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.

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 contains 3 SCCs with 13 less nodes.

↳ QTRS
  ↳ DependencyPairsProof
    ↳ QDP
      ↳ DependencyGraphProof
        ↳ AND
QDP
            ↳ QDPAfsSolverProof
          ↳ 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.
By using an argument filtering and a montonic ordering, at least one Dependency Pair of this SCC can be strictly oriented.

GEQ2(n__s1(X), n__s1(Y)) -> GEQ2(activate1(X), activate1(Y))
Used argument filtering: GEQ2(x1, x2)  =  x2
n__s1(x1)  =  n__s1(x1)
activate1(x1)  =  x1
n__0  =  n__0
0  =  0
s1(x1)  =  s1(x1)
Used ordering: Quasi Precedence: [n__s_1, s_1] [n__0, 0]


↳ QTRS
  ↳ DependencyPairsProof
    ↳ QDP
      ↳ DependencyGraphProof
        ↳ AND
          ↳ QDP
            ↳ QDPAfsSolverProof
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
            ↳ QDPAfsSolverProof
          ↳ 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.
By using an argument filtering and a montonic ordering, at least one Dependency Pair of this SCC can be strictly oriented.

MINUS2(n__s1(X), n__s1(Y)) -> MINUS2(activate1(X), activate1(Y))
Used argument filtering: MINUS2(x1, x2)  =  x2
n__s1(x1)  =  n__s1(x1)
activate1(x1)  =  x1
n__0  =  n__0
0  =  0
s1(x1)  =  s1(x1)
Used ordering: Quasi Precedence: [n__s_1, s_1] [n__0, 0]


↳ QTRS
  ↳ DependencyPairsProof
    ↳ QDP
      ↳ DependencyGraphProof
        ↳ AND
          ↳ QDP
          ↳ QDP
            ↳ QDPAfsSolverProof
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
            ↳ QDPAfsSolverProof

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.
By using an argument filtering and a montonic ordering, at least one Dependency Pair of this SCC can be strictly oriented.

DIV2(s1(X), n__s1(Y)) -> DIV2(minus2(X, activate1(Y)), n__s1(activate1(Y)))
Used argument filtering: DIV2(x1, x2)  =  x1
s1(x1)  =  s
minus2(x1, x2)  =  minus
0  =  0
n__0  =  n__0
Used ordering: Quasi Precedence: s > minus > 0 > n__0


↳ QTRS
  ↳ DependencyPairsProof
    ↳ QDP
      ↳ DependencyGraphProof
        ↳ AND
          ↳ QDP
          ↳ QDP
          ↳ QDP
            ↳ QDPAfsSolverProof
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.