Termination w.r.t. Q proof of /tmp/tmptIppDd/Ex5_Zan97

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(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

  ↳ RRRPoloQTRSProof

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.

The following Q TRS is given: 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.
The following rules can be removed by the rule removal processor [15] because they are oriented strictly by a polynomial ordering:

active(if(false, X, Y)) → mark(Y)

Used ordering:
Polynomial interpretation [25]:

POL(active(x₁)) = x₁
POL(c) = 0
POL(f(x₁)) = 2·x₁
POL(false) = 2
POL(if(x₁, x₂, x₃)) = 2·x₁ + 2·x₂ + 2·x₃
POL(mark(x₁)) = x₁
POL(true) = 0

↳ QTRS

  ↳ RRRPoloQTRSProof

    ↳ QTRS

      ↳ RRRPoloQTRSProof

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)
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.

The following Q TRS is given: 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)
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.
The following rules can be removed by the rule removal processor [15] because they are oriented strictly by a polynomial ordering:

mark(false) → active(false)

Used ordering:
Polynomial interpretation [25]:

POL(active(x₁)) = x₁
POL(c) = 0
POL(f(x₁)) = 2·x₁
POL(false) = 2
POL(if(x₁, x₂, x₃)) = x₁ + 2·x₂ + x₃
POL(mark(x₁)) = 2·x₁
POL(true) = 0

↳ QTRS

  ↳ RRRPoloQTRSProof

    ↳ QTRS

      ↳ RRRPoloQTRSProof

        ↳ 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)
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)
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)))
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)
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)
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

  ↳ RRRPoloQTRSProof

    ↳ QTRS

      ↳ RRRPoloQTRSProof

        ↳ 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)))
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)
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)
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 6 less nodes.

↳ QTRS

  ↳ RRRPoloQTRSProof

    ↳ QTRS

      ↳ RRRPoloQTRSProof

        ↳ QTRS

          ↳ DependencyPairsProof

            ↳ QDP

              ↳ DependencyGraphProof

                ↳ AND

                  ↳ QDP

                    ↳ UsableRulesProof

                  ↳ 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)
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)
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 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

  ↳ RRRPoloQTRSProof

    ↳ QTRS

      ↳ RRRPoloQTRSProof

        ↳ QTRS

          ↳ DependencyPairsProof

            ↳ QDP

              ↳ DependencyGraphProof

                ↳ AND

                  ↳ QDP

                    ↳ UsableRulesProof

                      ↳ QDP

                        ↳ QDPSizeChangeProof

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

IF(X1, mark(X2), X3) → IF(X1, X2, X3)
The graph contains the following edges 1 >= 1, 2 > 2, 3 >= 3
IF(X1, X2, active(X3)) → IF(X1, X2, X3)
The graph contains the following edges 1 >= 1, 2 >= 2, 3 > 3
IF(X1, active(X2), X3) → IF(X1, X2, X3)
The graph contains the following edges 1 >= 1, 2 > 2, 3 >= 3
IF(X1, X2, mark(X3)) → IF(X1, X2, X3)
The graph contains the following edges 1 >= 1, 2 >= 2, 3 > 3
IF(active(X1), X2, X3) → IF(X1, X2, X3)
The graph contains the following edges 1 > 1, 2 >= 2, 3 >= 3
IF(mark(X1), X2, X3) → IF(X1, X2, X3)
The graph contains the following edges 1 > 1, 2 >= 2, 3 >= 3

↳ QTRS

  ↳ RRRPoloQTRSProof

    ↳ QTRS

      ↳ RRRPoloQTRSProof

        ↳ QTRS

          ↳ DependencyPairsProof

            ↳ QDP

              ↳ DependencyGraphProof

                ↳ AND

                  ↳ QDP

                  ↳ QDP

                    ↳ UsableRulesProof

                  ↳ 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)
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)
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)

↳ QTRS

  ↳ RRRPoloQTRSProof

    ↳ QTRS

      ↳ RRRPoloQTRSProof

        ↳ QTRS

          ↳ DependencyPairsProof

            ↳ QDP

              ↳ DependencyGraphProof

                ↳ AND

                  ↳ QDP

                  ↳ QDP

                    ↳ UsableRulesProof

                      ↳ QDP

                        ↳ QDPSizeChangeProof

                  ↳ QDP

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

F(mark(X)) → F(X)
F(active(X)) → F(X)

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

F(mark(X)) → F(X)
The graph contains the following edges 1 > 1
F(active(X)) → F(X)
The graph contains the following edges 1 > 1

↳ QTRS

  ↳ RRRPoloQTRSProof

    ↳ QTRS

      ↳ RRRPoloQTRSProof

        ↳ QTRS

          ↳ DependencyPairsProof

            ↳ QDP

              ↳ DependencyGraphProof

                ↳ AND

                  ↳ QDP

                  ↳ QDP

                  ↳ QDP

                    ↳ RuleRemovalProof

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)) → 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)
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)
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.
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 dependency pairs:

MARK(f(X)) → MARK(X)

Used ordering: POLO with Polynomial interpretation [25]:

POL(ACTIVE(x₁)) = x₁
POL(MARK(x₁)) = x₁
POL(active(x₁)) = x₁
POL(c) = 0
POL(f(x₁)) = 2 + x₁
POL(if(x₁, x₂, x₃)) = x₁ + x₂ + x₃
POL(mark(x₁)) = x₁
POL(true) = 0

↳ QTRS

  ↳ RRRPoloQTRSProof

    ↳ QTRS

      ↳ RRRPoloQTRSProof

        ↳ QTRS

          ↳ DependencyPairsProof

            ↳ QDP

              ↳ DependencyGraphProof

                ↳ AND

                  ↳ QDP

                  ↳ QDP

                  ↳ QDP

                    ↳ RuleRemovalProof

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

Used ordering: Polynomial interpretation [25]:

POL(ACTIVE(x₁)) = x₁
POL(MARK(x₁)) = x₁
POL(active(x₁)) = x₁
POL(c) = 0
POL(f(x₁)) = x₁
POL(if(x₁, x₂, x₃)) = x₁ + x₂
POL(mark(x₁)) = x₁
POL(true) = 1

The following usable rules [17] were oriented:

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

↳ QTRS

  ↳ RRRPoloQTRSProof

    ↳ QTRS

      ↳ RRRPoloQTRSProof

        ↳ QTRS

          ↳ DependencyPairsProof

            ↳ QDP

              ↳ DependencyGraphProof

                ↳ AND

                  ↳ QDP

                  ↳ QDP

                  ↳ QDP

                    ↳ RuleRemovalProof

                      ↳ 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)
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)
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)
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(if(X1, X2, X3)) → ACTIVE(if(mark(X1), mark(X2), X3))

The remaining pairs can at least be oriented weakly.

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

Used ordering: Polynomial interpretation [25]:

POL(ACTIVE(x₁)) = x₁
POL(MARK(x₁)) = 1
POL(active(x₁)) = 0
POL(c) = 0
POL(f(x₁)) = 1
POL(if(x₁, x₂, x₃)) = 0
POL(mark(x₁)) = 0
POL(true) = 0

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)

↳ QTRS

  ↳ RRRPoloQTRSProof

    ↳ QTRS

      ↳ RRRPoloQTRSProof

        ↳ QTRS

          ↳ DependencyPairsProof

            ↳ QDP

              ↳ DependencyGraphProof

                ↳ AND

                  ↳ QDP

                  ↳ QDP

                  ↳ QDP

                    ↳ RuleRemovalProof

                      ↳ QDP

                        ↳ QDPOrderProof

                          ↳ QDP

                            ↳ QDPOrderProof

                              ↳ QDP

                                ↳ QDPOrderProof

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

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

The remaining pairs can at least be oriented weakly.

MARK(f(X)) → ACTIVE(f(mark(X)))

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

Recursive path order with status [2].
Quasi-Precedence:

[ACTIVE₁, MARK₁] > [f₁, true] > if₂
[ACTIVE₁, MARK₁] > [f₁, true] > c

Status:

c: multiset
f₁: [1]
MARK₁: [1]
if₂: multiset
true: multiset
ACTIVE₁: [1]

The following usable rules [17] were oriented:

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

↳ QTRS

  ↳ RRRPoloQTRSProof

    ↳ QTRS

      ↳ RRRPoloQTRSProof

        ↳ QTRS

          ↳ DependencyPairsProof

            ↳ QDP

              ↳ DependencyGraphProof

                ↳ AND

                  ↳ QDP

                  ↳ QDP

                  ↳ QDP

                    ↳ RuleRemovalProof

                      ↳ QDP

                        ↳ QDPOrderProof

                          ↳ QDP

                            ↳ QDPOrderProof

                              ↳ QDP

                                ↳ QDPOrderProof

                                  ↳ QDP

                                    ↳ DependencyGraphProof

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

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)
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)
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 0 SCCs with 1 less node.