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:

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
  ↳ Overlay + Local Confluence

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.

The TRS is overlay and locally confluent. By [19] we can switch to innermost.

↳ QTRS
  ↳ Overlay + Local Confluence
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)

The set Q consists of the following terms:

minus(x0, x0)
minus(s(x0), s(x1))
minus(0, x0)
minus(x0, 0)
div(s(x0), s(x1))
div(0, s(x0))
f(x0, 0, x1)
f(x0, s(x1), x2)


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)

The set Q consists of the following terms:

minus(x0, x0)
minus(s(x0), s(x1))
minus(0, x0)
minus(x0, 0)
div(s(x0), s(x1))
div(0, s(x0))
f(x0, 0, x1)
f(x0, s(x1), x2)

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

↳ QTRS
  ↳ Overlay + Local Confluence
    ↳ 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)

The set Q consists of the following terms:

minus(x0, x0)
minus(s(x0), s(x1))
minus(0, x0)
minus(x0, 0)
div(s(x0), s(x1))
div(0, s(x0))
f(x0, 0, x1)
f(x0, s(x1), x2)

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
  ↳ Overlay + Local Confluence
    ↳ QTRS
      ↳ DependencyPairsProof
        ↳ QDP
          ↳ DependencyGraphProof
            ↳ AND
QDP
                ↳ UsableRulesProof
              ↳ 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)

The set Q consists of the following terms:

minus(x0, x0)
minus(s(x0), s(x1))
minus(0, x0)
minus(x0, 0)
div(s(x0), s(x1))
div(0, s(x0))
f(x0, 0, x1)
f(x0, s(x1), x2)

We have to consider all minimal (P,Q,R)-chains.
As all Q-normal forms are R-normal forms we are in the innermost case. Hence, by the usable rules processor [15] we can delete all non-usable rules [17] from R.

↳ QTRS
  ↳ Overlay + Local Confluence
    ↳ QTRS
      ↳ DependencyPairsProof
        ↳ QDP
          ↳ DependencyGraphProof
            ↳ AND
              ↳ QDP
                ↳ UsableRulesProof
QDP
                    ↳ QReductionProof
              ↳ QDP
              ↳ QDP

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

MINUS(s(x), s(y)) → MINUS(x, y)

R is empty.
The set Q consists of the following terms:

minus(x0, x0)
minus(s(x0), s(x1))
minus(0, x0)
minus(x0, 0)
div(s(x0), s(x1))
div(0, s(x0))
f(x0, 0, x1)
f(x0, s(x1), x2)

We have to consider all minimal (P,Q,R)-chains.
We deleted the following terms from Q as each root-symbol of these terms does neither occur in P nor in R.

minus(x0, x0)
minus(s(x0), s(x1))
minus(0, x0)
minus(x0, 0)
div(s(x0), s(x1))
div(0, s(x0))
f(x0, 0, x1)
f(x0, s(x1), x2)



↳ QTRS
  ↳ Overlay + Local Confluence
    ↳ QTRS
      ↳ DependencyPairsProof
        ↳ QDP
          ↳ DependencyGraphProof
            ↳ AND
              ↳ QDP
                ↳ UsableRulesProof
                  ↳ QDP
                    ↳ QReductionProof
QDP
                        ↳ QDPSizeChangeProof
              ↳ QDP
              ↳ QDP

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

MINUS(s(x), s(y)) → MINUS(x, y)

R is empty.
Q is empty.
We have to consider all minimal (P,Q,R)-chains.
By using the subterm criterion [20] together with the size-change analysis [32] we have proven that there are no infinite chains for this DP problem.

From the DPs we obtained the following set of size-change graphs:



↳ QTRS
  ↳ Overlay + Local Confluence
    ↳ QTRS
      ↳ DependencyPairsProof
        ↳ QDP
          ↳ DependencyGraphProof
            ↳ AND
              ↳ QDP
QDP
                ↳ UsableRulesProof
              ↳ 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)

The set Q consists of the following terms:

minus(x0, x0)
minus(s(x0), s(x1))
minus(0, x0)
minus(x0, 0)
div(s(x0), s(x1))
div(0, s(x0))
f(x0, 0, x1)
f(x0, s(x1), x2)

We have to consider all minimal (P,Q,R)-chains.
As all Q-normal forms are R-normal forms we are in the innermost case. Hence, by the usable rules processor [15] we can delete all non-usable rules [17] from R.

↳ QTRS
  ↳ Overlay + Local Confluence
    ↳ QTRS
      ↳ DependencyPairsProof
        ↳ QDP
          ↳ DependencyGraphProof
            ↳ AND
              ↳ QDP
              ↳ QDP
                ↳ UsableRulesProof
QDP
                    ↳ QReductionProof
              ↳ 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

The set Q consists of the following terms:

minus(x0, x0)
minus(s(x0), s(x1))
minus(0, x0)
minus(x0, 0)
div(s(x0), s(x1))
div(0, s(x0))
f(x0, 0, x1)
f(x0, s(x1), x2)

We have to consider all minimal (P,Q,R)-chains.
We deleted the following terms from Q as each root-symbol of these terms does neither occur in P nor in R.

div(s(x0), s(x1))
div(0, s(x0))
f(x0, 0, x1)
f(x0, s(x1), x2)



↳ QTRS
  ↳ Overlay + Local Confluence
    ↳ QTRS
      ↳ DependencyPairsProof
        ↳ QDP
          ↳ DependencyGraphProof
            ↳ AND
              ↳ QDP
              ↳ QDP
                ↳ UsableRulesProof
                  ↳ QDP
                    ↳ QReductionProof
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

The set Q consists of the following terms:

minus(x0, x0)
minus(s(x0), s(x1))
minus(0, x0)
minus(x0, 0)

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]:

POL(0) = 0   
POL(DIV(x1, x2)) = x1   
POL(minus(x1, x2)) = x1   
POL(s(x1)) = 1 + x1   

The following usable rules [17] were oriented:

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



↳ QTRS
  ↳ Overlay + Local Confluence
    ↳ QTRS
      ↳ DependencyPairsProof
        ↳ QDP
          ↳ DependencyGraphProof
            ↳ AND
              ↳ QDP
              ↳ QDP
                ↳ UsableRulesProof
                  ↳ QDP
                    ↳ QReductionProof
                      ↳ 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

The set Q consists of the following terms:

minus(x0, x0)
minus(s(x0), s(x1))
minus(0, x0)
minus(x0, 0)

We have to consider all minimal (P,Q,R)-chains.
The TRS P is empty. Hence, there is no (P,Q,R) chain.

↳ QTRS
  ↳ Overlay + Local Confluence
    ↳ QTRS
      ↳ DependencyPairsProof
        ↳ QDP
          ↳ DependencyGraphProof
            ↳ AND
              ↳ QDP
              ↳ QDP
QDP
                ↳ UsableRulesProof

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)

The set Q consists of the following terms:

minus(x0, x0)
minus(s(x0), s(x1))
minus(0, x0)
minus(x0, 0)
div(s(x0), s(x1))
div(0, s(x0))
f(x0, 0, x1)
f(x0, s(x1), x2)

We have to consider all minimal (P,Q,R)-chains.
As all Q-normal forms are R-normal forms we are in the innermost case. Hence, by the usable rules processor [15] we can delete all non-usable rules [17] from R.

↳ QTRS
  ↳ Overlay + Local Confluence
    ↳ QTRS
      ↳ DependencyPairsProof
        ↳ QDP
          ↳ DependencyGraphProof
            ↳ AND
              ↳ QDP
              ↳ QDP
              ↳ QDP
                ↳ UsableRulesProof
QDP
                    ↳ QReductionProof

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

The set Q consists of the following terms:

minus(x0, x0)
minus(s(x0), s(x1))
minus(0, x0)
minus(x0, 0)
div(s(x0), s(x1))
div(0, s(x0))
f(x0, 0, x1)
f(x0, s(x1), x2)

We have to consider all minimal (P,Q,R)-chains.
We deleted the following terms from Q as each root-symbol of these terms does neither occur in P nor in R.

div(s(x0), s(x1))
div(0, s(x0))
f(x0, 0, x1)
f(x0, s(x1), x2)



↳ QTRS
  ↳ Overlay + Local Confluence
    ↳ QTRS
      ↳ DependencyPairsProof
        ↳ QDP
          ↳ DependencyGraphProof
            ↳ AND
              ↳ QDP
              ↳ QDP
              ↳ QDP
                ↳ UsableRulesProof
                  ↳ QDP
                    ↳ QReductionProof
QDP
                        ↳ Narrowing

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

The set Q consists of the following terms:

minus(x0, x0)
minus(s(x0), s(x1))
minus(0, x0)
minus(x0, 0)

We have to consider all minimal (P,Q,R)-chains.
By narrowing [15] the rule F(x, s(y), b) → F(x, minus(s(y), s(0)), b) at position [1] we obtained the following new rules:

F(y0, s(x0), y2) → F(y0, minus(x0, 0), y2)
F(y0, s(0), y2) → F(y0, 0, y2)



↳ QTRS
  ↳ Overlay + Local Confluence
    ↳ QTRS
      ↳ DependencyPairsProof
        ↳ QDP
          ↳ DependencyGraphProof
            ↳ AND
              ↳ QDP
              ↳ QDP
              ↳ QDP
                ↳ UsableRulesProof
                  ↳ QDP
                    ↳ QReductionProof
                      ↳ QDP
                        ↳ Narrowing
QDP
                            ↳ DependencyGraphProof

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

F(y0, s(x0), y2) → F(y0, minus(x0, 0), y2)
F(y0, s(0), y2) → F(y0, 0, y2)

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

The set Q consists of the following terms:

minus(x0, x0)
minus(s(x0), s(x1))
minus(0, x0)
minus(x0, 0)

We have to consider all minimal (P,Q,R)-chains.
The approximation of the Dependency Graph [15,17,22] contains 1 SCC with 1 less node.

↳ QTRS
  ↳ Overlay + Local Confluence
    ↳ QTRS
      ↳ DependencyPairsProof
        ↳ QDP
          ↳ DependencyGraphProof
            ↳ AND
              ↳ QDP
              ↳ QDP
              ↳ QDP
                ↳ UsableRulesProof
                  ↳ QDP
                    ↳ QReductionProof
                      ↳ QDP
                        ↳ Narrowing
                          ↳ QDP
                            ↳ DependencyGraphProof
QDP
                                ↳ UsableRulesProof

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

F(y0, s(x0), y2) → F(y0, minus(x0, 0), y2)

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

The set Q consists of the following terms:

minus(x0, x0)
minus(s(x0), s(x1))
minus(0, x0)
minus(x0, 0)

We have to consider all minimal (P,Q,R)-chains.
As all Q-normal forms are R-normal forms we are in the innermost case. Hence, by the usable rules processor [15] we can delete all non-usable rules [17] from R.

↳ QTRS
  ↳ Overlay + Local Confluence
    ↳ QTRS
      ↳ DependencyPairsProof
        ↳ QDP
          ↳ DependencyGraphProof
            ↳ AND
              ↳ QDP
              ↳ QDP
              ↳ QDP
                ↳ UsableRulesProof
                  ↳ QDP
                    ↳ QReductionProof
                      ↳ QDP
                        ↳ Narrowing
                          ↳ QDP
                            ↳ DependencyGraphProof
                              ↳ QDP
                                ↳ UsableRulesProof
QDP
                                    ↳ Rewriting

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

F(y0, s(x0), y2) → F(y0, minus(x0, 0), y2)

The TRS R consists of the following rules:

minus(x, x) → 0
minus(0, x) → 0
minus(x, 0) → x

The set Q consists of the following terms:

minus(x0, x0)
minus(s(x0), s(x1))
minus(0, x0)
minus(x0, 0)

We have to consider all minimal (P,Q,R)-chains.
By rewriting [15] the rule F(y0, s(x0), y2) → F(y0, minus(x0, 0), y2) at position [1] we obtained the following new rules:

F(y0, s(x0), y2) → F(y0, x0, y2)



↳ QTRS
  ↳ Overlay + Local Confluence
    ↳ QTRS
      ↳ DependencyPairsProof
        ↳ QDP
          ↳ DependencyGraphProof
            ↳ AND
              ↳ QDP
              ↳ QDP
              ↳ QDP
                ↳ UsableRulesProof
                  ↳ QDP
                    ↳ QReductionProof
                      ↳ QDP
                        ↳ Narrowing
                          ↳ QDP
                            ↳ DependencyGraphProof
                              ↳ QDP
                                ↳ UsableRulesProof
                                  ↳ QDP
                                    ↳ Rewriting
QDP
                                        ↳ UsableRulesProof

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

F(y0, s(x0), y2) → F(y0, x0, y2)

The TRS R consists of the following rules:

minus(x, x) → 0
minus(0, x) → 0
minus(x, 0) → x

The set Q consists of the following terms:

minus(x0, x0)
minus(s(x0), s(x1))
minus(0, x0)
minus(x0, 0)

We have to consider all minimal (P,Q,R)-chains.
As all Q-normal forms are R-normal forms we are in the innermost case. Hence, by the usable rules processor [15] we can delete all non-usable rules [17] from R.

↳ QTRS
  ↳ Overlay + Local Confluence
    ↳ QTRS
      ↳ DependencyPairsProof
        ↳ QDP
          ↳ DependencyGraphProof
            ↳ AND
              ↳ QDP
              ↳ QDP
              ↳ QDP
                ↳ UsableRulesProof
                  ↳ QDP
                    ↳ QReductionProof
                      ↳ QDP
                        ↳ Narrowing
                          ↳ QDP
                            ↳ DependencyGraphProof
                              ↳ QDP
                                ↳ UsableRulesProof
                                  ↳ QDP
                                    ↳ Rewriting
                                      ↳ QDP
                                        ↳ UsableRulesProof
QDP
                                            ↳ QReductionProof

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

F(y0, s(x0), y2) → F(y0, x0, y2)

R is empty.
The set Q consists of the following terms:

minus(x0, x0)
minus(s(x0), s(x1))
minus(0, x0)
minus(x0, 0)

We have to consider all minimal (P,Q,R)-chains.
We deleted the following terms from Q as each root-symbol of these terms does neither occur in P nor in R.

minus(x0, x0)
minus(s(x0), s(x1))
minus(0, x0)
minus(x0, 0)



↳ QTRS
  ↳ Overlay + Local Confluence
    ↳ QTRS
      ↳ DependencyPairsProof
        ↳ QDP
          ↳ DependencyGraphProof
            ↳ AND
              ↳ QDP
              ↳ QDP
              ↳ QDP
                ↳ UsableRulesProof
                  ↳ QDP
                    ↳ QReductionProof
                      ↳ QDP
                        ↳ Narrowing
                          ↳ QDP
                            ↳ DependencyGraphProof
                              ↳ QDP
                                ↳ UsableRulesProof
                                  ↳ QDP
                                    ↳ Rewriting
                                      ↳ QDP
                                        ↳ UsableRulesProof
                                          ↳ QDP
                                            ↳ QReductionProof
QDP
                                                ↳ ForwardInstantiation

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

F(y0, s(x0), y2) → F(y0, x0, y2)

R is empty.
Q is empty.
We have to consider all minimal (P,Q,R)-chains.
By forward instantiating [14] the rule F(y0, s(x0), y2) → F(y0, x0, y2) we obtained the following new rules:

F(x0, s(s(y_1)), x2) → F(x0, s(y_1), x2)



↳ QTRS
  ↳ Overlay + Local Confluence
    ↳ QTRS
      ↳ DependencyPairsProof
        ↳ QDP
          ↳ DependencyGraphProof
            ↳ AND
              ↳ QDP
              ↳ QDP
              ↳ QDP
                ↳ UsableRulesProof
                  ↳ QDP
                    ↳ QReductionProof
                      ↳ QDP
                        ↳ Narrowing
                          ↳ QDP
                            ↳ DependencyGraphProof
                              ↳ QDP
                                ↳ UsableRulesProof
                                  ↳ QDP
                                    ↳ Rewriting
                                      ↳ QDP
                                        ↳ UsableRulesProof
                                          ↳ QDP
                                            ↳ QReductionProof
                                              ↳ QDP
                                                ↳ ForwardInstantiation
QDP
                                                    ↳ QDPSizeChangeProof

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

F(x0, s(s(y_1)), x2) → F(x0, s(y_1), x2)

R is empty.
Q is empty.
We have to consider all minimal (P,Q,R)-chains.
By using the subterm criterion [20] together with the size-change analysis [32] we have proven that there are no infinite chains for this DP problem.

From the DPs we obtained the following set of size-change graphs: