Termination w.r.t. Q proof of TRS/TRCSREx49_GM04

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:

minus(n__0, Y) → 0
minus(n__s(X), n__s(Y)) → minus(activate(X), activate(Y))
geq(X, n__0) → true
geq(n__0, n__s(Y)) → false
geq(n__s(X), n__s(Y)) → geq(activate(X), activate(Y))
div(0, n__s(Y)) → 0
div(s(X), n__s(Y)) → if(geq(X, activate(Y)), n__s(div(minus(X, activate(Y)), n__s(activate(Y)))), n__0)
if(true, X, Y) → activate(X)
if(false, X, Y) → activate(Y)
0 → n__0
s(X) → n__s(X)
activate(n__0) → 0
activate(n__s(X)) → s(X)
activate(X) → X

Q is empty.

↳ QTRS

  ↳ DependencyPairsProof

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

minus(n__0, Y) → 0
minus(n__s(X), n__s(Y)) → minus(activate(X), activate(Y))
geq(X, n__0) → true
geq(n__0, n__s(Y)) → false
geq(n__s(X), n__s(Y)) → geq(activate(X), activate(Y))
div(0, n__s(Y)) → 0
div(s(X), n__s(Y)) → if(geq(X, activate(Y)), n__s(div(minus(X, activate(Y)), n__s(activate(Y)))), n__0)
if(true, X, Y) → activate(X)
if(false, X, Y) → activate(Y)
0 → n__0
s(X) → n__s(X)
activate(n__0) → 0
activate(n__s(X)) → s(X)
activate(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:

GEQ(n__s(X), n__s(Y)) → ACTIVATE(Y)
ACTIVATE(n__0) → 0¹
ACTIVATE(n__s(X)) → S(X)
DIV(s(X), n__s(Y)) → IF(geq(X, activate(Y)), n__s(div(minus(X, activate(Y)), n__s(activate(Y)))), n__0)
IF(true, X, Y) → ACTIVATE(X)
GEQ(n__s(X), n__s(Y)) → ACTIVATE(X)
MINUS(n__s(X), n__s(Y)) → ACTIVATE(Y)
DIV(s(X), n__s(Y)) → ACTIVATE(Y)
DIV(s(X), n__s(Y)) → GEQ(X, activate(Y))
MINUS(n__s(X), n__s(Y)) → MINUS(activate(X), activate(Y))
MINUS(n__s(X), n__s(Y)) → ACTIVATE(X)
IF(false, X, Y) → ACTIVATE(Y)
MINUS(n__0, Y) → 0¹
DIV(s(X), n__s(Y)) → MINUS(X, activate(Y))
GEQ(n__s(X), n__s(Y)) → GEQ(activate(X), activate(Y))
DIV(s(X), n__s(Y)) → DIV(minus(X, activate(Y)), n__s(activate(Y)))

The TRS R consists of the following rules:

minus(n__0, Y) → 0
minus(n__s(X), n__s(Y)) → minus(activate(X), activate(Y))
geq(X, n__0) → true
geq(n__0, n__s(Y)) → false
geq(n__s(X), n__s(Y)) → geq(activate(X), activate(Y))
div(0, n__s(Y)) → 0
div(s(X), n__s(Y)) → if(geq(X, activate(Y)), n__s(div(minus(X, activate(Y)), n__s(activate(Y)))), n__0)
if(true, X, Y) → activate(X)
if(false, X, Y) → activate(Y)
0 → n__0
s(X) → n__s(X)
activate(n__0) → 0
activate(n__s(X)) → s(X)
activate(X) → X

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

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ EdgeDeletionProof

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

GEQ(n__s(X), n__s(Y)) → ACTIVATE(Y)
ACTIVATE(n__0) → 0¹
ACTIVATE(n__s(X)) → S(X)
DIV(s(X), n__s(Y)) → IF(geq(X, activate(Y)), n__s(div(minus(X, activate(Y)), n__s(activate(Y)))), n__0)
IF(true, X, Y) → ACTIVATE(X)
GEQ(n__s(X), n__s(Y)) → ACTIVATE(X)
MINUS(n__s(X), n__s(Y)) → ACTIVATE(Y)
DIV(s(X), n__s(Y)) → ACTIVATE(Y)
DIV(s(X), n__s(Y)) → GEQ(X, activate(Y))
MINUS(n__s(X), n__s(Y)) → MINUS(activate(X), activate(Y))
MINUS(n__s(X), n__s(Y)) → ACTIVATE(X)
IF(false, X, Y) → ACTIVATE(Y)
MINUS(n__0, Y) → 0¹
DIV(s(X), n__s(Y)) → MINUS(X, activate(Y))
GEQ(n__s(X), n__s(Y)) → GEQ(activate(X), activate(Y))
DIV(s(X), n__s(Y)) → DIV(minus(X, activate(Y)), n__s(activate(Y)))

The TRS R consists of the following rules:

minus(n__0, Y) → 0
minus(n__s(X), n__s(Y)) → minus(activate(X), activate(Y))
geq(X, n__0) → true
geq(n__0, n__s(Y)) → false
geq(n__s(X), n__s(Y)) → geq(activate(X), activate(Y))
div(0, n__s(Y)) → 0
div(s(X), n__s(Y)) → if(geq(X, activate(Y)), n__s(div(minus(X, activate(Y)), n__s(activate(Y)))), n__0)
if(true, X, Y) → activate(X)
if(false, X, Y) → activate(Y)
0 → n__0
s(X) → n__s(X)
activate(n__0) → 0
activate(n__s(X)) → s(X)
activate(X) → X

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

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ EdgeDeletionProof

        ↳ QDP

          ↳ DependencyGraphProof

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

GEQ(n__s(X), n__s(Y)) → ACTIVATE(Y)
ACTIVATE(n__0) → 0¹
ACTIVATE(n__s(X)) → S(X)
DIV(s(X), n__s(Y)) → IF(geq(X, activate(Y)), n__s(div(minus(X, activate(Y)), n__s(activate(Y)))), n__0)
IF(true, X, Y) → ACTIVATE(X)
GEQ(n__s(X), n__s(Y)) → ACTIVATE(X)
MINUS(n__s(X), n__s(Y)) → ACTIVATE(Y)
DIV(s(X), n__s(Y)) → ACTIVATE(Y)
DIV(s(X), n__s(Y)) → GEQ(X, activate(Y))
MINUS(n__s(X), n__s(Y)) → MINUS(activate(X), activate(Y))
MINUS(n__s(X), n__s(Y)) → ACTIVATE(X)
MINUS(n__0, Y) → 0¹
IF(false, X, Y) → ACTIVATE(Y)
DIV(s(X), n__s(Y)) → MINUS(X, activate(Y))
GEQ(n__s(X), n__s(Y)) → GEQ(activate(X), activate(Y))
DIV(s(X), n__s(Y)) → DIV(minus(X, activate(Y)), n__s(activate(Y)))

The TRS R consists of the following rules:

minus(n__0, Y) → 0
minus(n__s(X), n__s(Y)) → minus(activate(X), activate(Y))
geq(X, n__0) → true
geq(n__0, n__s(Y)) → false
geq(n__s(X), n__s(Y)) → geq(activate(X), activate(Y))
div(0, n__s(Y)) → 0
div(s(X), n__s(Y)) → if(geq(X, activate(Y)), n__s(div(minus(X, activate(Y)), n__s(activate(Y)))), n__0)
if(true, X, Y) → activate(X)
if(false, X, Y) → activate(Y)
0 → n__0
s(X) → n__s(X)
activate(n__0) → 0
activate(n__s(X)) → s(X)
activate(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

      ↳ EdgeDeletionProof

        ↳ QDP

          ↳ DependencyGraphProof

            ↳ AND

              ↳ QDP

              ↳ QDP

              ↳ QDP

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

GEQ(n__s(X), n__s(Y)) → GEQ(activate(X), activate(Y))

The TRS R consists of the following rules:

minus(n__0, Y) → 0
minus(n__s(X), n__s(Y)) → minus(activate(X), activate(Y))
geq(X, n__0) → true
geq(n__0, n__s(Y)) → false
geq(n__s(X), n__s(Y)) → geq(activate(X), activate(Y))
div(0, n__s(Y)) → 0
div(s(X), n__s(Y)) → if(geq(X, activate(Y)), n__s(div(minus(X, activate(Y)), n__s(activate(Y)))), n__0)
if(true, X, Y) → activate(X)
if(false, X, Y) → activate(Y)
0 → n__0
s(X) → n__s(X)
activate(n__0) → 0
activate(n__s(X)) → s(X)
activate(X) → X

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

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ EdgeDeletionProof

        ↳ QDP

          ↳ DependencyGraphProof

            ↳ AND

              ↳ QDP

              ↳ QDP

              ↳ QDP

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

MINUS(n__s(X), n__s(Y)) → MINUS(activate(X), activate(Y))

The TRS R consists of the following rules:

minus(n__0, Y) → 0
minus(n__s(X), n__s(Y)) → minus(activate(X), activate(Y))
geq(X, n__0) → true
geq(n__0, n__s(Y)) → false
geq(n__s(X), n__s(Y)) → geq(activate(X), activate(Y))
div(0, n__s(Y)) → 0
div(s(X), n__s(Y)) → if(geq(X, activate(Y)), n__s(div(minus(X, activate(Y)), n__s(activate(Y)))), n__0)
if(true, X, Y) → activate(X)
if(false, X, Y) → activate(Y)
0 → n__0
s(X) → n__s(X)
activate(n__0) → 0
activate(n__s(X)) → s(X)
activate(X) → X

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

↳ 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), n__s(Y)) → DIV(minus(X, activate(Y)), n__s(activate(Y)))

The TRS R consists of the following rules:

minus(n__0, Y) → 0
minus(n__s(X), n__s(Y)) → minus(activate(X), activate(Y))
geq(X, n__0) → true
geq(n__0, n__s(Y)) → false
geq(n__s(X), n__s(Y)) → geq(activate(X), activate(Y))
div(0, n__s(Y)) → 0
div(s(X), n__s(Y)) → if(geq(X, activate(Y)), n__s(div(minus(X, activate(Y)), n__s(activate(Y)))), n__0)
if(true, X, Y) → activate(X)
if(false, X, Y) → activate(Y)
0 → n__0
s(X) → n__s(X)
activate(n__0) → 0
activate(n__s(X)) → s(X)
activate(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 oriented strictly and are deleted.

DIV(s(X), n__s(Y)) → DIV(minus(X, activate(Y)), n__s(activate(Y)))

The remaining pairs can at least be oriented weakly.
none
Used ordering: Combined order from the following AFS and order.
DIV(x1, x2) = x1
s(x1) = s
minus(x1, x2) = minus
0 = 0
n__0 = n__0

Lexicographic Path Order [19].
Precedence:

s > minus > 0 > n_₀

The following usable rules [14] were oriented:

minus(n__s(X), n__s(Y)) → minus(activate(X), activate(Y))
minus(n__0, Y) → 0
0 → n__0

↳ 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:

minus(n__0, Y) → 0
minus(n__s(X), n__s(Y)) → minus(activate(X), activate(Y))
geq(X, n__0) → true
geq(n__0, n__s(Y)) → false
geq(n__s(X), n__s(Y)) → geq(activate(X), activate(Y))
div(0, n__s(Y)) → 0
div(s(X), n__s(Y)) → if(geq(X, activate(Y)), n__s(div(minus(X, activate(Y)), n__s(activate(Y)))), n__0)
if(true, X, Y) → activate(X)
if(false, X, Y) → activate(Y)
0 → n__0
s(X) → n__s(X)
activate(n__0) → 0
activate(n__s(X)) → s(X)
activate(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.