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:

-(x, 0) → x
-(0, s(y)) → 0
-(s(x), s(y)) → -(x, y)
lt(x, 0) → false
lt(0, s(y)) → true
lt(s(x), s(y)) → lt(x, y)
if(true, x, y) → x
if(false, x, y) → y
div(x, 0) → 0
div(0, y) → 0
div(s(x), s(y)) → if(lt(x, y), 0, s(div(-(x, y), s(y))))

Q is empty.

↳ QTRS

  ↳ Overlay + Local Confluence

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

-(x, 0) → x
-(0, s(y)) → 0
-(s(x), s(y)) → -(x, y)
lt(x, 0) → false
lt(0, s(y)) → true
lt(s(x), s(y)) → lt(x, y)
if(true, x, y) → x
if(false, x, y) → y
div(x, 0) → 0
div(0, y) → 0
div(s(x), s(y)) → if(lt(x, y), 0, s(div(-(x, y), s(y))))

Q is empty.

The TRS is overlay and locally confluent. By [15] we can switch to innermost.

↳ QTRS

  ↳ Overlay + Local Confluence

    ↳ QTRS

      ↳ DependencyPairsProof

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

-(x, 0) → x
-(0, s(y)) → 0
-(s(x), s(y)) → -(x, y)
lt(x, 0) → false
lt(0, s(y)) → true
lt(s(x), s(y)) → lt(x, y)
if(true, x, y) → x
if(false, x, y) → y
div(x, 0) → 0
div(0, y) → 0
div(s(x), s(y)) → if(lt(x, y), 0, s(div(-(x, y), s(y))))

The set Q consists of the following terms:

-(x0, 0)
-(0, s(x0))
-(s(x0), s(x1))
lt(x0, 0)
lt(0, s(x0))
lt(s(x0), s(x1))
if(true, x0, x1)
if(false, x0, x1)
div(x0, 0)
div(0, x0)
div(s(x0), s(x1))

Using Dependency Pairs [1,13] we result in the following initial DP problem:
Q DP problem:
The TRS P consists of the following rules:

DIV(s(x), s(y)) → LT(x, y)
-¹(s(x), s(y)) → -¹(x, y)
DIV(s(x), s(y)) → -¹(x, y)
LT(s(x), s(y)) → LT(x, y)
DIV(s(x), s(y)) → IF(lt(x, y), 0, s(div(-(x, y), s(y))))
DIV(s(x), s(y)) → DIV(-(x, y), s(y))

The TRS R consists of the following rules:

-(x, 0) → x
-(0, s(y)) → 0
-(s(x), s(y)) → -(x, y)
lt(x, 0) → false
lt(0, s(y)) → true
lt(s(x), s(y)) → lt(x, y)
if(true, x, y) → x
if(false, x, y) → y
div(x, 0) → 0
div(0, y) → 0
div(s(x), s(y)) → if(lt(x, y), 0, s(div(-(x, y), s(y))))

The set Q consists of the following terms:

-(x0, 0)
-(0, s(x0))
-(s(x0), s(x1))
lt(x0, 0)
lt(0, s(x0))
lt(s(x0), s(x1))
if(true, x0, x1)
if(false, x0, x1)
div(x0, 0)
div(0, x0)
div(s(x0), s(x1))

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

↳ QTRS

  ↳ Overlay + Local Confluence

    ↳ QTRS

      ↳ DependencyPairsProof

        ↳ QDP

          ↳ EdgeDeletionProof

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

DIV(s(x), s(y)) → LT(x, y)
-¹(s(x), s(y)) → -¹(x, y)
DIV(s(x), s(y)) → -¹(x, y)
LT(s(x), s(y)) → LT(x, y)
DIV(s(x), s(y)) → IF(lt(x, y), 0, s(div(-(x, y), s(y))))
DIV(s(x), s(y)) → DIV(-(x, y), s(y))

The TRS R consists of the following rules:

-(x, 0) → x
-(0, s(y)) → 0
-(s(x), s(y)) → -(x, y)
lt(x, 0) → false
lt(0, s(y)) → true
lt(s(x), s(y)) → lt(x, y)
if(true, x, y) → x
if(false, x, y) → y
div(x, 0) → 0
div(0, y) → 0
div(s(x), s(y)) → if(lt(x, y), 0, s(div(-(x, y), s(y))))

The set Q consists of the following terms:

-(x0, 0)
-(0, s(x0))
-(s(x0), s(x1))
lt(x0, 0)
lt(0, s(x0))
lt(s(x0), s(x1))
if(true, x0, x1)
if(false, x0, x1)
div(x0, 0)
div(0, x0)
div(s(x0), s(x1))

We have to consider all minimal (P,Q,R)-chains.
We deleted some edges using various graph approximations

↳ QTRS

  ↳ Overlay + Local Confluence

    ↳ QTRS

      ↳ DependencyPairsProof

        ↳ QDP

          ↳ EdgeDeletionProof

            ↳ QDP

              ↳ DependencyGraphProof

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

DIV(s(x), s(y)) → LT(x, y)
-¹(s(x), s(y)) → -¹(x, y)
DIV(s(x), s(y)) → -¹(x, y)
LT(s(x), s(y)) → LT(x, y)
DIV(s(x), s(y)) → IF(lt(x, y), 0, s(div(-(x, y), s(y))))
DIV(s(x), s(y)) → DIV(-(x, y), s(y))

The TRS R consists of the following rules:

-(x, 0) → x
-(0, s(y)) → 0
-(s(x), s(y)) → -(x, y)
lt(x, 0) → false
lt(0, s(y)) → true
lt(s(x), s(y)) → lt(x, y)
if(true, x, y) → x
if(false, x, y) → y
div(x, 0) → 0
div(0, y) → 0
div(s(x), s(y)) → if(lt(x, y), 0, s(div(-(x, y), s(y))))

The set Q consists of the following terms:

-(x0, 0)
-(0, s(x0))
-(s(x0), s(x1))
lt(x0, 0)
lt(0, s(x0))
lt(s(x0), s(x1))
if(true, x0, x1)
if(false, x0, x1)
div(x0, 0)
div(0, x0)
div(s(x0), s(x1))

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

  ↳ Overlay + Local Confluence

    ↳ QTRS

      ↳ DependencyPairsProof

        ↳ QDP

          ↳ EdgeDeletionProof

            ↳ QDP

              ↳ DependencyGraphProof

                ↳ AND

                  ↳ QDP

                    ↳ QDPOrderProof

                  ↳ QDP

                  ↳ QDP

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

LT(s(x), s(y)) → LT(x, y)

The TRS R consists of the following rules:

-(x, 0) → x
-(0, s(y)) → 0
-(s(x), s(y)) → -(x, y)
lt(x, 0) → false
lt(0, s(y)) → true
lt(s(x), s(y)) → lt(x, y)
if(true, x, y) → x
if(false, x, y) → y
div(x, 0) → 0
div(0, y) → 0
div(s(x), s(y)) → if(lt(x, y), 0, s(div(-(x, y), s(y))))

The set Q consists of the following terms:

-(x0, 0)
-(0, s(x0))
-(s(x0), s(x1))
lt(x0, 0)
lt(0, s(x0))
lt(s(x0), s(x1))
if(true, x0, x1)
if(false, x0, x1)
div(x0, 0)
div(0, x0)
div(s(x0), s(x1))

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.

LT(s(x), s(y)) → LT(x, y)

The remaining pairs can at least be oriented weakly.
none
Used ordering: Combined order from the following AFS and order.
LT(x1, x2) = LT(x2)
s(x1) = s(x1)

Recursive Path Order [2].
Precedence:

s₁ > LT₁

The following usable rules [14] were oriented: none

↳ QTRS

  ↳ Overlay + Local Confluence

    ↳ QTRS

      ↳ DependencyPairsProof

        ↳ QDP

          ↳ EdgeDeletionProof

            ↳ QDP

              ↳ DependencyGraphProof

                ↳ AND

                  ↳ QDP

                    ↳ QDPOrderProof

                      ↳ QDP

                        ↳ PisEmptyProof

                  ↳ QDP

                  ↳ QDP

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

-(x, 0) → x
-(0, s(y)) → 0
-(s(x), s(y)) → -(x, y)
lt(x, 0) → false
lt(0, s(y)) → true
lt(s(x), s(y)) → lt(x, y)
if(true, x, y) → x
if(false, x, y) → y
div(x, 0) → 0
div(0, y) → 0
div(s(x), s(y)) → if(lt(x, y), 0, s(div(-(x, y), s(y))))

The set Q consists of the following terms:

-(x0, 0)
-(0, s(x0))
-(s(x0), s(x1))
lt(x0, 0)
lt(0, s(x0))
lt(s(x0), s(x1))
if(true, x0, x1)
if(false, x0, x1)
div(x0, 0)
div(0, x0)
div(s(x0), s(x1))

We have to consider all minimal (P,Q,R)-chains.
The TRS P is empty. Hence, there is no (P,Q,R) chain.

↳ QTRS

  ↳ Overlay + Local Confluence

    ↳ QTRS

      ↳ DependencyPairsProof

        ↳ QDP

          ↳ EdgeDeletionProof

            ↳ QDP

              ↳ DependencyGraphProof

                ↳ AND

                  ↳ QDP

                  ↳ QDP

                    ↳ QDPOrderProof

                  ↳ QDP

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

-¹(s(x), s(y)) → -¹(x, y)

The TRS R consists of the following rules:

-(x, 0) → x
-(0, s(y)) → 0
-(s(x), s(y)) → -(x, y)
lt(x, 0) → false
lt(0, s(y)) → true
lt(s(x), s(y)) → lt(x, y)
if(true, x, y) → x
if(false, x, y) → y
div(x, 0) → 0
div(0, y) → 0
div(s(x), s(y)) → if(lt(x, y), 0, s(div(-(x, y), s(y))))

The set Q consists of the following terms:

-(x0, 0)
-(0, s(x0))
-(s(x0), s(x1))
lt(x0, 0)
lt(0, s(x0))
lt(s(x0), s(x1))
if(true, x0, x1)
if(false, x0, x1)
div(x0, 0)
div(0, x0)
div(s(x0), s(x1))

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.

-¹(s(x), s(y)) → -¹(x, y)

The remaining pairs can at least be oriented weakly.
none
Used ordering: Combined order from the following AFS and order.
-¹(x1, x2) = -¹(x2)
s(x1) = s(x1)

Recursive Path Order [2].
Precedence:

s₁ > -^1₁

The following usable rules [14] were oriented: none

↳ QTRS

  ↳ Overlay + Local Confluence

    ↳ QTRS

      ↳ DependencyPairsProof

        ↳ QDP

          ↳ EdgeDeletionProof

            ↳ QDP

              ↳ DependencyGraphProof

                ↳ AND

                  ↳ QDP

                  ↳ QDP

                    ↳ QDPOrderProof

                      ↳ QDP

                        ↳ PisEmptyProof

                  ↳ QDP

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

-(x, 0) → x
-(0, s(y)) → 0
-(s(x), s(y)) → -(x, y)
lt(x, 0) → false
lt(0, s(y)) → true
lt(s(x), s(y)) → lt(x, y)
if(true, x, y) → x
if(false, x, y) → y
div(x, 0) → 0
div(0, y) → 0
div(s(x), s(y)) → if(lt(x, y), 0, s(div(-(x, y), s(y))))

The set Q consists of the following terms:

-(x0, 0)
-(0, s(x0))
-(s(x0), s(x1))
lt(x0, 0)
lt(0, s(x0))
lt(s(x0), s(x1))
if(true, x0, x1)
if(false, x0, x1)
div(x0, 0)
div(0, x0)
div(s(x0), s(x1))

We have to consider all minimal (P,Q,R)-chains.
The TRS P is empty. Hence, there is no (P,Q,R) chain.

↳ QTRS

  ↳ Overlay + Local Confluence

    ↳ QTRS

      ↳ DependencyPairsProof

        ↳ QDP

          ↳ EdgeDeletionProof

            ↳ QDP

              ↳ DependencyGraphProof

                ↳ AND

                  ↳ QDP

                  ↳ QDP

                  ↳ QDP

                    ↳ QDPOrderProof

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

DIV(s(x), s(y)) → DIV(-(x, y), s(y))

The TRS R consists of the following rules:

-(x, 0) → x
-(0, s(y)) → 0
-(s(x), s(y)) → -(x, y)
lt(x, 0) → false
lt(0, s(y)) → true
lt(s(x), s(y)) → lt(x, y)
if(true, x, y) → x
if(false, x, y) → y
div(x, 0) → 0
div(0, y) → 0
div(s(x), s(y)) → if(lt(x, y), 0, s(div(-(x, y), s(y))))

The set Q consists of the following terms:

-(x0, 0)
-(0, s(x0))
-(s(x0), s(x1))
lt(x0, 0)
lt(0, s(x0))
lt(s(x0), s(x1))
if(true, x0, x1)
if(false, x0, x1)
div(x0, 0)
div(0, x0)
div(s(x0), s(x1))

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.

DIV(s(x), s(y)) → DIV(-(x, y), s(y))

The remaining pairs can at least be oriented weakly.
none
Used ordering: Combined order from the following AFS and order.
DIV(x1, x2) = DIV(x1, x2)
s(x1) = s(x1)
-(x1, x2) = x1
0 = 0

Recursive Path Order [2].
Precedence:

s₁ > 0

The following usable rules [14] were oriented:

-(0, s(y)) → 0
-(s(x), s(y)) → -(x, y)
-(x, 0) → x

↳ QTRS

  ↳ Overlay + Local Confluence

    ↳ QTRS

      ↳ DependencyPairsProof

        ↳ QDP

          ↳ EdgeDeletionProof

            ↳ QDP

              ↳ DependencyGraphProof

                ↳ AND

                  ↳ QDP

                  ↳ QDP

                  ↳ QDP

                    ↳ QDPOrderProof

                      ↳ QDP

                        ↳ PisEmptyProof

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

-(x, 0) → x
-(0, s(y)) → 0
-(s(x), s(y)) → -(x, y)
lt(x, 0) → false
lt(0, s(y)) → true
lt(s(x), s(y)) → lt(x, y)
if(true, x, y) → x
if(false, x, y) → y
div(x, 0) → 0
div(0, y) → 0
div(s(x), s(y)) → if(lt(x, y), 0, s(div(-(x, y), s(y))))

The set Q consists of the following terms:

-(x0, 0)
-(0, s(x0))
-(s(x0), s(x1))
lt(x0, 0)
lt(0, s(x0))
lt(s(x0), s(x1))
if(true, x0, x1)
if(false, x0, x1)
div(x0, 0)
div(0, x0)
div(s(x0), s(x1))

We have to consider all minimal (P,Q,R)-chains.
The TRS P is empty. Hence, there is no (P,Q,R) chain.