Termination w.r.t. Q proof of /home/nowonder/forschung/aprove/satrung/aprove/lpar4/TRCSRSLASHEx5_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.

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 contains 5 SCCs with 9 less nodes.

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ DependencyGraphProof

        ↳ AND

          ↳ QDP

            ↳ QDPAfsSolverProof

          ↳ 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.
By using an argument filtering and a montonic ordering, at least one Dependency Pair of this SCC can be strictly oriented.

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

Used argument filtering: IF₃(x1, x2, x3) = x3
ok₁(x1) = ok₁(x1)
Used ordering: Quasi Precedence: trivial

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ DependencyGraphProof

        ↳ AND

          ↳ QDP

            ↳ QDPAfsSolverProof

              ↳ QDP

                ↳ QDPAfsSolverProof

          ↳ 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)

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.
By using an argument filtering and a montonic ordering, at least one Dependency Pair of this SCC can be strictly oriented.

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

Used argument filtering: IF₃(x1, x2, x3) = x2
mark₁(x1) = mark₁(x1)
Used ordering: Quasi Precedence: trivial

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ DependencyGraphProof

        ↳ AND

          ↳ QDP

            ↳ QDPAfsSolverProof

              ↳ QDP

                ↳ QDPAfsSolverProof

                  ↳ QDP

                    ↳ QDPAfsSolverProof

          ↳ QDP

          ↳ QDP

          ↳ QDP

          ↳ QDP

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

IF₃(mark₁(X1), X2, 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.
By using an argument filtering and a montonic ordering, at least one Dependency Pair of this SCC can be strictly oriented.

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

Used argument filtering: IF₃(x1, x2, x3) = x1
mark₁(x1) = mark₁(x1)
Used ordering: Quasi Precedence: trivial

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ DependencyGraphProof

        ↳ AND

          ↳ QDP

            ↳ QDPAfsSolverProof

              ↳ QDP

                ↳ QDPAfsSolverProof

                  ↳ QDP

                    ↳ QDPAfsSolverProof

                      ↳ 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

            ↳ QDPAfsSolverProof

          ↳ 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.
By using an argument filtering and a montonic ordering, at least one Dependency Pair of this SCC can be strictly oriented.

F₁(ok₁(X)) -> F₁(X)

Used argument filtering: F₁(x1) = x1
mark₁(x1) = x1
ok₁(x1) = ok₁(x1)
Used ordering: Quasi Precedence: trivial

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ DependencyGraphProof

        ↳ AND

          ↳ QDP

          ↳ QDP

            ↳ QDPAfsSolverProof

              ↳ QDP

                ↳ QDPAfsSolverProof

          ↳ QDP

          ↳ QDP

          ↳ QDP

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

F₁(mark₁(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.
By using an argument filtering and a montonic ordering, at least one Dependency Pair of this SCC can be strictly oriented.

F₁(mark₁(X)) -> F₁(X)

Used argument filtering: F₁(x1) = x1
mark₁(x1) = mark₁(x1)
Used ordering: Quasi Precedence: trivial

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ DependencyGraphProof

        ↳ AND

          ↳ QDP

          ↳ QDP

            ↳ QDPAfsSolverProof

              ↳ QDP

                ↳ QDPAfsSolverProof

                  ↳ 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

            ↳ QDPAfsSolverProof

          ↳ 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.
By using an argument filtering and a montonic ordering, at least one Dependency Pair of this SCC can be strictly oriented.

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

Used argument filtering: PROPER₁(x1) = x1
f₁(x1) = x1
if₃(x1, x2, x3) = if₃(x1, x2, x3)
Used ordering: Quasi Precedence: trivial

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ DependencyGraphProof

        ↳ AND

          ↳ QDP

          ↳ QDP

          ↳ QDP

            ↳ QDPAfsSolverProof

              ↳ QDP

                ↳ QDPAfsSolverProof

          ↳ QDP

          ↳ QDP

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

PROPER₁(f₁(X)) -> 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.
By using an argument filtering and a montonic ordering, at least one Dependency Pair of this SCC can be strictly oriented.

PROPER₁(f₁(X)) -> PROPER₁(X)

Used argument filtering: PROPER₁(x1) = x1
f₁(x1) = f₁(x1)
Used ordering: Quasi Precedence: trivial

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ DependencyGraphProof

        ↳ AND

          ↳ QDP

          ↳ QDP

          ↳ QDP

            ↳ QDPAfsSolverProof

              ↳ QDP

                ↳ QDPAfsSolverProof

                  ↳ 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

            ↳ QDPAfsSolverProof

          ↳ 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.
By using an argument filtering and a montonic ordering, at least one Dependency Pair of this SCC can be strictly oriented.

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

Used argument filtering: ACTIVE₁(x1) = x1
f₁(x1) = x1
if₃(x1, x2, x3) = if₂(x1, x2)
Used ordering: Quasi Precedence: trivial

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ DependencyGraphProof

        ↳ AND

          ↳ QDP

          ↳ QDP

          ↳ QDP

          ↳ QDP

            ↳ QDPAfsSolverProof

              ↳ QDP

                ↳ QDPAfsSolverProof

          ↳ QDP

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

ACTIVE₁(f₁(X)) -> ACTIVE₁(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.
By using an argument filtering and a montonic ordering, at least one Dependency Pair of this SCC can be strictly oriented.

ACTIVE₁(f₁(X)) -> ACTIVE₁(X)

Used argument filtering: ACTIVE₁(x1) = x1
f₁(x1) = f₁(x1)
Used ordering: Quasi Precedence: trivial

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ DependencyGraphProof

        ↳ AND

          ↳ QDP

          ↳ QDP

          ↳ QDP

          ↳ QDP

            ↳ QDPAfsSolverProof

              ↳ QDP

                ↳ QDPAfsSolverProof

                  ↳ 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.