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:

-2(x, 0) -> x
-2(0, s1(y)) -> 0
-2(s1(x), s1(y)) -> -2(x, y)
lt2(x, 0) -> false
lt2(0, s1(y)) -> true
lt2(s1(x), s1(y)) -> lt2(x, y)
if3(true, x, y) -> x
if3(false, x, y) -> y
div2(x, 0) -> 0
div2(0, y) -> 0
div2(s1(x), s1(y)) -> if3(lt2(x, y), 0, s1(div2(-2(x, y), s1(y))))

Q is empty.


↳ QTRS
  ↳ DependencyPairsProof

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

-2(x, 0) -> x
-2(0, s1(y)) -> 0
-2(s1(x), s1(y)) -> -2(x, y)
lt2(x, 0) -> false
lt2(0, s1(y)) -> true
lt2(s1(x), s1(y)) -> lt2(x, y)
if3(true, x, y) -> x
if3(false, x, y) -> y
div2(x, 0) -> 0
div2(0, y) -> 0
div2(s1(x), s1(y)) -> if3(lt2(x, y), 0, s1(div2(-2(x, y), s1(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)) -> LT2(x, y)
-12(s1(x), s1(y)) -> -12(x, y)
DIV2(s1(x), s1(y)) -> -12(x, y)
LT2(s1(x), s1(y)) -> LT2(x, y)
DIV2(s1(x), s1(y)) -> IF3(lt2(x, y), 0, s1(div2(-2(x, y), s1(y))))
DIV2(s1(x), s1(y)) -> DIV2(-2(x, y), s1(y))

The TRS R consists of the following rules:

-2(x, 0) -> x
-2(0, s1(y)) -> 0
-2(s1(x), s1(y)) -> -2(x, y)
lt2(x, 0) -> false
lt2(0, s1(y)) -> true
lt2(s1(x), s1(y)) -> lt2(x, y)
if3(true, x, y) -> x
if3(false, x, y) -> y
div2(x, 0) -> 0
div2(0, y) -> 0
div2(s1(x), s1(y)) -> if3(lt2(x, y), 0, s1(div2(-2(x, y), s1(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)) -> LT2(x, y)
-12(s1(x), s1(y)) -> -12(x, y)
DIV2(s1(x), s1(y)) -> -12(x, y)
LT2(s1(x), s1(y)) -> LT2(x, y)
DIV2(s1(x), s1(y)) -> IF3(lt2(x, y), 0, s1(div2(-2(x, y), s1(y))))
DIV2(s1(x), s1(y)) -> DIV2(-2(x, y), s1(y))

The TRS R consists of the following rules:

-2(x, 0) -> x
-2(0, s1(y)) -> 0
-2(s1(x), s1(y)) -> -2(x, y)
lt2(x, 0) -> false
lt2(0, s1(y)) -> true
lt2(s1(x), s1(y)) -> lt2(x, y)
if3(true, x, y) -> x
if3(false, x, y) -> y
div2(x, 0) -> 0
div2(0, y) -> 0
div2(s1(x), s1(y)) -> if3(lt2(x, y), 0, s1(div2(-2(x, y), s1(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:

LT2(s1(x), s1(y)) -> LT2(x, y)

The TRS R consists of the following rules:

-2(x, 0) -> x
-2(0, s1(y)) -> 0
-2(s1(x), s1(y)) -> -2(x, y)
lt2(x, 0) -> false
lt2(0, s1(y)) -> true
lt2(s1(x), s1(y)) -> lt2(x, y)
if3(true, x, y) -> x
if3(false, x, y) -> y
div2(x, 0) -> 0
div2(0, y) -> 0
div2(s1(x), s1(y)) -> if3(lt2(x, y), 0, s1(div2(-2(x, y), s1(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.


LT2(s1(x), s1(y)) -> LT2(x, y)
The remaining pairs can at least be oriented weakly.
none
Used ordering: Polynomial Order [17,21] with Interpretation:

POL( s1(x1) ) = x1 + 1


POL( LT2(x1, x2) ) = x1



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:

-2(x, 0) -> x
-2(0, s1(y)) -> 0
-2(s1(x), s1(y)) -> -2(x, y)
lt2(x, 0) -> false
lt2(0, s1(y)) -> true
lt2(s1(x), s1(y)) -> lt2(x, y)
if3(true, x, y) -> x
if3(false, x, y) -> y
div2(x, 0) -> 0
div2(0, y) -> 0
div2(s1(x), s1(y)) -> if3(lt2(x, y), 0, s1(div2(-2(x, y), s1(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:

-12(s1(x), s1(y)) -> -12(x, y)

The TRS R consists of the following rules:

-2(x, 0) -> x
-2(0, s1(y)) -> 0
-2(s1(x), s1(y)) -> -2(x, y)
lt2(x, 0) -> false
lt2(0, s1(y)) -> true
lt2(s1(x), s1(y)) -> lt2(x, y)
if3(true, x, y) -> x
if3(false, x, y) -> y
div2(x, 0) -> 0
div2(0, y) -> 0
div2(s1(x), s1(y)) -> if3(lt2(x, y), 0, s1(div2(-2(x, y), s1(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.


-12(s1(x), s1(y)) -> -12(x, y)
The remaining pairs can at least be oriented weakly.
none
Used ordering: Polynomial Order [17,21] with Interpretation:

POL( s1(x1) ) = x1 + 1


POL( -12(x1, x2) ) = x1



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:

-2(x, 0) -> x
-2(0, s1(y)) -> 0
-2(s1(x), s1(y)) -> -2(x, y)
lt2(x, 0) -> false
lt2(0, s1(y)) -> true
lt2(s1(x), s1(y)) -> lt2(x, y)
if3(true, x, y) -> x
if3(false, x, y) -> y
div2(x, 0) -> 0
div2(0, y) -> 0
div2(s1(x), s1(y)) -> if3(lt2(x, y), 0, s1(div2(-2(x, y), s1(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(-2(x, y), s1(y))

The TRS R consists of the following rules:

-2(x, 0) -> x
-2(0, s1(y)) -> 0
-2(s1(x), s1(y)) -> -2(x, y)
lt2(x, 0) -> false
lt2(0, s1(y)) -> true
lt2(s1(x), s1(y)) -> lt2(x, y)
if3(true, x, y) -> x
if3(false, x, y) -> y
div2(x, 0) -> 0
div2(0, y) -> 0
div2(s1(x), s1(y)) -> if3(lt2(x, y), 0, s1(div2(-2(x, y), s1(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(-2(x, y), s1(y))
The remaining pairs can at least be oriented weakly.
none
Used ordering: Polynomial Order [17,21] with Interpretation:

POL( -2(x1, x2) ) = x1


POL( s1(x1) ) = x1 + 1


POL( 0 ) = max{0, -1}


POL( DIV2(x1, x2) ) = x1 + x2 + 1



The following usable rules [14] were oriented:

-2(0, s1(y)) -> 0
-2(s1(x), s1(y)) -> -2(x, y)
-2(x, 0) -> x



↳ 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:

-2(x, 0) -> x
-2(0, s1(y)) -> 0
-2(s1(x), s1(y)) -> -2(x, y)
lt2(x, 0) -> false
lt2(0, s1(y)) -> true
lt2(s1(x), s1(y)) -> lt2(x, y)
if3(true, x, y) -> x
if3(false, x, y) -> y
div2(x, 0) -> 0
div2(0, y) -> 0
div2(s1(x), s1(y)) -> if3(lt2(x, y), 0, s1(div2(-2(x, y), s1(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.