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:

a__minus2(0, Y) -> 0
a__minus2(s1(X), s1(Y)) -> a__minus2(X, Y)
a__geq2(X, 0) -> true
a__geq2(0, s1(Y)) -> false
a__geq2(s1(X), s1(Y)) -> a__geq2(X, Y)
a__div2(0, s1(Y)) -> 0
a__div2(s1(X), s1(Y)) -> a__if3(a__geq2(X, Y), s1(div2(minus2(X, Y), s1(Y))), 0)
a__if3(true, X, Y) -> mark1(X)
a__if3(false, X, Y) -> mark1(Y)
mark1(minus2(X1, X2)) -> a__minus2(X1, X2)
mark1(geq2(X1, X2)) -> a__geq2(X1, X2)
mark1(div2(X1, X2)) -> a__div2(mark1(X1), X2)
mark1(if3(X1, X2, X3)) -> a__if3(mark1(X1), X2, X3)
mark1(0) -> 0
mark1(s1(X)) -> s1(mark1(X))
mark1(true) -> true
mark1(false) -> false
a__minus2(X1, X2) -> minus2(X1, X2)
a__geq2(X1, X2) -> geq2(X1, X2)
a__div2(X1, X2) -> div2(X1, X2)
a__if3(X1, X2, X3) -> if3(X1, X2, X3)

Q is empty.


↳ QTRS
  ↳ DependencyPairsProof

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

a__minus2(0, Y) -> 0
a__minus2(s1(X), s1(Y)) -> a__minus2(X, Y)
a__geq2(X, 0) -> true
a__geq2(0, s1(Y)) -> false
a__geq2(s1(X), s1(Y)) -> a__geq2(X, Y)
a__div2(0, s1(Y)) -> 0
a__div2(s1(X), s1(Y)) -> a__if3(a__geq2(X, Y), s1(div2(minus2(X, Y), s1(Y))), 0)
a__if3(true, X, Y) -> mark1(X)
a__if3(false, X, Y) -> mark1(Y)
mark1(minus2(X1, X2)) -> a__minus2(X1, X2)
mark1(geq2(X1, X2)) -> a__geq2(X1, X2)
mark1(div2(X1, X2)) -> a__div2(mark1(X1), X2)
mark1(if3(X1, X2, X3)) -> a__if3(mark1(X1), X2, X3)
mark1(0) -> 0
mark1(s1(X)) -> s1(mark1(X))
mark1(true) -> true
mark1(false) -> false
a__minus2(X1, X2) -> minus2(X1, X2)
a__geq2(X1, X2) -> geq2(X1, X2)
a__div2(X1, X2) -> div2(X1, X2)
a__if3(X1, X2, X3) -> if3(X1, X2, X3)

Q is empty.

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

MARK1(geq2(X1, X2)) -> A__GEQ2(X1, X2)
A__GEQ2(s1(X), s1(Y)) -> A__GEQ2(X, Y)
A__IF3(false, X, Y) -> MARK1(Y)
MARK1(minus2(X1, X2)) -> A__MINUS2(X1, X2)
MARK1(if3(X1, X2, X3)) -> MARK1(X1)
MARK1(s1(X)) -> MARK1(X)
A__DIV2(s1(X), s1(Y)) -> A__IF3(a__geq2(X, Y), s1(div2(minus2(X, Y), s1(Y))), 0)
A__DIV2(s1(X), s1(Y)) -> A__GEQ2(X, Y)
A__IF3(true, X, Y) -> MARK1(X)
MARK1(div2(X1, X2)) -> A__DIV2(mark1(X1), X2)
MARK1(div2(X1, X2)) -> MARK1(X1)
A__MINUS2(s1(X), s1(Y)) -> A__MINUS2(X, Y)
MARK1(if3(X1, X2, X3)) -> A__IF3(mark1(X1), X2, X3)

The TRS R consists of the following rules:

a__minus2(0, Y) -> 0
a__minus2(s1(X), s1(Y)) -> a__minus2(X, Y)
a__geq2(X, 0) -> true
a__geq2(0, s1(Y)) -> false
a__geq2(s1(X), s1(Y)) -> a__geq2(X, Y)
a__div2(0, s1(Y)) -> 0
a__div2(s1(X), s1(Y)) -> a__if3(a__geq2(X, Y), s1(div2(minus2(X, Y), s1(Y))), 0)
a__if3(true, X, Y) -> mark1(X)
a__if3(false, X, Y) -> mark1(Y)
mark1(minus2(X1, X2)) -> a__minus2(X1, X2)
mark1(geq2(X1, X2)) -> a__geq2(X1, X2)
mark1(div2(X1, X2)) -> a__div2(mark1(X1), X2)
mark1(if3(X1, X2, X3)) -> a__if3(mark1(X1), X2, X3)
mark1(0) -> 0
mark1(s1(X)) -> s1(mark1(X))
mark1(true) -> true
mark1(false) -> false
a__minus2(X1, X2) -> minus2(X1, X2)
a__geq2(X1, X2) -> geq2(X1, X2)
a__div2(X1, X2) -> div2(X1, X2)
a__if3(X1, X2, X3) -> if3(X1, X2, X3)

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:

MARK1(geq2(X1, X2)) -> A__GEQ2(X1, X2)
A__GEQ2(s1(X), s1(Y)) -> A__GEQ2(X, Y)
A__IF3(false, X, Y) -> MARK1(Y)
MARK1(minus2(X1, X2)) -> A__MINUS2(X1, X2)
MARK1(if3(X1, X2, X3)) -> MARK1(X1)
MARK1(s1(X)) -> MARK1(X)
A__DIV2(s1(X), s1(Y)) -> A__IF3(a__geq2(X, Y), s1(div2(minus2(X, Y), s1(Y))), 0)
A__DIV2(s1(X), s1(Y)) -> A__GEQ2(X, Y)
A__IF3(true, X, Y) -> MARK1(X)
MARK1(div2(X1, X2)) -> A__DIV2(mark1(X1), X2)
MARK1(div2(X1, X2)) -> MARK1(X1)
A__MINUS2(s1(X), s1(Y)) -> A__MINUS2(X, Y)
MARK1(if3(X1, X2, X3)) -> A__IF3(mark1(X1), X2, X3)

The TRS R consists of the following rules:

a__minus2(0, Y) -> 0
a__minus2(s1(X), s1(Y)) -> a__minus2(X, Y)
a__geq2(X, 0) -> true
a__geq2(0, s1(Y)) -> false
a__geq2(s1(X), s1(Y)) -> a__geq2(X, Y)
a__div2(0, s1(Y)) -> 0
a__div2(s1(X), s1(Y)) -> a__if3(a__geq2(X, Y), s1(div2(minus2(X, Y), s1(Y))), 0)
a__if3(true, X, Y) -> mark1(X)
a__if3(false, X, Y) -> mark1(Y)
mark1(minus2(X1, X2)) -> a__minus2(X1, X2)
mark1(geq2(X1, X2)) -> a__geq2(X1, X2)
mark1(div2(X1, X2)) -> a__div2(mark1(X1), X2)
mark1(if3(X1, X2, X3)) -> a__if3(mark1(X1), X2, X3)
mark1(0) -> 0
mark1(s1(X)) -> s1(mark1(X))
mark1(true) -> true
mark1(false) -> false
a__minus2(X1, X2) -> minus2(X1, X2)
a__geq2(X1, X2) -> geq2(X1, X2)
a__div2(X1, X2) -> div2(X1, X2)
a__if3(X1, X2, X3) -> if3(X1, X2, X3)

Q is empty.
We have to consider all minimal (P,Q,R)-chains.
The approximation of the Dependency Graph contains 3 SCCs with 3 less nodes.

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

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

A__GEQ2(s1(X), s1(Y)) -> A__GEQ2(X, Y)

The TRS R consists of the following rules:

a__minus2(0, Y) -> 0
a__minus2(s1(X), s1(Y)) -> a__minus2(X, Y)
a__geq2(X, 0) -> true
a__geq2(0, s1(Y)) -> false
a__geq2(s1(X), s1(Y)) -> a__geq2(X, Y)
a__div2(0, s1(Y)) -> 0
a__div2(s1(X), s1(Y)) -> a__if3(a__geq2(X, Y), s1(div2(minus2(X, Y), s1(Y))), 0)
a__if3(true, X, Y) -> mark1(X)
a__if3(false, X, Y) -> mark1(Y)
mark1(minus2(X1, X2)) -> a__minus2(X1, X2)
mark1(geq2(X1, X2)) -> a__geq2(X1, X2)
mark1(div2(X1, X2)) -> a__div2(mark1(X1), X2)
mark1(if3(X1, X2, X3)) -> a__if3(mark1(X1), X2, X3)
mark1(0) -> 0
mark1(s1(X)) -> s1(mark1(X))
mark1(true) -> true
mark1(false) -> false
a__minus2(X1, X2) -> minus2(X1, X2)
a__geq2(X1, X2) -> geq2(X1, X2)
a__div2(X1, X2) -> div2(X1, X2)
a__if3(X1, X2, X3) -> if3(X1, X2, X3)

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.

A__GEQ2(s1(X), s1(Y)) -> A__GEQ2(X, Y)
Used argument filtering: A__GEQ2(x1, x2)  =  x2
s1(x1)  =  s1(x1)
Used ordering: Quasi Precedence: trivial 


↳ 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:

a__minus2(0, Y) -> 0
a__minus2(s1(X), s1(Y)) -> a__minus2(X, Y)
a__geq2(X, 0) -> true
a__geq2(0, s1(Y)) -> false
a__geq2(s1(X), s1(Y)) -> a__geq2(X, Y)
a__div2(0, s1(Y)) -> 0
a__div2(s1(X), s1(Y)) -> a__if3(a__geq2(X, Y), s1(div2(minus2(X, Y), s1(Y))), 0)
a__if3(true, X, Y) -> mark1(X)
a__if3(false, X, Y) -> mark1(Y)
mark1(minus2(X1, X2)) -> a__minus2(X1, X2)
mark1(geq2(X1, X2)) -> a__geq2(X1, X2)
mark1(div2(X1, X2)) -> a__div2(mark1(X1), X2)
mark1(if3(X1, X2, X3)) -> a__if3(mark1(X1), X2, X3)
mark1(0) -> 0
mark1(s1(X)) -> s1(mark1(X))
mark1(true) -> true
mark1(false) -> false
a__minus2(X1, X2) -> minus2(X1, X2)
a__geq2(X1, X2) -> geq2(X1, X2)
a__div2(X1, X2) -> div2(X1, X2)
a__if3(X1, X2, X3) -> if3(X1, X2, X3)

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:

A__MINUS2(s1(X), s1(Y)) -> A__MINUS2(X, Y)

The TRS R consists of the following rules:

a__minus2(0, Y) -> 0
a__minus2(s1(X), s1(Y)) -> a__minus2(X, Y)
a__geq2(X, 0) -> true
a__geq2(0, s1(Y)) -> false
a__geq2(s1(X), s1(Y)) -> a__geq2(X, Y)
a__div2(0, s1(Y)) -> 0
a__div2(s1(X), s1(Y)) -> a__if3(a__geq2(X, Y), s1(div2(minus2(X, Y), s1(Y))), 0)
a__if3(true, X, Y) -> mark1(X)
a__if3(false, X, Y) -> mark1(Y)
mark1(minus2(X1, X2)) -> a__minus2(X1, X2)
mark1(geq2(X1, X2)) -> a__geq2(X1, X2)
mark1(div2(X1, X2)) -> a__div2(mark1(X1), X2)
mark1(if3(X1, X2, X3)) -> a__if3(mark1(X1), X2, X3)
mark1(0) -> 0
mark1(s1(X)) -> s1(mark1(X))
mark1(true) -> true
mark1(false) -> false
a__minus2(X1, X2) -> minus2(X1, X2)
a__geq2(X1, X2) -> geq2(X1, X2)
a__div2(X1, X2) -> div2(X1, X2)
a__if3(X1, X2, X3) -> if3(X1, X2, X3)

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.

A__MINUS2(s1(X), s1(Y)) -> A__MINUS2(X, Y)
Used argument filtering: A__MINUS2(x1, x2)  =  x2
s1(x1)  =  s1(x1)
Used ordering: Quasi Precedence: trivial 


↳ 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:

a__minus2(0, Y) -> 0
a__minus2(s1(X), s1(Y)) -> a__minus2(X, Y)
a__geq2(X, 0) -> true
a__geq2(0, s1(Y)) -> false
a__geq2(s1(X), s1(Y)) -> a__geq2(X, Y)
a__div2(0, s1(Y)) -> 0
a__div2(s1(X), s1(Y)) -> a__if3(a__geq2(X, Y), s1(div2(minus2(X, Y), s1(Y))), 0)
a__if3(true, X, Y) -> mark1(X)
a__if3(false, X, Y) -> mark1(Y)
mark1(minus2(X1, X2)) -> a__minus2(X1, X2)
mark1(geq2(X1, X2)) -> a__geq2(X1, X2)
mark1(div2(X1, X2)) -> a__div2(mark1(X1), X2)
mark1(if3(X1, X2, X3)) -> a__if3(mark1(X1), X2, X3)
mark1(0) -> 0
mark1(s1(X)) -> s1(mark1(X))
mark1(true) -> true
mark1(false) -> false
a__minus2(X1, X2) -> minus2(X1, X2)
a__geq2(X1, X2) -> geq2(X1, X2)
a__div2(X1, X2) -> div2(X1, X2)
a__if3(X1, X2, X3) -> if3(X1, X2, X3)

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

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

MARK1(if3(X1, X2, X3)) -> MARK1(X1)
MARK1(s1(X)) -> MARK1(X)
A__DIV2(s1(X), s1(Y)) -> A__IF3(a__geq2(X, Y), s1(div2(minus2(X, Y), s1(Y))), 0)
A__IF3(false, X, Y) -> MARK1(Y)
A__IF3(true, X, Y) -> MARK1(X)
MARK1(div2(X1, X2)) -> A__DIV2(mark1(X1), X2)
MARK1(div2(X1, X2)) -> MARK1(X1)
MARK1(if3(X1, X2, X3)) -> A__IF3(mark1(X1), X2, X3)

The TRS R consists of the following rules:

a__minus2(0, Y) -> 0
a__minus2(s1(X), s1(Y)) -> a__minus2(X, Y)
a__geq2(X, 0) -> true
a__geq2(0, s1(Y)) -> false
a__geq2(s1(X), s1(Y)) -> a__geq2(X, Y)
a__div2(0, s1(Y)) -> 0
a__div2(s1(X), s1(Y)) -> a__if3(a__geq2(X, Y), s1(div2(minus2(X, Y), s1(Y))), 0)
a__if3(true, X, Y) -> mark1(X)
a__if3(false, X, Y) -> mark1(Y)
mark1(minus2(X1, X2)) -> a__minus2(X1, X2)
mark1(geq2(X1, X2)) -> a__geq2(X1, X2)
mark1(div2(X1, X2)) -> a__div2(mark1(X1), X2)
mark1(if3(X1, X2, X3)) -> a__if3(mark1(X1), X2, X3)
mark1(0) -> 0
mark1(s1(X)) -> s1(mark1(X))
mark1(true) -> true
mark1(false) -> false
a__minus2(X1, X2) -> minus2(X1, X2)
a__geq2(X1, X2) -> geq2(X1, X2)
a__div2(X1, X2) -> div2(X1, X2)
a__if3(X1, X2, X3) -> if3(X1, X2, X3)

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