Termination w.r.t. Q proof of /tmp/tmpGLxWpx/Ex5_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)
mark(f(X)) → active(f(mark(X)))
mark(if(X1, X2, X3)) → active(if(mark(X1), mark(X2), X3))
mark(c) → active(c)
mark(true) → active(true)
mark(false) → active(false)
f(mark(X)) → f(X)
f(active(X)) → f(X)
if(mark(X1), X2, X3) → if(X1, X2, X3)
if(X1, mark(X2), X3) → if(X1, X2, X3)
if(X1, X2, mark(X3)) → if(X1, X2, X3)
if(active(X1), X2, X3) → if(X1, X2, X3)
if(X1, active(X2), X3) → if(X1, X2, X3)
if(X1, X2, active(X3)) → if(X1, X2, X3)

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

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:

IF(X1, mark(X2), X3) → IF(X1, X2, X3)
F(mark(X)) → F(X)
IF(X1, X2, active(X3)) → IF(X1, X2, X3)
ACTIVE(f(X)) → MARK(if(X, c, f(true)))
MARK(false) → ACTIVE(false)
ACTIVE(if(false, X, Y)) → MARK(Y)
IF(active(X1), X2, X3) → IF(X1, X2, X3)
IF(mark(X1), X2, X3) → IF(X1, X2, X3)
ACTIVE(f(X)) → F(true)
MARK(f(X)) → ACTIVE(f(mark(X)))
MARK(if(X1, X2, X3)) → ACTIVE(if(mark(X1), mark(X2), X3))
MARK(true) → ACTIVE(true)
MARK(c) → ACTIVE(c)
IF(X1, active(X2), X3) → IF(X1, X2, X3)
MARK(if(X1, X2, X3)) → MARK(X1)
IF(X1, X2, mark(X3)) → IF(X1, X2, X3)
MARK(if(X1, X2, X3)) → IF(mark(X1), mark(X2), X3)
MARK(f(X)) → F(mark(X))
ACTIVE(f(X)) → IF(X, c, f(true))
ACTIVE(if(true, X, Y)) → MARK(X)
F(active(X)) → F(X)
MARK(f(X)) → MARK(X)
MARK(if(X1, X2, X3)) → MARK(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)
mark(f(X)) → active(f(mark(X)))
mark(if(X1, X2, X3)) → active(if(mark(X1), mark(X2), X3))
mark(c) → active(c)
mark(true) → active(true)
mark(false) → active(false)
f(mark(X)) → f(X)
f(active(X)) → f(X)
if(mark(X1), X2, X3) → if(X1, X2, X3)
if(X1, mark(X2), X3) → if(X1, X2, X3)
if(X1, X2, mark(X3)) → if(X1, X2, X3)
if(active(X1), X2, X3) → if(X1, X2, X3)
if(X1, active(X2), X3) → if(X1, X2, X3)
if(X1, X2, active(X3)) → if(X1, X2, X3)

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:

IF(X1, mark(X2), X3) → IF(X1, X2, X3)
F(mark(X)) → F(X)
IF(X1, X2, active(X3)) → IF(X1, X2, X3)
ACTIVE(f(X)) → MARK(if(X, c, f(true)))
MARK(false) → ACTIVE(false)
ACTIVE(if(false, X, Y)) → MARK(Y)
IF(active(X1), X2, X3) → IF(X1, X2, X3)
IF(mark(X1), X2, X3) → IF(X1, X2, X3)
ACTIVE(f(X)) → F(true)
MARK(f(X)) → ACTIVE(f(mark(X)))
MARK(if(X1, X2, X3)) → ACTIVE(if(mark(X1), mark(X2), X3))
MARK(true) → ACTIVE(true)
MARK(c) → ACTIVE(c)
IF(X1, active(X2), X3) → IF(X1, X2, X3)
MARK(if(X1, X2, X3)) → MARK(X1)
IF(X1, X2, mark(X3)) → IF(X1, X2, X3)
MARK(if(X1, X2, X3)) → IF(mark(X1), mark(X2), X3)
MARK(f(X)) → F(mark(X))
ACTIVE(f(X)) → IF(X, c, f(true))
ACTIVE(if(true, X, Y)) → MARK(X)
F(active(X)) → F(X)
MARK(f(X)) → MARK(X)
MARK(if(X1, X2, X3)) → MARK(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)
mark(f(X)) → active(f(mark(X)))
mark(if(X1, X2, X3)) → active(if(mark(X1), mark(X2), X3))
mark(c) → active(c)
mark(true) → active(true)
mark(false) → active(false)
f(mark(X)) → f(X)
f(active(X)) → f(X)
if(mark(X1), X2, X3) → if(X1, X2, X3)
if(X1, mark(X2), X3) → if(X1, X2, X3)
if(X1, X2, mark(X3)) → if(X1, X2, X3)
if(active(X1), X2, X3) → if(X1, X2, X3)
if(X1, active(X2), X3) → if(X1, X2, X3)
if(X1, X2, active(X3)) → if(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 3 SCCs with 7 less nodes.

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ DependencyGraphProof

        ↳ AND

          ↳ QDP

            ↳ QDPOrderProof

          ↳ QDP

          ↳ QDP

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

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

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

The remaining pairs can at least be oriented weakly.

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

Used ordering: Polynomial interpretation [25,35]:

POL(active(x₁)) = 1 + (2)x_1
POL(mark(x₁)) = 1 + (4)x_1
POL(IF(x₁, x₂, x₃)) = x_1 + x_2

The value of delta used in the strict ordering is 1.
The following usable rules [17] were oriented: none

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ DependencyGraphProof

        ↳ AND

          ↳ QDP

            ↳ QDPOrderProof

              ↳ QDP

                ↳ QDPOrderProof

          ↳ QDP

          ↳ QDP

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

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

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

The remaining pairs can at least be oriented weakly.
none
Used ordering: Polynomial interpretation [25,35]:

POL(active(x₁)) = 4 + x_1
POL(mark(x₁)) = 4 + (4)x_1
POL(IF(x₁, x₂, x₃)) = (4)x_3

The value of delta used in the strict ordering is 16.
The following usable rules [17] were oriented: none

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ DependencyGraphProof

        ↳ AND

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

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

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

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

F(mark(X)) → F(X)
F(active(X)) → F(X)

The remaining pairs can at least be oriented weakly.
none
Used ordering: Polynomial interpretation [25,35]:

POL(active(x₁)) = 4 + x_1
POL(mark(x₁)) = 4 + (4)x_1
POL(F(x₁)) = (4)x_1

The value of delta used in the strict ordering is 16.
The following usable rules [17] were oriented: none

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ DependencyGraphProof

        ↳ AND

          ↳ QDP

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

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

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

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

ACTIVE(if(false, X, Y)) → MARK(Y)

The remaining pairs can at least be oriented weakly.

MARK(if(X1, X2, X3)) → ACTIVE(if(mark(X1), mark(X2), X3))
ACTIVE(f(X)) → MARK(if(X, c, f(true)))
MARK(if(X1, X2, X3)) → MARK(X1)
ACTIVE(if(true, X, Y)) → MARK(X)
MARK(if(X1, X2, X3)) → MARK(X2)
MARK(f(X)) → MARK(X)
MARK(f(X)) → ACTIVE(f(mark(X)))

Used ordering: Polynomial interpretation [25,35]:

POL(c) = 0
POL(active(x₁)) = x_1
POL(MARK(x₁)) = (4)x_1
POL(if(x₁, x₂, x₃)) = x_1 + (2)x_2 + (4)x_3
POL(f(x₁)) = (2)x_1
POL(true) = 0
POL(mark(x₁)) = (2)x_1
POL(false) = 4
POL(ACTIVE(x₁)) = (2)x_1

The value of delta used in the strict ordering is 8.
The following usable rules [17] were oriented:

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

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ DependencyGraphProof

        ↳ AND

          ↳ QDP

          ↳ QDP

          ↳ QDP

            ↳ QDPOrderProof

              ↳ QDP

                ↳ QDPOrderProof

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

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

MARK(f(X)) → MARK(X)

The remaining pairs can at least be oriented weakly.

MARK(if(X1, X2, X3)) → ACTIVE(if(mark(X1), mark(X2), X3))
ACTIVE(f(X)) → MARK(if(X, c, f(true)))
MARK(if(X1, X2, X3)) → MARK(X1)
ACTIVE(if(true, X, Y)) → MARK(X)
MARK(f(X)) → ACTIVE(f(mark(X)))
MARK(if(X1, X2, X3)) → MARK(X2)

Used ordering: Polynomial interpretation [25,35]:

POL(c) = 0
POL(active(x₁)) = x_1
POL(MARK(x₁)) = (4)x_1
POL(if(x₁, x₂, x₃)) = (4)x_1 + (4)x_2 + x_3
POL(f(x₁)) = 2 + (4)x_1
POL(true) = 0
POL(mark(x₁)) = x_1
POL(false) = 0
POL(ACTIVE(x₁)) = (4)x_1

The value of delta used in the strict ordering is 8.
The following usable rules [17] were oriented:

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

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ DependencyGraphProof

        ↳ AND

          ↳ QDP

          ↳ QDP

          ↳ QDP

            ↳ QDPOrderProof

              ↳ QDP

                ↳ QDPOrderProof

                  ↳ QDP

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

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

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