Termination w.r.t. Q proof of TRS/secret05aprove2.trs

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:

ge₂(x, 0) -> true
ge₂(0, s₁(x)) -> false
ge₂(s₁(x), s₁(y)) -> ge₂(x, y)
minus₂(x, 0) -> x
minus₂(s₁(x), s₁(y)) -> minus₂(x, y)
div₂(x, y) -> ify₃(ge₂(y, s₁(0)), x, y)
ify₃(false, x, y) -> divByZeroError
ify₃(true, x, y) -> if₃(ge₂(x, y), x, y)
if₃(false, x, y) -> 0
if₃(true, x, y) -> s₁(div₂(minus₂(x, y), y))

Q is empty.

↳ QTRS

  ↳ DependencyPairsProof

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

ge₂(x, 0) -> true
ge₂(0, s₁(x)) -> false
ge₂(s₁(x), s₁(y)) -> ge₂(x, y)
minus₂(x, 0) -> x
minus₂(s₁(x), s₁(y)) -> minus₂(x, y)
div₂(x, y) -> ify₃(ge₂(y, s₁(0)), x, y)
ify₃(false, x, y) -> divByZeroError
ify₃(true, x, y) -> if₃(ge₂(x, y), x, y)
if₃(false, x, y) -> 0
if₃(true, x, y) -> s₁(div₂(minus₂(x, y), y))

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:

IF₃(true, x, y) -> DIV₂(minus₂(x, y), y)
IF₃(true, x, y) -> MINUS₂(x, y)
IFY₃(true, x, y) -> GE₂(x, y)
IFY₃(true, x, y) -> IF₃(ge₂(x, y), x, y)
DIV₂(x, y) -> IFY₃(ge₂(y, s₁(0)), x, y)
MINUS₂(s₁(x), s₁(y)) -> MINUS₂(x, y)
DIV₂(x, y) -> GE₂(y, s₁(0))
GE₂(s₁(x), s₁(y)) -> GE₂(x, y)

The TRS R consists of the following rules:

ge₂(x, 0) -> true
ge₂(0, s₁(x)) -> false
ge₂(s₁(x), s₁(y)) -> ge₂(x, y)
minus₂(x, 0) -> x
minus₂(s₁(x), s₁(y)) -> minus₂(x, y)
div₂(x, y) -> ify₃(ge₂(y, s₁(0)), x, y)
ify₃(false, x, y) -> divByZeroError
ify₃(true, x, y) -> if₃(ge₂(x, y), x, y)
if₃(false, x, y) -> 0
if₃(true, x, y) -> s₁(div₂(minus₂(x, y), y))

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:

IF₃(true, x, y) -> DIV₂(minus₂(x, y), y)
IF₃(true, x, y) -> MINUS₂(x, y)
IFY₃(true, x, y) -> GE₂(x, y)
IFY₃(true, x, y) -> IF₃(ge₂(x, y), x, y)
DIV₂(x, y) -> IFY₃(ge₂(y, s₁(0)), x, y)
MINUS₂(s₁(x), s₁(y)) -> MINUS₂(x, y)
DIV₂(x, y) -> GE₂(y, s₁(0))
GE₂(s₁(x), s₁(y)) -> GE₂(x, y)

The TRS R consists of the following rules:

ge₂(x, 0) -> true
ge₂(0, s₁(x)) -> false
ge₂(s₁(x), s₁(y)) -> ge₂(x, y)
minus₂(x, 0) -> x
minus₂(s₁(x), s₁(y)) -> minus₂(x, y)
div₂(x, y) -> ify₃(ge₂(y, s₁(0)), x, y)
ify₃(false, x, y) -> divByZeroError
ify₃(true, x, y) -> if₃(ge₂(x, y), x, y)
if₃(false, x, y) -> 0
if₃(true, x, y) -> s₁(div₂(minus₂(x, y), y))

Q is empty.
We have to consider all minimal (P,Q,R)-chains.
The approximation of the Dependency Graph [13,14,18] contains 3 SCCs with 3 less nodes.

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ DependencyGraphProof

        ↳ AND

          ↳ QDP

            ↳ QDPOrderProof

          ↳ QDP

          ↳ QDP

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

MINUS₂(s₁(x), s₁(y)) -> MINUS₂(x, y)

The TRS R consists of the following rules:

ge₂(x, 0) -> true
ge₂(0, s₁(x)) -> false
ge₂(s₁(x), s₁(y)) -> ge₂(x, y)
minus₂(x, 0) -> x
minus₂(s₁(x), s₁(y)) -> minus₂(x, y)
div₂(x, y) -> ify₃(ge₂(y, s₁(0)), x, y)
ify₃(false, x, y) -> divByZeroError
ify₃(true, x, y) -> if₃(ge₂(x, y), x, y)
if₃(false, x, y) -> 0
if₃(true, x, y) -> s₁(div₂(minus₂(x, y), y))

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.

MINUS₂(s₁(x), s₁(y)) -> MINUS₂(x, y)

The remaining pairs can at least be oriented weakly.
none
Used ordering: Polynomial interpretation [21]:

POL(MINUS₂(x₁, x₂)) = 2·x₁ + 2·x₂
POL(s₁(x₁)) = 2 + x₁

The following usable rules [14] were oriented: none

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ DependencyGraphProof

        ↳ AND

          ↳ QDP

            ↳ QDPOrderProof

              ↳ QDP

                ↳ PisEmptyProof

          ↳ QDP

          ↳ QDP

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

ge₂(x, 0) -> true
ge₂(0, s₁(x)) -> false
ge₂(s₁(x), s₁(y)) -> ge₂(x, y)
minus₂(x, 0) -> x
minus₂(s₁(x), s₁(y)) -> minus₂(x, y)
div₂(x, y) -> ify₃(ge₂(y, s₁(0)), x, y)
ify₃(false, x, y) -> divByZeroError
ify₃(true, x, y) -> if₃(ge₂(x, y), x, y)
if₃(false, x, y) -> 0
if₃(true, x, y) -> s₁(div₂(minus₂(x, y), y))

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

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

GE₂(s₁(x), s₁(y)) -> GE₂(x, y)

The TRS R consists of the following rules:

ge₂(x, 0) -> true
ge₂(0, s₁(x)) -> false
ge₂(s₁(x), s₁(y)) -> ge₂(x, y)
minus₂(x, 0) -> x
minus₂(s₁(x), s₁(y)) -> minus₂(x, y)
div₂(x, y) -> ify₃(ge₂(y, s₁(0)), x, y)
ify₃(false, x, y) -> divByZeroError
ify₃(true, x, y) -> if₃(ge₂(x, y), x, y)
if₃(false, x, y) -> 0
if₃(true, x, y) -> s₁(div₂(minus₂(x, y), y))

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.

GE₂(s₁(x), s₁(y)) -> GE₂(x, y)

The remaining pairs can at least be oriented weakly.
none
Used ordering: Polynomial interpretation [21]:

POL(GE₂(x₁, x₂)) = 2·x₁ + 2·x₂
POL(s₁(x₁)) = 2 + x₁

The following usable rules [14] were oriented: none

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ DependencyGraphProof

        ↳ AND

          ↳ QDP

          ↳ QDP

            ↳ QDPOrderProof

              ↳ QDP

                ↳ PisEmptyProof

          ↳ QDP

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

ge₂(x, 0) -> true
ge₂(0, s₁(x)) -> false
ge₂(s₁(x), s₁(y)) -> ge₂(x, y)
minus₂(x, 0) -> x
minus₂(s₁(x), s₁(y)) -> minus₂(x, y)
div₂(x, y) -> ify₃(ge₂(y, s₁(0)), x, y)
ify₃(false, x, y) -> divByZeroError
ify₃(true, x, y) -> if₃(ge₂(x, y), x, y)
if₃(false, x, y) -> 0
if₃(true, x, y) -> s₁(div₂(minus₂(x, y), y))

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

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

IF₃(true, x, y) -> DIV₂(minus₂(x, y), y)
IFY₃(true, x, y) -> IF₃(ge₂(x, y), x, y)
DIV₂(x, y) -> IFY₃(ge₂(y, s₁(0)), x, y)

The TRS R consists of the following rules:

ge₂(x, 0) -> true
ge₂(0, s₁(x)) -> false
ge₂(s₁(x), s₁(y)) -> ge₂(x, y)
minus₂(x, 0) -> x
minus₂(s₁(x), s₁(y)) -> minus₂(x, y)
div₂(x, y) -> ify₃(ge₂(y, s₁(0)), x, y)
ify₃(false, x, y) -> divByZeroError
ify₃(true, x, y) -> if₃(ge₂(x, y), x, y)
if₃(false, x, y) -> 0
if₃(true, x, y) -> s₁(div₂(minus₂(x, y), y))

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