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:

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(n__div₂(n__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)
div₂(X1, X2) -> n__div₂(X1, X2)
minus₂(X1, X2) -> n__minus₂(X1, X2)
activate₁(n__0) -> 0
activate₁(n__s₁(X)) -> s₁(activate₁(X))
activate₁(n__div₂(X1, X2)) -> div₂(activate₁(X1), X2)
activate₁(n__minus₂(X1, X2)) -> minus₂(X1, X2)
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(n__div₂(n__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)
div₂(X1, X2) -> n__div₂(X1, X2)
minus₂(X1, X2) -> n__minus₂(X1, X2)
activate₁(n__0) -> 0
activate₁(n__s₁(X)) -> s₁(activate₁(X))
activate₁(n__div₂(X1, X2)) -> div₂(activate₁(X1), X2)
activate₁(n__minus₂(X1, X2)) -> minus₂(X1, X2)
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:

DIV₂(s₁(X), n__s₁(Y)) -> IF₃(geq₂(X, activate₁(Y)), n__s₁(n__div₂(n__minus₂(X, activate₁(Y)), n__s₁(activate₁(Y)))), n__0)
GEQ₂(n__s₁(X), n__s₁(Y)) -> ACTIVATE₁(Y)
ACTIVATE₁(n__0) -> 0¹
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)
ACTIVATE₁(n__div₂(X1, X2)) -> ACTIVATE₁(X1)
DIV₂(s₁(X), n__s₁(Y)) -> GEQ₂(X, activate₁(Y))
ACTIVATE₁(n__s₁(X)) -> S₁(activate₁(X))
MINUS₂(n__s₁(X), n__s₁(Y)) -> MINUS₂(activate₁(X), activate₁(Y))
MINUS₂(n__s₁(X), n__s₁(Y)) -> ACTIVATE₁(X)
MINUS₂(n__0, Y) -> 0¹
IF₃(false, X, Y) -> ACTIVATE₁(Y)
ACTIVATE₁(n__div₂(X1, X2)) -> DIV₂(activate₁(X1), X2)
GEQ₂(n__s₁(X), n__s₁(Y)) -> GEQ₂(activate₁(X), activate₁(Y))
ACTIVATE₁(n__minus₂(X1, X2)) -> MINUS₂(X1, X2)
ACTIVATE₁(n__s₁(X)) -> ACTIVATE₁(X)

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(n__div₂(n__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)
div₂(X1, X2) -> n__div₂(X1, X2)
minus₂(X1, X2) -> n__minus₂(X1, X2)
activate₁(n__0) -> 0
activate₁(n__s₁(X)) -> s₁(activate₁(X))
activate₁(n__div₂(X1, X2)) -> div₂(activate₁(X1), X2)
activate₁(n__minus₂(X1, X2)) -> minus₂(X1, X2)
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:

DIV₂(s₁(X), n__s₁(Y)) -> IF₃(geq₂(X, activate₁(Y)), n__s₁(n__div₂(n__minus₂(X, activate₁(Y)), n__s₁(activate₁(Y)))), n__0)
GEQ₂(n__s₁(X), n__s₁(Y)) -> ACTIVATE₁(Y)
ACTIVATE₁(n__0) -> 0¹
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)
ACTIVATE₁(n__div₂(X1, X2)) -> ACTIVATE₁(X1)
DIV₂(s₁(X), n__s₁(Y)) -> GEQ₂(X, activate₁(Y))
ACTIVATE₁(n__s₁(X)) -> S₁(activate₁(X))
MINUS₂(n__s₁(X), n__s₁(Y)) -> MINUS₂(activate₁(X), activate₁(Y))
MINUS₂(n__s₁(X), n__s₁(Y)) -> ACTIVATE₁(X)
MINUS₂(n__0, Y) -> 0¹
IF₃(false, X, Y) -> ACTIVATE₁(Y)
ACTIVATE₁(n__div₂(X1, X2)) -> DIV₂(activate₁(X1), X2)
GEQ₂(n__s₁(X), n__s₁(Y)) -> GEQ₂(activate₁(X), activate₁(Y))
ACTIVATE₁(n__minus₂(X1, X2)) -> MINUS₂(X1, X2)
ACTIVATE₁(n__s₁(X)) -> ACTIVATE₁(X)

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(n__div₂(n__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)
div₂(X1, X2) -> n__div₂(X1, X2)
minus₂(X1, X2) -> n__minus₂(X1, X2)
activate₁(n__0) -> 0
activate₁(n__s₁(X)) -> s₁(activate₁(X))
activate₁(n__div₂(X1, X2)) -> div₂(activate₁(X1), X2)
activate₁(n__minus₂(X1, X2)) -> minus₂(X1, X2)
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 1 SCC with 3 less nodes.

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ DependencyGraphProof

        ↳ QDP

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

DIV₂(s₁(X), n__s₁(Y)) -> IF₃(geq₂(X, activate₁(Y)), n__s₁(n__div₂(n__minus₂(X, activate₁(Y)), n__s₁(activate₁(Y)))), n__0)
GEQ₂(n__s₁(X), n__s₁(Y)) -> ACTIVATE₁(Y)
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)
ACTIVATE₁(n__div₂(X1, X2)) -> ACTIVATE₁(X1)
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)
ACTIVATE₁(n__div₂(X1, X2)) -> DIV₂(activate₁(X1), X2)
GEQ₂(n__s₁(X), n__s₁(Y)) -> GEQ₂(activate₁(X), activate₁(Y))
ACTIVATE₁(n__s₁(X)) -> ACTIVATE₁(X)
ACTIVATE₁(n__minus₂(X1, X2)) -> MINUS₂(X1, X2)

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(n__div₂(n__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)
div₂(X1, X2) -> n__div₂(X1, X2)
minus₂(X1, X2) -> n__minus₂(X1, X2)
activate₁(n__0) -> 0
activate₁(n__s₁(X)) -> s₁(activate₁(X))
activate₁(n__div₂(X1, X2)) -> div₂(activate₁(X1), X2)
activate₁(n__minus₂(X1, X2)) -> minus₂(X1, X2)
activate₁(X) -> X

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