Termination w.r.t. Q proof of /home/nowonder/forschung/aprove/satrung/aprove/lpar1/HMSLASHt014.trs

Termination w.r.t. Q of the following Term Rewriting System could be proven:

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

-₂(x, 0) -> x
-₂(0, s₁(y)) -> 0
-₂(s₁(x), s₁(y)) -> -₂(x, y)
lt₂(x, 0) -> false
lt₂(0, s₁(y)) -> true
lt₂(s₁(x), s₁(y)) -> lt₂(x, y)
if₃(true, x, y) -> x
if₃(false, x, y) -> y
div₂(x, 0) -> 0
div₂(0, y) -> 0
div₂(s₁(x), s₁(y)) -> if₃(lt₂(x, y), 0, s₁(div₂(-₂(x, y), s₁(y))))

Q is empty.

↳ QTRS

  ↳ DependencyPairsProof

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

-₂(x, 0) -> x
-₂(0, s₁(y)) -> 0
-₂(s₁(x), s₁(y)) -> -₂(x, y)
lt₂(x, 0) -> false
lt₂(0, s₁(y)) -> true
lt₂(s₁(x), s₁(y)) -> lt₂(x, y)
if₃(true, x, y) -> x
if₃(false, x, y) -> y
div₂(x, 0) -> 0
div₂(0, y) -> 0
div₂(s₁(x), s₁(y)) -> if₃(lt₂(x, y), 0, s₁(div₂(-₂(x, y), s₁(y))))

Q is empty.

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

DIV₂(s₁(x), s₁(y)) -> LT₂(x, y)
-¹₂(s₁(x), s₁(y)) -> -¹₂(x, y)
DIV₂(s₁(x), s₁(y)) -> -¹₂(x, y)
LT₂(s₁(x), s₁(y)) -> LT₂(x, y)
DIV₂(s₁(x), s₁(y)) -> IF₃(lt₂(x, y), 0, s₁(div₂(-₂(x, y), s₁(y))))
DIV₂(s₁(x), s₁(y)) -> DIV₂(-₂(x, y), s₁(y))

The TRS R consists of the following rules:

-₂(x, 0) -> x
-₂(0, s₁(y)) -> 0
-₂(s₁(x), s₁(y)) -> -₂(x, y)
lt₂(x, 0) -> false
lt₂(0, s₁(y)) -> true
lt₂(s₁(x), s₁(y)) -> lt₂(x, y)
if₃(true, x, y) -> x
if₃(false, x, y) -> y
div₂(x, 0) -> 0
div₂(0, y) -> 0
div₂(s₁(x), s₁(y)) -> if₃(lt₂(x, y), 0, s₁(div₂(-₂(x, y), s₁(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:

DIV₂(s₁(x), s₁(y)) -> LT₂(x, y)
-¹₂(s₁(x), s₁(y)) -> -¹₂(x, y)
DIV₂(s₁(x), s₁(y)) -> -¹₂(x, y)
LT₂(s₁(x), s₁(y)) -> LT₂(x, y)
DIV₂(s₁(x), s₁(y)) -> IF₃(lt₂(x, y), 0, s₁(div₂(-₂(x, y), s₁(y))))
DIV₂(s₁(x), s₁(y)) -> DIV₂(-₂(x, y), s₁(y))

The TRS R consists of the following rules:

-₂(x, 0) -> x
-₂(0, s₁(y)) -> 0
-₂(s₁(x), s₁(y)) -> -₂(x, y)
lt₂(x, 0) -> false
lt₂(0, s₁(y)) -> true
lt₂(s₁(x), s₁(y)) -> lt₂(x, y)
if₃(true, x, y) -> x
if₃(false, x, y) -> y
div₂(x, 0) -> 0
div₂(0, y) -> 0
div₂(s₁(x), s₁(y)) -> if₃(lt₂(x, y), 0, s₁(div₂(-₂(x, y), s₁(y))))

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

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ DependencyGraphProof

        ↳ AND

          ↳ QDP

            ↳ QDPAfsSolverProof

          ↳ 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:

-₂(x, 0) -> x
-₂(0, s₁(y)) -> 0
-₂(s₁(x), s₁(y)) -> -₂(x, y)
lt₂(x, 0) -> false
lt₂(0, s₁(y)) -> true
lt₂(s₁(x), s₁(y)) -> lt₂(x, y)
if₃(true, x, y) -> x
if₃(false, x, y) -> y
div₂(x, 0) -> 0
div₂(0, y) -> 0
div₂(s₁(x), s₁(y)) -> if₃(lt₂(x, y), 0, s₁(div₂(-₂(x, y), s₁(y))))

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.

LT₂(s₁(x), s₁(y)) -> LT₂(x, y)

Used argument filtering: LT₂(x1, x2) = x2
s₁(x1) = s₁(x1)
Used ordering: Precedence:

trivial

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ DependencyGraphProof

        ↳ AND

          ↳ QDP

            ↳ QDPAfsSolverProof

              ↳ QDP

                ↳ PisEmptyProof

          ↳ QDP

          ↳ QDP

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

-₂(x, 0) -> x
-₂(0, s₁(y)) -> 0
-₂(s₁(x), s₁(y)) -> -₂(x, y)
lt₂(x, 0) -> false
lt₂(0, s₁(y)) -> true
lt₂(s₁(x), s₁(y)) -> lt₂(x, y)
if₃(true, x, y) -> x
if₃(false, x, y) -> y
div₂(x, 0) -> 0
div₂(0, y) -> 0
div₂(s₁(x), s₁(y)) -> if₃(lt₂(x, y), 0, s₁(div₂(-₂(x, y), s₁(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

            ↳ QDPAfsSolverProof

          ↳ QDP

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

-¹₂(s₁(x), s₁(y)) -> -¹₂(x, y)

The TRS R consists of the following rules:

-₂(x, 0) -> x
-₂(0, s₁(y)) -> 0
-₂(s₁(x), s₁(y)) -> -₂(x, y)
lt₂(x, 0) -> false
lt₂(0, s₁(y)) -> true
lt₂(s₁(x), s₁(y)) -> lt₂(x, y)
if₃(true, x, y) -> x
if₃(false, x, y) -> y
div₂(x, 0) -> 0
div₂(0, y) -> 0
div₂(s₁(x), s₁(y)) -> if₃(lt₂(x, y), 0, s₁(div₂(-₂(x, y), s₁(y))))

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.

-¹₂(s₁(x), s₁(y)) -> -¹₂(x, y)

Used argument filtering: -¹₂(x1, x2) = x2
s₁(x1) = s₁(x1)
Used ordering: Precedence:

trivial

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ DependencyGraphProof

        ↳ AND

          ↳ QDP

          ↳ QDP

            ↳ QDPAfsSolverProof

              ↳ QDP

                ↳ PisEmptyProof

          ↳ QDP

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

-₂(x, 0) -> x
-₂(0, s₁(y)) -> 0
-₂(s₁(x), s₁(y)) -> -₂(x, y)
lt₂(x, 0) -> false
lt₂(0, s₁(y)) -> true
lt₂(s₁(x), s₁(y)) -> lt₂(x, y)
if₃(true, x, y) -> x
if₃(false, x, y) -> y
div₂(x, 0) -> 0
div₂(0, y) -> 0
div₂(s₁(x), s₁(y)) -> if₃(lt₂(x, y), 0, s₁(div₂(-₂(x, y), s₁(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

            ↳ QDPAfsSolverProof

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

DIV₂(s₁(x), s₁(y)) -> DIV₂(-₂(x, y), s₁(y))

The TRS R consists of the following rules:

-₂(x, 0) -> x
-₂(0, s₁(y)) -> 0
-₂(s₁(x), s₁(y)) -> -₂(x, y)
lt₂(x, 0) -> false
lt₂(0, s₁(y)) -> true
lt₂(s₁(x), s₁(y)) -> lt₂(x, y)
if₃(true, x, y) -> x
if₃(false, x, y) -> y
div₂(x, 0) -> 0
div₂(0, y) -> 0
div₂(s₁(x), s₁(y)) -> if₃(lt₂(x, y), 0, s₁(div₂(-₂(x, y), s₁(y))))

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.

DIV₂(s₁(x), s₁(y)) -> DIV₂(-₂(x, y), s₁(y))

Used argument filtering: DIV₂(x1, x2) = x1
s₁(x1) = s₁(x1)
-₂(x1, x2) = x1
0 = 0
Used ordering: Precedence:

trivial

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ DependencyGraphProof

        ↳ AND

          ↳ QDP

          ↳ QDP

          ↳ QDP

            ↳ QDPAfsSolverProof

              ↳ QDP

                ↳ PisEmptyProof

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

-₂(x, 0) -> x
-₂(0, s₁(y)) -> 0
-₂(s₁(x), s₁(y)) -> -₂(x, y)
lt₂(x, 0) -> false
lt₂(0, s₁(y)) -> true
lt₂(s₁(x), s₁(y)) -> lt₂(x, y)
if₃(true, x, y) -> x
if₃(false, x, y) -> y
div₂(x, 0) -> 0
div₂(0, y) -> 0
div₂(s₁(x), s₁(y)) -> if₃(lt₂(x, y), 0, s₁(div₂(-₂(x, y), s₁(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.