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:

division(x, y) → div(x, y, 0)
div(x, y, z) → if(lt(x, y), x, y, inc(z))
if(true, x, y, z) → z
if(false, x, s(y), z) → div(minus(x, s(y)), s(y), z)
minus(x, 0) → x
minus(s(x), s(y)) → minus(x, y)
lt(x, 0) → false
lt(0, s(y)) → true
lt(s(x), s(y)) → lt(x, y)
inc(0) → s(0)
inc(s(x)) → s(inc(x))

Q is empty.

↳ QTRS

  ↳ Overlay + Local Confluence

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

division(x, y) → div(x, y, 0)
div(x, y, z) → if(lt(x, y), x, y, inc(z))
if(true, x, y, z) → z
if(false, x, s(y), z) → div(minus(x, s(y)), s(y), z)
minus(x, 0) → x
minus(s(x), s(y)) → minus(x, y)
lt(x, 0) → false
lt(0, s(y)) → true
lt(s(x), s(y)) → lt(x, y)
inc(0) → s(0)
inc(s(x)) → s(inc(x))

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:

division(x, y) → div(x, y, 0)
div(x, y, z) → if(lt(x, y), x, y, inc(z))
if(true, x, y, z) → z
if(false, x, s(y), z) → div(minus(x, s(y)), s(y), z)
minus(x, 0) → x
minus(s(x), s(y)) → minus(x, y)
lt(x, 0) → false
lt(0, s(y)) → true
lt(s(x), s(y)) → lt(x, y)
inc(0) → s(0)
inc(s(x)) → s(inc(x))

The set Q consists of the following terms:

division(x0, x1)
div(x0, x1, x2)
if(true, x0, x1, x2)
if(false, x0, s(x1), x2)
minus(x0, 0)
minus(s(x0), s(x1))
lt(x0, 0)
lt(0, s(x0))
lt(s(x0), s(x1))
inc(0)
inc(s(x0))

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

DIVISION(x, y) → DIV(x, y, 0)
DIV(x, y, z) → LT(x, y)
LT(s(x), s(y)) → LT(x, y)
MINUS(s(x), s(y)) → MINUS(x, y)
INC(s(x)) → INC(x)
DIV(x, y, z) → INC(z)
IF(false, x, s(y), z) → DIV(minus(x, s(y)), s(y), z)
DIV(x, y, z) → IF(lt(x, y), x, y, inc(z))
IF(false, x, s(y), z) → MINUS(x, s(y))

The TRS R consists of the following rules:

division(x, y) → div(x, y, 0)
div(x, y, z) → if(lt(x, y), x, y, inc(z))
if(true, x, y, z) → z
if(false, x, s(y), z) → div(minus(x, s(y)), s(y), z)
minus(x, 0) → x
minus(s(x), s(y)) → minus(x, y)
lt(x, 0) → false
lt(0, s(y)) → true
lt(s(x), s(y)) → lt(x, y)
inc(0) → s(0)
inc(s(x)) → s(inc(x))

The set Q consists of the following terms:

division(x0, x1)
div(x0, x1, x2)
if(true, x0, x1, x2)
if(false, x0, s(x1), x2)
minus(x0, 0)
minus(s(x0), s(x1))
lt(x0, 0)
lt(0, s(x0))
lt(s(x0), s(x1))
inc(0)
inc(s(x0))

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:

DIVISION(x, y) → DIV(x, y, 0)
DIV(x, y, z) → LT(x, y)
LT(s(x), s(y)) → LT(x, y)
MINUS(s(x), s(y)) → MINUS(x, y)
INC(s(x)) → INC(x)
DIV(x, y, z) → INC(z)
IF(false, x, s(y), z) → DIV(minus(x, s(y)), s(y), z)
DIV(x, y, z) → IF(lt(x, y), x, y, inc(z))
IF(false, x, s(y), z) → MINUS(x, s(y))

The TRS R consists of the following rules:

division(x, y) → div(x, y, 0)
div(x, y, z) → if(lt(x, y), x, y, inc(z))
if(true, x, y, z) → z
if(false, x, s(y), z) → div(minus(x, s(y)), s(y), z)
minus(x, 0) → x
minus(s(x), s(y)) → minus(x, y)
lt(x, 0) → false
lt(0, s(y)) → true
lt(s(x), s(y)) → lt(x, y)
inc(0) → s(0)
inc(s(x)) → s(inc(x))

The set Q consists of the following terms:

division(x0, x1)
div(x0, x1, x2)
if(true, x0, x1, x2)
if(false, x0, s(x1), x2)
minus(x0, 0)
minus(s(x0), s(x1))
lt(x0, 0)
lt(0, s(x0))
lt(s(x0), s(x1))
inc(0)
inc(s(x0))

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:

DIVISION(x, y) → DIV(x, y, 0)
DIV(x, y, z) → LT(x, y)
MINUS(s(x), s(y)) → MINUS(x, y)
LT(s(x), s(y)) → LT(x, y)
INC(s(x)) → INC(x)
DIV(x, y, z) → INC(z)
DIV(x, y, z) → IF(lt(x, y), x, y, inc(z))
IF(false, x, s(y), z) → DIV(minus(x, s(y)), s(y), z)
IF(false, x, s(y), z) → MINUS(x, s(y))

The TRS R consists of the following rules:

division(x, y) → div(x, y, 0)
div(x, y, z) → if(lt(x, y), x, y, inc(z))
if(true, x, y, z) → z
if(false, x, s(y), z) → div(minus(x, s(y)), s(y), z)
minus(x, 0) → x
minus(s(x), s(y)) → minus(x, y)
lt(x, 0) → false
lt(0, s(y)) → true
lt(s(x), s(y)) → lt(x, y)
inc(0) → s(0)
inc(s(x)) → s(inc(x))

The set Q consists of the following terms:

division(x0, x1)
div(x0, x1, x2)
if(true, x0, x1, x2)
if(false, x0, s(x1), x2)
minus(x0, 0)
minus(s(x0), s(x1))
lt(x0, 0)
lt(0, s(x0))
lt(s(x0), s(x1))
inc(0)
inc(s(x0))

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

  ↳ Overlay + Local Confluence

    ↳ QTRS

      ↳ DependencyPairsProof

        ↳ QDP

          ↳ EdgeDeletionProof

            ↳ QDP

              ↳ DependencyGraphProof

                ↳ AND

                  ↳ QDP

                    ↳ QDPOrderProof

                  ↳ QDP

                  ↳ QDP

                  ↳ QDP

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

INC(s(x)) → INC(x)

The TRS R consists of the following rules:

division(x, y) → div(x, y, 0)
div(x, y, z) → if(lt(x, y), x, y, inc(z))
if(true, x, y, z) → z
if(false, x, s(y), z) → div(minus(x, s(y)), s(y), z)
minus(x, 0) → x
minus(s(x), s(y)) → minus(x, y)
lt(x, 0) → false
lt(0, s(y)) → true
lt(s(x), s(y)) → lt(x, y)
inc(0) → s(0)
inc(s(x)) → s(inc(x))

The set Q consists of the following terms:

division(x0, x1)
div(x0, x1, x2)
if(true, x0, x1, x2)
if(false, x0, s(x1), x2)
minus(x0, 0)
minus(s(x0), s(x1))
lt(x0, 0)
lt(0, s(x0))
lt(s(x0), s(x1))
inc(0)
inc(s(x0))

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.

INC(s(x)) → INC(x)

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

Lexicographic Path Order [19].
Precedence:

s₁ > INC₁

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

                  ↳ QDP

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

division(x, y) → div(x, y, 0)
div(x, y, z) → if(lt(x, y), x, y, inc(z))
if(true, x, y, z) → z
if(false, x, s(y), z) → div(minus(x, s(y)), s(y), z)
minus(x, 0) → x
minus(s(x), s(y)) → minus(x, y)
lt(x, 0) → false
lt(0, s(y)) → true
lt(s(x), s(y)) → lt(x, y)
inc(0) → s(0)
inc(s(x)) → s(inc(x))

The set Q consists of the following terms:

division(x0, x1)
div(x0, x1, x2)
if(true, x0, x1, x2)
if(false, x0, s(x1), x2)
minus(x0, 0)
minus(s(x0), s(x1))
lt(x0, 0)
lt(0, s(x0))
lt(s(x0), s(x1))
inc(0)
inc(s(x0))

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

                  ↳ 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:

division(x, y) → div(x, y, 0)
div(x, y, z) → if(lt(x, y), x, y, inc(z))
if(true, x, y, z) → z
if(false, x, s(y), z) → div(minus(x, s(y)), s(y), z)
minus(x, 0) → x
minus(s(x), s(y)) → minus(x, y)
lt(x, 0) → false
lt(0, s(y)) → true
lt(s(x), s(y)) → lt(x, y)
inc(0) → s(0)
inc(s(x)) → s(inc(x))

The set Q consists of the following terms:

division(x0, x1)
div(x0, x1, x2)
if(true, x0, x1, x2)
if(false, x0, s(x1), x2)
minus(x0, 0)
minus(s(x0), s(x1))
lt(x0, 0)
lt(0, s(x0))
lt(s(x0), s(x1))
inc(0)
inc(s(x0))

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)

Lexicographic Path Order [19].
Precedence:

s₁ > LT₁

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

                  ↳ QDP

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

division(x, y) → div(x, y, 0)
div(x, y, z) → if(lt(x, y), x, y, inc(z))
if(true, x, y, z) → z
if(false, x, s(y), z) → div(minus(x, s(y)), s(y), z)
minus(x, 0) → x
minus(s(x), s(y)) → minus(x, y)
lt(x, 0) → false
lt(0, s(y)) → true
lt(s(x), s(y)) → lt(x, y)
inc(0) → s(0)
inc(s(x)) → s(inc(x))

The set Q consists of the following terms:

division(x0, x1)
div(x0, x1, x2)
if(true, x0, x1, x2)
if(false, x0, s(x1), x2)
minus(x0, 0)
minus(s(x0), s(x1))
lt(x0, 0)
lt(0, s(x0))
lt(s(x0), s(x1))
inc(0)
inc(s(x0))

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

                  ↳ QDP

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

MINUS(s(x), s(y)) → MINUS(x, y)

The TRS R consists of the following rules:

division(x, y) → div(x, y, 0)
div(x, y, z) → if(lt(x, y), x, y, inc(z))
if(true, x, y, z) → z
if(false, x, s(y), z) → div(minus(x, s(y)), s(y), z)
minus(x, 0) → x
minus(s(x), s(y)) → minus(x, y)
lt(x, 0) → false
lt(0, s(y)) → true
lt(s(x), s(y)) → lt(x, y)
inc(0) → s(0)
inc(s(x)) → s(inc(x))

The set Q consists of the following terms:

division(x0, x1)
div(x0, x1, x2)
if(true, x0, x1, x2)
if(false, x0, s(x1), x2)
minus(x0, 0)
minus(s(x0), s(x1))
lt(x0, 0)
lt(0, s(x0))
lt(s(x0), s(x1))
inc(0)
inc(s(x0))

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.

MINUS(s(x), s(y)) → MINUS(x, y)

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

Lexicographic Path Order [19].
Precedence:

s₁ > MINUS₁

The following usable rules [14] were oriented: none

↳ QTRS

  ↳ Overlay + Local Confluence

    ↳ QTRS

      ↳ DependencyPairsProof

        ↳ QDP

          ↳ EdgeDeletionProof

            ↳ QDP

              ↳ DependencyGraphProof

                ↳ AND

                  ↳ QDP

                  ↳ QDP

                  ↳ QDP

                    ↳ QDPOrderProof

                      ↳ QDP

                        ↳ PisEmptyProof

                  ↳ QDP

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

division(x, y) → div(x, y, 0)
div(x, y, z) → if(lt(x, y), x, y, inc(z))
if(true, x, y, z) → z
if(false, x, s(y), z) → div(minus(x, s(y)), s(y), z)
minus(x, 0) → x
minus(s(x), s(y)) → minus(x, y)
lt(x, 0) → false
lt(0, s(y)) → true
lt(s(x), s(y)) → lt(x, y)
inc(0) → s(0)
inc(s(x)) → s(inc(x))

The set Q consists of the following terms:

division(x0, x1)
div(x0, x1, x2)
if(true, x0, x1, x2)
if(false, x0, s(x1), x2)
minus(x0, 0)
minus(s(x0), s(x1))
lt(x0, 0)
lt(0, s(x0))
lt(s(x0), s(x1))
inc(0)
inc(s(x0))

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

                  ↳ QDP

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

DIV(x, y, z) → IF(lt(x, y), x, y, inc(z))
IF(false, x, s(y), z) → DIV(minus(x, s(y)), s(y), z)

The TRS R consists of the following rules:

division(x, y) → div(x, y, 0)
div(x, y, z) → if(lt(x, y), x, y, inc(z))
if(true, x, y, z) → z
if(false, x, s(y), z) → div(minus(x, s(y)), s(y), z)
minus(x, 0) → x
minus(s(x), s(y)) → minus(x, y)
lt(x, 0) → false
lt(0, s(y)) → true
lt(s(x), s(y)) → lt(x, y)
inc(0) → s(0)
inc(s(x)) → s(inc(x))

The set Q consists of the following terms:

division(x0, x1)
div(x0, x1, x2)
if(true, x0, x1, x2)
if(false, x0, s(x1), x2)
minus(x0, 0)
minus(s(x0), s(x1))
lt(x0, 0)
lt(0, s(x0))
lt(s(x0), s(x1))
inc(0)
inc(s(x0))

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