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(0, Y) -> 0
minus2(s1(X), s1(Y)) -> minus2(X, Y)
geq2(X, 0) -> true
geq2(0, s1(Y)) -> false
geq2(s1(X), s1(Y)) -> geq2(X, Y)
div2(0, s1(Y)) -> 0
div2(s1(X), s1(Y)) -> if3(geq2(X, Y), s1(div2(minus2(X, Y), s1(Y))), 0)
if3(true, X, Y) -> X
if3(false, X, Y) -> Y

Q is empty.


↳ QTRS
  ↳ DependencyPairsProof

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

minus2(0, Y) -> 0
minus2(s1(X), s1(Y)) -> minus2(X, Y)
geq2(X, 0) -> true
geq2(0, s1(Y)) -> false
geq2(s1(X), s1(Y)) -> geq2(X, Y)
div2(0, s1(Y)) -> 0
div2(s1(X), s1(Y)) -> if3(geq2(X, Y), s1(div2(minus2(X, Y), s1(Y))), 0)
if3(true, X, Y) -> X
if3(false, X, Y) -> Y

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), s1(Y)) -> IF3(geq2(X, Y), s1(div2(minus2(X, Y), s1(Y))), 0)
DIV2(s1(X), s1(Y)) -> MINUS2(X, Y)
DIV2(s1(X), s1(Y)) -> DIV2(minus2(X, Y), s1(Y))
DIV2(s1(X), s1(Y)) -> GEQ2(X, Y)
GEQ2(s1(X), s1(Y)) -> GEQ2(X, Y)
MINUS2(s1(X), s1(Y)) -> MINUS2(X, Y)

The TRS R consists of the following rules:

minus2(0, Y) -> 0
minus2(s1(X), s1(Y)) -> minus2(X, Y)
geq2(X, 0) -> true
geq2(0, s1(Y)) -> false
geq2(s1(X), s1(Y)) -> geq2(X, Y)
div2(0, s1(Y)) -> 0
div2(s1(X), s1(Y)) -> if3(geq2(X, Y), s1(div2(minus2(X, Y), s1(Y))), 0)
if3(true, X, Y) -> X
if3(false, X, Y) -> Y

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), s1(Y)) -> IF3(geq2(X, Y), s1(div2(minus2(X, Y), s1(Y))), 0)
DIV2(s1(X), s1(Y)) -> MINUS2(X, Y)
DIV2(s1(X), s1(Y)) -> DIV2(minus2(X, Y), s1(Y))
DIV2(s1(X), s1(Y)) -> GEQ2(X, Y)
GEQ2(s1(X), s1(Y)) -> GEQ2(X, Y)
MINUS2(s1(X), s1(Y)) -> MINUS2(X, Y)

The TRS R consists of the following rules:

minus2(0, Y) -> 0
minus2(s1(X), s1(Y)) -> minus2(X, Y)
geq2(X, 0) -> true
geq2(0, s1(Y)) -> false
geq2(s1(X), s1(Y)) -> geq2(X, Y)
div2(0, s1(Y)) -> 0
div2(s1(X), s1(Y)) -> if3(geq2(X, Y), s1(div2(minus2(X, Y), s1(Y))), 0)
if3(true, X, Y) -> X
if3(false, X, Y) -> Y

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 3 less nodes.

↳ QTRS
  ↳ DependencyPairsProof
    ↳ QDP
      ↳ DependencyGraphProof
        ↳ AND
          ↳ QDP
            ↳ QDPOrderProof
          ↳ QDP
          ↳ QDP

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

GEQ2(s1(X), s1(Y)) -> GEQ2(X, Y)

The TRS R consists of the following rules:

minus2(0, Y) -> 0
minus2(s1(X), s1(Y)) -> minus2(X, Y)
geq2(X, 0) -> true
geq2(0, s1(Y)) -> false
geq2(s1(X), s1(Y)) -> geq2(X, Y)
div2(0, s1(Y)) -> 0
div2(s1(X), s1(Y)) -> if3(geq2(X, Y), s1(div2(minus2(X, Y), s1(Y))), 0)
if3(true, X, Y) -> X
if3(false, X, Y) -> Y

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(s1(X), s1(Y)) -> GEQ2(X, Y)
The remaining pairs can at least be oriented weakly.
none
Used ordering: Polynomial Order [17,21] with Interpretation:

POL( s1(x1) ) = x1 + 3


POL( GEQ2(x1, x2) ) = 3x2 + 3



The following usable rules [14] were oriented: none



↳ 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(0, Y) -> 0
minus2(s1(X), s1(Y)) -> minus2(X, Y)
geq2(X, 0) -> true
geq2(0, s1(Y)) -> false
geq2(s1(X), s1(Y)) -> geq2(X, Y)
div2(0, s1(Y)) -> 0
div2(s1(X), s1(Y)) -> if3(geq2(X, Y), s1(div2(minus2(X, Y), s1(Y))), 0)
if3(true, X, Y) -> X
if3(false, X, Y) -> Y

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(s1(X), s1(Y)) -> MINUS2(X, Y)

The TRS R consists of the following rules:

minus2(0, Y) -> 0
minus2(s1(X), s1(Y)) -> minus2(X, Y)
geq2(X, 0) -> true
geq2(0, s1(Y)) -> false
geq2(s1(X), s1(Y)) -> geq2(X, Y)
div2(0, s1(Y)) -> 0
div2(s1(X), s1(Y)) -> if3(geq2(X, Y), s1(div2(minus2(X, Y), s1(Y))), 0)
if3(true, X, Y) -> X
if3(false, X, Y) -> Y

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(s1(X), s1(Y)) -> MINUS2(X, Y)
The remaining pairs can at least be oriented weakly.
none
Used ordering: Polynomial Order [17,21] with Interpretation:

POL( MINUS2(x1, x2) ) = 3x2 + 3


POL( s1(x1) ) = x1 + 3



The following usable rules [14] were oriented: none



↳ 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(0, Y) -> 0
minus2(s1(X), s1(Y)) -> minus2(X, Y)
geq2(X, 0) -> true
geq2(0, s1(Y)) -> false
geq2(s1(X), s1(Y)) -> geq2(X, Y)
div2(0, s1(Y)) -> 0
div2(s1(X), s1(Y)) -> if3(geq2(X, Y), s1(div2(minus2(X, Y), s1(Y))), 0)
if3(true, X, Y) -> X
if3(false, X, Y) -> Y

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), s1(Y)) -> DIV2(minus2(X, Y), s1(Y))

The TRS R consists of the following rules:

minus2(0, Y) -> 0
minus2(s1(X), s1(Y)) -> minus2(X, Y)
geq2(X, 0) -> true
geq2(0, s1(Y)) -> false
geq2(s1(X), s1(Y)) -> geq2(X, Y)
div2(0, s1(Y)) -> 0
div2(s1(X), s1(Y)) -> if3(geq2(X, Y), s1(div2(minus2(X, Y), s1(Y))), 0)
if3(true, X, Y) -> X
if3(false, X, Y) -> Y

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), s1(Y)) -> DIV2(minus2(X, Y), s1(Y))
The remaining pairs can at least be oriented weakly.
none
Used ordering: Polynomial Order [17,21] with Interpretation:

POL( minus2(x1, x2) ) = 2x1 + 1


POL( 0 ) = 2


POL( s1(x1) ) = 3x1 + 3


POL( DIV2(x1, x2) ) = 3x1 + 3



The following usable rules [14] were oriented:

minus2(s1(X), s1(Y)) -> minus2(X, Y)
minus2(0, Y) -> 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(0, Y) -> 0
minus2(s1(X), s1(Y)) -> minus2(X, Y)
geq2(X, 0) -> true
geq2(0, s1(Y)) -> false
geq2(s1(X), s1(Y)) -> geq2(X, Y)
div2(0, s1(Y)) -> 0
div2(s1(X), s1(Y)) -> if3(geq2(X, Y), s1(div2(minus2(X, Y), s1(Y))), 0)
if3(true, X, Y) -> X
if3(false, X, Y) -> Y

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.