Termination w.r.t. Q proof of /home/nowonder/forschung/aprove/satrung/aprove/lpar14/TRCSRSLASHEx49_GM04

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(n__0, Y) -> 0
minus₂(n__s₁(X), n__s₁(Y)) -> minus₂(activate₁(X), activate₁(Y))
geq₂(X, n__0) -> true
geq₂(n__0, n__s₁(Y)) -> false
geq₂(n__s₁(X), n__s₁(Y)) -> geq₂(activate₁(X), activate₁(Y))
div₂(0, n__s₁(Y)) -> 0
div₂(s₁(X), n__s₁(Y)) -> if₃(geq₂(X, activate₁(Y)), n__s₁(div₂(minus₂(X, activate₁(Y)), n__s₁(activate₁(Y)))), n__0)
if₃(true, X, Y) -> activate₁(X)
if₃(false, X, Y) -> activate₁(Y)
0 -> n__0
s₁(X) -> n__s₁(X)
activate₁(n__0) -> 0
activate₁(n__s₁(X)) -> s₁(X)
activate₁(X) -> X

Q is empty.

↳ QTRS

  ↳ DependencyPairsProof

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

minus₂(n__0, Y) -> 0
minus₂(n__s₁(X), n__s₁(Y)) -> minus₂(activate₁(X), activate₁(Y))
geq₂(X, n__0) -> true
geq₂(n__0, n__s₁(Y)) -> false
geq₂(n__s₁(X), n__s₁(Y)) -> geq₂(activate₁(X), activate₁(Y))
div₂(0, n__s₁(Y)) -> 0
div₂(s₁(X), n__s₁(Y)) -> if₃(geq₂(X, activate₁(Y)), n__s₁(div₂(minus₂(X, activate₁(Y)), n__s₁(activate₁(Y)))), n__0)
if₃(true, X, Y) -> activate₁(X)
if₃(false, X, Y) -> activate₁(Y)
0 -> n__0
s₁(X) -> n__s₁(X)
activate₁(n__0) -> 0
activate₁(n__s₁(X)) -> s₁(X)
activate₁(X) -> X

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:

GEQ₂(n__s₁(X), n__s₁(Y)) -> ACTIVATE₁(Y)
ACTIVATE₁(n__0) -> 0¹
ACTIVATE₁(n__s₁(X)) -> S₁(X)
DIV₂(s₁(X), n__s₁(Y)) -> IF₃(geq₂(X, activate₁(Y)), n__s₁(div₂(minus₂(X, activate₁(Y)), n__s₁(activate₁(Y)))), n__0)
IF₃(true, X, Y) -> ACTIVATE₁(X)
GEQ₂(n__s₁(X), n__s₁(Y)) -> ACTIVATE₁(X)
MINUS₂(n__s₁(X), n__s₁(Y)) -> ACTIVATE₁(Y)
DIV₂(s₁(X), n__s₁(Y)) -> ACTIVATE₁(Y)
DIV₂(s₁(X), n__s₁(Y)) -> GEQ₂(X, activate₁(Y))
MINUS₂(n__s₁(X), n__s₁(Y)) -> MINUS₂(activate₁(X), activate₁(Y))
MINUS₂(n__s₁(X), n__s₁(Y)) -> ACTIVATE₁(X)
IF₃(false, X, Y) -> ACTIVATE₁(Y)
MINUS₂(n__0, Y) -> 0¹
DIV₂(s₁(X), n__s₁(Y)) -> MINUS₂(X, activate₁(Y))
GEQ₂(n__s₁(X), n__s₁(Y)) -> GEQ₂(activate₁(X), activate₁(Y))
DIV₂(s₁(X), n__s₁(Y)) -> DIV₂(minus₂(X, activate₁(Y)), n__s₁(activate₁(Y)))

The TRS R consists of the following rules:

minus₂(n__0, Y) -> 0
minus₂(n__s₁(X), n__s₁(Y)) -> minus₂(activate₁(X), activate₁(Y))
geq₂(X, n__0) -> true
geq₂(n__0, n__s₁(Y)) -> false
geq₂(n__s₁(X), n__s₁(Y)) -> geq₂(activate₁(X), activate₁(Y))
div₂(0, n__s₁(Y)) -> 0
div₂(s₁(X), n__s₁(Y)) -> if₃(geq₂(X, activate₁(Y)), n__s₁(div₂(minus₂(X, activate₁(Y)), n__s₁(activate₁(Y)))), n__0)
if₃(true, X, Y) -> activate₁(X)
if₃(false, X, Y) -> activate₁(Y)
0 -> n__0
s₁(X) -> n__s₁(X)
activate₁(n__0) -> 0
activate₁(n__s₁(X)) -> s₁(X)
activate₁(X) -> X

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:

GEQ₂(n__s₁(X), n__s₁(Y)) -> ACTIVATE₁(Y)
ACTIVATE₁(n__0) -> 0¹
ACTIVATE₁(n__s₁(X)) -> S₁(X)
DIV₂(s₁(X), n__s₁(Y)) -> IF₃(geq₂(X, activate₁(Y)), n__s₁(div₂(minus₂(X, activate₁(Y)), n__s₁(activate₁(Y)))), n__0)
IF₃(true, X, Y) -> ACTIVATE₁(X)
GEQ₂(n__s₁(X), n__s₁(Y)) -> ACTIVATE₁(X)
MINUS₂(n__s₁(X), n__s₁(Y)) -> ACTIVATE₁(Y)
DIV₂(s₁(X), n__s₁(Y)) -> ACTIVATE₁(Y)
DIV₂(s₁(X), n__s₁(Y)) -> GEQ₂(X, activate₁(Y))
MINUS₂(n__s₁(X), n__s₁(Y)) -> MINUS₂(activate₁(X), activate₁(Y))
MINUS₂(n__s₁(X), n__s₁(Y)) -> ACTIVATE₁(X)
IF₃(false, X, Y) -> ACTIVATE₁(Y)
MINUS₂(n__0, Y) -> 0¹
DIV₂(s₁(X), n__s₁(Y)) -> MINUS₂(X, activate₁(Y))
GEQ₂(n__s₁(X), n__s₁(Y)) -> GEQ₂(activate₁(X), activate₁(Y))
DIV₂(s₁(X), n__s₁(Y)) -> DIV₂(minus₂(X, activate₁(Y)), n__s₁(activate₁(Y)))

The TRS R consists of the following rules:

minus₂(n__0, Y) -> 0
minus₂(n__s₁(X), n__s₁(Y)) -> minus₂(activate₁(X), activate₁(Y))
geq₂(X, n__0) -> true
geq₂(n__0, n__s₁(Y)) -> false
geq₂(n__s₁(X), n__s₁(Y)) -> geq₂(activate₁(X), activate₁(Y))
div₂(0, n__s₁(Y)) -> 0
div₂(s₁(X), n__s₁(Y)) -> if₃(geq₂(X, activate₁(Y)), n__s₁(div₂(minus₂(X, activate₁(Y)), n__s₁(activate₁(Y)))), n__0)
if₃(true, X, Y) -> activate₁(X)
if₃(false, X, Y) -> activate₁(Y)
0 -> n__0
s₁(X) -> n__s₁(X)
activate₁(n__0) -> 0
activate₁(n__s₁(X)) -> s₁(X)
activate₁(X) -> X

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 13 less nodes.

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ DependencyGraphProof

        ↳ AND

          ↳ QDP

            ↳ QDPOrderProof

          ↳ QDP

          ↳ QDP

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

GEQ₂(n__s₁(X), n__s₁(Y)) -> GEQ₂(activate₁(X), activate₁(Y))

The TRS R consists of the following rules:

minus₂(n__0, Y) -> 0
minus₂(n__s₁(X), n__s₁(Y)) -> minus₂(activate₁(X), activate₁(Y))
geq₂(X, n__0) -> true
geq₂(n__0, n__s₁(Y)) -> false
geq₂(n__s₁(X), n__s₁(Y)) -> geq₂(activate₁(X), activate₁(Y))
div₂(0, n__s₁(Y)) -> 0
div₂(s₁(X), n__s₁(Y)) -> if₃(geq₂(X, activate₁(Y)), n__s₁(div₂(minus₂(X, activate₁(Y)), n__s₁(activate₁(Y)))), n__0)
if₃(true, X, Y) -> activate₁(X)
if₃(false, X, Y) -> activate₁(Y)
0 -> n__0
s₁(X) -> n__s₁(X)
activate₁(n__0) -> 0
activate₁(n__s₁(X)) -> s₁(X)
activate₁(X) -> X

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 strictly oriented and are deleted.

GEQ₂(n__s₁(X), n__s₁(Y)) -> GEQ₂(activate₁(X), activate₁(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(x2)
n__s₁(x1) = n__s₁(x1)
activate₁(x1) = x1
s₁(x1) = s₁(x1)
n__0 = n__0
0 = 0

Lexicographic Path Order [19].
Precedence:

[n_{_s₁}, s_1]
[n_₀, 0]

The following usable rules [14] were oriented:

activate₁(n__0) -> 0
activate₁(n__s₁(X)) -> s₁(X)
activate₁(X) -> X
s₁(X) -> n__s₁(X)
0 -> n__0

↳ 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:

minus₂(n__0, Y) -> 0
minus₂(n__s₁(X), n__s₁(Y)) -> minus₂(activate₁(X), activate₁(Y))
geq₂(X, n__0) -> true
geq₂(n__0, n__s₁(Y)) -> false
geq₂(n__s₁(X), n__s₁(Y)) -> geq₂(activate₁(X), activate₁(Y))
div₂(0, n__s₁(Y)) -> 0
div₂(s₁(X), n__s₁(Y)) -> if₃(geq₂(X, activate₁(Y)), n__s₁(div₂(minus₂(X, activate₁(Y)), n__s₁(activate₁(Y)))), n__0)
if₃(true, X, Y) -> activate₁(X)
if₃(false, X, Y) -> activate₁(Y)
0 -> n__0
s₁(X) -> n__s₁(X)
activate₁(n__0) -> 0
activate₁(n__s₁(X)) -> s₁(X)
activate₁(X) -> X

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:

MINUS₂(n__s₁(X), n__s₁(Y)) -> MINUS₂(activate₁(X), activate₁(Y))

The TRS R consists of the following rules:

minus₂(n__0, Y) -> 0
minus₂(n__s₁(X), n__s₁(Y)) -> minus₂(activate₁(X), activate₁(Y))
geq₂(X, n__0) -> true
geq₂(n__0, n__s₁(Y)) -> false
geq₂(n__s₁(X), n__s₁(Y)) -> geq₂(activate₁(X), activate₁(Y))
div₂(0, n__s₁(Y)) -> 0
div₂(s₁(X), n__s₁(Y)) -> if₃(geq₂(X, activate₁(Y)), n__s₁(div₂(minus₂(X, activate₁(Y)), n__s₁(activate₁(Y)))), n__0)
if₃(true, X, Y) -> activate₁(X)
if₃(false, X, Y) -> activate₁(Y)
0 -> n__0
s₁(X) -> n__s₁(X)
activate₁(n__0) -> 0
activate₁(n__s₁(X)) -> s₁(X)
activate₁(X) -> X

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 strictly oriented and are deleted.

MINUS₂(n__s₁(X), n__s₁(Y)) -> MINUS₂(activate₁(X), activate₁(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(x2)
n__s₁(x1) = n__s₁(x1)
activate₁(x1) = x1
s₁(x1) = s₁(x1)
n__0 = n__0
0 = 0

Lexicographic Path Order [19].
Precedence:

[n_{_s₁}, s_1]
[n_₀, 0]

The following usable rules [14] were oriented:

activate₁(n__0) -> 0
activate₁(n__s₁(X)) -> s₁(X)
activate₁(X) -> X
s₁(X) -> n__s₁(X)
0 -> n__0

↳ 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:

minus₂(n__0, Y) -> 0
minus₂(n__s₁(X), n__s₁(Y)) -> minus₂(activate₁(X), activate₁(Y))
geq₂(X, n__0) -> true
geq₂(n__0, n__s₁(Y)) -> false
geq₂(n__s₁(X), n__s₁(Y)) -> geq₂(activate₁(X), activate₁(Y))
div₂(0, n__s₁(Y)) -> 0
div₂(s₁(X), n__s₁(Y)) -> if₃(geq₂(X, activate₁(Y)), n__s₁(div₂(minus₂(X, activate₁(Y)), n__s₁(activate₁(Y)))), n__0)
if₃(true, X, Y) -> activate₁(X)
if₃(false, X, Y) -> activate₁(Y)
0 -> n__0
s₁(X) -> n__s₁(X)
activate₁(n__0) -> 0
activate₁(n__s₁(X)) -> s₁(X)
activate₁(X) -> X

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), n__s₁(Y)) -> DIV₂(minus₂(X, activate₁(Y)), n__s₁(activate₁(Y)))

The TRS R consists of the following rules:

minus₂(n__0, Y) -> 0
minus₂(n__s₁(X), n__s₁(Y)) -> minus₂(activate₁(X), activate₁(Y))
geq₂(X, n__0) -> true
geq₂(n__0, n__s₁(Y)) -> false
geq₂(n__s₁(X), n__s₁(Y)) -> geq₂(activate₁(X), activate₁(Y))
div₂(0, n__s₁(Y)) -> 0
div₂(s₁(X), n__s₁(Y)) -> if₃(geq₂(X, activate₁(Y)), n__s₁(div₂(minus₂(X, activate₁(Y)), n__s₁(activate₁(Y)))), n__0)
if₃(true, X, Y) -> activate₁(X)
if₃(false, X, Y) -> activate₁(Y)
0 -> n__0
s₁(X) -> n__s₁(X)
activate₁(n__0) -> 0
activate₁(n__s₁(X)) -> s₁(X)
activate₁(X) -> X

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 strictly oriented and are deleted.

DIV₂(s₁(X), n__s₁(Y)) -> DIV₂(minus₂(X, activate₁(Y)), n__s₁(activate₁(Y)))

The remaining pairs can at least by weakly be oriented.
none
Used ordering: Combined order from the following AFS and order.
DIV₂(x1, x2) = DIV₁(x1)
s₁(x1) = s₁(x1)
n__s₁(x1) = x1
minus₂(x1, x2) = minus
activate₁(x1) = activate₁(x1)
n__0 = n__0
0 = 0

Lexicographic Path Order [19].
Precedence:

[DIV₁, s₁, minus] > activate₁ > [n_₀, 0]

The following usable rules [14] were oriented:

minus₂(n__0, Y) -> 0
minus₂(n__s₁(X), n__s₁(Y)) -> minus₂(activate₁(X), activate₁(Y))
0 -> n__0

↳ 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:

minus₂(n__0, Y) -> 0
minus₂(n__s₁(X), n__s₁(Y)) -> minus₂(activate₁(X), activate₁(Y))
geq₂(X, n__0) -> true
geq₂(n__0, n__s₁(Y)) -> false
geq₂(n__s₁(X), n__s₁(Y)) -> geq₂(activate₁(X), activate₁(Y))
div₂(0, n__s₁(Y)) -> 0
div₂(s₁(X), n__s₁(Y)) -> if₃(geq₂(X, activate₁(Y)), n__s₁(div₂(minus₂(X, activate₁(Y)), n__s₁(activate₁(Y)))), n__0)
if₃(true, X, Y) -> activate₁(X)
if₃(false, X, Y) -> activate₁(Y)
0 -> n__0
s₁(X) -> n__s₁(X)
activate₁(n__0) -> 0
activate₁(n__s₁(X)) -> s₁(X)
activate₁(X) -> X

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.