Termination w.r.t. Q proof of TRS/Thiemanndiv

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(0, 0) -> true
ge₂(s₁(x), 0) -> ge₂(x, 0)
ge₂(0, s₁(0)) -> false
ge₂(0, s₁(s₁(x))) -> ge₂(0, s₁(x))
ge₂(s₁(x), s₁(y)) -> ge₂(x, y)
minus₂(0, 0) -> 0
minus₂(0, s₁(x)) -> minus₂(0, x)
minus₂(s₁(x), 0) -> s₁(minus₂(x, 0))
minus₂(s₁(x), s₁(y)) -> minus₂(x, y)
plus₂(0, 0) -> 0
plus₂(0, s₁(x)) -> s₁(plus₂(0, x))
plus₂(s₁(x), y) -> s₁(plus₂(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))
div₂(plus₂(x, y), z) -> plus₂(div₂(x, z), div₂(y, z))

Q is empty.

↳ QTRS

  ↳ DependencyPairsProof

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

ge₂(0, 0) -> true
ge₂(s₁(x), 0) -> ge₂(x, 0)
ge₂(0, s₁(0)) -> false
ge₂(0, s₁(s₁(x))) -> ge₂(0, s₁(x))
ge₂(s₁(x), s₁(y)) -> ge₂(x, y)
minus₂(0, 0) -> 0
minus₂(0, s₁(x)) -> minus₂(0, x)
minus₂(s₁(x), 0) -> s₁(minus₂(x, 0))
minus₂(s₁(x), s₁(y)) -> minus₂(x, y)
plus₂(0, 0) -> 0
plus₂(0, s₁(x)) -> s₁(plus₂(0, x))
plus₂(s₁(x), y) -> s₁(plus₂(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))
div₂(plus₂(x, y), z) -> plus₂(div₂(x, z), div₂(y, z))

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:

DIV₂(plus₂(x, y), z) -> PLUS₂(div₂(x, z), div₂(y, z))
PLUS₂(0, s₁(x)) -> PLUS₂(0, x)
GE₂(s₁(x), 0) -> GE₂(x, 0)
IFY₃(true, x, y) -> GE₂(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(0, s₁(s₁(x))) -> GE₂(0, s₁(x))
IF₃(true, x, y) -> DIV₂(minus₂(x, y), y)
PLUS₂(s₁(x), y) -> PLUS₂(x, y)
MINUS₂(0, s₁(x)) -> MINUS₂(0, x)
MINUS₂(s₁(x), 0) -> MINUS₂(x, 0)
IF₃(true, x, y) -> MINUS₂(x, y)
IFY₃(true, x, y) -> IF₃(ge₂(x, y), x, y)
DIV₂(plus₂(x, y), z) -> DIV₂(x, z)
DIV₂(plus₂(x, y), z) -> DIV₂(y, z)
GE₂(s₁(x), s₁(y)) -> GE₂(x, y)

The TRS R consists of the following rules:

ge₂(0, 0) -> true
ge₂(s₁(x), 0) -> ge₂(x, 0)
ge₂(0, s₁(0)) -> false
ge₂(0, s₁(s₁(x))) -> ge₂(0, s₁(x))
ge₂(s₁(x), s₁(y)) -> ge₂(x, y)
minus₂(0, 0) -> 0
minus₂(0, s₁(x)) -> minus₂(0, x)
minus₂(s₁(x), 0) -> s₁(minus₂(x, 0))
minus₂(s₁(x), s₁(y)) -> minus₂(x, y)
plus₂(0, 0) -> 0
plus₂(0, s₁(x)) -> s₁(plus₂(0, x))
plus₂(s₁(x), y) -> s₁(plus₂(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))
div₂(plus₂(x, y), z) -> plus₂(div₂(x, z), div₂(y, z))

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:

DIV₂(plus₂(x, y), z) -> PLUS₂(div₂(x, z), div₂(y, z))
PLUS₂(0, s₁(x)) -> PLUS₂(0, x)
GE₂(s₁(x), 0) -> GE₂(x, 0)
IFY₃(true, x, y) -> GE₂(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(0, s₁(s₁(x))) -> GE₂(0, s₁(x))
IF₃(true, x, y) -> DIV₂(minus₂(x, y), y)
PLUS₂(s₁(x), y) -> PLUS₂(x, y)
MINUS₂(0, s₁(x)) -> MINUS₂(0, x)
MINUS₂(s₁(x), 0) -> MINUS₂(x, 0)
IF₃(true, x, y) -> MINUS₂(x, y)
IFY₃(true, x, y) -> IF₃(ge₂(x, y), x, y)
DIV₂(plus₂(x, y), z) -> DIV₂(x, z)
DIV₂(plus₂(x, y), z) -> DIV₂(y, z)
GE₂(s₁(x), s₁(y)) -> GE₂(x, y)

The TRS R consists of the following rules:

ge₂(0, 0) -> true
ge₂(s₁(x), 0) -> ge₂(x, 0)
ge₂(0, s₁(0)) -> false
ge₂(0, s₁(s₁(x))) -> ge₂(0, s₁(x))
ge₂(s₁(x), s₁(y)) -> ge₂(x, y)
minus₂(0, 0) -> 0
minus₂(0, s₁(x)) -> minus₂(0, x)
minus₂(s₁(x), 0) -> s₁(minus₂(x, 0))
minus₂(s₁(x), s₁(y)) -> minus₂(x, y)
plus₂(0, 0) -> 0
plus₂(0, s₁(x)) -> s₁(plus₂(0, x))
plus₂(s₁(x), y) -> s₁(plus₂(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))
div₂(plus₂(x, y), z) -> plus₂(div₂(x, z), div₂(y, z))

Q is empty.
We have to consider all minimal (P,Q,R)-chains.
The approximation of the Dependency Graph [13,14,18] contains 9 SCCs with 4 less nodes.

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ DependencyGraphProof

        ↳ AND

          ↳ QDP

            ↳ QDPOrderProof

          ↳ QDP

          ↳ QDP

          ↳ QDP

          ↳ QDP

          ↳ QDP

          ↳ QDP

          ↳ QDP

          ↳ QDP

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

PLUS₂(0, s₁(x)) -> PLUS₂(0, x)

The TRS R consists of the following rules:

ge₂(0, 0) -> true
ge₂(s₁(x), 0) -> ge₂(x, 0)
ge₂(0, s₁(0)) -> false
ge₂(0, s₁(s₁(x))) -> ge₂(0, s₁(x))
ge₂(s₁(x), s₁(y)) -> ge₂(x, y)
minus₂(0, 0) -> 0
minus₂(0, s₁(x)) -> minus₂(0, x)
minus₂(s₁(x), 0) -> s₁(minus₂(x, 0))
minus₂(s₁(x), s₁(y)) -> minus₂(x, y)
plus₂(0, 0) -> 0
plus₂(0, s₁(x)) -> s₁(plus₂(0, x))
plus₂(s₁(x), y) -> s₁(plus₂(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))
div₂(plus₂(x, y), z) -> plus₂(div₂(x, z), div₂(y, z))

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.

PLUS₂(0, s₁(x)) -> PLUS₂(0, x)

The remaining pairs can at least be oriented weakly.
none
Used ordering: Polynomial Order [17,21] with Interpretation:

POL( 0 ) = 3

POL( s₁(x₁) ) = x₁ + 2

POL( PLUS₂(x₁, x₂) ) = 2x₁ + 3x₂

The following usable rules [14] were oriented: none

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ DependencyGraphProof

        ↳ AND

          ↳ QDP

            ↳ QDPOrderProof

              ↳ QDP

                ↳ PisEmptyProof

          ↳ QDP

          ↳ QDP

          ↳ QDP

          ↳ QDP

          ↳ QDP

          ↳ QDP

          ↳ QDP

          ↳ QDP

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

ge₂(0, 0) -> true
ge₂(s₁(x), 0) -> ge₂(x, 0)
ge₂(0, s₁(0)) -> false
ge₂(0, s₁(s₁(x))) -> ge₂(0, s₁(x))
ge₂(s₁(x), s₁(y)) -> ge₂(x, y)
minus₂(0, 0) -> 0
minus₂(0, s₁(x)) -> minus₂(0, x)
minus₂(s₁(x), 0) -> s₁(minus₂(x, 0))
minus₂(s₁(x), s₁(y)) -> minus₂(x, y)
plus₂(0, 0) -> 0
plus₂(0, s₁(x)) -> s₁(plus₂(0, x))
plus₂(s₁(x), y) -> s₁(plus₂(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))
div₂(plus₂(x, y), z) -> plus₂(div₂(x, z), div₂(y, z))

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

          ↳ QDP

          ↳ QDP

          ↳ QDP

          ↳ QDP

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

PLUS₂(s₁(x), y) -> PLUS₂(x, y)

The TRS R consists of the following rules:

ge₂(0, 0) -> true
ge₂(s₁(x), 0) -> ge₂(x, 0)
ge₂(0, s₁(0)) -> false
ge₂(0, s₁(s₁(x))) -> ge₂(0, s₁(x))
ge₂(s₁(x), s₁(y)) -> ge₂(x, y)
minus₂(0, 0) -> 0
minus₂(0, s₁(x)) -> minus₂(0, x)
minus₂(s₁(x), 0) -> s₁(minus₂(x, 0))
minus₂(s₁(x), s₁(y)) -> minus₂(x, y)
plus₂(0, 0) -> 0
plus₂(0, s₁(x)) -> s₁(plus₂(0, x))
plus₂(s₁(x), y) -> s₁(plus₂(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))
div₂(plus₂(x, y), z) -> plus₂(div₂(x, z), div₂(y, z))

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.

PLUS₂(s₁(x), y) -> PLUS₂(x, y)

The remaining pairs can at least be oriented weakly.
none
Used ordering: Polynomial Order [17,21] with Interpretation:

POL( s₁(x₁) ) = x₁ + 1

POL( PLUS₂(x₁, x₂) ) = 2x₁ + 3x₂ + 2

The following usable rules [14] were oriented: none

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ DependencyGraphProof

        ↳ AND

          ↳ QDP

          ↳ QDP

            ↳ QDPOrderProof

              ↳ QDP

                ↳ PisEmptyProof

          ↳ QDP

          ↳ QDP

          ↳ QDP

          ↳ QDP

          ↳ QDP

          ↳ QDP

          ↳ QDP

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

ge₂(0, 0) -> true
ge₂(s₁(x), 0) -> ge₂(x, 0)
ge₂(0, s₁(0)) -> false
ge₂(0, s₁(s₁(x))) -> ge₂(0, s₁(x))
ge₂(s₁(x), s₁(y)) -> ge₂(x, y)
minus₂(0, 0) -> 0
minus₂(0, s₁(x)) -> minus₂(0, x)
minus₂(s₁(x), 0) -> s₁(minus₂(x, 0))
minus₂(s₁(x), s₁(y)) -> minus₂(x, y)
plus₂(0, 0) -> 0
plus₂(0, s₁(x)) -> s₁(plus₂(0, x))
plus₂(s₁(x), y) -> s₁(plus₂(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))
div₂(plus₂(x, y), z) -> plus₂(div₂(x, z), div₂(y, z))

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

          ↳ QDP

          ↳ QDP

          ↳ QDP

          ↳ QDP

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

MINUS₂(s₁(x), 0) -> MINUS₂(x, 0)

The TRS R consists of the following rules:

ge₂(0, 0) -> true
ge₂(s₁(x), 0) -> ge₂(x, 0)
ge₂(0, s₁(0)) -> false
ge₂(0, s₁(s₁(x))) -> ge₂(0, s₁(x))
ge₂(s₁(x), s₁(y)) -> ge₂(x, y)
minus₂(0, 0) -> 0
minus₂(0, s₁(x)) -> minus₂(0, x)
minus₂(s₁(x), 0) -> s₁(minus₂(x, 0))
minus₂(s₁(x), s₁(y)) -> minus₂(x, y)
plus₂(0, 0) -> 0
plus₂(0, s₁(x)) -> s₁(plus₂(0, x))
plus₂(s₁(x), y) -> s₁(plus₂(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))
div₂(plus₂(x, y), z) -> plus₂(div₂(x, z), div₂(y, z))

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), 0) -> MINUS₂(x, 0)

The remaining pairs can at least be oriented weakly.
none
Used ordering: Polynomial Order [17,21] with Interpretation:

POL( MINUS₂(x₁, x₂) ) = 2x₁ + 3x₂

POL( s₁(x₁) ) = 3x₁ + 3

POL( 0 ) = 2

The following usable rules [14] were oriented: none

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ DependencyGraphProof

        ↳ AND

          ↳ QDP

          ↳ QDP

          ↳ QDP

            ↳ QDPOrderProof

              ↳ QDP

                ↳ PisEmptyProof

          ↳ QDP

          ↳ QDP

          ↳ QDP

          ↳ QDP

          ↳ QDP

          ↳ QDP

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

ge₂(0, 0) -> true
ge₂(s₁(x), 0) -> ge₂(x, 0)
ge₂(0, s₁(0)) -> false
ge₂(0, s₁(s₁(x))) -> ge₂(0, s₁(x))
ge₂(s₁(x), s₁(y)) -> ge₂(x, y)
minus₂(0, 0) -> 0
minus₂(0, s₁(x)) -> minus₂(0, x)
minus₂(s₁(x), 0) -> s₁(minus₂(x, 0))
minus₂(s₁(x), s₁(y)) -> minus₂(x, y)
plus₂(0, 0) -> 0
plus₂(0, s₁(x)) -> s₁(plus₂(0, x))
plus₂(s₁(x), y) -> s₁(plus₂(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))
div₂(plus₂(x, y), z) -> plus₂(div₂(x, z), div₂(y, z))

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

          ↳ QDP

          ↳ QDP

          ↳ QDP

          ↳ QDP

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

MINUS₂(0, s₁(x)) -> MINUS₂(0, x)

The TRS R consists of the following rules:

ge₂(0, 0) -> true
ge₂(s₁(x), 0) -> ge₂(x, 0)
ge₂(0, s₁(0)) -> false
ge₂(0, s₁(s₁(x))) -> ge₂(0, s₁(x))
ge₂(s₁(x), s₁(y)) -> ge₂(x, y)
minus₂(0, 0) -> 0
minus₂(0, s₁(x)) -> minus₂(0, x)
minus₂(s₁(x), 0) -> s₁(minus₂(x, 0))
minus₂(s₁(x), s₁(y)) -> minus₂(x, y)
plus₂(0, 0) -> 0
plus₂(0, s₁(x)) -> s₁(plus₂(0, x))
plus₂(s₁(x), y) -> s₁(plus₂(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))
div₂(plus₂(x, y), z) -> plus₂(div₂(x, z), div₂(y, z))

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(0, s₁(x)) -> MINUS₂(0, x)

The remaining pairs can at least be oriented weakly.
none
Used ordering: Polynomial Order [17,21] with Interpretation:

POL( MINUS₂(x₁, x₂) ) = 2x₁ + 3x₂

POL( 0 ) = 3

POL( s₁(x₁) ) = x₁ + 2

The following usable rules [14] were oriented: none

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ DependencyGraphProof

        ↳ AND

          ↳ QDP

          ↳ QDP

          ↳ QDP

          ↳ QDP

            ↳ QDPOrderProof

              ↳ QDP

                ↳ PisEmptyProof

          ↳ QDP

          ↳ QDP

          ↳ QDP

          ↳ QDP

          ↳ QDP

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

ge₂(0, 0) -> true
ge₂(s₁(x), 0) -> ge₂(x, 0)
ge₂(0, s₁(0)) -> false
ge₂(0, s₁(s₁(x))) -> ge₂(0, s₁(x))
ge₂(s₁(x), s₁(y)) -> ge₂(x, y)
minus₂(0, 0) -> 0
minus₂(0, s₁(x)) -> minus₂(0, x)
minus₂(s₁(x), 0) -> s₁(minus₂(x, 0))
minus₂(s₁(x), s₁(y)) -> minus₂(x, y)
plus₂(0, 0) -> 0
plus₂(0, s₁(x)) -> s₁(plus₂(0, x))
plus₂(s₁(x), y) -> s₁(plus₂(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))
div₂(plus₂(x, y), z) -> plus₂(div₂(x, z), div₂(y, z))

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

            ↳ QDPOrderProof

          ↳ QDP

          ↳ QDP

          ↳ 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(0, 0) -> true
ge₂(s₁(x), 0) -> ge₂(x, 0)
ge₂(0, s₁(0)) -> false
ge₂(0, s₁(s₁(x))) -> ge₂(0, s₁(x))
ge₂(s₁(x), s₁(y)) -> ge₂(x, y)
minus₂(0, 0) -> 0
minus₂(0, s₁(x)) -> minus₂(0, x)
minus₂(s₁(x), 0) -> s₁(minus₂(x, 0))
minus₂(s₁(x), s₁(y)) -> minus₂(x, y)
plus₂(0, 0) -> 0
plus₂(0, s₁(x)) -> s₁(plus₂(0, x))
plus₂(s₁(x), y) -> s₁(plus₂(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))
div₂(plus₂(x, y), z) -> plus₂(div₂(x, z), div₂(y, z))

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 Order [17,21] with Interpretation:

POL( MINUS₂(x₁, x₂) ) = 3x₂ + 3

POL( s₁(x₁) ) = x₁ + 3

The following usable rules [14] were oriented: none

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ DependencyGraphProof

        ↳ AND

          ↳ QDP

          ↳ QDP

          ↳ QDP

          ↳ QDP

          ↳ QDP

            ↳ QDPOrderProof

              ↳ QDP

                ↳ PisEmptyProof

          ↳ QDP

          ↳ QDP

          ↳ QDP

          ↳ QDP

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

ge₂(0, 0) -> true
ge₂(s₁(x), 0) -> ge₂(x, 0)
ge₂(0, s₁(0)) -> false
ge₂(0, s₁(s₁(x))) -> ge₂(0, s₁(x))
ge₂(s₁(x), s₁(y)) -> ge₂(x, y)
minus₂(0, 0) -> 0
minus₂(0, s₁(x)) -> minus₂(0, x)
minus₂(s₁(x), 0) -> s₁(minus₂(x, 0))
minus₂(s₁(x), s₁(y)) -> minus₂(x, y)
plus₂(0, 0) -> 0
plus₂(0, s₁(x)) -> s₁(plus₂(0, x))
plus₂(s₁(x), y) -> s₁(plus₂(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))
div₂(plus₂(x, y), z) -> plus₂(div₂(x, z), div₂(y, z))

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

          ↳ QDP

            ↳ QDPOrderProof

          ↳ QDP

          ↳ QDP

          ↳ QDP

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

GE₂(0, s₁(s₁(x))) -> GE₂(0, s₁(x))

The TRS R consists of the following rules:

ge₂(0, 0) -> true
ge₂(s₁(x), 0) -> ge₂(x, 0)
ge₂(0, s₁(0)) -> false
ge₂(0, s₁(s₁(x))) -> ge₂(0, s₁(x))
ge₂(s₁(x), s₁(y)) -> ge₂(x, y)
minus₂(0, 0) -> 0
minus₂(0, s₁(x)) -> minus₂(0, x)
minus₂(s₁(x), 0) -> s₁(minus₂(x, 0))
minus₂(s₁(x), s₁(y)) -> minus₂(x, y)
plus₂(0, 0) -> 0
plus₂(0, s₁(x)) -> s₁(plus₂(0, x))
plus₂(s₁(x), y) -> s₁(plus₂(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))
div₂(plus₂(x, y), z) -> plus₂(div₂(x, z), div₂(y, z))

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(0, s₁(s₁(x))) -> GE₂(0, s₁(x))

The remaining pairs can at least be oriented weakly.
none
Used ordering: Polynomial Order [17,21] with Interpretation:

POL( 0 ) = 1

POL( s₁(x₁) ) = 2x₁ + 2

POL( GE₂(x₁, x₂) ) = max{0, 2x₁ + x₂ - 3}

The following usable rules [14] were oriented: none

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ DependencyGraphProof

        ↳ AND

          ↳ QDP

          ↳ QDP

          ↳ QDP

          ↳ QDP

          ↳ QDP

          ↳ QDP

            ↳ QDPOrderProof

              ↳ QDP

                ↳ PisEmptyProof

          ↳ QDP

          ↳ QDP

          ↳ QDP

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

ge₂(0, 0) -> true
ge₂(s₁(x), 0) -> ge₂(x, 0)
ge₂(0, s₁(0)) -> false
ge₂(0, s₁(s₁(x))) -> ge₂(0, s₁(x))
ge₂(s₁(x), s₁(y)) -> ge₂(x, y)
minus₂(0, 0) -> 0
minus₂(0, s₁(x)) -> minus₂(0, x)
minus₂(s₁(x), 0) -> s₁(minus₂(x, 0))
minus₂(s₁(x), s₁(y)) -> minus₂(x, y)
plus₂(0, 0) -> 0
plus₂(0, s₁(x)) -> s₁(plus₂(0, x))
plus₂(s₁(x), y) -> s₁(plus₂(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))
div₂(plus₂(x, y), z) -> plus₂(div₂(x, z), div₂(y, z))

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

          ↳ QDP

          ↳ QDP

            ↳ QDPOrderProof

          ↳ QDP

          ↳ QDP

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

GE₂(s₁(x), 0) -> GE₂(x, 0)

The TRS R consists of the following rules:

ge₂(0, 0) -> true
ge₂(s₁(x), 0) -> ge₂(x, 0)
ge₂(0, s₁(0)) -> false
ge₂(0, s₁(s₁(x))) -> ge₂(0, s₁(x))
ge₂(s₁(x), s₁(y)) -> ge₂(x, y)
minus₂(0, 0) -> 0
minus₂(0, s₁(x)) -> minus₂(0, x)
minus₂(s₁(x), 0) -> s₁(minus₂(x, 0))
minus₂(s₁(x), s₁(y)) -> minus₂(x, y)
plus₂(0, 0) -> 0
plus₂(0, s₁(x)) -> s₁(plus₂(0, x))
plus₂(s₁(x), y) -> s₁(plus₂(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))
div₂(plus₂(x, y), z) -> plus₂(div₂(x, z), div₂(y, z))

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), 0) -> GE₂(x, 0)

The remaining pairs can at least be oriented weakly.
none
Used ordering: Polynomial Order [17,21] with Interpretation:

POL( s₁(x₁) ) = 3x₁ + 3

POL( 0 ) = 2

POL( GE₂(x₁, x₂) ) = 2x₁ + 3x₂

The following usable rules [14] were oriented: none

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ DependencyGraphProof

        ↳ AND

          ↳ QDP

          ↳ QDP

          ↳ QDP

          ↳ QDP

          ↳ QDP

          ↳ QDP

          ↳ QDP

            ↳ QDPOrderProof

              ↳ QDP

                ↳ PisEmptyProof

          ↳ QDP

          ↳ QDP

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

ge₂(0, 0) -> true
ge₂(s₁(x), 0) -> ge₂(x, 0)
ge₂(0, s₁(0)) -> false
ge₂(0, s₁(s₁(x))) -> ge₂(0, s₁(x))
ge₂(s₁(x), s₁(y)) -> ge₂(x, y)
minus₂(0, 0) -> 0
minus₂(0, s₁(x)) -> minus₂(0, x)
minus₂(s₁(x), 0) -> s₁(minus₂(x, 0))
minus₂(s₁(x), s₁(y)) -> minus₂(x, y)
plus₂(0, 0) -> 0
plus₂(0, s₁(x)) -> s₁(plus₂(0, x))
plus₂(s₁(x), y) -> s₁(plus₂(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))
div₂(plus₂(x, y), z) -> plus₂(div₂(x, z), div₂(y, z))

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

          ↳ QDP

          ↳ 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(0, 0) -> true
ge₂(s₁(x), 0) -> ge₂(x, 0)
ge₂(0, s₁(0)) -> false
ge₂(0, s₁(s₁(x))) -> ge₂(0, s₁(x))
ge₂(s₁(x), s₁(y)) -> ge₂(x, y)
minus₂(0, 0) -> 0
minus₂(0, s₁(x)) -> minus₂(0, x)
minus₂(s₁(x), 0) -> s₁(minus₂(x, 0))
minus₂(s₁(x), s₁(y)) -> minus₂(x, y)
plus₂(0, 0) -> 0
plus₂(0, s₁(x)) -> s₁(plus₂(0, x))
plus₂(s₁(x), y) -> s₁(plus₂(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))
div₂(plus₂(x, y), z) -> plus₂(div₂(x, z), div₂(y, z))

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 Order [17,21] with Interpretation:

POL( s₁(x₁) ) = x₁ + 3

POL( GE₂(x₁, x₂) ) = 3x₂ + 3

The following usable rules [14] were oriented: none

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ DependencyGraphProof

        ↳ AND

          ↳ QDP

          ↳ QDP

          ↳ QDP

          ↳ QDP

          ↳ QDP

          ↳ QDP

          ↳ QDP

          ↳ QDP

            ↳ QDPOrderProof

              ↳ QDP

                ↳ PisEmptyProof

          ↳ QDP

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

ge₂(0, 0) -> true
ge₂(s₁(x), 0) -> ge₂(x, 0)
ge₂(0, s₁(0)) -> false
ge₂(0, s₁(s₁(x))) -> ge₂(0, s₁(x))
ge₂(s₁(x), s₁(y)) -> ge₂(x, y)
minus₂(0, 0) -> 0
minus₂(0, s₁(x)) -> minus₂(0, x)
minus₂(s₁(x), 0) -> s₁(minus₂(x, 0))
minus₂(s₁(x), s₁(y)) -> minus₂(x, y)
plus₂(0, 0) -> 0
plus₂(0, s₁(x)) -> s₁(plus₂(0, x))
plus₂(s₁(x), y) -> s₁(plus₂(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))
div₂(plus₂(x, y), z) -> plus₂(div₂(x, z), div₂(y, z))

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

          ↳ QDP

          ↳ QDP

          ↳ QDP

          ↳ QDP

            ↳ QDPOrderProof

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)
DIV₂(plus₂(x, y), z) -> DIV₂(x, z)
DIV₂(plus₂(x, y), z) -> DIV₂(y, z)

The TRS R consists of the following rules:

ge₂(0, 0) -> true
ge₂(s₁(x), 0) -> ge₂(x, 0)
ge₂(0, s₁(0)) -> false
ge₂(0, s₁(s₁(x))) -> ge₂(0, s₁(x))
ge₂(s₁(x), s₁(y)) -> ge₂(x, y)
minus₂(0, 0) -> 0
minus₂(0, s₁(x)) -> minus₂(0, x)
minus₂(s₁(x), 0) -> s₁(minus₂(x, 0))
minus₂(s₁(x), s₁(y)) -> minus₂(x, y)
plus₂(0, 0) -> 0
plus₂(0, s₁(x)) -> s₁(plus₂(0, x))
plus₂(s₁(x), y) -> s₁(plus₂(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))
div₂(plus₂(x, y), z) -> plus₂(div₂(x, z), div₂(y, z))

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.

DIV₂(plus₂(x, y), z) -> DIV₂(x, z)
DIV₂(plus₂(x, y), z) -> DIV₂(y, z)

The remaining pairs can at least be oriented weakly.

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)

Used ordering: Polynomial Order [17,21] with Interpretation:

POL( IFY₃(x₁, ..., x₃) ) = max{0, -3}

POL( minus₂(x₁, x₂) ) = max{0, -3}

POL( true ) = 1

POL( false ) = 1

POL( IF₃(x₁, ..., x₃) ) = max{0, -3}

POL( plus₂(x₁, x₂) ) = 3x₁ + 3x₂ + 1

POL( 0 ) = 0

POL( s₁(x₁) ) = max{0, -3}

POL( DIV₂(x₁, x₂) ) = 2x₁

POL( ge₂(x₁, x₂) ) = max{0, -3}

The following usable rules [14] were oriented:

minus₂(s₁(x), s₁(y)) -> minus₂(x, y)
minus₂(0, s₁(x)) -> minus₂(0, x)
minus₂(s₁(x), 0) -> s₁(minus₂(x, 0))
minus₂(0, 0) -> 0

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ DependencyGraphProof

        ↳ AND

          ↳ QDP

          ↳ QDP

          ↳ QDP

          ↳ QDP

          ↳ QDP

          ↳ QDP

          ↳ QDP

          ↳ QDP

          ↳ QDP

            ↳ QDPOrderProof

              ↳ 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(0, 0) -> true
ge₂(s₁(x), 0) -> ge₂(x, 0)
ge₂(0, s₁(0)) -> false
ge₂(0, s₁(s₁(x))) -> ge₂(0, s₁(x))
ge₂(s₁(x), s₁(y)) -> ge₂(x, y)
minus₂(0, 0) -> 0
minus₂(0, s₁(x)) -> minus₂(0, x)
minus₂(s₁(x), 0) -> s₁(minus₂(x, 0))
minus₂(s₁(x), s₁(y)) -> minus₂(x, y)
plus₂(0, 0) -> 0
plus₂(0, s₁(x)) -> s₁(plus₂(0, x))
plus₂(s₁(x), y) -> s₁(plus₂(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))
div₂(plus₂(x, y), z) -> plus₂(div₂(x, z), div₂(y, z))

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