Termination w.r.t. Q proof of /home/nowonder/forschung/aprove/satrung/aprove/lpar16/nonterminSLASHCSRSLASHEx49

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:

minus₂(0, Y) -> 0
minus₂(s₁(X), s₁(Y)) -> minus₂(X, Y)
geq₂(X, 0) -> true
geq₂(0, s₁(Y)) -> false
geq₂(s₁(X), s₁(Y)) -> geq₂(X, Y)
div₂(0, s₁(Y)) -> 0
div₂(s₁(X), s₁(Y)) -> if₃(geq₂(X, Y), s₁(div₂(minus₂(X, Y), s₁(Y))), 0)
if₃(true, X, Y) -> X
if₃(false, X, Y) -> Y

Q is empty.

↳ QTRS

  ↳ Non-Overlap Check

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

minus₂(0, Y) -> 0
minus₂(s₁(X), s₁(Y)) -> minus₂(X, Y)
geq₂(X, 0) -> true
geq₂(0, s₁(Y)) -> false
geq₂(s₁(X), s₁(Y)) -> geq₂(X, Y)
div₂(0, s₁(Y)) -> 0
div₂(s₁(X), s₁(Y)) -> if₃(geq₂(X, Y), s₁(div₂(minus₂(X, Y), s₁(Y))), 0)
if₃(true, X, Y) -> X
if₃(false, 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:

minus₂(0, Y) -> 0
minus₂(s₁(X), s₁(Y)) -> minus₂(X, Y)
geq₂(X, 0) -> true
geq₂(0, s₁(Y)) -> false
geq₂(s₁(X), s₁(Y)) -> geq₂(X, Y)
div₂(0, s₁(Y)) -> 0
div₂(s₁(X), s₁(Y)) -> if₃(geq₂(X, Y), s₁(div₂(minus₂(X, Y), s₁(Y))), 0)
if₃(true, X, Y) -> X
if₃(false, X, Y) -> Y

The set Q consists of the following terms:

minus₂(0, x0)
minus₂(s₁(x0), s₁(x1))
geq₂(x0, 0)
geq₂(0, s₁(x0))
geq₂(s₁(x0), s₁(x1))
div₂(0, s₁(x0))
div₂(s₁(x0), s₁(x1))
if₃(true, x0, x1)
if₃(false, 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:

DIV₂(s₁(X), s₁(Y)) -> IF₃(geq₂(X, Y), s₁(div₂(minus₂(X, Y), s₁(Y))), 0)
DIV₂(s₁(X), s₁(Y)) -> MINUS₂(X, Y)
DIV₂(s₁(X), s₁(Y)) -> DIV₂(minus₂(X, Y), s₁(Y))
DIV₂(s₁(X), s₁(Y)) -> GEQ₂(X, Y)
GEQ₂(s₁(X), s₁(Y)) -> GEQ₂(X, Y)
MINUS₂(s₁(X), s₁(Y)) -> MINUS₂(X, Y)

The TRS R consists of the following rules:

minus₂(0, Y) -> 0
minus₂(s₁(X), s₁(Y)) -> minus₂(X, Y)
geq₂(X, 0) -> true
geq₂(0, s₁(Y)) -> false
geq₂(s₁(X), s₁(Y)) -> geq₂(X, Y)
div₂(0, s₁(Y)) -> 0
div₂(s₁(X), s₁(Y)) -> if₃(geq₂(X, Y), s₁(div₂(minus₂(X, Y), s₁(Y))), 0)
if₃(true, X, Y) -> X
if₃(false, X, Y) -> Y

The set Q consists of the following terms:

minus₂(0, x0)
minus₂(s₁(x0), s₁(x1))
geq₂(x0, 0)
geq₂(0, s₁(x0))
geq₂(s₁(x0), s₁(x1))
div₂(0, s₁(x0))
div₂(s₁(x0), s₁(x1))
if₃(true, x0, x1)
if₃(false, 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:

DIV₂(s₁(X), s₁(Y)) -> IF₃(geq₂(X, Y), s₁(div₂(minus₂(X, Y), s₁(Y))), 0)
DIV₂(s₁(X), s₁(Y)) -> MINUS₂(X, Y)
DIV₂(s₁(X), s₁(Y)) -> DIV₂(minus₂(X, Y), s₁(Y))
DIV₂(s₁(X), s₁(Y)) -> GEQ₂(X, Y)
GEQ₂(s₁(X), s₁(Y)) -> GEQ₂(X, Y)
MINUS₂(s₁(X), s₁(Y)) -> MINUS₂(X, Y)

The TRS R consists of the following rules:

minus₂(0, Y) -> 0
minus₂(s₁(X), s₁(Y)) -> minus₂(X, Y)
geq₂(X, 0) -> true
geq₂(0, s₁(Y)) -> false
geq₂(s₁(X), s₁(Y)) -> geq₂(X, Y)
div₂(0, s₁(Y)) -> 0
div₂(s₁(X), s₁(Y)) -> if₃(geq₂(X, Y), s₁(div₂(minus₂(X, Y), s₁(Y))), 0)
if₃(true, X, Y) -> X
if₃(false, X, Y) -> Y

The set Q consists of the following terms:

minus₂(0, x0)
minus₂(s₁(x0), s₁(x1))
geq₂(x0, 0)
geq₂(0, s₁(x0))
geq₂(s₁(x0), s₁(x1))
div₂(0, s₁(x0))
div₂(s₁(x0), s₁(x1))
if₃(true, x0, x1)
if₃(false, x0, x1)

We have to consider all minimal (P,Q,R)-chains.
The approximation of the Dependency Graph [13,14,18] contains 2 SCCs with 4 less nodes.

↳ QTRS

  ↳ Non-Overlap Check

    ↳ QTRS

      ↳ DependencyPairsProof

        ↳ QDP

          ↳ DependencyGraphProof

            ↳ AND

              ↳ QDP

                ↳ QDPOrderProof

              ↳ QDP

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

GEQ₂(s₁(X), s₁(Y)) -> GEQ₂(X, Y)

The TRS R consists of the following rules:

minus₂(0, Y) -> 0
minus₂(s₁(X), s₁(Y)) -> minus₂(X, Y)
geq₂(X, 0) -> true
geq₂(0, s₁(Y)) -> false
geq₂(s₁(X), s₁(Y)) -> geq₂(X, Y)
div₂(0, s₁(Y)) -> 0
div₂(s₁(X), s₁(Y)) -> if₃(geq₂(X, Y), s₁(div₂(minus₂(X, Y), s₁(Y))), 0)
if₃(true, X, Y) -> X
if₃(false, X, Y) -> Y

The set Q consists of the following terms:

minus₂(0, x0)
minus₂(s₁(x0), s₁(x1))
geq₂(x0, 0)
geq₂(0, s₁(x0))
geq₂(s₁(x0), s₁(x1))
div₂(0, s₁(x0))
div₂(s₁(x0), s₁(x1))
if₃(true, x0, x1)
if₃(false, 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.

GEQ₂(s₁(X), s₁(Y)) -> GEQ₂(X, Y)

The remaining pairs can at least by weakly be oriented.
none
Used ordering: Combined order from the following AFS and order.
GEQ₂(x1, x2) = GEQ₁(x1)
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

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

minus₂(0, Y) -> 0
minus₂(s₁(X), s₁(Y)) -> minus₂(X, Y)
geq₂(X, 0) -> true
geq₂(0, s₁(Y)) -> false
geq₂(s₁(X), s₁(Y)) -> geq₂(X, Y)
div₂(0, s₁(Y)) -> 0
div₂(s₁(X), s₁(Y)) -> if₃(geq₂(X, Y), s₁(div₂(minus₂(X, Y), s₁(Y))), 0)
if₃(true, X, Y) -> X
if₃(false, X, Y) -> Y

The set Q consists of the following terms:

minus₂(0, x0)
minus₂(s₁(x0), s₁(x1))
geq₂(x0, 0)
geq₂(0, s₁(x0))
geq₂(s₁(x0), s₁(x1))
div₂(0, s₁(x0))
div₂(s₁(x0), s₁(x1))
if₃(true, x0, x1)
if₃(false, 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

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:

minus₂(0, Y) -> 0
minus₂(s₁(X), s₁(Y)) -> minus₂(X, Y)
geq₂(X, 0) -> true
geq₂(0, s₁(Y)) -> false
geq₂(s₁(X), s₁(Y)) -> geq₂(X, Y)
div₂(0, s₁(Y)) -> 0
div₂(s₁(X), s₁(Y)) -> if₃(geq₂(X, Y), s₁(div₂(minus₂(X, Y), s₁(Y))), 0)
if₃(true, X, Y) -> X
if₃(false, X, Y) -> Y

The set Q consists of the following terms:

minus₂(0, x0)
minus₂(s₁(x0), s₁(x1))
geq₂(x0, 0)
geq₂(0, s₁(x0))
geq₂(s₁(x0), s₁(x1))
div₂(0, s₁(x0))
div₂(s₁(x0), s₁(x1))
if₃(true, x0, x1)
if₃(false, 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) = MINUS₁(x1)
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

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

minus₂(0, Y) -> 0
minus₂(s₁(X), s₁(Y)) -> minus₂(X, Y)
geq₂(X, 0) -> true
geq₂(0, s₁(Y)) -> false
geq₂(s₁(X), s₁(Y)) -> geq₂(X, Y)
div₂(0, s₁(Y)) -> 0
div₂(s₁(X), s₁(Y)) -> if₃(geq₂(X, Y), s₁(div₂(minus₂(X, Y), s₁(Y))), 0)
if₃(true, X, Y) -> X
if₃(false, X, Y) -> Y

The set Q consists of the following terms:

minus₂(0, x0)
minus₂(s₁(x0), s₁(x1))
geq₂(x0, 0)
geq₂(0, s₁(x0))
geq₂(s₁(x0), s₁(x1))
div₂(0, s₁(x0))
div₂(s₁(x0), s₁(x1))
if₃(true, x0, x1)
if₃(false, 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.