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:

active1(f1(X)) -> mark1(if3(X, c, f1(true)))
active1(if3(true, X, Y)) -> mark1(X)
active1(if3(false, X, Y)) -> mark1(Y)
active1(f1(X)) -> f1(active1(X))
active1(if3(X1, X2, X3)) -> if3(active1(X1), X2, X3)
active1(if3(X1, X2, X3)) -> if3(X1, active1(X2), X3)
f1(mark1(X)) -> mark1(f1(X))
if3(mark1(X1), X2, X3) -> mark1(if3(X1, X2, X3))
if3(X1, mark1(X2), X3) -> mark1(if3(X1, X2, X3))
proper1(f1(X)) -> f1(proper1(X))
proper1(if3(X1, X2, X3)) -> if3(proper1(X1), proper1(X2), proper1(X3))
proper1(c) -> ok1(c)
proper1(true) -> ok1(true)
proper1(false) -> ok1(false)
f1(ok1(X)) -> ok1(f1(X))
if3(ok1(X1), ok1(X2), ok1(X3)) -> ok1(if3(X1, X2, X3))
top1(mark1(X)) -> top1(proper1(X))
top1(ok1(X)) -> top1(active1(X))

Q is empty.


↳ QTRS
  ↳ DependencyPairsProof

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

active1(f1(X)) -> mark1(if3(X, c, f1(true)))
active1(if3(true, X, Y)) -> mark1(X)
active1(if3(false, X, Y)) -> mark1(Y)
active1(f1(X)) -> f1(active1(X))
active1(if3(X1, X2, X3)) -> if3(active1(X1), X2, X3)
active1(if3(X1, X2, X3)) -> if3(X1, active1(X2), X3)
f1(mark1(X)) -> mark1(f1(X))
if3(mark1(X1), X2, X3) -> mark1(if3(X1, X2, X3))
if3(X1, mark1(X2), X3) -> mark1(if3(X1, X2, X3))
proper1(f1(X)) -> f1(proper1(X))
proper1(if3(X1, X2, X3)) -> if3(proper1(X1), proper1(X2), proper1(X3))
proper1(c) -> ok1(c)
proper1(true) -> ok1(true)
proper1(false) -> ok1(false)
f1(ok1(X)) -> ok1(f1(X))
if3(ok1(X1), ok1(X2), ok1(X3)) -> ok1(if3(X1, X2, X3))
top1(mark1(X)) -> top1(proper1(X))
top1(ok1(X)) -> top1(active1(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:

TOP1(mark1(X)) -> PROPER1(X)
ACTIVE1(f1(X)) -> ACTIVE1(X)
IF3(X1, mark1(X2), X3) -> IF3(X1, X2, X3)
ACTIVE1(f1(X)) -> F1(active1(X))
PROPER1(if3(X1, X2, X3)) -> PROPER1(X1)
ACTIVE1(if3(X1, X2, X3)) -> ACTIVE1(X2)
TOP1(ok1(X)) -> ACTIVE1(X)
IF3(ok1(X1), ok1(X2), ok1(X3)) -> IF3(X1, X2, X3)
ACTIVE1(if3(X1, X2, X3)) -> IF3(X1, active1(X2), X3)
F1(mark1(X)) -> F1(X)
ACTIVE1(f1(X)) -> IF3(X, c, f1(true))
ACTIVE1(f1(X)) -> F1(true)
PROPER1(if3(X1, X2, X3)) -> IF3(proper1(X1), proper1(X2), proper1(X3))
ACTIVE1(if3(X1, X2, X3)) -> IF3(active1(X1), X2, X3)
PROPER1(f1(X)) -> PROPER1(X)
TOP1(ok1(X)) -> TOP1(active1(X))
ACTIVE1(if3(X1, X2, X3)) -> ACTIVE1(X1)
F1(ok1(X)) -> F1(X)
TOP1(mark1(X)) -> TOP1(proper1(X))
PROPER1(if3(X1, X2, X3)) -> PROPER1(X3)
IF3(mark1(X1), X2, X3) -> IF3(X1, X2, X3)
PROPER1(f1(X)) -> F1(proper1(X))
PROPER1(if3(X1, X2, X3)) -> PROPER1(X2)

The TRS R consists of the following rules:

active1(f1(X)) -> mark1(if3(X, c, f1(true)))
active1(if3(true, X, Y)) -> mark1(X)
active1(if3(false, X, Y)) -> mark1(Y)
active1(f1(X)) -> f1(active1(X))
active1(if3(X1, X2, X3)) -> if3(active1(X1), X2, X3)
active1(if3(X1, X2, X3)) -> if3(X1, active1(X2), X3)
f1(mark1(X)) -> mark1(f1(X))
if3(mark1(X1), X2, X3) -> mark1(if3(X1, X2, X3))
if3(X1, mark1(X2), X3) -> mark1(if3(X1, X2, X3))
proper1(f1(X)) -> f1(proper1(X))
proper1(if3(X1, X2, X3)) -> if3(proper1(X1), proper1(X2), proper1(X3))
proper1(c) -> ok1(c)
proper1(true) -> ok1(true)
proper1(false) -> ok1(false)
f1(ok1(X)) -> ok1(f1(X))
if3(ok1(X1), ok1(X2), ok1(X3)) -> ok1(if3(X1, X2, X3))
top1(mark1(X)) -> top1(proper1(X))
top1(ok1(X)) -> top1(active1(X))

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:

TOP1(mark1(X)) -> PROPER1(X)
ACTIVE1(f1(X)) -> ACTIVE1(X)
IF3(X1, mark1(X2), X3) -> IF3(X1, X2, X3)
ACTIVE1(f1(X)) -> F1(active1(X))
PROPER1(if3(X1, X2, X3)) -> PROPER1(X1)
ACTIVE1(if3(X1, X2, X3)) -> ACTIVE1(X2)
TOP1(ok1(X)) -> ACTIVE1(X)
IF3(ok1(X1), ok1(X2), ok1(X3)) -> IF3(X1, X2, X3)
ACTIVE1(if3(X1, X2, X3)) -> IF3(X1, active1(X2), X3)
F1(mark1(X)) -> F1(X)
ACTIVE1(f1(X)) -> IF3(X, c, f1(true))
ACTIVE1(f1(X)) -> F1(true)
PROPER1(if3(X1, X2, X3)) -> IF3(proper1(X1), proper1(X2), proper1(X3))
ACTIVE1(if3(X1, X2, X3)) -> IF3(active1(X1), X2, X3)
PROPER1(f1(X)) -> PROPER1(X)
TOP1(ok1(X)) -> TOP1(active1(X))
ACTIVE1(if3(X1, X2, X3)) -> ACTIVE1(X1)
F1(ok1(X)) -> F1(X)
TOP1(mark1(X)) -> TOP1(proper1(X))
PROPER1(if3(X1, X2, X3)) -> PROPER1(X3)
IF3(mark1(X1), X2, X3) -> IF3(X1, X2, X3)
PROPER1(f1(X)) -> F1(proper1(X))
PROPER1(if3(X1, X2, X3)) -> PROPER1(X2)

The TRS R consists of the following rules:

active1(f1(X)) -> mark1(if3(X, c, f1(true)))
active1(if3(true, X, Y)) -> mark1(X)
active1(if3(false, X, Y)) -> mark1(Y)
active1(f1(X)) -> f1(active1(X))
active1(if3(X1, X2, X3)) -> if3(active1(X1), X2, X3)
active1(if3(X1, X2, X3)) -> if3(X1, active1(X2), X3)
f1(mark1(X)) -> mark1(f1(X))
if3(mark1(X1), X2, X3) -> mark1(if3(X1, X2, X3))
if3(X1, mark1(X2), X3) -> mark1(if3(X1, X2, X3))
proper1(f1(X)) -> f1(proper1(X))
proper1(if3(X1, X2, X3)) -> if3(proper1(X1), proper1(X2), proper1(X3))
proper1(c) -> ok1(c)
proper1(true) -> ok1(true)
proper1(false) -> ok1(false)
f1(ok1(X)) -> ok1(f1(X))
if3(ok1(X1), ok1(X2), ok1(X3)) -> ok1(if3(X1, X2, X3))
top1(mark1(X)) -> top1(proper1(X))
top1(ok1(X)) -> top1(active1(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
      ↳ DependencyGraphProof
        ↳ AND
          ↳ QDP
            ↳ QDPOrderProof
          ↳ QDP
          ↳ QDP
          ↳ QDP
          ↳ QDP

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

IF3(X1, mark1(X2), X3) -> IF3(X1, X2, X3)
IF3(mark1(X1), X2, X3) -> IF3(X1, X2, X3)
IF3(ok1(X1), ok1(X2), ok1(X3)) -> IF3(X1, X2, X3)

The TRS R consists of the following rules:

active1(f1(X)) -> mark1(if3(X, c, f1(true)))
active1(if3(true, X, Y)) -> mark1(X)
active1(if3(false, X, Y)) -> mark1(Y)
active1(f1(X)) -> f1(active1(X))
active1(if3(X1, X2, X3)) -> if3(active1(X1), X2, X3)
active1(if3(X1, X2, X3)) -> if3(X1, active1(X2), X3)
f1(mark1(X)) -> mark1(f1(X))
if3(mark1(X1), X2, X3) -> mark1(if3(X1, X2, X3))
if3(X1, mark1(X2), X3) -> mark1(if3(X1, X2, X3))
proper1(f1(X)) -> f1(proper1(X))
proper1(if3(X1, X2, X3)) -> if3(proper1(X1), proper1(X2), proper1(X3))
proper1(c) -> ok1(c)
proper1(true) -> ok1(true)
proper1(false) -> ok1(false)
f1(ok1(X)) -> ok1(f1(X))
if3(ok1(X1), ok1(X2), ok1(X3)) -> ok1(if3(X1, X2, X3))
top1(mark1(X)) -> top1(proper1(X))
top1(ok1(X)) -> top1(active1(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.


IF3(X1, mark1(X2), X3) -> IF3(X1, X2, X3)
IF3(mark1(X1), X2, X3) -> IF3(X1, X2, X3)
IF3(ok1(X1), ok1(X2), ok1(X3)) -> IF3(X1, X2, X3)
The remaining pairs can at least be oriented weakly.
none
Used ordering: Polynomial Order [17,21] with Interpretation:

POL( IF3(x1, ..., x3) ) = 2x1 + 2x2 + x3 + 3


POL( mark1(x1) ) = 3x1 + 1


POL( ok1(x1) ) = x1 + 3



The following usable rules [14] were oriented: none



↳ QTRS
  ↳ DependencyPairsProof
    ↳ QDP
      ↳ DependencyGraphProof
        ↳ AND
          ↳ QDP
            ↳ QDPOrderProof
              ↳ QDP
                ↳ PisEmptyProof
          ↳ QDP
          ↳ QDP
          ↳ QDP
          ↳ QDP

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

active1(f1(X)) -> mark1(if3(X, c, f1(true)))
active1(if3(true, X, Y)) -> mark1(X)
active1(if3(false, X, Y)) -> mark1(Y)
active1(f1(X)) -> f1(active1(X))
active1(if3(X1, X2, X3)) -> if3(active1(X1), X2, X3)
active1(if3(X1, X2, X3)) -> if3(X1, active1(X2), X3)
f1(mark1(X)) -> mark1(f1(X))
if3(mark1(X1), X2, X3) -> mark1(if3(X1, X2, X3))
if3(X1, mark1(X2), X3) -> mark1(if3(X1, X2, X3))
proper1(f1(X)) -> f1(proper1(X))
proper1(if3(X1, X2, X3)) -> if3(proper1(X1), proper1(X2), proper1(X3))
proper1(c) -> ok1(c)
proper1(true) -> ok1(true)
proper1(false) -> ok1(false)
f1(ok1(X)) -> ok1(f1(X))
if3(ok1(X1), ok1(X2), ok1(X3)) -> ok1(if3(X1, X2, X3))
top1(mark1(X)) -> top1(proper1(X))
top1(ok1(X)) -> top1(active1(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
      ↳ DependencyGraphProof
        ↳ AND
          ↳ QDP
          ↳ QDP
            ↳ QDPOrderProof
          ↳ QDP
          ↳ QDP
          ↳ QDP

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

F1(mark1(X)) -> F1(X)
F1(ok1(X)) -> F1(X)

The TRS R consists of the following rules:

active1(f1(X)) -> mark1(if3(X, c, f1(true)))
active1(if3(true, X, Y)) -> mark1(X)
active1(if3(false, X, Y)) -> mark1(Y)
active1(f1(X)) -> f1(active1(X))
active1(if3(X1, X2, X3)) -> if3(active1(X1), X2, X3)
active1(if3(X1, X2, X3)) -> if3(X1, active1(X2), X3)
f1(mark1(X)) -> mark1(f1(X))
if3(mark1(X1), X2, X3) -> mark1(if3(X1, X2, X3))
if3(X1, mark1(X2), X3) -> mark1(if3(X1, X2, X3))
proper1(f1(X)) -> f1(proper1(X))
proper1(if3(X1, X2, X3)) -> if3(proper1(X1), proper1(X2), proper1(X3))
proper1(c) -> ok1(c)
proper1(true) -> ok1(true)
proper1(false) -> ok1(false)
f1(ok1(X)) -> ok1(f1(X))
if3(ok1(X1), ok1(X2), ok1(X3)) -> ok1(if3(X1, X2, X3))
top1(mark1(X)) -> top1(proper1(X))
top1(ok1(X)) -> top1(active1(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.


F1(mark1(X)) -> F1(X)
F1(ok1(X)) -> F1(X)
The remaining pairs can at least be oriented weakly.
none
Used ordering: Polynomial Order [17,21] with Interpretation:

POL( F1(x1) ) = 2x1


POL( mark1(x1) ) = 3x1 + 1


POL( ok1(x1) ) = 2x1 + 2



The following usable rules [14] were oriented: none



↳ QTRS
  ↳ DependencyPairsProof
    ↳ QDP
      ↳ DependencyGraphProof
        ↳ AND
          ↳ QDP
          ↳ QDP
            ↳ QDPOrderProof
              ↳ QDP
                ↳ PisEmptyProof
          ↳ QDP
          ↳ QDP
          ↳ QDP

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

active1(f1(X)) -> mark1(if3(X, c, f1(true)))
active1(if3(true, X, Y)) -> mark1(X)
active1(if3(false, X, Y)) -> mark1(Y)
active1(f1(X)) -> f1(active1(X))
active1(if3(X1, X2, X3)) -> if3(active1(X1), X2, X3)
active1(if3(X1, X2, X3)) -> if3(X1, active1(X2), X3)
f1(mark1(X)) -> mark1(f1(X))
if3(mark1(X1), X2, X3) -> mark1(if3(X1, X2, X3))
if3(X1, mark1(X2), X3) -> mark1(if3(X1, X2, X3))
proper1(f1(X)) -> f1(proper1(X))
proper1(if3(X1, X2, X3)) -> if3(proper1(X1), proper1(X2), proper1(X3))
proper1(c) -> ok1(c)
proper1(true) -> ok1(true)
proper1(false) -> ok1(false)
f1(ok1(X)) -> ok1(f1(X))
if3(ok1(X1), ok1(X2), ok1(X3)) -> ok1(if3(X1, X2, X3))
top1(mark1(X)) -> top1(proper1(X))
top1(ok1(X)) -> top1(active1(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
      ↳ DependencyGraphProof
        ↳ AND
          ↳ QDP
          ↳ QDP
          ↳ QDP
            ↳ QDPOrderProof
          ↳ QDP
          ↳ QDP

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

PROPER1(f1(X)) -> PROPER1(X)
PROPER1(if3(X1, X2, X3)) -> PROPER1(X3)
PROPER1(if3(X1, X2, X3)) -> PROPER1(X1)
PROPER1(if3(X1, X2, X3)) -> PROPER1(X2)

The TRS R consists of the following rules:

active1(f1(X)) -> mark1(if3(X, c, f1(true)))
active1(if3(true, X, Y)) -> mark1(X)
active1(if3(false, X, Y)) -> mark1(Y)
active1(f1(X)) -> f1(active1(X))
active1(if3(X1, X2, X3)) -> if3(active1(X1), X2, X3)
active1(if3(X1, X2, X3)) -> if3(X1, active1(X2), X3)
f1(mark1(X)) -> mark1(f1(X))
if3(mark1(X1), X2, X3) -> mark1(if3(X1, X2, X3))
if3(X1, mark1(X2), X3) -> mark1(if3(X1, X2, X3))
proper1(f1(X)) -> f1(proper1(X))
proper1(if3(X1, X2, X3)) -> if3(proper1(X1), proper1(X2), proper1(X3))
proper1(c) -> ok1(c)
proper1(true) -> ok1(true)
proper1(false) -> ok1(false)
f1(ok1(X)) -> ok1(f1(X))
if3(ok1(X1), ok1(X2), ok1(X3)) -> ok1(if3(X1, X2, X3))
top1(mark1(X)) -> top1(proper1(X))
top1(ok1(X)) -> top1(active1(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.


PROPER1(f1(X)) -> PROPER1(X)
PROPER1(if3(X1, X2, X3)) -> PROPER1(X3)
PROPER1(if3(X1, X2, X3)) -> PROPER1(X1)
PROPER1(if3(X1, X2, X3)) -> PROPER1(X2)
The remaining pairs can at least be oriented weakly.
none
Used ordering: Polynomial Order [17,21] with Interpretation:

POL( PROPER1(x1) ) = max{0, 2x1 - 1}


POL( if3(x1, ..., x3) ) = x1 + x2 + 3x3 + 3


POL( f1(x1) ) = 2x1 + 3



The following usable rules [14] were oriented: none



↳ QTRS
  ↳ DependencyPairsProof
    ↳ QDP
      ↳ DependencyGraphProof
        ↳ AND
          ↳ QDP
          ↳ QDP
          ↳ QDP
            ↳ QDPOrderProof
              ↳ QDP
                ↳ PisEmptyProof
          ↳ QDP
          ↳ QDP

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

active1(f1(X)) -> mark1(if3(X, c, f1(true)))
active1(if3(true, X, Y)) -> mark1(X)
active1(if3(false, X, Y)) -> mark1(Y)
active1(f1(X)) -> f1(active1(X))
active1(if3(X1, X2, X3)) -> if3(active1(X1), X2, X3)
active1(if3(X1, X2, X3)) -> if3(X1, active1(X2), X3)
f1(mark1(X)) -> mark1(f1(X))
if3(mark1(X1), X2, X3) -> mark1(if3(X1, X2, X3))
if3(X1, mark1(X2), X3) -> mark1(if3(X1, X2, X3))
proper1(f1(X)) -> f1(proper1(X))
proper1(if3(X1, X2, X3)) -> if3(proper1(X1), proper1(X2), proper1(X3))
proper1(c) -> ok1(c)
proper1(true) -> ok1(true)
proper1(false) -> ok1(false)
f1(ok1(X)) -> ok1(f1(X))
if3(ok1(X1), ok1(X2), ok1(X3)) -> ok1(if3(X1, X2, X3))
top1(mark1(X)) -> top1(proper1(X))
top1(ok1(X)) -> top1(active1(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
      ↳ DependencyGraphProof
        ↳ AND
          ↳ QDP
          ↳ QDP
          ↳ QDP
          ↳ QDP
            ↳ QDPOrderProof
          ↳ QDP

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

ACTIVE1(f1(X)) -> ACTIVE1(X)
ACTIVE1(if3(X1, X2, X3)) -> ACTIVE1(X1)
ACTIVE1(if3(X1, X2, X3)) -> ACTIVE1(X2)

The TRS R consists of the following rules:

active1(f1(X)) -> mark1(if3(X, c, f1(true)))
active1(if3(true, X, Y)) -> mark1(X)
active1(if3(false, X, Y)) -> mark1(Y)
active1(f1(X)) -> f1(active1(X))
active1(if3(X1, X2, X3)) -> if3(active1(X1), X2, X3)
active1(if3(X1, X2, X3)) -> if3(X1, active1(X2), X3)
f1(mark1(X)) -> mark1(f1(X))
if3(mark1(X1), X2, X3) -> mark1(if3(X1, X2, X3))
if3(X1, mark1(X2), X3) -> mark1(if3(X1, X2, X3))
proper1(f1(X)) -> f1(proper1(X))
proper1(if3(X1, X2, X3)) -> if3(proper1(X1), proper1(X2), proper1(X3))
proper1(c) -> ok1(c)
proper1(true) -> ok1(true)
proper1(false) -> ok1(false)
f1(ok1(X)) -> ok1(f1(X))
if3(ok1(X1), ok1(X2), ok1(X3)) -> ok1(if3(X1, X2, X3))
top1(mark1(X)) -> top1(proper1(X))
top1(ok1(X)) -> top1(active1(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.


ACTIVE1(f1(X)) -> ACTIVE1(X)
ACTIVE1(if3(X1, X2, X3)) -> ACTIVE1(X1)
ACTIVE1(if3(X1, X2, X3)) -> ACTIVE1(X2)
The remaining pairs can at least be oriented weakly.
none
Used ordering: Polynomial Order [17,21] with Interpretation:

POL( ACTIVE1(x1) ) = x1 + 2


POL( if3(x1, ..., x3) ) = 2x1 + x2 + 3x3 + 3


POL( f1(x1) ) = 2x1 + 3



The following usable rules [14] were oriented: none



↳ QTRS
  ↳ DependencyPairsProof
    ↳ QDP
      ↳ DependencyGraphProof
        ↳ AND
          ↳ QDP
          ↳ QDP
          ↳ QDP
          ↳ QDP
            ↳ QDPOrderProof
              ↳ QDP
                ↳ PisEmptyProof
          ↳ QDP

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

active1(f1(X)) -> mark1(if3(X, c, f1(true)))
active1(if3(true, X, Y)) -> mark1(X)
active1(if3(false, X, Y)) -> mark1(Y)
active1(f1(X)) -> f1(active1(X))
active1(if3(X1, X2, X3)) -> if3(active1(X1), X2, X3)
active1(if3(X1, X2, X3)) -> if3(X1, active1(X2), X3)
f1(mark1(X)) -> mark1(f1(X))
if3(mark1(X1), X2, X3) -> mark1(if3(X1, X2, X3))
if3(X1, mark1(X2), X3) -> mark1(if3(X1, X2, X3))
proper1(f1(X)) -> f1(proper1(X))
proper1(if3(X1, X2, X3)) -> if3(proper1(X1), proper1(X2), proper1(X3))
proper1(c) -> ok1(c)
proper1(true) -> ok1(true)
proper1(false) -> ok1(false)
f1(ok1(X)) -> ok1(f1(X))
if3(ok1(X1), ok1(X2), ok1(X3)) -> ok1(if3(X1, X2, X3))
top1(mark1(X)) -> top1(proper1(X))
top1(ok1(X)) -> top1(active1(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
      ↳ DependencyGraphProof
        ↳ AND
          ↳ QDP
          ↳ QDP
          ↳ QDP
          ↳ QDP
          ↳ QDP

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

TOP1(ok1(X)) -> TOP1(active1(X))
TOP1(mark1(X)) -> TOP1(proper1(X))

The TRS R consists of the following rules:

active1(f1(X)) -> mark1(if3(X, c, f1(true)))
active1(if3(true, X, Y)) -> mark1(X)
active1(if3(false, X, Y)) -> mark1(Y)
active1(f1(X)) -> f1(active1(X))
active1(if3(X1, X2, X3)) -> if3(active1(X1), X2, X3)
active1(if3(X1, X2, X3)) -> if3(X1, active1(X2), X3)
f1(mark1(X)) -> mark1(f1(X))
if3(mark1(X1), X2, X3) -> mark1(if3(X1, X2, X3))
if3(X1, mark1(X2), X3) -> mark1(if3(X1, X2, X3))
proper1(f1(X)) -> f1(proper1(X))
proper1(if3(X1, X2, X3)) -> if3(proper1(X1), proper1(X2), proper1(X3))
proper1(c) -> ok1(c)
proper1(true) -> ok1(true)
proper1(false) -> ok1(false)
f1(ok1(X)) -> ok1(f1(X))
if3(ok1(X1), ok1(X2), ok1(X3)) -> ok1(if3(X1, X2, X3))
top1(mark1(X)) -> top1(proper1(X))
top1(ok1(X)) -> top1(active1(X))

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