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:



↳ 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)

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
                  ↳ 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)

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
  ↳ 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)))
The remaining pairs can at least be oriented weakly.

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)) = 1 + x1   
POL(if(x1, x2, x3)) = x1 + x2   
POL(mark(x1)) = x1   
POL(true) = 0   

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(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.
The approximation of the Dependency Graph [15,17,22] contains 1 SCC with 1 less node.

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

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

MARK(if(X1, X2, X3)) → MARK(X1)
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 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
                  ↳ QDP
                  ↳ QDP
                    ↳ RuleRemovalProof
                      ↳ QDP
                        ↳ QDPOrderProof
                          ↳ QDP
                            ↳ QDPOrderProof
                              ↳ QDP
                                ↳ QDPOrderProof
                                  ↳ QDP
                                    ↳ DependencyGraphProof
                                      ↳ QDP
                                        ↳ UsableRulesProof
QDP
                                            ↳ QDPSizeChangeProof

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

MARK(if(X1, X2, X3)) → MARK(X1)
MARK(if(X1, X2, X3)) → MARK(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: