Termination w.r.t. Q proof of /tmp/tmptu-1ob/kabasci06.trs

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, app(p, app(s, x))), app(p, app(s, y)))
app(p, app(s, x)) → x
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), app(id, y))), app(s, y)))
app(id, x) → x
app(id, x) → app(s, app(s, app(s, x)))
app(id, app(p, x)) → app(id, app(s, app(id, x)))

Q is empty.

↳ 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, app(p, app(s, x))), app(p, app(s, y)))
app(p, app(s, x)) → x
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), app(id, y))), app(s, y)))
app(id, x) → x
app(id, x) → app(s, app(s, app(s, x)))
app(id, app(p, x)) → app(id, app(s, app(id, x)))

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:

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

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:

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

Q is empty.
We have to consider all minimal (P,Q,R)-chains.
The approximation of the Dependency Graph [15,17,22] contains 4 SCCs with 15 less nodes.

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ DependencyGraphProof

        ↳ AND

          ↳ QDP

            ↳ UsableRulesProof

          ↳ QDP

          ↳ QDP

          ↳ QDP

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

APP(id, app(p, x)) → APP(id, x)

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, app(p, app(s, x))), app(p, app(s, y)))
app(p, app(s, x)) → x
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), app(id, y))), app(s, y)))
app(id, x) → x
app(id, x) → app(s, app(s, app(s, x)))
app(id, app(p, x)) → app(id, app(s, app(id, x)))

Q is empty.
We have to consider all minimal (P,Q,R)-chains.
We can use the usable rules and reduction pair processor [15] with the Ce-compatible extension of the polynomial order that maps every function symbol to the sum of its argument. Then, we can delete all non-usable rules [17] from R.

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ DependencyGraphProof

        ↳ AND

          ↳ QDP

            ↳ UsableRulesProof

              ↳ QDP

                ↳ ATransformationProof

          ↳ QDP

          ↳ QDP

          ↳ QDP

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

APP(id, app(p, x)) → APP(id, x)

R is empty.
Q is empty.
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

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ DependencyGraphProof

        ↳ AND

          ↳ QDP

            ↳ UsableRulesProof

              ↳ QDP

                ↳ ATransformationProof

                  ↳ QDP

                    ↳ QDPSizeChangeProof

          ↳ QDP

          ↳ QDP

          ↳ QDP

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

id(p(x)) → id(x)

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:

id(p(x)) → id(x)
The graph contains the following edges 1 > 1

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ DependencyGraphProof

        ↳ AND

          ↳ QDP

          ↳ 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, app(p, app(s, x))), app(p, 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, app(p, app(s, x))), app(p, app(s, y)))
app(p, app(s, x)) → x
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), app(id, y))), app(s, y)))
app(id, x) → x
app(id, x) → app(s, app(s, app(s, x)))
app(id, app(p, x)) → app(id, app(s, app(id, x)))

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ DependencyGraphProof

        ↳ AND

          ↳ QDP

          ↳ QDP

            ↳ UsableRulesProof

              ↳ QDP

                ↳ MNOCProof

          ↳ 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, app(p, app(s, x))), app(p, app(s, y)))

The TRS R consists of the following rules:

app(p, app(s, x)) → x

Q is empty.
We have to consider all minimal (P,Q,R)-chains.
We use the modular non-overlap check [15] to enlarge Q to all left-hand sides of R.

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ DependencyGraphProof

        ↳ AND

          ↳ QDP

          ↳ QDP

            ↳ UsableRulesProof

              ↳ QDP

                ↳ MNOCProof

                  ↳ 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, app(p, app(s, x))), app(p, app(s, y)))

The TRS R consists of the following rules:

app(p, app(s, x)) → x

The set Q consists of the following terms:

app(p, app(s, x0))

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

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ DependencyGraphProof

        ↳ AND

          ↳ QDP

          ↳ QDP

            ↳ UsableRulesProof

              ↳ QDP

                ↳ MNOCProof

                  ↳ QDP

                    ↳ ATransformationProof

                      ↳ QDP

                        ↳ RuleRemovalProof

          ↳ QDP

          ↳ QDP

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

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

The TRS R consists of the following rules:

p(s(x)) → x

The set Q consists of the following terms:

p(s(x0))

We have to consider all minimal (P,Q,R)-chains.
By using the rule removal processor [15] with the following polynomial ordering [25], at least one Dependency Pair or term rewrite system rule of this QDP problem can be strictly oriented.

Strictly oriented rules of the TRS R:

p(s(x)) → x

Used ordering: POLO with Polynomial interpretation [25]:

POL(minus(x₁, x₂)) = 2·x₁ + 2·x₂
POL(p(x₁)) = x₁
POL(s(x₁)) = 2 + x₁

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ DependencyGraphProof

        ↳ AND

          ↳ QDP

          ↳ QDP

            ↳ UsableRulesProof

              ↳ QDP

                ↳ MNOCProof

                  ↳ QDP

                    ↳ ATransformationProof

                      ↳ QDP

                        ↳ RuleRemovalProof

                          ↳ QDP

                            ↳ DependencyGraphProof

          ↳ QDP

          ↳ QDP

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

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

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

p(s(x0))

We have to consider all minimal (P,Q,R)-chains.
The approximation of the Dependency Graph [15,17,22] contains 0 SCCs with 1 less node.

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ DependencyGraphProof

        ↳ AND

          ↳ QDP

          ↳ QDP

          ↳ QDP

            ↳ QDPApplicativeOrderProof

          ↳ 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), app(id, 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, app(p, app(s, x))), app(p, app(s, y)))
app(p, app(s, x)) → x
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), app(id, y))), app(s, y)))
app(id, x) → x
app(id, x) → app(s, app(s, app(s, x)))
app(id, app(p, x)) → app(id, app(s, app(id, x)))

Q is empty.
We have to consider all minimal (P,Q,R)-chains.
We use the reduction pair processor [15].Here, we combined the reduction pair processor with the A-transformation [17] which results in the following intermediate Q-DP Problem.
The a-transformed P is

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

The a-transformed usable rules are

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

The following pairs can be oriented strictly and are deleted.

APP(app(div, app(s, x)), app(s, y)) → APP(app(div, app(app(minus, x), app(id, y))), app(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(id(x₁)) = x₁
POL(minus(x₁, x₂)) = x₁
POL(p(x₁)) = x₁
POL(s(x₁)) = 1 + x₁

The following usable rules [17] were oriented:

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

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ DependencyGraphProof

        ↳ AND

          ↳ QDP

          ↳ QDP

          ↳ QDP

            ↳ QDPApplicativeOrderProof

              ↳ QDP

                ↳ PisEmptyProof

          ↳ QDP

Q DP problem:
P is empty.
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, app(p, app(s, x))), app(p, app(s, y)))
app(p, app(s, x)) → x
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), app(id, y))), app(s, y)))
app(id, x) → x
app(id, x) → app(s, app(s, app(s, x)))
app(id, app(p, x)) → app(id, app(s, app(id, x)))

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

          ↳ 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, app(p, app(s, x))), app(p, app(s, y)))
app(p, app(s, x)) → x
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), app(id, y))), app(s, y)))
app(id, x) → x
app(id, x) → app(s, app(s, app(s, x)))
app(id, app(p, x)) → app(id, app(s, app(id, x)))

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ DependencyGraphProof

        ↳ AND

          ↳ QDP

          ↳ 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)

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

APP(app(map, f), app(app(cons, x), xs)) → APP(f, x)
The graph contains the following edges 1 > 1, 2 > 2
APP(app(map, f), app(app(cons, x), xs)) → APP(app(map, f), xs)
The graph contains the following edges 1 >= 1, 2 > 2