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:

app(app(map, f), nil) → nil
app(app(map, f), app(app(cons, x), xs)) → app(app(cons, app(f, x)), app(app(map, f), xs))
app(app(minus, x), 0) → x
app(app(minus, app(s, x)), app(s, y)) → app(app(minus, x), y)
app(app(div, 0), app(s, y)) → 0
app(app(div, app(s, x)), app(s, y)) → app(s, app(app(div, app(app(minus, x), y)), app(s, y)))

Q is empty.


QTRS
  ↳ Overlay + Local Confluence

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

app(app(map, f), nil) → nil
app(app(map, f), app(app(cons, x), xs)) → app(app(cons, app(f, x)), app(app(map, f), xs))
app(app(minus, x), 0) → x
app(app(minus, app(s, x)), app(s, y)) → app(app(minus, x), y)
app(app(div, 0), app(s, y)) → 0
app(app(div, app(s, x)), app(s, y)) → app(s, app(app(div, app(app(minus, x), y)), app(s, y)))

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:

app(app(map, f), nil) → nil
app(app(map, f), app(app(cons, x), xs)) → app(app(cons, app(f, x)), app(app(map, f), xs))
app(app(minus, x), 0) → x
app(app(minus, app(s, x)), app(s, y)) → app(app(minus, x), y)
app(app(div, 0), app(s, y)) → 0
app(app(div, app(s, x)), app(s, y)) → app(s, app(app(div, app(app(minus, x), y)), app(s, y)))

The set Q consists of the following terms:

app(app(map, x0), nil)
app(app(map, x0), app(app(cons, x1), x2))
app(app(minus, x0), 0)
app(app(minus, app(s, x0)), app(s, x1))
app(app(div, 0), app(s, x0))
app(app(div, app(s, x0)), app(s, x1))


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

APP(app(minus, app(s, x)), app(s, y)) → APP(app(minus, x), y)
APP(app(div, app(s, x)), app(s, y)) → APP(minus, x)
APP(app(map, f), app(app(cons, x), xs)) → APP(cons, app(f, x))
APP(app(div, app(s, x)), app(s, y)) → APP(app(div, app(app(minus, x), y)), app(s, y))
APP(app(minus, app(s, x)), app(s, y)) → APP(minus, x)
APP(app(div, app(s, x)), app(s, y)) → APP(app(minus, x), y)
APP(app(div, app(s, x)), app(s, y)) → APP(s, app(app(div, app(app(minus, x), y)), app(s, y)))
APP(app(map, f), app(app(cons, x), xs)) → APP(f, x)
APP(app(map, f), app(app(cons, x), xs)) → APP(app(map, f), xs)
APP(app(div, app(s, x)), app(s, y)) → APP(div, app(app(minus, x), y))
APP(app(map, f), app(app(cons, x), xs)) → APP(app(cons, app(f, x)), app(app(map, f), xs))

The TRS R consists of the following rules:

app(app(map, f), nil) → nil
app(app(map, f), app(app(cons, x), xs)) → app(app(cons, app(f, x)), app(app(map, f), xs))
app(app(minus, x), 0) → x
app(app(minus, app(s, x)), app(s, y)) → app(app(minus, x), y)
app(app(div, 0), app(s, y)) → 0
app(app(div, app(s, x)), app(s, y)) → app(s, app(app(div, app(app(minus, x), y)), app(s, y)))

The set Q consists of the following terms:

app(app(map, x0), nil)
app(app(map, x0), app(app(cons, x1), x2))
app(app(minus, x0), 0)
app(app(minus, app(s, x0)), app(s, x1))
app(app(div, 0), app(s, x0))
app(app(div, app(s, x0)), app(s, x1))

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:

APP(app(minus, app(s, x)), app(s, y)) → APP(app(minus, x), y)
APP(app(div, app(s, x)), app(s, y)) → APP(minus, x)
APP(app(map, f), app(app(cons, x), xs)) → APP(cons, app(f, x))
APP(app(div, app(s, x)), app(s, y)) → APP(app(div, app(app(minus, x), y)), app(s, y))
APP(app(minus, app(s, x)), app(s, y)) → APP(minus, x)
APP(app(div, app(s, x)), app(s, y)) → APP(app(minus, x), y)
APP(app(div, app(s, x)), app(s, y)) → APP(s, app(app(div, app(app(minus, x), y)), app(s, y)))
APP(app(map, f), app(app(cons, x), xs)) → APP(f, x)
APP(app(map, f), app(app(cons, x), xs)) → APP(app(map, f), xs)
APP(app(div, app(s, x)), app(s, y)) → APP(div, app(app(minus, x), y))
APP(app(map, f), app(app(cons, x), xs)) → APP(app(cons, app(f, x)), app(app(map, f), xs))

The TRS R consists of the following rules:

app(app(map, f), nil) → nil
app(app(map, f), app(app(cons, x), xs)) → app(app(cons, app(f, x)), app(app(map, f), xs))
app(app(minus, x), 0) → x
app(app(minus, app(s, x)), app(s, y)) → app(app(minus, x), y)
app(app(div, 0), app(s, y)) → 0
app(app(div, app(s, x)), app(s, y)) → app(s, app(app(div, app(app(minus, x), y)), app(s, y)))

The set Q consists of the following terms:

app(app(map, x0), nil)
app(app(map, x0), app(app(cons, x1), x2))
app(app(minus, x0), 0)
app(app(minus, app(s, x0)), app(s, x1))
app(app(div, 0), app(s, x0))
app(app(div, app(s, x0)), app(s, x1))

We have to consider all minimal (P,Q,R)-chains.
The approximation of the Dependency Graph [15,17,22] contains 3 SCCs with 7 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:

APP(app(minus, app(s, x)), app(s, y)) → APP(app(minus, x), y)

The TRS R consists of the following rules:

app(app(map, f), nil) → nil
app(app(map, f), app(app(cons, x), xs)) → app(app(cons, app(f, x)), app(app(map, f), xs))
app(app(minus, x), 0) → x
app(app(minus, app(s, x)), app(s, y)) → app(app(minus, x), y)
app(app(div, 0), app(s, y)) → 0
app(app(div, app(s, x)), app(s, y)) → app(s, app(app(div, app(app(minus, x), y)), app(s, y)))

The set Q consists of the following terms:

app(app(map, x0), nil)
app(app(map, x0), app(app(cons, x1), x2))
app(app(minus, x0), 0)
app(app(minus, app(s, x0)), app(s, x1))
app(app(div, 0), app(s, x0))
app(app(div, app(s, x0)), app(s, x1))

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
                    ↳ ATransformationProof
              ↳ QDP
              ↳ QDP

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

APP(app(minus, app(s, x)), app(s, y)) → APP(app(minus, x), y)

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

app(app(map, x0), nil)
app(app(map, x0), app(app(cons, x1), x2))
app(app(minus, x0), 0)
app(app(minus, app(s, x0)), app(s, x1))
app(app(div, 0), app(s, x0))
app(app(div, app(s, x0)), app(s, x1))

We have to consider all minimal (P,Q,R)-chains.
We have applied the A-Transformation [17] to get from an applicative problem to a standard problem.

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

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

minus1(s(x), s(y)) → minus1(x, y)

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

map(x0, nil)
map(x0, cons(x1, x2))
minus(x0, 0)
minus(s(x0), s(x1))
div(0, s(x0))
div(s(x0), s(x1))

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.

map(x0, nil)
map(x0, cons(x1, x2))
minus(x0, 0)
minus(s(x0), s(x1))
div(0, s(x0))
div(s(x0), s(x1))



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

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

minus1(s(x), s(y)) → minus1(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:

APP(app(div, app(s, x)), app(s, y)) → APP(app(div, app(app(minus, x), y)), app(s, y))

The TRS R consists of the following rules:

app(app(map, f), nil) → nil
app(app(map, f), app(app(cons, x), xs)) → app(app(cons, app(f, x)), app(app(map, f), xs))
app(app(minus, x), 0) → x
app(app(minus, app(s, x)), app(s, y)) → app(app(minus, x), y)
app(app(div, 0), app(s, y)) → 0
app(app(div, app(s, x)), app(s, y)) → app(s, app(app(div, app(app(minus, x), y)), app(s, y)))

The set Q consists of the following terms:

app(app(map, x0), nil)
app(app(map, x0), app(app(cons, x1), x2))
app(app(minus, x0), 0)
app(app(minus, app(s, x0)), app(s, x1))
app(app(div, 0), app(s, x0))
app(app(div, app(s, x0)), app(s, x1))

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
                    ↳ ATransformationProof
              ↳ QDP

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

APP(app(div, app(s, x)), app(s, y)) → APP(app(div, app(app(minus, x), y)), app(s, y))

The TRS R consists of the following rules:

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

The set Q consists of the following terms:

app(app(map, x0), nil)
app(app(map, x0), app(app(cons, x1), x2))
app(app(minus, x0), 0)
app(app(minus, app(s, x0)), app(s, x1))
app(app(div, 0), app(s, x0))
app(app(div, app(s, x0)), app(s, x1))

We have to consider all minimal (P,Q,R)-chains.
We have applied the A-Transformation [17] to get from an applicative problem to a standard problem.

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

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

div1(s(x), s(y)) → div1(minus(x, y), s(y))

The TRS R consists of the following rules:

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

The set Q consists of the following terms:

map(x0, nil)
map(x0, cons(x1, x2))
minus(x0, 0)
minus(s(x0), s(x1))
div(0, s(x0))
div(s(x0), s(x1))

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.

map(x0, nil)
map(x0, cons(x1, x2))
div(0, s(x0))
div(s(x0), s(x1))



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

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

div1(s(x), s(y)) → div1(minus(x, y), s(y))

The TRS R consists of the following rules:

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

The set Q consists of the following terms:

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

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.


div1(s(x), s(y)) → div1(minus(x, y), s(y))
The remaining pairs can at least be oriented weakly.
none
Used ordering: Polynomial interpretation [25]:

POL(0) = 0   
POL(div1(x1, x2)) = x1   
POL(minus(x1, x2)) = x1   
POL(s(x1)) = 1 + x1   

The following usable rules [17] were oriented:

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



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

Q DP problem:
P is empty.
The TRS R consists of the following rules:

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

The set Q consists of the following terms:

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

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:

APP(app(map, f), app(app(cons, x), xs)) → APP(f, x)
APP(app(map, f), app(app(cons, x), xs)) → APP(app(map, f), xs)

The TRS R consists of the following rules:

app(app(map, f), nil) → nil
app(app(map, f), app(app(cons, x), xs)) → app(app(cons, app(f, x)), app(app(map, f), xs))
app(app(minus, x), 0) → x
app(app(minus, app(s, x)), app(s, y)) → app(app(minus, x), y)
app(app(div, 0), app(s, y)) → 0
app(app(div, app(s, x)), app(s, y)) → app(s, app(app(div, app(app(minus, x), y)), app(s, y)))

The set Q consists of the following terms:

app(app(map, x0), nil)
app(app(map, x0), app(app(cons, x1), x2))
app(app(minus, x0), 0)
app(app(minus, app(s, x0)), app(s, x1))
app(app(div, 0), app(s, x0))
app(app(div, app(s, x0)), app(s, x1))

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
                    ↳ QDPSizeChangeProof

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

APP(app(map, f), app(app(cons, x), xs)) → APP(f, x)
APP(app(map, f), app(app(cons, x), xs)) → APP(app(map, f), xs)

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

app(app(map, x0), nil)
app(app(map, x0), app(app(cons, x1), x2))
app(app(minus, x0), 0)
app(app(minus, app(s, x0)), app(s, x1))
app(app(div, 0), app(s, x0))
app(app(div, app(s, x0)), app(s, x1))

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: