Termination w.r.t. Q proof of /home/nowonder/forschung/aprove/satrung/aprove/lpar18/secret2005SLASHaprove2.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:

ge₂(x, 0) -> true
ge₂(0, s₁(x)) -> false
ge₂(s₁(x), s₁(y)) -> ge₂(x, y)
minus₂(x, 0) -> x
minus₂(s₁(x), s₁(y)) -> minus₂(x, y)
div₂(x, y) -> ify₃(ge₂(y, s₁(0)), x, y)
ify₃(false, x, y) -> divByZeroError
ify₃(true, x, y) -> if₃(ge₂(x, y), x, y)
if₃(false, x, y) -> 0
if₃(true, x, y) -> s₁(div₂(minus₂(x, y), y))

Q is empty.

↳ QTRS

  ↳ Non-Overlap Check

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

ge₂(x, 0) -> true
ge₂(0, s₁(x)) -> false
ge₂(s₁(x), s₁(y)) -> ge₂(x, y)
minus₂(x, 0) -> x
minus₂(s₁(x), s₁(y)) -> minus₂(x, y)
div₂(x, y) -> ify₃(ge₂(y, s₁(0)), x, y)
ify₃(false, x, y) -> divByZeroError
ify₃(true, x, y) -> if₃(ge₂(x, y), x, y)
if₃(false, x, y) -> 0
if₃(true, x, y) -> s₁(div₂(minus₂(x, y), y))

Q is empty.

The TRS is non-overlapping. Hence, we can switch to innermost.

↳ QTRS

  ↳ Non-Overlap Check

    ↳ QTRS

      ↳ DependencyPairsProof

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

ge₂(x, 0) -> true
ge₂(0, s₁(x)) -> false
ge₂(s₁(x), s₁(y)) -> ge₂(x, y)
minus₂(x, 0) -> x
minus₂(s₁(x), s₁(y)) -> minus₂(x, y)
div₂(x, y) -> ify₃(ge₂(y, s₁(0)), x, y)
ify₃(false, x, y) -> divByZeroError
ify₃(true, x, y) -> if₃(ge₂(x, y), x, y)
if₃(false, x, y) -> 0
if₃(true, x, y) -> s₁(div₂(minus₂(x, y), y))

The set Q consists of the following terms:

ge₂(x0, 0)
ge₂(0, s₁(x0))
ge₂(s₁(x0), s₁(x1))
minus₂(x0, 0)
minus₂(s₁(x0), s₁(x1))
div₂(x0, x1)
ify₃(false, x0, x1)
ify₃(true, x0, x1)
if₃(false, x0, x1)
if₃(true, x0, 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:

IF₃(true, x, y) -> DIV₂(minus₂(x, y), y)
IF₃(true, x, y) -> MINUS₂(x, y)
IFY₃(true, x, y) -> GE₂(x, y)
IFY₃(true, x, y) -> IF₃(ge₂(x, y), x, y)
DIV₂(x, y) -> IFY₃(ge₂(y, s₁(0)), x, y)
MINUS₂(s₁(x), s₁(y)) -> MINUS₂(x, y)
DIV₂(x, y) -> GE₂(y, s₁(0))
GE₂(s₁(x), s₁(y)) -> GE₂(x, y)

The TRS R consists of the following rules:

ge₂(x, 0) -> true
ge₂(0, s₁(x)) -> false
ge₂(s₁(x), s₁(y)) -> ge₂(x, y)
minus₂(x, 0) -> x
minus₂(s₁(x), s₁(y)) -> minus₂(x, y)
div₂(x, y) -> ify₃(ge₂(y, s₁(0)), x, y)
ify₃(false, x, y) -> divByZeroError
ify₃(true, x, y) -> if₃(ge₂(x, y), x, y)
if₃(false, x, y) -> 0
if₃(true, x, y) -> s₁(div₂(minus₂(x, y), y))

The set Q consists of the following terms:

ge₂(x0, 0)
ge₂(0, s₁(x0))
ge₂(s₁(x0), s₁(x1))
minus₂(x0, 0)
minus₂(s₁(x0), s₁(x1))
div₂(x0, x1)
ify₃(false, x0, x1)
ify₃(true, x0, x1)
if₃(false, x0, x1)
if₃(true, x0, x1)

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

↳ QTRS

  ↳ Non-Overlap Check

    ↳ QTRS

      ↳ DependencyPairsProof

        ↳ QDP

          ↳ DependencyGraphProof

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

IF₃(true, x, y) -> DIV₂(minus₂(x, y), y)
IF₃(true, x, y) -> MINUS₂(x, y)
IFY₃(true, x, y) -> GE₂(x, y)
IFY₃(true, x, y) -> IF₃(ge₂(x, y), x, y)
DIV₂(x, y) -> IFY₃(ge₂(y, s₁(0)), x, y)
MINUS₂(s₁(x), s₁(y)) -> MINUS₂(x, y)
DIV₂(x, y) -> GE₂(y, s₁(0))
GE₂(s₁(x), s₁(y)) -> GE₂(x, y)

The TRS R consists of the following rules:

ge₂(x, 0) -> true
ge₂(0, s₁(x)) -> false
ge₂(s₁(x), s₁(y)) -> ge₂(x, y)
minus₂(x, 0) -> x
minus₂(s₁(x), s₁(y)) -> minus₂(x, y)
div₂(x, y) -> ify₃(ge₂(y, s₁(0)), x, y)
ify₃(false, x, y) -> divByZeroError
ify₃(true, x, y) -> if₃(ge₂(x, y), x, y)
if₃(false, x, y) -> 0
if₃(true, x, y) -> s₁(div₂(minus₂(x, y), y))

The set Q consists of the following terms:

ge₂(x0, 0)
ge₂(0, s₁(x0))
ge₂(s₁(x0), s₁(x1))
minus₂(x0, 0)
minus₂(s₁(x0), s₁(x1))
div₂(x0, x1)
ify₃(false, x0, x1)
ify₃(true, x0, x1)
if₃(false, x0, x1)
if₃(true, x0, 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

  ↳ Non-Overlap Check

    ↳ QTRS

      ↳ DependencyPairsProof

        ↳ QDP

          ↳ DependencyGraphProof

            ↳ AND

              ↳ QDP

                ↳ QDPOrderProof

              ↳ QDP

              ↳ 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:

ge₂(x, 0) -> true
ge₂(0, s₁(x)) -> false
ge₂(s₁(x), s₁(y)) -> ge₂(x, y)
minus₂(x, 0) -> x
minus₂(s₁(x), s₁(y)) -> minus₂(x, y)
div₂(x, y) -> ify₃(ge₂(y, s₁(0)), x, y)
ify₃(false, x, y) -> divByZeroError
ify₃(true, x, y) -> if₃(ge₂(x, y), x, y)
if₃(false, x, y) -> 0
if₃(true, x, y) -> s₁(div₂(minus₂(x, y), y))

The set Q consists of the following terms:

ge₂(x0, 0)
ge₂(0, s₁(x0))
ge₂(s₁(x0), s₁(x1))
minus₂(x0, 0)
minus₂(s₁(x0), s₁(x1))
div₂(x0, x1)
ify₃(false, x0, x1)
ify₃(true, x0, x1)
if₃(false, x0, x1)
if₃(true, x0, x1)

We have to consider all minimal (P,Q,R)-chains.
We use the reduction pair processor [13].

The following pairs can be strictly oriented and are deleted.

MINUS₂(s₁(x), s₁(y)) -> MINUS₂(x, y)

The remaining pairs can at least by weakly be oriented.
none
Used ordering: Combined order from the following AFS and order.
MINUS₂(x1, x2) = x2
s₁(x1) = s₁(x1)

Lexicographic Path Order [19].
Precedence:

trivial

The following usable rules [14] were oriented: none

↳ QTRS

  ↳ Non-Overlap Check

    ↳ 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:

ge₂(x, 0) -> true
ge₂(0, s₁(x)) -> false
ge₂(s₁(x), s₁(y)) -> ge₂(x, y)
minus₂(x, 0) -> x
minus₂(s₁(x), s₁(y)) -> minus₂(x, y)
div₂(x, y) -> ify₃(ge₂(y, s₁(0)), x, y)
ify₃(false, x, y) -> divByZeroError
ify₃(true, x, y) -> if₃(ge₂(x, y), x, y)
if₃(false, x, y) -> 0
if₃(true, x, y) -> s₁(div₂(minus₂(x, y), y))

The set Q consists of the following terms:

ge₂(x0, 0)
ge₂(0, s₁(x0))
ge₂(s₁(x0), s₁(x1))
minus₂(x0, 0)
minus₂(s₁(x0), s₁(x1))
div₂(x0, x1)
ify₃(false, x0, x1)
ify₃(true, x0, x1)
if₃(false, x0, x1)
if₃(true, x0, 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

  ↳ Non-Overlap Check

    ↳ QTRS

      ↳ DependencyPairsProof

        ↳ QDP

          ↳ DependencyGraphProof

            ↳ AND

              ↳ QDP

              ↳ QDP

                ↳ QDPOrderProof

              ↳ QDP

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

GE₂(s₁(x), s₁(y)) -> GE₂(x, y)

The TRS R consists of the following rules:

ge₂(x, 0) -> true
ge₂(0, s₁(x)) -> false
ge₂(s₁(x), s₁(y)) -> ge₂(x, y)
minus₂(x, 0) -> x
minus₂(s₁(x), s₁(y)) -> minus₂(x, y)
div₂(x, y) -> ify₃(ge₂(y, s₁(0)), x, y)
ify₃(false, x, y) -> divByZeroError
ify₃(true, x, y) -> if₃(ge₂(x, y), x, y)
if₃(false, x, y) -> 0
if₃(true, x, y) -> s₁(div₂(minus₂(x, y), y))

The set Q consists of the following terms:

ge₂(x0, 0)
ge₂(0, s₁(x0))
ge₂(s₁(x0), s₁(x1))
minus₂(x0, 0)
minus₂(s₁(x0), s₁(x1))
div₂(x0, x1)
ify₃(false, x0, x1)
ify₃(true, x0, x1)
if₃(false, x0, x1)
if₃(true, x0, x1)

We have to consider all minimal (P,Q,R)-chains.
We use the reduction pair processor [13].

The following pairs can be strictly oriented and are deleted.

GE₂(s₁(x), s₁(y)) -> GE₂(x, y)

The remaining pairs can at least by weakly be oriented.
none
Used ordering: Combined order from the following AFS and order.
GE₂(x1, x2) = x2
s₁(x1) = s₁(x1)

Lexicographic Path Order [19].
Precedence:

trivial

The following usable rules [14] were oriented: none

↳ QTRS

  ↳ Non-Overlap Check

    ↳ 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:

ge₂(x, 0) -> true
ge₂(0, s₁(x)) -> false
ge₂(s₁(x), s₁(y)) -> ge₂(x, y)
minus₂(x, 0) -> x
minus₂(s₁(x), s₁(y)) -> minus₂(x, y)
div₂(x, y) -> ify₃(ge₂(y, s₁(0)), x, y)
ify₃(false, x, y) -> divByZeroError
ify₃(true, x, y) -> if₃(ge₂(x, y), x, y)
if₃(false, x, y) -> 0
if₃(true, x, y) -> s₁(div₂(minus₂(x, y), y))

The set Q consists of the following terms:

ge₂(x0, 0)
ge₂(0, s₁(x0))
ge₂(s₁(x0), s₁(x1))
minus₂(x0, 0)
minus₂(s₁(x0), s₁(x1))
div₂(x0, x1)
ify₃(false, x0, x1)
ify₃(true, x0, x1)
if₃(false, x0, x1)
if₃(true, x0, 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

  ↳ Non-Overlap Check

    ↳ QTRS

      ↳ DependencyPairsProof

        ↳ QDP

          ↳ DependencyGraphProof

            ↳ AND

              ↳ QDP

              ↳ QDP

              ↳ QDP

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

IF₃(true, x, y) -> DIV₂(minus₂(x, y), y)
IFY₃(true, x, y) -> IF₃(ge₂(x, y), x, y)
DIV₂(x, y) -> IFY₃(ge₂(y, s₁(0)), x, y)

The TRS R consists of the following rules:

ge₂(x, 0) -> true
ge₂(0, s₁(x)) -> false
ge₂(s₁(x), s₁(y)) -> ge₂(x, y)
minus₂(x, 0) -> x
minus₂(s₁(x), s₁(y)) -> minus₂(x, y)
div₂(x, y) -> ify₃(ge₂(y, s₁(0)), x, y)
ify₃(false, x, y) -> divByZeroError
ify₃(true, x, y) -> if₃(ge₂(x, y), x, y)
if₃(false, x, y) -> 0
if₃(true, x, y) -> s₁(div₂(minus₂(x, y), y))

The set Q consists of the following terms:

ge₂(x0, 0)
ge₂(0, s₁(x0))
ge₂(s₁(x0), s₁(x1))
minus₂(x0, 0)
minus₂(s₁(x0), s₁(x1))
div₂(x0, x1)
ify₃(false, x0, x1)
ify₃(true, x0, x1)
if₃(false, x0, x1)
if₃(true, x0, x1)

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