Termination w.r.t. Q proof of TRS/TRCSREx49_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:

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

                ↳ QDPOrderProof

              ↳ 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.
We use the reduction pair processor [13].

The following pairs can be oriented strictly and are deleted.

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

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

Lexicographic path order with status [19].
Quasi-Precedence:

[n_{_s₁}, s_1]
[n_₀, 0]

Status:

n_₀: multiset
n_{_s₁}: [1]
s₁: [1]
0: multiset

The following usable rules [14] were oriented:

activate(n__0) → 0
activate(X) → X
activate(n__s(X)) → s(X)
s(X) → n__s(X)
0 → n__0

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ EdgeDeletionProof

        ↳ QDP

          ↳ DependencyGraphProof

            ↳ AND

              ↳ QDP

                ↳ QDPOrderProof

                  ↳ QDP

                    ↳ PisEmptyProof

              ↳ QDP

              ↳ QDP

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.

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ EdgeDeletionProof

        ↳ QDP

          ↳ DependencyGraphProof

            ↳ AND

              ↳ QDP

              ↳ QDP

                ↳ QDPOrderProof

              ↳ 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.
We use the reduction pair processor [13].

The following pairs can be oriented strictly and are deleted.

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

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

Lexicographic path order with status [19].
Quasi-Precedence:

[n_{_s₁}, s_1]
[n_₀, 0]

Status:

n_₀: multiset
n_{_s₁}: [1]
s₁: [1]
0: multiset

The following usable rules [14] were oriented:

activate(n__0) → 0
activate(X) → X
activate(n__s(X)) → s(X)
s(X) → n__s(X)
0 → n__0

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ EdgeDeletionProof

        ↳ QDP

          ↳ DependencyGraphProof

            ↳ AND

              ↳ QDP

              ↳ QDP

                ↳ QDPOrderProof

                  ↳ QDP

                    ↳ PisEmptyProof

              ↳ QDP

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.

↳ 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) = DIV(x1, x2)
s(x1) = s(x1)
n__s(x1) = n__s(x1)
minus(x1, x2) = x1
activate(x1) = x1
n__0 = n__0
0 = 0

Lexicographic path order with status [19].
Quasi-Precedence:

[s₁, n_{_{s_1]}} > DIV₂
[n_₀, 0] > DIV₂

Status:

n_₀: multiset
n_{_s₁}: [1]
s₁: [1]
0: multiset
DIV₂: [2,1]

The following usable rules [14] were oriented:

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