Termination w.r.t. Q proof of /tmp/tmp2f0TGB/#3.36.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(minus, x), 0) → x
app(app(minus, app(s, x)), app(s, y)) → app(app(minus, x), y)
app(f, 0) → app(s, 0)
app(f, app(s, x)) → app(app(minus, app(s, x)), app(g, app(f, x)))
app(g, 0) → 0
app(g, app(s, x)) → app(app(minus, app(s, x)), app(f, app(g, x)))
app(app(map, fun), nil) → nil
app(app(map, fun), app(app(cons, x), xs)) → app(app(cons, app(fun, x)), app(app(map, fun), xs))
app(app(filter, fun), nil) → nil
app(app(filter, fun), app(app(cons, x), xs)) → app(app(app(app(filter2, app(fun, x)), fun), x), xs)
app(app(app(app(filter2, true), fun), x), xs) → app(app(cons, x), app(app(filter, fun), xs))
app(app(app(app(filter2, false), fun), x), xs) → app(app(filter, fun), xs)

Q is empty.

↳ QTRS

  ↳ Overlay + Local Confluence

Q restricted rewrite system:
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)
app(f, 0) → app(s, 0)
app(f, app(s, x)) → app(app(minus, app(s, x)), app(g, app(f, x)))
app(g, 0) → 0
app(g, app(s, x)) → app(app(minus, app(s, x)), app(f, app(g, x)))
app(app(map, fun), nil) → nil
app(app(map, fun), app(app(cons, x), xs)) → app(app(cons, app(fun, x)), app(app(map, fun), xs))
app(app(filter, fun), nil) → nil
app(app(filter, fun), app(app(cons, x), xs)) → app(app(app(app(filter2, app(fun, x)), fun), x), xs)
app(app(app(app(filter2, true), fun), x), xs) → app(app(cons, x), app(app(filter, fun), xs))
app(app(app(app(filter2, false), fun), x), xs) → app(app(filter, fun), xs)

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(minus, x), 0) → x
app(app(minus, app(s, x)), app(s, y)) → app(app(minus, x), y)
app(f, 0) → app(s, 0)
app(f, app(s, x)) → app(app(minus, app(s, x)), app(g, app(f, x)))
app(g, 0) → 0
app(g, app(s, x)) → app(app(minus, app(s, x)), app(f, app(g, x)))
app(app(map, fun), nil) → nil
app(app(map, fun), app(app(cons, x), xs)) → app(app(cons, app(fun, x)), app(app(map, fun), xs))
app(app(filter, fun), nil) → nil
app(app(filter, fun), app(app(cons, x), xs)) → app(app(app(app(filter2, app(fun, x)), fun), x), xs)
app(app(app(app(filter2, true), fun), x), xs) → app(app(cons, x), app(app(filter, fun), xs))
app(app(app(app(filter2, false), fun), x), xs) → app(app(filter, fun), xs)

The set Q consists of the following terms:

app(app(minus, x0), 0)
app(app(minus, app(s, x0)), app(s, x1))
app(f, 0)
app(f, app(s, x0))
app(g, 0)
app(g, app(s, x0))
app(app(map, x0), nil)
app(app(map, x0), app(app(cons, x1), x2))
app(app(filter, x0), nil)
app(app(filter, x0), app(app(cons, x1), x2))
app(app(app(app(filter2, true), x0), x1), x2)
app(app(app(app(filter2, false), x0), 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:

APP(app(filter, fun), app(app(cons, x), xs)) → APP(app(filter2, app(fun, x)), fun)
APP(app(filter, fun), app(app(cons, x), xs)) → APP(fun, x)
APP(app(minus, app(s, x)), app(s, y)) → APP(app(minus, x), y)
APP(app(filter, fun), app(app(cons, x), xs)) → APP(app(app(app(filter2, app(fun, x)), fun), x), xs)
APP(f, app(s, x)) → APP(app(minus, app(s, x)), app(g, app(f, x)))
APP(app(map, fun), app(app(cons, x), xs)) → APP(app(map, fun), xs)
APP(app(app(app(filter2, true), fun), x), xs) → APP(app(filter, fun), xs)
APP(app(app(app(filter2, false), fun), x), xs) → APP(app(filter, fun), xs)
APP(app(app(app(filter2, false), fun), x), xs) → APP(filter, fun)
APP(f, 0) → APP(s, 0)
APP(app(filter, fun), app(app(cons, x), xs)) → APP(app(app(filter2, app(fun, x)), fun), x)
APP(app(minus, app(s, x)), app(s, y)) → APP(minus, x)
APP(app(map, fun), app(app(cons, x), xs)) → APP(fun, x)
APP(g, app(s, x)) → APP(f, app(g, x))
APP(g, app(s, x)) → APP(minus, app(s, x))
APP(f, app(s, x)) → APP(minus, app(s, x))
APP(app(app(app(filter2, true), fun), x), xs) → APP(cons, x)
APP(app(filter, fun), app(app(cons, x), xs)) → APP(filter2, app(fun, x))
APP(g, app(s, x)) → APP(g, x)
APP(app(map, fun), app(app(cons, x), xs)) → APP(cons, app(fun, x))
APP(g, app(s, x)) → APP(app(minus, app(s, x)), app(f, app(g, x)))
APP(app(app(app(filter2, true), fun), x), xs) → APP(filter, fun)
APP(f, app(s, x)) → APP(f, x)
APP(f, app(s, x)) → APP(g, app(f, x))
APP(app(app(app(filter2, true), fun), x), xs) → APP(app(cons, x), app(app(filter, fun), xs))
APP(app(map, fun), app(app(cons, x), xs)) → APP(app(cons, app(fun, x)), app(app(map, fun), xs))

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)
app(f, 0) → app(s, 0)
app(f, app(s, x)) → app(app(minus, app(s, x)), app(g, app(f, x)))
app(g, 0) → 0
app(g, app(s, x)) → app(app(minus, app(s, x)), app(f, app(g, x)))
app(app(map, fun), nil) → nil
app(app(map, fun), app(app(cons, x), xs)) → app(app(cons, app(fun, x)), app(app(map, fun), xs))
app(app(filter, fun), nil) → nil
app(app(filter, fun), app(app(cons, x), xs)) → app(app(app(app(filter2, app(fun, x)), fun), x), xs)
app(app(app(app(filter2, true), fun), x), xs) → app(app(cons, x), app(app(filter, fun), xs))
app(app(app(app(filter2, false), fun), x), xs) → app(app(filter, fun), xs)

The set Q consists of the following terms:

app(app(minus, x0), 0)
app(app(minus, app(s, x0)), app(s, x1))
app(f, 0)
app(f, app(s, x0))
app(g, 0)
app(g, app(s, x0))
app(app(map, x0), nil)
app(app(map, x0), app(app(cons, x1), x2))
app(app(filter, x0), nil)
app(app(filter, x0), app(app(cons, x1), x2))
app(app(app(app(filter2, true), x0), x1), x2)
app(app(app(app(filter2, false), x0), 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:

APP(app(filter, fun), app(app(cons, x), xs)) → APP(app(filter2, app(fun, x)), fun)
APP(app(filter, fun), app(app(cons, x), xs)) → APP(fun, x)
APP(app(minus, app(s, x)), app(s, y)) → APP(app(minus, x), y)
APP(app(filter, fun), app(app(cons, x), xs)) → APP(app(app(app(filter2, app(fun, x)), fun), x), xs)
APP(f, app(s, x)) → APP(app(minus, app(s, x)), app(g, app(f, x)))
APP(app(map, fun), app(app(cons, x), xs)) → APP(app(map, fun), xs)
APP(app(app(app(filter2, true), fun), x), xs) → APP(app(filter, fun), xs)
APP(app(app(app(filter2, false), fun), x), xs) → APP(app(filter, fun), xs)
APP(app(app(app(filter2, false), fun), x), xs) → APP(filter, fun)
APP(f, 0) → APP(s, 0)
APP(app(filter, fun), app(app(cons, x), xs)) → APP(app(app(filter2, app(fun, x)), fun), x)
APP(app(minus, app(s, x)), app(s, y)) → APP(minus, x)
APP(app(map, fun), app(app(cons, x), xs)) → APP(fun, x)
APP(g, app(s, x)) → APP(f, app(g, x))
APP(g, app(s, x)) → APP(minus, app(s, x))
APP(f, app(s, x)) → APP(minus, app(s, x))
APP(app(app(app(filter2, true), fun), x), xs) → APP(cons, x)
APP(app(filter, fun), app(app(cons, x), xs)) → APP(filter2, app(fun, x))
APP(g, app(s, x)) → APP(g, x)
APP(app(map, fun), app(app(cons, x), xs)) → APP(cons, app(fun, x))
APP(g, app(s, x)) → APP(app(minus, app(s, x)), app(f, app(g, x)))
APP(app(app(app(filter2, true), fun), x), xs) → APP(filter, fun)
APP(f, app(s, x)) → APP(f, x)
APP(f, app(s, x)) → APP(g, app(f, x))
APP(app(app(app(filter2, true), fun), x), xs) → APP(app(cons, x), app(app(filter, fun), xs))
APP(app(map, fun), app(app(cons, x), xs)) → APP(app(cons, app(fun, x)), app(app(map, fun), xs))

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)
app(f, 0) → app(s, 0)
app(f, app(s, x)) → app(app(minus, app(s, x)), app(g, app(f, x)))
app(g, 0) → 0
app(g, app(s, x)) → app(app(minus, app(s, x)), app(f, app(g, x)))
app(app(map, fun), nil) → nil
app(app(map, fun), app(app(cons, x), xs)) → app(app(cons, app(fun, x)), app(app(map, fun), xs))
app(app(filter, fun), nil) → nil
app(app(filter, fun), app(app(cons, x), xs)) → app(app(app(app(filter2, app(fun, x)), fun), x), xs)
app(app(app(app(filter2, true), fun), x), xs) → app(app(cons, x), app(app(filter, fun), xs))
app(app(app(app(filter2, false), fun), x), xs) → app(app(filter, fun), xs)

The set Q consists of the following terms:

app(app(minus, x0), 0)
app(app(minus, app(s, x0)), app(s, x1))
app(f, 0)
app(f, app(s, x0))
app(g, 0)
app(g, app(s, x0))
app(app(map, x0), nil)
app(app(map, x0), app(app(cons, x1), x2))
app(app(filter, x0), nil)
app(app(filter, x0), app(app(cons, x1), x2))
app(app(app(app(filter2, true), x0), x1), x2)
app(app(app(app(filter2, false), x0), 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 15 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(minus, x), 0) → x
app(app(minus, app(s, x)), app(s, y)) → app(app(minus, x), y)
app(f, 0) → app(s, 0)
app(f, app(s, x)) → app(app(minus, app(s, x)), app(g, app(f, x)))
app(g, 0) → 0
app(g, app(s, x)) → app(app(minus, app(s, x)), app(f, app(g, x)))
app(app(map, fun), nil) → nil
app(app(map, fun), app(app(cons, x), xs)) → app(app(cons, app(fun, x)), app(app(map, fun), xs))
app(app(filter, fun), nil) → nil
app(app(filter, fun), app(app(cons, x), xs)) → app(app(app(app(filter2, app(fun, x)), fun), x), xs)
app(app(app(app(filter2, true), fun), x), xs) → app(app(cons, x), app(app(filter, fun), xs))
app(app(app(app(filter2, false), fun), x), xs) → app(app(filter, fun), xs)

The set Q consists of the following terms:

app(app(minus, x0), 0)
app(app(minus, app(s, x0)), app(s, x1))
app(f, 0)
app(f, app(s, x0))
app(g, 0)
app(g, app(s, x0))
app(app(map, x0), nil)
app(app(map, x0), app(app(cons, x1), x2))
app(app(filter, x0), nil)
app(app(filter, x0), app(app(cons, x1), x2))
app(app(app(app(filter2, true), x0), x1), x2)
app(app(app(app(filter2, false), x0), 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

                    ↳ 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(minus, x0), 0)
app(app(minus, app(s, x0)), app(s, x1))
app(f, 0)
app(f, app(s, x0))
app(g, 0)
app(g, app(s, x0))
app(app(map, x0), nil)
app(app(map, x0), app(app(cons, x1), x2))
app(app(filter, x0), nil)
app(app(filter, x0), app(app(cons, x1), x2))
app(app(app(app(filter2, true), x0), x1), x2)
app(app(app(app(filter2, false), x0), x1), x2)

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:

minus(x0, 0)
minus(s(x0), s(x1))
f(0)
f(s(x0))
g(0)
g(s(x0))
map(x0, nil)
map(x0, cons(x1, x2))
filter(x0, nil)
filter(x0, cons(x1, x2))
filter2(true, x0, x1, x2)
filter2(false, x0, 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, 0)
minus(s(x0), s(x1))
f(0)
f(s(x0))
g(0)
g(s(x0))
map(x0, nil)
map(x0, cons(x1, x2))
filter(x0, nil)
filter(x0, cons(x1, x2))
filter2(true, x0, x1, x2)
filter2(false, x0, x1, x2)

↳ 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:

minus1(s(x), s(y)) → minus1(x, y)
The graph contains the following edges 1 > 1, 2 > 2

↳ 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(g, app(s, x)) → APP(f, app(g, x))
APP(g, app(s, x)) → APP(g, x)
APP(f, app(s, x)) → APP(f, x)
APP(f, app(s, x)) → APP(g, app(f, x))

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)
app(f, 0) → app(s, 0)
app(f, app(s, x)) → app(app(minus, app(s, x)), app(g, app(f, x)))
app(g, 0) → 0
app(g, app(s, x)) → app(app(minus, app(s, x)), app(f, app(g, x)))
app(app(map, fun), nil) → nil
app(app(map, fun), app(app(cons, x), xs)) → app(app(cons, app(fun, x)), app(app(map, fun), xs))
app(app(filter, fun), nil) → nil
app(app(filter, fun), app(app(cons, x), xs)) → app(app(app(app(filter2, app(fun, x)), fun), x), xs)
app(app(app(app(filter2, true), fun), x), xs) → app(app(cons, x), app(app(filter, fun), xs))
app(app(app(app(filter2, false), fun), x), xs) → app(app(filter, fun), xs)

The set Q consists of the following terms:

app(app(minus, x0), 0)
app(app(minus, app(s, x0)), app(s, x1))
app(f, 0)
app(f, app(s, x0))
app(g, 0)
app(g, app(s, x0))
app(app(map, x0), nil)
app(app(map, x0), app(app(cons, x1), x2))
app(app(filter, x0), nil)
app(app(filter, x0), app(app(cons, x1), x2))
app(app(app(app(filter2, true), x0), x1), x2)
app(app(app(app(filter2, false), x0), x1), x2)

↳ 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(g, app(s, x)) → APP(f, app(g, x))
APP(g, app(s, x)) → APP(g, x)
APP(f, app(s, x)) → APP(f, x)
APP(f, app(s, x)) → APP(g, app(f, x))

The TRS R consists of the following rules:

app(g, 0) → 0
app(g, app(s, x)) → app(app(minus, app(s, x)), app(f, app(g, x)))
app(f, 0) → app(s, 0)
app(f, app(s, x)) → app(app(minus, app(s, x)), app(g, app(f, x)))
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(minus, x0), 0)
app(app(minus, app(s, x0)), app(s, x1))
app(f, 0)
app(f, app(s, x0))
app(g, 0)
app(g, app(s, x0))
app(app(map, x0), nil)
app(app(map, x0), app(app(cons, x1), x2))
app(app(filter, x0), nil)
app(app(filter, x0), app(app(cons, x1), x2))
app(app(app(app(filter2, true), x0), x1), x2)
app(app(app(app(filter2, false), x0), x1), x2)

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:

f1(s(x)) → f1(x)
g1(s(x)) → f1(g(x))
f1(s(x)) → g1(f(x))
g1(s(x)) → g1(x)

The TRS R consists of the following rules:

g(0) → 0
g(s(x)) → minus(s(x), f(g(x)))
f(0) → s(0)
f(s(x)) → minus(s(x), g(f(x)))
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))
f(0)
f(s(x0))
g(0)
g(s(x0))
map(x0, nil)
map(x0, cons(x1, x2))
filter(x0, nil)
filter(x0, cons(x1, x2))
filter2(true, x0, x1, x2)
filter2(false, x0, 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.

map(x0, nil)
map(x0, cons(x1, x2))
filter(x0, nil)
filter(x0, cons(x1, x2))
filter2(true, x0, x1, x2)
filter2(false, x0, x1, x2)

↳ 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:

g1(s(x)) → f1(g(x))
f1(s(x)) → f1(x)
f1(s(x)) → g1(f(x))
g1(s(x)) → g1(x)

The TRS R consists of the following rules:

g(0) → 0
g(s(x)) → minus(s(x), f(g(x)))
f(0) → s(0)
f(s(x)) → minus(s(x), g(f(x)))
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))
f(0)
f(s(x0))
g(0)
g(s(x0))

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.

f1(s(x)) → f1(x)
f1(s(x)) → g1(f(x))
g1(s(x)) → g1(x)

The remaining pairs can at least be oriented weakly.

g1(s(x)) → f1(g(x))

Used ordering: Polynomial interpretation [25]:

POL(0) = 0
POL(f(x₁)) = 1 + x₁
POL(f1(x₁)) = 1 + x₁
POL(g(x₁)) = x₁
POL(g1(x₁)) = x₁
POL(minus(x₁, x₂)) = x₁
POL(s(x₁)) = 1 + x₁

The following usable rules [17] were oriented:

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

↳ QTRS

  ↳ Overlay + Local Confluence

    ↳ QTRS

      ↳ DependencyPairsProof

        ↳ QDP

          ↳ DependencyGraphProof

            ↳ AND

              ↳ QDP

              ↳ QDP

                ↳ UsableRulesProof

                  ↳ QDP

                    ↳ ATransformationProof

                      ↳ QDP

                        ↳ QReductionProof

                          ↳ QDP

                            ↳ QDPOrderProof

                              ↳ QDP

                                ↳ DependencyGraphProof

              ↳ QDP

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

g1(s(x)) → f1(g(x))

The TRS R consists of the following rules:

g(0) → 0
g(s(x)) → minus(s(x), f(g(x)))
f(0) → s(0)
f(s(x)) → minus(s(x), g(f(x)))
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))
f(0)
f(s(x0))
g(0)
g(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

  ↳ Overlay + Local Confluence

    ↳ QTRS

      ↳ DependencyPairsProof

        ↳ QDP

          ↳ DependencyGraphProof

            ↳ AND

              ↳ QDP

              ↳ QDP

              ↳ QDP

                ↳ QDPSizeChangeProof

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

APP(app(filter, fun), app(app(cons, x), xs)) → APP(fun, x)
APP(app(app(app(filter2, false), fun), x), xs) → APP(app(filter, fun), xs)
APP(app(app(app(filter2, true), fun), x), xs) → APP(app(filter, fun), xs)
APP(app(filter, fun), app(app(cons, x), xs)) → APP(app(app(app(filter2, app(fun, x)), fun), x), xs)
APP(app(map, fun), app(app(cons, x), xs)) → APP(fun, x)
APP(app(map, fun), app(app(cons, x), xs)) → APP(app(map, fun), xs)

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)
app(f, 0) → app(s, 0)
app(f, app(s, x)) → app(app(minus, app(s, x)), app(g, app(f, x)))
app(g, 0) → 0
app(g, app(s, x)) → app(app(minus, app(s, x)), app(f, app(g, x)))
app(app(map, fun), nil) → nil
app(app(map, fun), app(app(cons, x), xs)) → app(app(cons, app(fun, x)), app(app(map, fun), xs))
app(app(filter, fun), nil) → nil
app(app(filter, fun), app(app(cons, x), xs)) → app(app(app(app(filter2, app(fun, x)), fun), x), xs)
app(app(app(app(filter2, true), fun), x), xs) → app(app(cons, x), app(app(filter, fun), xs))
app(app(app(app(filter2, false), fun), x), xs) → app(app(filter, fun), xs)

The set Q consists of the following terms:

app(app(minus, x0), 0)
app(app(minus, app(s, x0)), app(s, x1))
app(f, 0)
app(f, app(s, x0))
app(g, 0)
app(g, app(s, x0))
app(app(map, x0), nil)
app(app(map, x0), app(app(cons, x1), x2))
app(app(filter, x0), nil)
app(app(filter, x0), app(app(cons, x1), x2))
app(app(app(app(filter2, true), x0), x1), x2)
app(app(app(app(filter2, false), x0), x1), x2)

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:

APP(app(filter, fun), app(app(cons, x), xs)) → APP(fun, x)
The graph contains the following edges 1 > 1, 2 > 2
APP(app(map, fun), app(app(cons, x), xs)) → APP(fun, x)
The graph contains the following edges 1 > 1, 2 > 2
APP(app(map, fun), app(app(cons, x), xs)) → APP(app(map, fun), xs)
The graph contains the following edges 1 >= 1, 2 > 2
APP(app(filter, fun), app(app(cons, x), xs)) → APP(app(app(app(filter2, app(fun, x)), fun), x), xs)
The graph contains the following edges 2 > 2
APP(app(app(app(filter2, true), fun), x), xs) → APP(app(filter, fun), xs)
The graph contains the following edges 2 >= 2
APP(app(app(app(filter2, false), fun), x), xs) → APP(app(filter, fun), xs)
The graph contains the following edges 2 >= 2