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(x, x) -> 0
minus2(s1(x), s1(y)) -> minus2(x, y)
minus2(0, x) -> 0
minus2(x, 0) -> x
div2(s1(x), s1(y)) -> s1(div2(minus2(x, y), s1(y)))
div2(0, s1(y)) -> 0
f3(x, 0, b) -> x
f3(x, s1(y), b) -> div2(f3(x, minus2(s1(y), s1(0)), b), b)

Q is empty.


↳ QTRS
  ↳ DependencyPairsProof

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

minus2(x, x) -> 0
minus2(s1(x), s1(y)) -> minus2(x, y)
minus2(0, x) -> 0
minus2(x, 0) -> x
div2(s1(x), s1(y)) -> s1(div2(minus2(x, y), s1(y)))
div2(0, s1(y)) -> 0
f3(x, 0, b) -> x
f3(x, s1(y), b) -> div2(f3(x, minus2(s1(y), s1(0)), b), b)

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)) -> MINUS2(x, y)
DIV2(s1(x), s1(y)) -> DIV2(minus2(x, y), s1(y))
F3(x, s1(y), b) -> DIV2(f3(x, minus2(s1(y), s1(0)), b), b)
F3(x, s1(y), b) -> F3(x, minus2(s1(y), s1(0)), b)
F3(x, s1(y), b) -> MINUS2(s1(y), s1(0))
MINUS2(s1(x), s1(y)) -> MINUS2(x, y)

The TRS R consists of the following rules:

minus2(x, x) -> 0
minus2(s1(x), s1(y)) -> minus2(x, y)
minus2(0, x) -> 0
minus2(x, 0) -> x
div2(s1(x), s1(y)) -> s1(div2(minus2(x, y), s1(y)))
div2(0, s1(y)) -> 0
f3(x, 0, b) -> x
f3(x, s1(y), b) -> div2(f3(x, minus2(s1(y), s1(0)), b), b)

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)) -> MINUS2(x, y)
DIV2(s1(x), s1(y)) -> DIV2(minus2(x, y), s1(y))
F3(x, s1(y), b) -> DIV2(f3(x, minus2(s1(y), s1(0)), b), b)
F3(x, s1(y), b) -> F3(x, minus2(s1(y), s1(0)), b)
F3(x, s1(y), b) -> MINUS2(s1(y), s1(0))
MINUS2(s1(x), s1(y)) -> MINUS2(x, y)

The TRS R consists of the following rules:

minus2(x, x) -> 0
minus2(s1(x), s1(y)) -> minus2(x, y)
minus2(0, x) -> 0
minus2(x, 0) -> x
div2(s1(x), s1(y)) -> s1(div2(minus2(x, y), s1(y)))
div2(0, s1(y)) -> 0
f3(x, 0, b) -> x
f3(x, s1(y), b) -> div2(f3(x, minus2(s1(y), s1(0)), b), b)

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:

MINUS2(s1(x), s1(y)) -> MINUS2(x, y)

The TRS R consists of the following rules:

minus2(x, x) -> 0
minus2(s1(x), s1(y)) -> minus2(x, y)
minus2(0, x) -> 0
minus2(x, 0) -> x
div2(s1(x), s1(y)) -> s1(div2(minus2(x, y), s1(y)))
div2(0, s1(y)) -> 0
f3(x, 0, b) -> x
f3(x, s1(y), b) -> div2(f3(x, minus2(s1(y), s1(0)), b), b)

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) ) = x1


POL( s1(x1) ) = x1 + 1



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(x, x) -> 0
minus2(s1(x), s1(y)) -> minus2(x, y)
minus2(0, x) -> 0
minus2(x, 0) -> x
div2(s1(x), s1(y)) -> s1(div2(minus2(x, y), s1(y)))
div2(0, s1(y)) -> 0
f3(x, 0, b) -> x
f3(x, s1(y), b) -> div2(f3(x, minus2(s1(y), s1(0)), b), b)

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:

DIV2(s1(x), s1(y)) -> DIV2(minus2(x, y), s1(y))

The TRS R consists of the following rules:

minus2(x, x) -> 0
minus2(s1(x), s1(y)) -> minus2(x, y)
minus2(0, x) -> 0
minus2(x, 0) -> x
div2(s1(x), s1(y)) -> s1(div2(minus2(x, y), s1(y)))
div2(0, s1(y)) -> 0
f3(x, 0, b) -> x
f3(x, s1(y), b) -> div2(f3(x, minus2(s1(y), s1(0)), b), b)

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) ) = x1


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


POL( s1(x1) ) = x1 + 1


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



The following usable rules [14] were oriented:

minus2(s1(x), s1(y)) -> minus2(x, y)
minus2(x, 0) -> x
minus2(0, x) -> 0
minus2(x, x) -> 0



↳ 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(x, x) -> 0
minus2(s1(x), s1(y)) -> minus2(x, y)
minus2(0, x) -> 0
minus2(x, 0) -> x
div2(s1(x), s1(y)) -> s1(div2(minus2(x, y), s1(y)))
div2(0, s1(y)) -> 0
f3(x, 0, b) -> x
f3(x, s1(y), b) -> div2(f3(x, minus2(s1(y), s1(0)), b), b)

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:

F3(x, s1(y), b) -> F3(x, minus2(s1(y), s1(0)), b)

The TRS R consists of the following rules:

minus2(x, x) -> 0
minus2(s1(x), s1(y)) -> minus2(x, y)
minus2(0, x) -> 0
minus2(x, 0) -> x
div2(s1(x), s1(y)) -> s1(div2(minus2(x, y), s1(y)))
div2(0, s1(y)) -> 0
f3(x, 0, b) -> x
f3(x, s1(y), b) -> div2(f3(x, minus2(s1(y), s1(0)), b), b)

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