Termination w.r.t. Q proof of TRS/secret06division.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:

division₂(x, y) -> div₃(x, y, 0)
div₃(x, y, z) -> if₄(lt₂(x, y), x, y, inc₁(z))
if₄(true, x, y, z) -> z
if₄(false, x, s₁(y), z) -> div₃(minus₂(x, s₁(y)), s₁(y), z)
minus₂(x, 0) -> x
minus₂(s₁(x), s₁(y)) -> minus₂(x, y)
lt₂(x, 0) -> false
lt₂(0, s₁(y)) -> true
lt₂(s₁(x), s₁(y)) -> lt₂(x, y)
inc₁(0) -> s₁(0)
inc₁(s₁(x)) -> s₁(inc₁(x))

Q is empty.

↳ QTRS

  ↳ DependencyPairsProof

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

division₂(x, y) -> div₃(x, y, 0)
div₃(x, y, z) -> if₄(lt₂(x, y), x, y, inc₁(z))
if₄(true, x, y, z) -> z
if₄(false, x, s₁(y), z) -> div₃(minus₂(x, s₁(y)), s₁(y), z)
minus₂(x, 0) -> x
minus₂(s₁(x), s₁(y)) -> minus₂(x, y)
lt₂(x, 0) -> false
lt₂(0, s₁(y)) -> true
lt₂(s₁(x), s₁(y)) -> lt₂(x, y)
inc₁(0) -> s₁(0)
inc₁(s₁(x)) -> s₁(inc₁(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:

DIVISION₂(x, y) -> DIV₃(x, y, 0)
DIV₃(x, y, z) -> LT₂(x, y)
LT₂(s₁(x), s₁(y)) -> LT₂(x, y)
MINUS₂(s₁(x), s₁(y)) -> MINUS₂(x, y)
INC₁(s₁(x)) -> INC₁(x)
DIV₃(x, y, z) -> INC₁(z)
IF₄(false, x, s₁(y), z) -> DIV₃(minus₂(x, s₁(y)), s₁(y), z)
DIV₃(x, y, z) -> IF₄(lt₂(x, y), x, y, inc₁(z))
IF₄(false, x, s₁(y), z) -> MINUS₂(x, s₁(y))

The TRS R consists of the following rules:

division₂(x, y) -> div₃(x, y, 0)
div₃(x, y, z) -> if₄(lt₂(x, y), x, y, inc₁(z))
if₄(true, x, y, z) -> z
if₄(false, x, s₁(y), z) -> div₃(minus₂(x, s₁(y)), s₁(y), z)
minus₂(x, 0) -> x
minus₂(s₁(x), s₁(y)) -> minus₂(x, y)
lt₂(x, 0) -> false
lt₂(0, s₁(y)) -> true
lt₂(s₁(x), s₁(y)) -> lt₂(x, y)
inc₁(0) -> s₁(0)
inc₁(s₁(x)) -> s₁(inc₁(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:

DIVISION₂(x, y) -> DIV₃(x, y, 0)
DIV₃(x, y, z) -> LT₂(x, y)
LT₂(s₁(x), s₁(y)) -> LT₂(x, y)
MINUS₂(s₁(x), s₁(y)) -> MINUS₂(x, y)
INC₁(s₁(x)) -> INC₁(x)
DIV₃(x, y, z) -> INC₁(z)
IF₄(false, x, s₁(y), z) -> DIV₃(minus₂(x, s₁(y)), s₁(y), z)
DIV₃(x, y, z) -> IF₄(lt₂(x, y), x, y, inc₁(z))
IF₄(false, x, s₁(y), z) -> MINUS₂(x, s₁(y))

The TRS R consists of the following rules:

division₂(x, y) -> div₃(x, y, 0)
div₃(x, y, z) -> if₄(lt₂(x, y), x, y, inc₁(z))
if₄(true, x, y, z) -> z
if₄(false, x, s₁(y), z) -> div₃(minus₂(x, s₁(y)), s₁(y), z)
minus₂(x, 0) -> x
minus₂(s₁(x), s₁(y)) -> minus₂(x, y)
lt₂(x, 0) -> false
lt₂(0, s₁(y)) -> true
lt₂(s₁(x), s₁(y)) -> lt₂(x, y)
inc₁(0) -> s₁(0)
inc₁(s₁(x)) -> s₁(inc₁(x))

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

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ DependencyGraphProof

        ↳ AND

          ↳ QDP

            ↳ QDPOrderProof

          ↳ QDP

          ↳ QDP

          ↳ QDP

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

INC₁(s₁(x)) -> INC₁(x)

The TRS R consists of the following rules:

division₂(x, y) -> div₃(x, y, 0)
div₃(x, y, z) -> if₄(lt₂(x, y), x, y, inc₁(z))
if₄(true, x, y, z) -> z
if₄(false, x, s₁(y), z) -> div₃(minus₂(x, s₁(y)), s₁(y), z)
minus₂(x, 0) -> x
minus₂(s₁(x), s₁(y)) -> minus₂(x, y)
lt₂(x, 0) -> false
lt₂(0, s₁(y)) -> true
lt₂(s₁(x), s₁(y)) -> lt₂(x, y)
inc₁(0) -> s₁(0)
inc₁(s₁(x)) -> s₁(inc₁(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.

INC₁(s₁(x)) -> INC₁(x)

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

POL(INC₁(x₁)) = 1 + x₁²
POL(s₁(x₁)) = 1 + x₁²

The following usable rules [14] were oriented: none

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ DependencyGraphProof

        ↳ AND

          ↳ QDP

            ↳ QDPOrderProof

              ↳ QDP

                ↳ PisEmptyProof

          ↳ QDP

          ↳ QDP

          ↳ QDP

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

division₂(x, y) -> div₃(x, y, 0)
div₃(x, y, z) -> if₄(lt₂(x, y), x, y, inc₁(z))
if₄(true, x, y, z) -> z
if₄(false, x, s₁(y), z) -> div₃(minus₂(x, s₁(y)), s₁(y), z)
minus₂(x, 0) -> x
minus₂(s₁(x), s₁(y)) -> minus₂(x, y)
lt₂(x, 0) -> false
lt₂(0, s₁(y)) -> true
lt₂(s₁(x), s₁(y)) -> lt₂(x, y)
inc₁(0) -> s₁(0)
inc₁(s₁(x)) -> s₁(inc₁(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

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

LT₂(s₁(x), s₁(y)) -> LT₂(x, y)

The TRS R consists of the following rules:

division₂(x, y) -> div₃(x, y, 0)
div₃(x, y, z) -> if₄(lt₂(x, y), x, y, inc₁(z))
if₄(true, x, y, z) -> z
if₄(false, x, s₁(y), z) -> div₃(minus₂(x, s₁(y)), s₁(y), z)
minus₂(x, 0) -> x
minus₂(s₁(x), s₁(y)) -> minus₂(x, y)
lt₂(x, 0) -> false
lt₂(0, s₁(y)) -> true
lt₂(s₁(x), s₁(y)) -> lt₂(x, y)
inc₁(0) -> s₁(0)
inc₁(s₁(x)) -> s₁(inc₁(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.

LT₂(s₁(x), s₁(y)) -> LT₂(x, y)

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

POL(LT₂(x₁, x₂)) = x₂
POL(s₁(x₁)) = 1 + x₁

The following usable rules [14] were oriented: none

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ DependencyGraphProof

        ↳ AND

          ↳ QDP

          ↳ QDP

            ↳ QDPOrderProof

              ↳ QDP

                ↳ PisEmptyProof

          ↳ QDP

          ↳ QDP

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

division₂(x, y) -> div₃(x, y, 0)
div₃(x, y, z) -> if₄(lt₂(x, y), x, y, inc₁(z))
if₄(true, x, y, z) -> z
if₄(false, x, s₁(y), z) -> div₃(minus₂(x, s₁(y)), s₁(y), z)
minus₂(x, 0) -> x
minus₂(s₁(x), s₁(y)) -> minus₂(x, y)
lt₂(x, 0) -> false
lt₂(0, s₁(y)) -> true
lt₂(s₁(x), s₁(y)) -> lt₂(x, y)
inc₁(0) -> s₁(0)
inc₁(s₁(x)) -> s₁(inc₁(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

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:

division₂(x, y) -> div₃(x, y, 0)
div₃(x, y, z) -> if₄(lt₂(x, y), x, y, inc₁(z))
if₄(true, x, y, z) -> z
if₄(false, x, s₁(y), z) -> div₃(minus₂(x, s₁(y)), s₁(y), z)
minus₂(x, 0) -> x
minus₂(s₁(x), s₁(y)) -> minus₂(x, y)
lt₂(x, 0) -> false
lt₂(0, s₁(y)) -> true
lt₂(s₁(x), s₁(y)) -> lt₂(x, y)
inc₁(0) -> s₁(0)
inc₁(s₁(x)) -> s₁(inc₁(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.

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)) = x₂
POL(s₁(x₁)) = 1 + x₁

The following usable rules [14] were oriented: none

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ DependencyGraphProof

        ↳ AND

          ↳ QDP

          ↳ QDP

          ↳ QDP

            ↳ QDPOrderProof

              ↳ QDP

                ↳ PisEmptyProof

          ↳ QDP

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

division₂(x, y) -> div₃(x, y, 0)
div₃(x, y, z) -> if₄(lt₂(x, y), x, y, inc₁(z))
if₄(true, x, y, z) -> z
if₄(false, x, s₁(y), z) -> div₃(minus₂(x, s₁(y)), s₁(y), z)
minus₂(x, 0) -> x
minus₂(s₁(x), s₁(y)) -> minus₂(x, y)
lt₂(x, 0) -> false
lt₂(0, s₁(y)) -> true
lt₂(s₁(x), s₁(y)) -> lt₂(x, y)
inc₁(0) -> s₁(0)
inc₁(s₁(x)) -> s₁(inc₁(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

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

DIV₃(x, y, z) -> IF₄(lt₂(x, y), x, y, inc₁(z))
IF₄(false, x, s₁(y), z) -> DIV₃(minus₂(x, s₁(y)), s₁(y), z)

The TRS R consists of the following rules:

division₂(x, y) -> div₃(x, y, 0)
div₃(x, y, z) -> if₄(lt₂(x, y), x, y, inc₁(z))
if₄(true, x, y, z) -> z
if₄(false, x, s₁(y), z) -> div₃(minus₂(x, s₁(y)), s₁(y), z)
minus₂(x, 0) -> x
minus₂(s₁(x), s₁(y)) -> minus₂(x, y)
lt₂(x, 0) -> false
lt₂(0, s₁(y)) -> true
lt₂(s₁(x), s₁(y)) -> lt₂(x, y)
inc₁(0) -> s₁(0)
inc₁(s₁(x)) -> s₁(inc₁(x))

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