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:

minus(x, x) → 0
minus(s(x), s(y)) → minus(x, y)
minus(0, x) → 0
minus(x, 0) → x
div(s(x), s(y)) → s(div(minus(x, y), s(y)))
div(0, s(y)) → 0
f(x, 0, b) → x
f(x, s(y), b) → div(f(x, minus(s(y), s(0)), b), b)

Q is empty.


↳ QTRS
  ↳ DependencyPairsProof

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

minus(x, x) → 0
minus(s(x), s(y)) → minus(x, y)
minus(0, x) → 0
minus(x, 0) → x
div(s(x), s(y)) → s(div(minus(x, y), s(y)))
div(0, s(y)) → 0
f(x, 0, b) → x
f(x, s(y), b) → div(f(x, minus(s(y), s(0)), b), b)

Q is empty.

Using Dependency Pairs [1,15] we result in the following initial DP problem:
Q DP problem:
The TRS P consists of the following rules:

DIV(s(x), s(y)) → DIV(minus(x, y), s(y))
MINUS(s(x), s(y)) → MINUS(x, y)
F(x, s(y), b) → MINUS(s(y), s(0))
F(x, s(y), b) → F(x, minus(s(y), s(0)), b)
DIV(s(x), s(y)) → MINUS(x, y)
F(x, s(y), b) → DIV(f(x, minus(s(y), s(0)), b), b)

The TRS R consists of the following rules:

minus(x, x) → 0
minus(s(x), s(y)) → minus(x, y)
minus(0, x) → 0
minus(x, 0) → x
div(s(x), s(y)) → s(div(minus(x, y), s(y)))
div(0, s(y)) → 0
f(x, 0, b) → x
f(x, s(y), b) → div(f(x, minus(s(y), s(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:

DIV(s(x), s(y)) → DIV(minus(x, y), s(y))
MINUS(s(x), s(y)) → MINUS(x, y)
F(x, s(y), b) → MINUS(s(y), s(0))
F(x, s(y), b) → F(x, minus(s(y), s(0)), b)
DIV(s(x), s(y)) → MINUS(x, y)
F(x, s(y), b) → DIV(f(x, minus(s(y), s(0)), b), b)

The TRS R consists of the following rules:

minus(x, x) → 0
minus(s(x), s(y)) → minus(x, y)
minus(0, x) → 0
minus(x, 0) → x
div(s(x), s(y)) → s(div(minus(x, y), s(y)))
div(0, s(y)) → 0
f(x, 0, b) → x
f(x, s(y), b) → div(f(x, minus(s(y), s(0)), b), b)

Q is empty.
We have to consider all minimal (P,Q,R)-chains.
The approximation of the Dependency Graph [15,17,22] 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:

MINUS(s(x), s(y)) → MINUS(x, y)

The TRS R consists of the following rules:

minus(x, x) → 0
minus(s(x), s(y)) → minus(x, y)
minus(0, x) → 0
minus(x, 0) → x
div(s(x), s(y)) → s(div(minus(x, y), s(y)))
div(0, s(y)) → 0
f(x, 0, b) → x
f(x, s(y), b) → div(f(x, minus(s(y), s(0)), b), b)

Q is empty.
We have to consider all minimal (P,Q,R)-chains.
We use the reduction pair processor [15].


The following pairs can be oriented strictly and are deleted.


MINUS(s(x), s(y)) → MINUS(x, y)
The remaining pairs can at least be oriented weakly.
none
Used ordering: Polynomial interpretation [25,35]:

POL(MINUS(x1, x2)) = (15/4)x_2   
POL(s(x1)) = 1/2 + (13/4)x_1   
The value of delta used in the strict ordering is 15/8.
The following usable rules [17] 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:

minus(x, x) → 0
minus(s(x), s(y)) → minus(x, y)
minus(0, x) → 0
minus(x, 0) → x
div(s(x), s(y)) → s(div(minus(x, y), s(y)))
div(0, s(y)) → 0
f(x, 0, b) → x
f(x, s(y), b) → div(f(x, minus(s(y), s(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:

DIV(s(x), s(y)) → DIV(minus(x, y), s(y))

The TRS R consists of the following rules:

minus(x, x) → 0
minus(s(x), s(y)) → minus(x, y)
minus(0, x) → 0
minus(x, 0) → x
div(s(x), s(y)) → s(div(minus(x, y), s(y)))
div(0, s(y)) → 0
f(x, 0, b) → x
f(x, s(y), b) → div(f(x, minus(s(y), s(0)), b), b)

Q is empty.
We have to consider all minimal (P,Q,R)-chains.
We use the reduction pair processor [15].


The following pairs can be oriented strictly and are deleted.


DIV(s(x), s(y)) → DIV(minus(x, y), s(y))
The remaining pairs can at least be oriented weakly.
none
Used ordering: Polynomial interpretation [25,35]:

POL(minus(x1, x2)) = 5/4 + (2)x_1   
POL(DIV(x1, x2)) = (9/4)x_1   
POL(s(x1)) = 4 + (15/4)x_1   
POL(0) = 1/4   
The value of delta used in the strict ordering is 99/16.
The following usable rules [17] were oriented:

minus(x, x) → 0
minus(s(x), s(y)) → minus(x, y)
minus(0, x) → 0
minus(x, 0) → x



↳ 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:

minus(x, x) → 0
minus(s(x), s(y)) → minus(x, y)
minus(0, x) → 0
minus(x, 0) → x
div(s(x), s(y)) → s(div(minus(x, y), s(y)))
div(0, s(y)) → 0
f(x, 0, b) → x
f(x, s(y), b) → div(f(x, minus(s(y), s(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:

F(x, s(y), b) → F(x, minus(s(y), s(0)), b)

The TRS R consists of the following rules:

minus(x, x) → 0
minus(s(x), s(y)) → minus(x, y)
minus(0, x) → 0
minus(x, 0) → x
div(s(x), s(y)) → s(div(minus(x, y), s(y)))
div(0, s(y)) → 0
f(x, 0, b) → x
f(x, s(y), b) → div(f(x, minus(s(y), s(0)), b), b)

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