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:

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

Q is empty.


↳ QTRS
  ↳ DependencyPairsProof

Q restricted rewrite system:
The TRS R consists of the following rules:

a__f(X) → a__if(mark(X), c, f(true))
a__if(true, X, Y) → mark(X)
a__if(false, X, Y) → mark(Y)
mark(f(X)) → a__f(mark(X))
mark(if(X1, X2, X3)) → a__if(mark(X1), mark(X2), X3)
mark(c) → c
mark(true) → true
mark(false) → false
a__f(X) → f(X)
a__if(X1, X2, 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:

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

The TRS R consists of the following rules:

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

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

↳ QTRS
  ↳ DependencyPairsProof
    ↳ QDP
      ↳ QDPOrderProof

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

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

The TRS R consists of the following rules:

a__f(X) → a__if(mark(X), c, f(true))
a__if(true, X, Y) → mark(X)
a__if(false, X, Y) → mark(Y)
mark(f(X)) → a__f(mark(X))
mark(if(X1, X2, X3)) → a__if(mark(X1), mark(X2), X3)
mark(c) → c
mark(true) → true
mark(false) → false
a__f(X) → f(X)
a__if(X1, X2, 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.


A__IF(false, X, Y) → MARK(Y)
The remaining pairs can at least be oriented weakly.

MARK(if(X1, X2, X3)) → A__IF(mark(X1), mark(X2), X3)
MARK(if(X1, X2, X3)) → MARK(X1)
A__F(X) → A__IF(mark(X), c, f(true))
MARK(f(X)) → A__F(mark(X))
A__IF(true, X, Y) → MARK(X)
A__F(X) → MARK(X)
MARK(if(X1, X2, X3)) → MARK(X2)
MARK(f(X)) → MARK(X)
Used ordering: Polynomial interpretation [25,35]:

POL(A__IF(x1, x2, x3)) = x_1 + (1/4)x_2 + (1/4)x_3   
POL(c) = 0   
POL(MARK(x1)) = (1/4)x_1   
POL(if(x1, x2, x3)) = (4)x_1 + (4)x_2 + (2)x_3   
POL(f(x1)) = (4)x_1   
POL(a__f(x1)) = (4)x_1   
POL(true) = 0   
POL(mark(x1)) = x_1   
POL(false) = 2   
POL(A__F(x1)) = x_1   
POL(a__if(x1, x2, x3)) = (4)x_1 + (4)x_2 + (2)x_3   
The value of delta used in the strict ordering is 2.
The following usable rules [17] were oriented:

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



↳ QTRS
  ↳ DependencyPairsProof
    ↳ QDP
      ↳ QDPOrderProof
        ↳ QDP
          ↳ QDPOrderProof

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

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

The TRS R consists of the following rules:

a__f(X) → a__if(mark(X), c, f(true))
a__if(true, X, Y) → mark(X)
a__if(false, X, Y) → mark(Y)
mark(f(X)) → a__f(mark(X))
mark(if(X1, X2, X3)) → a__if(mark(X1), mark(X2), X3)
mark(c) → c
mark(true) → true
mark(false) → false
a__f(X) → f(X)
a__if(X1, X2, 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.


A__F(X) → A__IF(mark(X), c, f(true))
A__F(X) → MARK(X)
MARK(f(X)) → MARK(X)
The remaining pairs can at least be oriented weakly.

MARK(if(X1, X2, X3)) → A__IF(mark(X1), mark(X2), X3)
MARK(if(X1, X2, X3)) → MARK(X1)
MARK(f(X)) → A__F(mark(X))
A__IF(true, X, Y) → MARK(X)
MARK(if(X1, X2, X3)) → MARK(X2)
Used ordering: Polynomial interpretation [25,35]:

POL(A__IF(x1, x2, x3)) = 1 + (4)x_2   
POL(c) = 0   
POL(MARK(x1)) = 1 + (4)x_1   
POL(if(x1, x2, x3)) = x_1 + (4)x_2 + (2)x_3   
POL(f(x1)) = 1/4 + (4)x_1   
POL(a__f(x1)) = 1 + (4)x_1   
POL(true) = 0   
POL(mark(x1)) = (4)x_1   
POL(A__F(x1)) = 2 + (4)x_1   
POL(false) = 2   
POL(a__if(x1, x2, x3)) = x_1 + (4)x_2 + (4)x_3   
The value of delta used in the strict ordering is 1.
The following usable rules [17] were oriented:

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



↳ QTRS
  ↳ DependencyPairsProof
    ↳ QDP
      ↳ QDPOrderProof
        ↳ QDP
          ↳ QDPOrderProof
            ↳ QDP
              ↳ DependencyGraphProof

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

MARK(if(X1, X2, X3)) → A__IF(mark(X1), mark(X2), X3)
MARK(if(X1, X2, X3)) → MARK(X1)
MARK(f(X)) → A__F(mark(X))
A__IF(true, X, Y) → MARK(X)
MARK(if(X1, X2, X3)) → MARK(X2)

The TRS R consists of the following rules:

a__f(X) → a__if(mark(X), c, f(true))
a__if(true, X, Y) → mark(X)
a__if(false, X, Y) → mark(Y)
mark(f(X)) → a__f(mark(X))
mark(if(X1, X2, X3)) → a__if(mark(X1), mark(X2), X3)
mark(c) → c
mark(true) → true
mark(false) → false
a__f(X) → f(X)
a__if(X1, X2, 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
  ↳ DependencyPairsProof
    ↳ QDP
      ↳ QDPOrderProof
        ↳ QDP
          ↳ QDPOrderProof
            ↳ QDP
              ↳ DependencyGraphProof
                ↳ QDP

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

MARK(if(X1, X2, X3)) → A__IF(mark(X1), mark(X2), X3)
MARK(if(X1, X2, X3)) → MARK(X1)
A__IF(true, X, Y) → MARK(X)
MARK(if(X1, X2, X3)) → MARK(X2)

The TRS R consists of the following rules:

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

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