Termination w.r.t. Q proof of TRS/HMt014.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.

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(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 [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:

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.
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)) = 2·x₁·x₂
POL(s₁(x₁)) = 2 + 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:

-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

            ↳ QDPOrderProof

          ↳ 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.
We use the reduction pair processor [13].

The following pairs can be oriented strictly and are deleted.

-¹₂(s₁(x), s₁(y)) -> -¹₂(x, y)

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

POL(-¹₂(x₁, x₂)) = 2·x₁·x₂
POL(s₁(x₁)) = 2 + 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:

-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

            ↳ QDPOrderProof

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.
We use the reduction pair processor [13].

The following pairs can be oriented strictly and are deleted.

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

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

POL(-₂(x₁, x₂)) = x₁
POL(0) = 0
POL(DIV₂(x₁, x₂)) = 2·x₁·x₂
POL(s₁(x₁)) = 2 + 2·x₁

The following usable rules [14] were oriented:

-₂(0, s₁(y)) -> 0
-₂(x, 0) -> x
-₂(s₁(x), s₁(y)) -> -₂(x, y)

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ DependencyGraphProof

        ↳ AND

          ↳ QDP

          ↳ QDP

          ↳ QDP

            ↳ QDPOrderProof

              ↳ 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.