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 oriented strictly and are deleted.


GEQ2(n__s1(X), n__s1(Y)) -> GEQ2(activate1(X), activate1(Y))
The remaining pairs can at least be oriented weakly.
none
Used ordering: Polynomial interpretation [21]:

POL(0) = 0   
POL(GEQ2(x1, x2)) = 3·x2   
POL(activate1(x1)) = 2 + x1   
POL(n__0) = 0   
POL(n__s1(x1)) = 3 + 2·x1   
POL(s1(x1)) = 3 + 2·x1   

The following usable rules [14] were oriented:

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



↳ 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 oriented strictly and are deleted.


MINUS2(n__s1(X), n__s1(Y)) -> MINUS2(activate1(X), activate1(Y))
The remaining pairs can at least be oriented weakly.
none
Used ordering: Polynomial interpretation [21]:

POL(0) = 0   
POL(MINUS2(x1, x2)) = 3·x2   
POL(activate1(x1)) = 2 + x1   
POL(n__0) = 0   
POL(n__s1(x1)) = 3 + 2·x1   
POL(s1(x1)) = 3 + 2·x1   

The following usable rules [14] were oriented:

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



↳ 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 oriented strictly and are deleted.


DIV2(s1(X), n__s1(Y)) -> DIV2(minus2(X, activate1(Y)), n__s1(activate1(Y)))
The remaining pairs can at least be oriented weakly.
none
Used ordering: Polynomial interpretation [21]:

POL(0) = 0   
POL(DIV2(x1, x2)) = 2·x1   
POL(activate1(x1)) = 0   
POL(minus2(x1, x2)) = 0   
POL(n__0) = 0   
POL(n__s1(x1)) = 0   
POL(s1(x1)) = 2   

The following usable rules [14] were oriented:

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