Termination w.r.t. Q proof of /tmp/aprove_benchmark_example

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(app(quot, 0), app(s, y)) → 0
app(app(quot, app(s, x)), app(s, y)) → app(s, app(app(quot, app(app(minus, x), y)), app(s, y)))
app(app(plus, 0), y) → y
app(app(plus, app(s, x)), y) → app(s, app(app(plus, x), y))
app(app(plus, app(app(minus, x), app(s, 0))), app(app(minus, y), app(s, app(s, z)))) → app(app(plus, app(app(minus, y), app(s, app(s, z)))), app(app(minus, x), app(s, 0)))
app(app(plus, app(app(plus, x), app(s, 0))), app(app(plus, y), app(s, app(s, z)))) → app(app(plus, app(app(plus, y), app(s, app(s, z)))), app(app(plus, x), app(s, 0)))
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(filter, f), nil) → nil
app(app(filter, f), app(app(cons, x), xs)) → app(app(app(app(filter2, app(f, x)), f), x), xs)
app(app(app(app(filter2, true), f), x), xs) → app(app(cons, x), app(app(filter, f), xs))
app(app(app(app(filter2, false), f), x), xs) → app(app(filter, f), xs)

Q is empty.

↳ 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(app(quot, 0), app(s, y)) → 0
app(app(quot, app(s, x)), app(s, y)) → app(s, app(app(quot, app(app(minus, x), y)), app(s, y)))
app(app(plus, 0), y) → y
app(app(plus, app(s, x)), y) → app(s, app(app(plus, x), y))
app(app(plus, app(app(minus, x), app(s, 0))), app(app(minus, y), app(s, app(s, z)))) → app(app(plus, app(app(minus, y), app(s, app(s, z)))), app(app(minus, x), app(s, 0)))
app(app(plus, app(app(plus, x), app(s, 0))), app(app(plus, y), app(s, app(s, z)))) → app(app(plus, app(app(plus, y), app(s, app(s, z)))), app(app(plus, x), app(s, 0)))
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(filter, f), nil) → nil
app(app(filter, f), app(app(cons, x), xs)) → app(app(app(app(filter2, app(f, x)), f), x), xs)
app(app(app(app(filter2, true), f), x), xs) → app(app(cons, x), app(app(filter, f), xs))
app(app(app(app(filter2, false), f), x), xs) → app(app(filter, f), xs)

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(app(plus, app(s, x)), y) → APP(app(plus, x), y)
APP(app(filter, f), app(app(cons, x), xs)) → APP(app(filter2, app(f, x)), f)
APP(app(filter, f), app(app(cons, x), xs)) → APP(f, x)
APP(app(minus, app(s, x)), app(s, y)) → APP(app(minus, x), y)
APP(app(quot, app(s, x)), app(s, y)) → APP(app(minus, x), y)
APP(app(quot, app(s, x)), app(s, y)) → APP(minus, x)
APP(app(quot, app(s, x)), app(s, y)) → APP(s, app(app(quot, app(app(minus, x), y)), app(s, y)))
APP(app(filter, f), app(app(cons, x), xs)) → APP(app(app(app(filter2, app(f, x)), f), x), xs)
APP(app(map, f), app(app(cons, x), xs)) → APP(app(map, f), xs)
APP(app(plus, app(s, x)), y) → APP(s, app(app(plus, x), y))
APP(app(app(app(filter2, false), f), x), xs) → APP(app(filter, f), xs)
APP(app(app(app(filter2, true), f), x), xs) → APP(app(filter, f), xs)
APP(app(plus, app(app(plus, x), app(s, 0))), app(app(plus, y), app(s, app(s, z)))) → APP(app(plus, app(app(plus, y), app(s, app(s, z)))), app(app(plus, x), app(s, 0)))
APP(app(app(app(filter2, false), f), x), xs) → APP(filter, f)
APP(app(quot, app(s, x)), app(s, y)) → APP(quot, app(app(minus, x), y))
APP(app(filter, f), app(app(cons, x), xs)) → APP(app(app(filter2, app(f, x)), f), x)
APP(app(minus, app(s, x)), app(s, y)) → APP(minus, x)
APP(app(plus, app(app(minus, x), app(s, 0))), app(app(minus, y), app(s, app(s, z)))) → APP(plus, app(app(minus, y), app(s, app(s, z))))
APP(app(plus, app(app(plus, x), app(s, 0))), app(app(plus, y), app(s, app(s, z)))) → APP(plus, app(app(plus, y), app(s, app(s, z))))
APP(app(plus, app(s, x)), y) → APP(plus, x)
APP(app(map, f), app(app(cons, x), xs)) → APP(f, x)
APP(app(quot, app(s, x)), app(s, y)) → APP(app(quot, app(app(minus, x), y)), app(s, y))
APP(app(app(app(filter2, true), f), x), xs) → APP(cons, x)
APP(app(filter, f), app(app(cons, x), xs)) → APP(filter2, app(f, x))
APP(app(map, f), app(app(cons, x), xs)) → APP(cons, app(f, x))
APP(app(app(app(filter2, true), f), x), xs) → APP(filter, f)
APP(app(plus, app(app(minus, x), app(s, 0))), app(app(minus, y), app(s, app(s, z)))) → APP(app(plus, app(app(minus, y), app(s, app(s, z)))), app(app(minus, x), app(s, 0)))
APP(app(app(app(filter2, true), f), x), xs) → APP(app(cons, x), app(app(filter, f), xs))
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(minus, x), 0) → x
app(app(minus, app(s, x)), app(s, y)) → app(app(minus, x), y)
app(app(quot, 0), app(s, y)) → 0
app(app(quot, app(s, x)), app(s, y)) → app(s, app(app(quot, app(app(minus, x), y)), app(s, y)))
app(app(plus, 0), y) → y
app(app(plus, app(s, x)), y) → app(s, app(app(plus, x), y))
app(app(plus, app(app(minus, x), app(s, 0))), app(app(minus, y), app(s, app(s, z)))) → app(app(plus, app(app(minus, y), app(s, app(s, z)))), app(app(minus, x), app(s, 0)))
app(app(plus, app(app(plus, x), app(s, 0))), app(app(plus, y), app(s, app(s, z)))) → app(app(plus, app(app(plus, y), app(s, app(s, z)))), app(app(plus, x), app(s, 0)))
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(filter, f), nil) → nil
app(app(filter, f), app(app(cons, x), xs)) → app(app(app(app(filter2, app(f, x)), f), x), xs)
app(app(app(app(filter2, true), f), x), xs) → app(app(cons, x), app(app(filter, f), xs))
app(app(app(app(filter2, false), f), x), xs) → app(app(filter, f), xs)

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(app(plus, app(s, x)), y) → APP(app(plus, x), y)
APP(app(filter, f), app(app(cons, x), xs)) → APP(app(filter2, app(f, x)), f)
APP(app(filter, f), app(app(cons, x), xs)) → APP(f, x)
APP(app(minus, app(s, x)), app(s, y)) → APP(app(minus, x), y)
APP(app(quot, app(s, x)), app(s, y)) → APP(app(minus, x), y)
APP(app(quot, app(s, x)), app(s, y)) → APP(minus, x)
APP(app(quot, app(s, x)), app(s, y)) → APP(s, app(app(quot, app(app(minus, x), y)), app(s, y)))
APP(app(filter, f), app(app(cons, x), xs)) → APP(app(app(app(filter2, app(f, x)), f), x), xs)
APP(app(map, f), app(app(cons, x), xs)) → APP(app(map, f), xs)
APP(app(plus, app(s, x)), y) → APP(s, app(app(plus, x), y))
APP(app(app(app(filter2, false), f), x), xs) → APP(app(filter, f), xs)
APP(app(app(app(filter2, true), f), x), xs) → APP(app(filter, f), xs)
APP(app(plus, app(app(plus, x), app(s, 0))), app(app(plus, y), app(s, app(s, z)))) → APP(app(plus, app(app(plus, y), app(s, app(s, z)))), app(app(plus, x), app(s, 0)))
APP(app(app(app(filter2, false), f), x), xs) → APP(filter, f)
APP(app(quot, app(s, x)), app(s, y)) → APP(quot, app(app(minus, x), y))
APP(app(filter, f), app(app(cons, x), xs)) → APP(app(app(filter2, app(f, x)), f), x)
APP(app(minus, app(s, x)), app(s, y)) → APP(minus, x)
APP(app(plus, app(app(minus, x), app(s, 0))), app(app(minus, y), app(s, app(s, z)))) → APP(plus, app(app(minus, y), app(s, app(s, z))))
APP(app(plus, app(app(plus, x), app(s, 0))), app(app(plus, y), app(s, app(s, z)))) → APP(plus, app(app(plus, y), app(s, app(s, z))))
APP(app(plus, app(s, x)), y) → APP(plus, x)
APP(app(map, f), app(app(cons, x), xs)) → APP(f, x)
APP(app(quot, app(s, x)), app(s, y)) → APP(app(quot, app(app(minus, x), y)), app(s, y))
APP(app(app(app(filter2, true), f), x), xs) → APP(cons, x)
APP(app(filter, f), app(app(cons, x), xs)) → APP(filter2, app(f, x))
APP(app(map, f), app(app(cons, x), xs)) → APP(cons, app(f, x))
APP(app(app(app(filter2, true), f), x), xs) → APP(filter, f)
APP(app(plus, app(app(minus, x), app(s, 0))), app(app(minus, y), app(s, app(s, z)))) → APP(app(plus, app(app(minus, y), app(s, app(s, z)))), app(app(minus, x), app(s, 0)))
APP(app(app(app(filter2, true), f), x), xs) → APP(app(cons, x), app(app(filter, f), xs))
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(minus, x), 0) → x
app(app(minus, app(s, x)), app(s, y)) → app(app(minus, x), y)
app(app(quot, 0), app(s, y)) → 0
app(app(quot, app(s, x)), app(s, y)) → app(s, app(app(quot, app(app(minus, x), y)), app(s, y)))
app(app(plus, 0), y) → y
app(app(plus, app(s, x)), y) → app(s, app(app(plus, x), y))
app(app(plus, app(app(minus, x), app(s, 0))), app(app(minus, y), app(s, app(s, z)))) → app(app(plus, app(app(minus, y), app(s, app(s, z)))), app(app(minus, x), app(s, 0)))
app(app(plus, app(app(plus, x), app(s, 0))), app(app(plus, y), app(s, app(s, z)))) → app(app(plus, app(app(plus, y), app(s, app(s, z)))), app(app(plus, x), app(s, 0)))
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(filter, f), nil) → nil
app(app(filter, f), app(app(cons, x), xs)) → app(app(app(app(filter2, app(f, x)), f), x), xs)
app(app(app(app(filter2, true), f), x), xs) → app(app(cons, x), app(app(filter, f), xs))
app(app(app(app(filter2, false), f), x), xs) → app(app(filter, f), xs)

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 18 less nodes.

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ DependencyGraphProof

        ↳ AND

          ↳ QDP

            ↳ UsableRulesProof

          ↳ QDP

          ↳ QDP

          ↳ QDP

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

APP(app(plus, app(s, x)), y) → APP(app(plus, x), y)
APP(app(plus, app(app(minus, x), app(s, 0))), app(app(minus, y), app(s, app(s, z)))) → APP(app(plus, app(app(minus, y), app(s, app(s, z)))), app(app(minus, x), app(s, 0)))
APP(app(plus, app(app(plus, x), app(s, 0))), app(app(plus, y), app(s, app(s, z)))) → APP(app(plus, app(app(plus, y), app(s, app(s, z)))), app(app(plus, x), app(s, 0)))

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(app(quot, 0), app(s, y)) → 0
app(app(quot, app(s, x)), app(s, y)) → app(s, app(app(quot, app(app(minus, x), y)), app(s, y)))
app(app(plus, 0), y) → y
app(app(plus, app(s, x)), y) → app(s, app(app(plus, x), y))
app(app(plus, app(app(minus, x), app(s, 0))), app(app(minus, y), app(s, app(s, z)))) → app(app(plus, app(app(minus, y), app(s, app(s, z)))), app(app(minus, x), app(s, 0)))
app(app(plus, app(app(plus, x), app(s, 0))), app(app(plus, y), app(s, app(s, z)))) → app(app(plus, app(app(plus, y), app(s, app(s, z)))), app(app(plus, x), app(s, 0)))
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(filter, f), nil) → nil
app(app(filter, f), app(app(cons, x), xs)) → app(app(app(app(filter2, app(f, x)), f), x), xs)
app(app(app(app(filter2, true), f), x), xs) → app(app(cons, x), app(app(filter, f), xs))
app(app(app(app(filter2, false), f), x), xs) → app(app(filter, f), xs)

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

                ↳ UsableRulesReductionPairsProof

          ↳ QDP

          ↳ QDP

          ↳ QDP

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

APP(app(plus, app(s, x)), y) → APP(app(plus, x), y)
APP(app(plus, app(app(minus, x), app(s, 0))), app(app(minus, y), app(s, app(s, z)))) → APP(app(plus, app(app(minus, y), app(s, app(s, z)))), app(app(minus, x), app(s, 0)))
APP(app(plus, app(app(plus, x), app(s, 0))), app(app(plus, y), app(s, app(s, z)))) → APP(app(plus, app(app(plus, y), app(s, app(s, z)))), app(app(plus, x), app(s, 0)))

The TRS R consists of the following rules:

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

Q is empty.
We have to consider all minimal (P,Q,R)-chains.
First, we A-transformed [17] the QDP-Problem. Then we obtain the following A-transformed DP problem.
The pairs P are:

plus1(s(x), y) → plus1(x, y)
plus1(minus(x, s(0)), minus(y, s(s(z)))) → plus1(minus(y, s(s(z))), minus(x, s(0)))
plus1(plus(x, s(0)), plus(y, s(s(z)))) → plus1(plus(y, s(s(z))), plus(x, s(0)))

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

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

Q is empty.

By using the usable rules with reduction pair processor [15] with a polynomial ordering [25], all dependency pairs and the corresponding usable rules [17] can be oriented non-strictly. All non-usable rules are removed, and those dependency pairs and usable rules that have been oriented strictly or contain non-usable symbols in their left-hand side are removed as well.

No dependency pairs are removed.

No rules are removed from R.

Used ordering: POLO with Polynomial interpretation [25]:

POL(0) = 0
POL(minus(x₁, x₂)) = x₁ + 2·x₂
POL(plus(x₁, x₂)) = 2·x₁ + 2·x₂
POL(plus1(x₁, x₂)) = x₁ + x₂
POL(s(x₁)) = x₁

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ DependencyGraphProof

        ↳ AND

          ↳ QDP

            ↳ UsableRulesProof

              ↳ QDP

                ↳ UsableRulesReductionPairsProof

                  ↳ QDP

                    ↳ RuleRemovalProof

          ↳ QDP

          ↳ QDP

          ↳ QDP

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

plus1(minus(x, s(0)), minus(y, s(s(z)))) → plus1(minus(y, s(s(z))), minus(x, s(0)))
plus1(plus(x, s(0)), plus(y, s(s(z)))) → plus1(plus(y, s(s(z))), plus(x, s(0)))
plus1(s(x), y) → plus1(x, y)

The TRS R consists of the following rules:

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

Q is empty.
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:

minus(x, 0) → x

Used ordering: POLO with Polynomial interpretation [25]:

POL(0) = 0
POL(minus(x₁, x₂)) = 1 + x₁ + 2·x₂
POL(plus(x₁, x₂)) = 2·x₁ + 2·x₂
POL(plus1(x₁, x₂)) = 2·x₁ + 2·x₂
POL(s(x₁)) = x₁

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ DependencyGraphProof

        ↳ AND

          ↳ QDP

            ↳ UsableRulesProof

              ↳ QDP

                ↳ UsableRulesReductionPairsProof

                  ↳ QDP

                    ↳ RuleRemovalProof

                      ↳ QDP

                        ↳ RuleRemovalProof

          ↳ QDP

          ↳ QDP

          ↳ QDP

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

plus1(minus(x, s(0)), minus(y, s(s(z)))) → plus1(minus(y, s(s(z))), minus(x, s(0)))
plus1(plus(x, s(0)), plus(y, s(s(z)))) → plus1(plus(y, s(s(z))), plus(x, s(0)))
plus1(s(x), y) → plus1(x, y)

The TRS R consists of the following rules:

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

plus(0, y) → y

Used ordering: POLO with Polynomial interpretation [25]:

POL(0) = 0
POL(minus(x₁, x₂)) = 2·x₁ + x₂
POL(plus(x₁, x₂)) = 1 + 2·x₁ + 2·x₂
POL(plus1(x₁, x₂)) = 2·x₁ + 2·x₂
POL(s(x₁)) = x₁

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ DependencyGraphProof

        ↳ AND

          ↳ QDP

            ↳ UsableRulesProof

              ↳ QDP

                ↳ UsableRulesReductionPairsProof

                  ↳ QDP

                    ↳ RuleRemovalProof

                      ↳ QDP

                        ↳ RuleRemovalProof

                          ↳ QDP

                            ↳ RuleRemovalProof

          ↳ QDP

          ↳ QDP

          ↳ QDP

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

plus1(minus(x, s(0)), minus(y, s(s(z)))) → plus1(minus(y, s(s(z))), minus(x, s(0)))
plus1(plus(x, s(0)), plus(y, s(s(z)))) → plus1(plus(y, s(s(z))), plus(x, s(0)))
plus1(s(x), y) → plus1(x, y)

The TRS R consists of the following rules:

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

plus1(s(x), y) → plus1(x, y)

Strictly oriented rules of the TRS R:

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

Used ordering: POLO with Polynomial interpretation [25]:

POL(0) = 0
POL(minus(x₁, x₂)) = 2·x₁ + x₂
POL(plus(x₁, x₂)) = 2 + 2·x₁ + 2·x₂
POL(plus1(x₁, x₂)) = x₁ + x₂
POL(s(x₁)) = 1 + x₁

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ DependencyGraphProof

        ↳ AND

          ↳ QDP

            ↳ UsableRulesProof

              ↳ QDP

                ↳ UsableRulesReductionPairsProof

                  ↳ QDP

                    ↳ RuleRemovalProof

                      ↳ QDP

                        ↳ RuleRemovalProof

                          ↳ QDP

                            ↳ RuleRemovalProof

                              ↳ QDP

                                ↳ DependencyGraphProof

          ↳ QDP

          ↳ QDP

          ↳ QDP

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

plus1(minus(x, s(0)), minus(y, s(s(z)))) → plus1(minus(y, s(s(z))), minus(x, s(0)))
plus1(plus(x, s(0)), plus(y, s(s(z)))) → plus1(plus(y, s(s(z))), plus(x, s(0)))

The TRS R consists of the following rules:

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

Q is empty.
We have to consider all minimal (P,Q,R)-chains.
The approximation of the Dependency Graph [15,17,22] contains 0 SCCs with 2 less nodes.

↳ 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, 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(app(quot, 0), app(s, y)) → 0
app(app(quot, app(s, x)), app(s, y)) → app(s, app(app(quot, app(app(minus, x), y)), app(s, y)))
app(app(plus, 0), y) → y
app(app(plus, app(s, x)), y) → app(s, app(app(plus, x), y))
app(app(plus, app(app(minus, x), app(s, 0))), app(app(minus, y), app(s, app(s, z)))) → app(app(plus, app(app(minus, y), app(s, app(s, z)))), app(app(minus, x), app(s, 0)))
app(app(plus, app(app(plus, x), app(s, 0))), app(app(plus, y), app(s, app(s, z)))) → app(app(plus, app(app(plus, y), app(s, app(s, z)))), app(app(plus, x), app(s, 0)))
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(filter, f), nil) → nil
app(app(filter, f), app(app(cons, x), xs)) → app(app(app(app(filter2, app(f, x)), f), x), xs)
app(app(app(app(filter2, true), f), x), xs) → app(app(cons, x), app(app(filter, f), xs))
app(app(app(app(filter2, false), f), x), xs) → app(app(filter, f), xs)

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ DependencyGraphProof

        ↳ AND

          ↳ QDP

          ↳ 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.
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

          ↳ QDP

            ↳ UsableRulesProof

              ↳ QDP

                ↳ ATransformationProof

                  ↳ 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:

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

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ DependencyGraphProof

        ↳ AND

          ↳ QDP

          ↳ QDP

          ↳ QDP

            ↳ QDPApplicativeOrderProof

          ↳ QDP

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

APP(app(quot, app(s, x)), app(s, y)) → APP(app(quot, 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)
app(app(quot, 0), app(s, y)) → 0
app(app(quot, app(s, x)), app(s, y)) → app(s, app(app(quot, app(app(minus, x), y)), app(s, y)))
app(app(plus, 0), y) → y
app(app(plus, app(s, x)), y) → app(s, app(app(plus, x), y))
app(app(plus, app(app(minus, x), app(s, 0))), app(app(minus, y), app(s, app(s, z)))) → app(app(plus, app(app(minus, y), app(s, app(s, z)))), app(app(minus, x), app(s, 0)))
app(app(plus, app(app(plus, x), app(s, 0))), app(app(plus, y), app(s, app(s, z)))) → app(app(plus, app(app(plus, y), app(s, app(s, z)))), app(app(plus, x), app(s, 0)))
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(filter, f), nil) → nil
app(app(filter, f), app(app(cons, x), xs)) → app(app(app(app(filter2, app(f, x)), f), x), xs)
app(app(app(app(filter2, true), f), x), xs) → app(app(cons, x), app(app(filter, f), xs))
app(app(app(app(filter2, false), f), x), xs) → app(app(filter, f), xs)

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

quot1(s(x), s(y)) → quot1(minus(x, y), s(y))

The a-transformed usable rules are

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

The following pairs can be oriented strictly and are deleted.

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

The remaining pairs can at least be oriented weakly.
none
Used ordering: Polynomial interpretation [25]:

POL(0) = 0
POL(minus(x₁, x₂)) = x₁
POL(quot1(x₁, x₂)) = x₁
POL(s(x₁)) = 1 + x₁

The following usable rules [17] were oriented:

app(app(minus, app(s, x)), app(s, y)) → app(app(minus, x), y)
app(app(minus, x), 0) → 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(minus, x), 0) → x
app(app(minus, app(s, x)), app(s, y)) → app(app(minus, x), y)
app(app(quot, 0), app(s, y)) → 0
app(app(quot, app(s, x)), app(s, y)) → app(s, app(app(quot, app(app(minus, x), y)), app(s, y)))
app(app(plus, 0), y) → y
app(app(plus, app(s, x)), y) → app(s, app(app(plus, x), y))
app(app(plus, app(app(minus, x), app(s, 0))), app(app(minus, y), app(s, app(s, z)))) → app(app(plus, app(app(minus, y), app(s, app(s, z)))), app(app(minus, x), app(s, 0)))
app(app(plus, app(app(plus, x), app(s, 0))), app(app(plus, y), app(s, app(s, z)))) → app(app(plus, app(app(plus, y), app(s, app(s, z)))), app(app(plus, x), app(s, 0)))
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(filter, f), nil) → nil
app(app(filter, f), app(app(cons, x), xs)) → app(app(app(app(filter2, app(f, x)), f), x), xs)
app(app(app(app(filter2, true), f), x), xs) → app(app(cons, x), app(app(filter, f), xs))
app(app(app(app(filter2, false), f), x), xs) → app(app(filter, f), xs)

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

            ↳ QDPSizeChangeProof

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

APP(app(filter, f), app(app(cons, x), xs)) → APP(f, x)
APP(app(app(app(filter2, false), f), x), xs) → APP(app(filter, f), xs)
APP(app(app(app(filter2, true), f), x), xs) → APP(app(filter, f), xs)
APP(app(filter, f), app(app(cons, x), xs)) → APP(app(app(app(filter2, app(f, x)), f), x), xs)
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(minus, x), 0) → x
app(app(minus, app(s, x)), app(s, y)) → app(app(minus, x), y)
app(app(quot, 0), app(s, y)) → 0
app(app(quot, app(s, x)), app(s, y)) → app(s, app(app(quot, app(app(minus, x), y)), app(s, y)))
app(app(plus, 0), y) → y
app(app(plus, app(s, x)), y) → app(s, app(app(plus, x), y))
app(app(plus, app(app(minus, x), app(s, 0))), app(app(minus, y), app(s, app(s, z)))) → app(app(plus, app(app(minus, y), app(s, app(s, z)))), app(app(minus, x), app(s, 0)))
app(app(plus, app(app(plus, x), app(s, 0))), app(app(plus, y), app(s, app(s, z)))) → app(app(plus, app(app(plus, y), app(s, app(s, z)))), app(app(plus, x), app(s, 0)))
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(filter, f), nil) → nil
app(app(filter, f), app(app(cons, x), xs)) → app(app(app(app(filter2, app(f, x)), f), x), xs)
app(app(app(app(filter2, true), f), x), xs) → app(app(cons, x), app(app(filter, f), xs))
app(app(app(app(filter2, false), f), x), xs) → app(app(filter, f), xs)

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:

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