1, X2, X3)) → ACTIVE(X2)
TOP(ok(X)) → ACTIVE(X)
IF(ok(X1), ok(X2), ok(X3)) → IF(X1, X2, X3)
ACTIVE(if(X1, X2, X3)) → IF(X1, active(X2), X3)
F(mark(X)) → F(X)
ACTIVE(f(X)) → IF(X, c, f(true))
ACTIVE(f(X)) → F(true)
PROPER(if(X1, X2, X3)) → IF(proper(X1), proper(X2), proper(X3))
ACTIVE(if(X1, X2, X3)) → IF(active(X1), X2, X3)
PROPER(f(X)) → PROPER(X)
TOP(ok(X)) → TOP(active(X))
ACTIVE(if(X1, X2, X3)) → ACTIVE(X1)
F(ok(X)) → F(X)
TOP(mark(X)) → TOP(proper(X))
PROPER(if(X1, X2, X3)) → PROPER(X3)
IF(mark(X1), X2, X3) → IF(X1, X2, X3)
PROPER(f(X)) → F(proper(X))
PROPER(if(X1, X2, X3)) → PROPER(X2)

The TRS R consists of the following rules:

active(f(X)) → mark(if(X, c, f(true)))
active(if(true, X, Y)) → mark(X)
active(if(false, X, Y)) → mark(Y)
active(f(X)) → f(active(X))
active(if(X1, X2, X3)) → if(active(X1), X2, X3)
active(if(X1, X2, X3)) → if(X1, active(X2), X3)
f(mark(X)) → mark(f(X))
if(mark(X1), X2, X3) → mark(if(X1, X2, X3))
if(X1, mark(X2), X3) → mark(if(X1, X2, X3))
proper(f(X)) → f(proper(X))
proper(if(X1, X2, X3)) → if(proper(X1), proper(X2), proper(X3))
proper(c) → ok(c)
proper(true) → ok(true)
proper(false) → ok(false)
f(ok(X)) → ok(f(X))
if(ok(X1), ok(X2), ok(X3)) → ok(if(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:

TOP(mark(X)) → PROPER(X)
ACTIVE(f(X)) → ACTIVE(X)
IF(X1, mark(X2), X3) → IF(X1, X2, X3)
ACTIVE(f(X)) → F(active(X))
PROPER(if(X1, X2, X3)) → PROPER(X1)
ACTIVE(if(X1, X2, X3)) → ACTIVE(X2)
TOP(ok(X)) → ACTIVE(X)
IF(ok(X1), ok(X2), ok(X3)) → IF(X1, X2, X3)
ACTIVE(if(X1, X2, X3)) → IF(X1, active(X2), X3)
F(mark(X)) → F(X)
ACTIVE(f(X)) → IF(X, c, f(true))
ACTIVE(f(X)) → F(true)
PROPER(if(X1, X2, X3)) → IF(proper(X1), proper(X2), proper(X3))
ACTIVE(if(X1, X2, X3)) → IF(active(X1), X2, X3)
PROPER(f(X)) → PROPER(X)
TOP(ok(X)) → TOP(active(X))
ACTIVE(if(X1, X2, X3)) → ACTIVE(X1)
F(ok(X)) → F(X)
TOP(mark(X)) → TOP(proper(X))
PROPER(if(X1, X2, X3)) → PROPER(X3)
IF(mark(X1), X2, X3) → IF(X1, X2, X3)
PROPER(f(X)) → F(proper(X))
PROPER(if(X1, X2, X3)) → PROPER(X2)

The TRS R consists of the following rules:

active(f(X)) → mark(if(X, c, f(true)))
active(if(true, X, Y)) → mark(X)
active(if(false, X, Y)) → mark(Y)
active(f(X)) → f(active(X))
active(if(X1, X2, X3)) → if(active(X1), X2, X3)
active(if(X1, X2, X3)) → if(X1, active(X2), X3)
f(mark(X)) → mark(f(X))
if(mark(X1), X2, X3) → mark(if(X1, X2, X3))
if(X1, mark(X2), X3) → mark(if(X1, X2, X3))
proper(f(X)) → f(proper(X))
proper(if(X1, X2, X3)) → if(proper(X1), proper(X2), proper(X3))
proper(c) → ok(c)
proper(true) → ok(true)
proper(false) → ok(false)
f(ok(X)) → ok(f(X))
if(ok(X1), ok(X2), ok(X3)) → ok(if(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:

TOP(mark(X)) → PROPER(X)
ACTIVE(f(X)) → ACTIVE(X)
IF(X1, mark(X2), X3) → IF(X1, X2, X3)
ACTIVE(f(X)) → F(active(X))
ACTIVE(if(X1, X2, X3)) → ACTIVE(X2)
PROPER(if(X1, X2, X3)) → PROPER(X1)
TOP(ok(X)) → ACTIVE(X)
IF(ok(X1), ok(X2), ok(X3)) → IF(X1, X2, X3)
F(mark(X)) → F(X)
ACTIVE(if(X1, X2, X3)) → IF(X1, active(X2), X3)
ACTIVE(f(X)) → F(true)
ACTIVE(f(X)) → IF(X, c, f(true))
PROPER(if(X1, X2, X3)) → IF(proper(X1), proper(X2), proper(X3))
ACTIVE(if(X1, X2, X3)) → IF(active(X1), X2, X3)
PROPER(f(X)) → PROPER(X)
ACTIVE(if(X1, X2, X3)) → ACTIVE(X1)
TOP(ok(X)) → TOP(active(X))
F(ok(X)) → F(X)
PROPER(if(X1, X2, X3)) → PROPER(X3)
TOP(mark(X)) → TOP(proper(X))
IF(mark(X1), X2, X3) → IF(X1, X2, X3)
PROPER(f(X)) → F(proper(X))
PROPER(if(X1, X2, X3)) → PROPER(X2)

The TRS R consists of the following rules:

active(f(X)) → mark(if(X, c, f(true)))
active(if(true, X, Y)) → mark(X)
active(if(false, X, Y)) → mark(Y)
active(f(X)) → f(active(X))
active(if(X1, X2, X3)) → if(active(X1), X2, X3)
active(if(X1, X2, X3)) → if(X1, active(X2), X3)
f(mark(X)) → mark(f(X))
if(mark(X1), X2, X3) → mark(if(X1, X2, X3))
if(X1, mark(X2), X3) → mark(if(X1, X2, X3))
proper(f(X)) → f(proper(X))
proper(if(X1, X2, X3)) → if(proper(X1), proper(X2), proper(X3))
proper(c) → ok(c)
proper(true) → ok(true)
proper(false) → ok(false)
f(ok(X)) → ok(f(X))
if(ok(X1), ok(X2), ok(X3)) → ok(if(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 5 SCCs with 9 less nodes.

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ EdgeDeletionProof

        ↳ QDP

          ↳ DependencyGraphProof

            ↳ AND

              ↳ QDP

                ↳ QDPOrderProof

              ↳ QDP

              ↳ QDP

              ↳ QDP

              ↳ QDP

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

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

The TRS R consists of the following rules:

active(f(X)) → mark(if(X, c, f(true)))
active(if(true, X, Y)) → mark(X)
active(if(false, X, Y)) → mark(Y)
active(f(X)) → f(active(X))
active(if(X1, X2, X3)) → if(active(X1), X2, X3)
active(if(X1, X2, X3)) → if(X1, active(X2), X3)
f(mark(X)) → mark(f(X))
if(mark(X1), X2, X3) → mark(if(X1, X2, X3))
if(X1, mark(X2), X3) → mark(if(X1, X2, X3))
proper(f(X)) → f(proper(X))
proper(if(X1, X2, X3)) → if(proper(X1), proper(X2), proper(X3))
proper(c) → ok(c)
proper(true) → ok(true)
proper(false) → ok(false)
f(ok(X)) → ok(f(X))
if(ok(X1), ok(X2), ok(X3)) → ok(if(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.

IF(ok(X1), ok(X2), ok(X3)) → IF(X1, X2, X3)

The remaining pairs can at least be oriented weakly.

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

Used ordering: Combined order from the following AFS and order.
IF(x1, x2, x3) = x3
ok(x1) = ok(x1)

Lexicographic path order with status [19].
Precedence:

trivial

Status:

trivial

The following usable rules [14] were oriented: none

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ EdgeDeletionProof

        ↳ QDP

          ↳ DependencyGraphProof

            ↳ AND

              ↳ QDP

                ↳ QDPOrderProof

                  ↳ QDP

                    ↳ QDPOrderProof

              ↳ QDP

              ↳ QDP

              ↳ QDP

              ↳ QDP

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

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

The TRS R consists of the following rules:

active(f(X)) → mark(if(X, c, f(true)))
active(if(true, X, Y)) → mark(X)
active(if(false, X, Y)) → mark(Y)
active(f(X)) → f(active(X))
active(if(X1, X2, X3)) → if(active(X1), X2, X3)
active(if(X1, X2, X3)) → if(X1, active(X2), X3)
f(mark(X)) → mark(f(X))
if(mark(X1), X2, X3) → mark(if(X1, X2, X3))
if(X1, mark(X2), X3) → mark(if(X1, X2, X3))
proper(f(X)) → f(proper(X))
proper(if(X1, X2, X3)) → if(proper(X1), proper(X2), proper(X3))
proper(c) → ok(c)
proper(true) → ok(true)
proper(false) → ok(false)
f(ok(X)) → ok(f(X))
if(ok(X1), ok(X2), ok(X3)) → ok(if(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.

IF(X1, mark(X2), X3) → IF(X1, X2, X3)

The remaining pairs can at least be oriented weakly.

IF(mark(X1), X2, X3) → IF(X1, X2, X3)

Used ordering: Combined order from the following AFS and order.
IF(x1, x2, x3) = x2
mark(x1) = mark(x1)

Lexicographic path order with status [19].
Precedence:

trivial

Status:

trivial

The following usable rules [14] were oriented: none

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ EdgeDeletionProof

        ↳ QDP

          ↳ DependencyGraphProof

            ↳ AND

              ↳ QDP

                ↳ QDPOrderProof

                  ↳ QDP

                    ↳ QDPOrderProof

                      ↳ QDP

                        ↳ QDPOrderProof

              ↳ QDP

              ↳ QDP

              ↳ QDP

              ↳ QDP

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

IF(mark(X1), X2, X3) → IF(X1, X2, X3)

The TRS R consists of the following rules:

active(f(X)) → mark(if(X, c, f(true)))
active(if(true, X, Y)) → mark(X)
active(if(false, X, Y)) → mark(Y)
active(f(X)) → f(active(X))
active(if(X1, X2, X3)) → if(active(X1), X2, X3)
active(if(X1, X2, X3)) → if(X1, active(X2), X3)
f(mark(X)) → mark(f(X))
if(mark(X1), X2, X3) → mark(if(X1, X2, X3))
if(X1, mark(X2), X3) → mark(if(X1, X2, X3))
proper(f(X)) → f(proper(X))
proper(if(X1, X2, X3)) → if(proper(X1), proper(X2), proper(X3))
proper(c) → ok(c)
proper(true) → ok(true)
proper(false) → ok(false)
f(ok(X)) → ok(f(X))
if(ok(X1), ok(X2), ok(X3)) → ok(if(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.

IF(mark(X1), X2, X3) → IF(X1, X2, X3)

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

Lexicographic path order with status [19].
Precedence:

trivial

Status:

trivial

The following usable rules [14] were oriented: none

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ EdgeDeletionProof

        ↳ QDP

          ↳ DependencyGraphProof

            ↳ AND

              ↳ QDP

                ↳ QDPOrderProof

                  ↳ QDP

                    ↳ QDPOrderProof

                      ↳ QDP

                        ↳ QDPOrderProof

                          ↳ QDP

                            ↳ PisEmptyProof

              ↳ QDP

              ↳ QDP

              ↳ QDP

              ↳ QDP

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

active(f(X)) → mark(if(X, c, f(true)))
active(if(true, X, Y)) → mark(X)
active(if(false, X, Y)) → mark(Y)
active(f(X)) → f(active(X))
active(if(X1, X2, X3)) → if(active(X1), X2, X3)
active(if(X1, X2, X3)) → if(X1, active(X2), X3)
f(mark(X)) → mark(f(X))
if(mark(X1), X2, X3) → mark(if(X1, X2, X3))
if(X1, mark(X2), X3) → mark(if(X1, X2, X3))
proper(f(X)) → f(proper(X))
proper(if(X1, X2, X3)) → if(proper(X1), proper(X2), proper(X3))
proper(c) → ok(c)
proper(true) → ok(true)
proper(false) → ok(false)
f(ok(X)) → ok(f(X))
if(ok(X1), ok(X2), ok(X3)) → ok(if(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

              ↳ QDP

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

F(mark(X)) → F(X)
F(ok(X)) → F(X)

The TRS R consists of the following rules:

active(f(X)) → mark(if(X, c, f(true)))
active(if(true, X, Y)) → mark(X)
active(if(false, X, Y)) → mark(Y)
active(f(X)) → f(active(X))
active(if(X1, X2, X3)) → if(active(X1), X2, X3)
active(if(X1, X2, X3)) → if(X1, active(X2), X3)
f(mark(X)) → mark(f(X))
if(mark(X1), X2, X3) → mark(if(X1, X2, X3))
if(X1, mark(X2), X3) → mark(if(X1, X2, X3))
proper(f(X)) → f(proper(X))
proper(if(X1, X2, X3)) → if(proper(X1), proper(X2), proper(X3))
proper(c) → ok(c)
proper(true) → ok(true)
proper(false) → ok(false)
f(ok(X)) → ok(f(X))
if(ok(X1), ok(X2), ok(X3)) → ok(if(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(X)) → F(X)

The remaining pairs can at least be oriented weakly.

F(mark(X)) → F(X)

Used ordering: Combined order from the following AFS and order.
F(x1) = x1
mark(x1) = x1
ok(x1) = ok(x1)

Lexicographic path order with status [19].
Precedence:

trivial

Status:

trivial

The following usable rules [14] were oriented: none

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ EdgeDeletionProof

        ↳ QDP

          ↳ DependencyGraphProof

            ↳ AND

              ↳ QDP

              ↳ QDP

                ↳ QDPOrderProof

                  ↳ QDP

                    ↳ QDPOrderProof

              ↳ QDP

              ↳ QDP

              ↳ QDP

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

F(mark(X)) → F(X)

The TRS R consists of the following rules:

active(f(X)) → mark(if(X, c, f(true)))
active(if(true, X, Y)) → mark(X)
active(if(false, X, Y)) → mark(Y)
active(f(X)) → f(active(X))
active(if(X1, X2, X3)) → if(active(X1), X2, X3)
active(if(X1, X2, X3)) → if(X1, active(X2), X3)
f(mark(X)) → mark(f(X))
if(mark(X1), X2, X3) → mark(if(X1, X2, X3))
if(X1, mark(X2), X3) → mark(if(X1, X2, X3))
proper(f(X)) → f(proper(X))
proper(if(X1, X2, X3)) → if(proper(X1), proper(X2), proper(X3))
proper(c) → ok(c)
proper(true) → ok(true)
proper(false) → ok(false)
f(ok(X)) → ok(f(X))
if(ok(X1), ok(X2), ok(X3)) → ok(if(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(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) = x1
mark(x1) = mark(x1)

Lexicographic path order with status [19].
Precedence:

trivial

Status:

trivial

The following usable rules [14] were oriented: none

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ EdgeDeletionProof

        ↳ QDP

          ↳ DependencyGraphProof

            ↳ AND

              ↳ QDP

              ↳ 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(X)) → mark(if(X, c, f(true)))
active(if(true, X, Y)) → mark(X)
active(if(false, X, Y)) → mark(Y)
active(f(X)) → f(active(X))
active(if(X1, X2, X3)) → if(active(X1), X2, X3)
active(if(X1, X2, X3)) → if(X1, active(X2), X3)
f(mark(X)) → mark(f(X))
if(mark(X1), X2, X3) → mark(if(X1, X2, X3))
if(X1, mark(X2), X3) → mark(if(X1, X2, X3))
proper(f(X)) → f(proper(X))
proper(if(X1, X2, X3)) → if(proper(X1), proper(X2), proper(X3))
proper(c) → ok(c)
proper(true) → ok(true)
proper(false) → ok(false)
f(ok(X)) → ok(f(X))
if(ok(X1), ok(X2), ok(X3)) → ok(if(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

              ↳ QDP

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

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

The TRS R consists of the following rules:

active(f(X)) → mark(if(X, c, f(true)))
active(if(true, X, Y)) → mark(X)
active(if(false, X, Y)) → mark(Y)
active(f(X)) → f(active(X))
active(if(X1, X2, X3)) → if(active(X1), X2, X3)
active(if(X1, X2, X3)) → if(X1, active(X2), X3)
f(mark(X)) → mark(f(X))
if(mark(X1), X2, X3) → mark(if(X1, X2, X3))
if(X1, mark(X2), X3) → mark(if(X1, X2, X3))
proper(f(X)) → f(proper(X))
proper(if(X1, X2, X3)) → if(proper(X1), proper(X2), proper(X3))
proper(c) → ok(c)
proper(true) → ok(true)
proper(false) → ok(false)
f(ok(X)) → ok(f(X))
if(ok(X1), ok(X2), ok(X3)) → ok(if(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(if(X1, X2, X3)) → PROPER(X3)
PROPER(if(X1, X2, X3)) → PROPER(X1)
PROPER(if(X1, X2, X3)) → PROPER(X2)

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) = x1
f(x1) = x1
if(x1, x2, x3) = if(x1, x2, x3)

Lexicographic path order with status [19].
Precedence:

trivial

Status:

trivial

The following usable rules [14] were oriented: none

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ EdgeDeletionProof

        ↳ QDP

          ↳ DependencyGraphProof

            ↳ AND

              ↳ QDP

              ↳ QDP

              ↳ QDP

                ↳ QDPOrderProof

                  ↳ QDP

                    ↳ QDPOrderProof

              ↳ QDP

              ↳ 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(f(X)) → mark(if(X, c, f(true)))
active(if(true, X, Y)) → mark(X)
active(if(false, X, Y)) → mark(Y)
active(f(X)) → f(active(X))
active(if(X1, X2, X3)) → if(active(X1), X2, X3)
active(if(X1, X2, X3)) → if(X1, active(X2), X3)
f(mark(X)) → mark(f(X))
if(mark(X1), X2, X3) → mark(if(X1, X2, X3))
if(X1, mark(X2), X3) → mark(if(X1, X2, X3))
proper(f(X)) → f(proper(X))
proper(if(X1, X2, X3)) → if(proper(X1), proper(X2), proper(X3))
proper(c) → ok(c)
proper(true) → ok(true)
proper(false) → ok(false)
f(ok(X)) → ok(f(X))
if(ok(X1), ok(X2), ok(X3)) → ok(if(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(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) = x1
f(x1) = f(x1)

Lexicographic path order with status [19].
Precedence:

trivial

Status:

trivial

The following usable rules [14] were oriented: none

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ EdgeDeletionProof

        ↳ QDP

          ↳ DependencyGraphProof

            ↳ AND

              ↳ QDP

              ↳ QDP

              ↳ QDP

                ↳ QDPOrderProof

                  ↳ QDP

                    ↳ QDPOrderProof

                      ↳ QDP

                        ↳ PisEmptyProof

              ↳ QDP

              ↳ QDP

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

active(f(X)) → mark(if(X, c, f(true)))
active(if(true, X, Y)) → mark(X)
active(if(false, X, Y)) → mark(Y)
active(f(X)) → f(active(X))
active(if(X1, X2, X3)) → if(active(X1), X2, X3)
active(if(X1, X2, X3)) → if(X1, active(X2), X3)
f(mark(X)) → mark(f(X))
if(mark(X1), X2, X3) → mark(if(X1, X2, X3))
if(X1, mark(X2), X3) → mark(if(X1, X2, X3))
proper(f(X)) → f(proper(X))
proper(if(X1, X2, X3)) → if(proper(X1), proper(X2), proper(X3))
proper(c) → ok(c)
proper(true) → ok(true)
proper(false) → ok(false)
f(ok(X)) → ok(f(X))
if(ok(X1), ok(X2), ok(X3)) → ok(if(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

                ↳ QDPOrderProof

              ↳ QDP

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

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

The TRS R consists of the following rules:

active(f(X)) → mark(if(X, c, f(true)))
active(if(true, X, Y)) → mark(X)
active(if(false, X, Y)) → mark(Y)
active(f(X)) → f(active(X))
active(if(X1, X2, X3)) → if(active(X1), X2, X3)
active(if(X1, X2, X3)) → if(X1, active(X2), X3)
f(mark(X)) → mark(f(X))
if(mark(X1), X2, X3) → mark(if(X1, X2, X3))
if(X1, mark(X2), X3) → mark(if(X1, X2, X3))
proper(f(X)) → f(proper(X))
proper(if(X1, X2, X3)) → if(proper(X1), proper(X2), proper(X3))
proper(c) → ok(c)
proper(true) → ok(true)
proper(false) → ok(false)
f(ok(X)) → ok(f(X))
if(ok(X1), ok(X2), ok(X3)) → ok(if(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(if(X1, X2, X3)) → ACTIVE(X1)
ACTIVE(if(X1, X2, X3)) → ACTIVE(X2)

The remaining pairs can at least be oriented weakly.

ACTIVE(f(X)) → ACTIVE(X)

Used ordering: Combined order from the following AFS and order.
ACTIVE(x1) = x1
f(x1) = x1
if(x1, x2, x3) = if(x1, x2)

Lexicographic path order with status [19].
Precedence:

trivial

Status:

trivial

The following usable rules [14] were oriented: none

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ EdgeDeletionProof

        ↳ QDP

          ↳ DependencyGraphProof

            ↳ AND

              ↳ QDP

              ↳ QDP

              ↳ QDP

              ↳ QDP

                ↳ QDPOrderProof

                  ↳ QDP

                    ↳ QDPOrderProof

              ↳ QDP

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

ACTIVE(f(X)) → ACTIVE(X)

The TRS R consists of the following rules:

active(f(X)) → mark(if(X, c, f(true)))
active(if(true, X, Y)) → mark(X)
active(if(false, X, Y)) → mark(Y)
active(f(X)) → f(active(X))
active(if(X1, X2, X3)) → if(active(X1), X2, X3)
active(if(X1, X2, X3)) → if(X1, active(X2), X3)
f(mark(X)) → mark(f(X))
if(mark(X1), X2, X3) → mark(if(X1, X2, X3))
if(X1, mark(X2), X3) → mark(if(X1, X2, X3))
proper(f(X)) → f(proper(X))
proper(if(X1, X2, X3)) → if(proper(X1), proper(X2), proper(X3))
proper(c) → ok(c)
proper(true) → ok(true)
proper(false) → ok(false)
f(ok(X)) → ok(f(X))
if(ok(X1), ok(X2), ok(X3)) → ok(if(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(X)) → ACTIVE(X)

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

Lexicographic path order with status [19].
Precedence:

trivial

Status:

trivial

The following usable rules [14] were oriented: none

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ EdgeDeletionProof

        ↳ QDP

          ↳ DependencyGraphProof

            ↳ AND

              ↳ QDP

              ↳ QDP

              ↳ QDP

              ↳ QDP

                ↳ QDPOrderProof

                  ↳ QDP

                    ↳ QDPOrderProof

                      ↳ QDP

                        ↳ PisEmptyProof

              ↳ QDP

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

active(f(X)) → mark(if(X, c, f(true)))
active(if(true, X, Y)) → mark(X)
active(if(false, X, Y)) → mark(Y)
active(f(X)) → f(active(X))
active(if(X1, X2, X3)) → if(active(X1), X2, X3)
active(if(X1, X2, X3)) → if(X1, active(X2), X3)
f(mark(X)) → mark(f(X))
if(mark(X1), X2, X3) → mark(if(X1, X2, X3))
if(X1, mark(X2), X3) → mark(if(X1, X2, X3))
proper(f(X)) → f(proper(X))
proper(if(X1, X2, X3)) → if(proper(X1), proper(X2), proper(X3))
proper(c) → ok(c)
proper(true) → ok(true)
proper(false) → ok(false)
f(ok(X)) → ok(f(X))
if(ok(X1), ok(X2), ok(X3)) → ok(if(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

              ↳ 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(X)) → mark(if(X, c, f(true)))
active(if(true, X, Y)) → mark(X)
active(if(false, X, Y)) → mark(Y)
active(f(X)) → f(active(X))
active(if(X1, X2, X3)) → if(active(X1), X2, X3)
active(if(X1, X2, X3)) → if(X1, active(X2), X3)
f(mark(X)) → mark(f(X))
if(mark(X1), X2, X3) → mark(if(X1, X2, X3))
if(X1, mark(X2), X3) → mark(if(X1, X2, X3))
proper(f(X)) → f(proper(X))
proper(if(X1, X2, X3)) → if(proper(X1), proper(X2), proper(X3))
proper(c) → ok(c)
proper(true) → ok(true)
proper(false) → ok(false)
f(ok(X)) → ok(f(X))
if(ok(X1), ok(X2), ok(X3)) → ok(if(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.