Termination w.r.t. Q proof of TRS/TRCSREx5_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:

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 interpretation [21]:

POL(IF₃(x₁, x₂, x₃)) = x₁ + 2·x₂ + x₃
POL(mark₁(x₁)) = 1 + 2·x₁
POL(ok₁(x₁)) = 1 + 2·x₁

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 interpretation [21]:

POL(F₁(x₁)) = 2·x₁
POL(mark₁(x₁)) = 2 + 2·x₁
POL(ok₁(x₁)) = 2 + 2·x₁

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 interpretation [21]:

POL(PROPER₁(x₁)) = 2·x₁
POL(f₁(x₁)) = 2 + 2·x₁
POL(if₃(x₁, x₂, x₃)) = 2 + 2·x₁ + 2·x₂ + 2·x₃

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 interpretation [21]:

POL(ACTIVE₁(x₁)) = 2·x₁
POL(f₁(x₁)) = 2 + 2·x₁
POL(if₃(x₁, x₂, x₃)) = 2 + 2·x₁ + 2·x₂

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.