Termination w.r.t. Q proof of /tmp/tmpsiTEML/#3.57.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'(app'(minus, app'(app'(minus, x), y)), z) → app'(app'(minus, x), app'(app'(plus, y), z))
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'(app, nil), k) → k
app'(app'(app, l), nil) → l
app'(app'(app, app'(app'(cons, x), l)), k) → app'(app'(cons, x), app'(app'(app, l), k))
app'(sum, app'(app'(cons, x), nil)) → app'(app'(cons, x), nil)
app'(sum, app'(app'(cons, x), app'(app'(cons, y), l))) → app'(sum, app'(app'(cons, app'(app'(plus, x), y)), l))
app'(sum, app'(app'(app, l), app'(app'(cons, x), app'(app'(cons, y), k)))) → app'(sum, app'(app'(app, l), app'(sum, app'(app'(cons, x), app'(app'(cons, y), k)))))
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'(minus, app'(app'(minus, x), y)), z) → app'(app'(minus, x), app'(app'(plus, y), z))
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'(app, nil), k) → k
app'(app'(app, l), nil) → l
app'(app'(app, app'(app'(cons, x), l)), k) → app'(app'(cons, x), app'(app'(app, l), k))
app'(sum, app'(app'(cons, x), nil)) → app'(app'(cons, x), nil)
app'(sum, app'(app'(cons, x), app'(app'(cons, y), l))) → app'(sum, app'(app'(cons, app'(app'(plus, x), y)), l))
app'(sum, app'(app'(app, l), app'(app'(cons, x), app'(app'(cons, y), k)))) → app'(sum, app'(app'(app, l), app'(sum, app'(app'(cons, x), app'(app'(cons, y), k)))))
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'(plus, app'(s, x)), y) → APP'(s, app'(app'(plus, x), y))
APP'(app'(filter, f), app'(app'(cons, x), xs)) → APP'(filter2, app'(f, x))
APP'(sum, app'(app'(cons, x), app'(app'(cons, y), l))) → APP'(cons, app'(app'(plus, x), y))
APP'(app'(minus, app'(s, x)), app'(s, y)) → APP'(minus, x)
APP'(app'(app, app'(app'(cons, x), l)), k) → APP'(app'(app, l), k)
APP'(sum, app'(app'(app, l), app'(app'(cons, x), app'(app'(cons, y), k)))) → APP'(sum, app'(app'(app, l), app'(sum, app'(app'(cons, x), app'(app'(cons, y), k)))))
APP'(app'(minus, app'(app'(minus, x), y)), z) → APP'(app'(minus, x), app'(app'(plus, y), z))
APP'(app'(quot, app'(s, x)), app'(s, y)) → APP'(quot, app'(app'(minus, x), y))
APP'(app'(map, f), app'(app'(cons, x), xs)) → APP'(app'(map, f), xs)
APP'(app'(quot, app'(s, x)), app'(s, y)) → APP'(s, app'(app'(quot, app'(app'(minus, x), y)), app'(s, y)))
APP'(app'(minus, app'(app'(minus, x), y)), z) → APP'(app'(plus, y), z)
APP'(sum, app'(app'(app, l), app'(app'(cons, x), app'(app'(cons, y), k)))) → APP'(sum, app'(app'(cons, x), app'(app'(cons, y), k)))
APP'(app'(filter, f), app'(app'(cons, x), xs)) → APP'(app'(filter2, app'(f, x)), f)
APP'(app'(plus, app'(s, x)), y) → APP'(plus, x)
APP'(sum, app'(app'(cons, x), app'(app'(cons, y), l))) → APP'(sum, app'(app'(cons, app'(app'(plus, x), y)), l))
APP'(app'(filter, f), app'(app'(cons, x), xs)) → APP'(app'(app'(filter2, app'(f, x)), f), x)
APP'(app'(quot, app'(s, x)), app'(s, y)) → APP'(app'(minus, x), y)
APP'(sum, app'(app'(cons, x), app'(app'(cons, y), l))) → APP'(plus, x)
APP'(app'(app'(app'(filter2, true), f), x), xs) → APP'(app'(cons, x), app'(app'(filter, f), xs))
APP'(sum, app'(app'(cons, x), app'(app'(cons, y), l))) → APP'(app'(plus, x), y)
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'(sum, app'(app'(cons, x), app'(app'(cons, y), l))) → APP'(app'(cons, app'(app'(plus, x), y)), l)
APP'(app'(app'(app'(filter2, true), f), x), xs) → APP'(app'(filter, f), xs)
APP'(app'(app'(app'(filter2, false), 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'(app'(app'(filter2, true), f), x), xs) → APP'(cons, x)
APP'(app'(minus, app'(app'(minus, x), y)), z) → APP'(plus, y)
APP'(app'(quot, app'(s, x)), app'(s, y)) → APP'(app'(quot, app'(app'(minus, x), y)), app'(s, y))
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'(app'(minus, x), y)
APP'(app'(map, f), app'(app'(cons, x), xs)) → APP'(f, x)
APP'(app'(filter, f), app'(app'(cons, x), xs)) → APP'(f, x)
APP'(app'(quot, app'(s, x)), app'(s, y)) → APP'(minus, x)
APP'(sum, app'(app'(app, l), app'(app'(cons, x), app'(app'(cons, y), k)))) → APP'(app'(app, l), app'(sum, app'(app'(cons, x), app'(app'(cons, y), k))))
APP'(app'(app, app'(app'(cons, x), l)), k) → APP'(app, l)
APP'(app'(app, app'(app'(cons, x), l)), k) → APP'(app'(cons, x), app'(app'(app, l), k))
APP'(app'(app'(app'(filter2, false), f), x), xs) → APP'(filter, f)

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

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'(minus, app'(app'(minus, x), y)), z) → app'(app'(minus, x), app'(app'(plus, y), z))
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'(app, nil), k) → k
app'(app'(app, l), nil) → l
app'(app'(app, app'(app'(cons, x), l)), k) → app'(app'(cons, x), app'(app'(app, l), k))
app'(sum, app'(app'(cons, x), nil)) → app'(app'(cons, x), nil)
app'(sum, app'(app'(cons, x), app'(app'(cons, y), l))) → app'(sum, app'(app'(cons, app'(app'(plus, x), y)), l))
app'(sum, app'(app'(app, l), app'(app'(cons, x), app'(app'(cons, y), k)))) → app'(sum, app'(app'(app, l), app'(sum, app'(app'(cons, x), app'(app'(cons, y), k)))))
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 7 SCCs with 26 less nodes.

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ DependencyGraphProof

        ↳ AND

          ↳ QDP

            ↳ UsableRulesProof

          ↳ QDP

          ↳ QDP

          ↳ QDP

          ↳ QDP

          ↳ QDP

          ↳ QDP

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

APP'(app'(app, app'(app'(cons, x), l)), k) → APP'(app'(app, l), k)

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'(minus, app'(app'(minus, x), y)), z) → app'(app'(minus, x), app'(app'(plus, y), z))
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'(app, nil), k) → k
app'(app'(app, l), nil) → l
app'(app'(app, app'(app'(cons, x), l)), k) → app'(app'(cons, x), app'(app'(app, l), k))
app'(sum, app'(app'(cons, x), nil)) → app'(app'(cons, x), nil)
app'(sum, app'(app'(cons, x), app'(app'(cons, y), l))) → app'(sum, app'(app'(cons, app'(app'(plus, x), y)), l))
app'(sum, app'(app'(app, l), app'(app'(cons, x), app'(app'(cons, y), k)))) → app'(sum, app'(app'(app, l), app'(sum, app'(app'(cons, x), app'(app'(cons, y), k)))))
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

                ↳ ATransformationProof

          ↳ QDP

          ↳ QDP

          ↳ QDP

          ↳ QDP

          ↳ QDP

          ↳ QDP

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

APP'(app'(app, app'(app'(cons, x), l)), k) → APP'(app'(app, l), k)

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

          ↳ QDP

          ↳ QDP

          ↳ QDP

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

app(cons(x, l), k) → app(l, k)

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:

app(cons(x, l), k) → app(l, k)
The graph contains the following edges 1 > 1, 2 >= 2

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ DependencyGraphProof

        ↳ AND

          ↳ QDP

          ↳ QDP

            ↳ UsableRulesProof

          ↳ QDP

          ↳ QDP

          ↳ 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)

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'(minus, app'(app'(minus, x), y)), z) → app'(app'(minus, x), app'(app'(plus, y), z))
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'(app, nil), k) → k
app'(app'(app, l), nil) → l
app'(app'(app, app'(app'(cons, x), l)), k) → app'(app'(cons, x), app'(app'(app, l), k))
app'(sum, app'(app'(cons, x), nil)) → app'(app'(cons, x), nil)
app'(sum, app'(app'(cons, x), app'(app'(cons, y), l))) → app'(sum, app'(app'(cons, app'(app'(plus, x), y)), l))
app'(sum, app'(app'(app, l), app'(app'(cons, x), app'(app'(cons, y), k)))) → app'(sum, app'(app'(app, l), app'(sum, app'(app'(cons, x), app'(app'(cons, y), k)))))
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

          ↳ 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)

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

          ↳ QDP

          ↳ QDP

          ↳ QDP

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

plus(s(x), y) → plus(x, y)

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

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

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ DependencyGraphProof

        ↳ AND

          ↳ QDP

          ↳ QDP

          ↳ QDP

            ↳ UsableRulesProof

          ↳ QDP

          ↳ QDP

          ↳ QDP

          ↳ QDP

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

APP'(sum, app'(app'(cons, x), app'(app'(cons, y), l))) → APP'(sum, app'(app'(cons, app'(app'(plus, x), y)), l))

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'(minus, app'(app'(minus, x), y)), z) → app'(app'(minus, x), app'(app'(plus, y), z))
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'(app, nil), k) → k
app'(app'(app, l), nil) → l
app'(app'(app, app'(app'(cons, x), l)), k) → app'(app'(cons, x), app'(app'(app, l), k))
app'(sum, app'(app'(cons, x), nil)) → app'(app'(cons, x), nil)
app'(sum, app'(app'(cons, x), app'(app'(cons, y), l))) → app'(sum, app'(app'(cons, app'(app'(plus, x), y)), l))
app'(sum, app'(app'(app, l), app'(app'(cons, x), app'(app'(cons, y), k)))) → app'(sum, app'(app'(app, l), app'(sum, app'(app'(cons, x), app'(app'(cons, y), k)))))
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

          ↳ QDP

            ↳ UsableRulesProof

              ↳ QDP

                ↳ MNOCProof

          ↳ QDP

          ↳ QDP

          ↳ QDP

          ↳ QDP

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

APP'(sum, app'(app'(cons, x), app'(app'(cons, y), l))) → APP'(sum, app'(app'(cons, app'(app'(plus, x), y)), l))

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))

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

          ↳ QDP

            ↳ UsableRulesProof

              ↳ QDP

                ↳ MNOCProof

                  ↳ QDP

                    ↳ ATransformationProof

          ↳ QDP

          ↳ QDP

          ↳ QDP

          ↳ QDP

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

APP'(sum, app'(app'(cons, x), app'(app'(cons, y), l))) → APP'(sum, app'(app'(cons, app'(app'(plus, x), y)), l))

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))

The set Q consists of the following terms:

app'(app'(plus, 0), x0)
app'(app'(plus, app'(s, x0)), 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

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ DependencyGraphProof

        ↳ AND

          ↳ QDP

          ↳ QDP

          ↳ QDP

            ↳ UsableRulesProof

              ↳ QDP

                ↳ MNOCProof

                  ↳ QDP

                    ↳ ATransformationProof

                      ↳ QDP

                        ↳ UsableRulesReductionPairsProof

          ↳ QDP

          ↳ QDP

          ↳ QDP

          ↳ QDP

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

sum(cons(x, cons(y, l))) → sum(cons(plus(x, y), l))

The TRS R consists of the following rules:

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

The set Q consists of the following terms:

plus(0, x0)
plus(s(x0), x1)

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

The following dependency pairs can be deleted:

sum(cons(x, cons(y, l))) → sum(cons(plus(x, y), l))

The following rules are removed from R:

plus(0, y) → y

Used ordering: POLO with Polynomial interpretation [25]:

POL(0) = 0
POL(cons(x₁, x₂)) = 2 + 2·x₁ + 2·x₂
POL(plus(x₁, x₂)) = x₁ + x₂
POL(s(x₁)) = x₁
POL(sum(x₁)) = 2·x₁

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ DependencyGraphProof

        ↳ AND

          ↳ QDP

          ↳ QDP

          ↳ QDP

            ↳ UsableRulesProof

              ↳ QDP

                ↳ MNOCProof

                  ↳ QDP

                    ↳ ATransformationProof

                      ↳ QDP

                        ↳ UsableRulesReductionPairsProof

                          ↳ QDP

                            ↳ PisEmptyProof

          ↳ QDP

          ↳ QDP

          ↳ QDP

          ↳ QDP

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

plus(s(x), y) → s(plus(x, y))

The set Q consists of the following terms:

plus(0, x0)
plus(s(x0), 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

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ DependencyGraphProof

        ↳ AND

          ↳ QDP

          ↳ QDP

          ↳ QDP

          ↳ QDP

            ↳ UsableRulesProof

          ↳ QDP

          ↳ QDP

          ↳ QDP

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

APP'(sum, app'(app'(app, l), app'(app'(cons, x), app'(app'(cons, y), k)))) → APP'(sum, app'(app'(app, l), app'(sum, app'(app'(cons, x), app'(app'(cons, y), k)))))

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'(minus, app'(app'(minus, x), y)), z) → app'(app'(minus, x), app'(app'(plus, y), z))
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'(app, nil), k) → k
app'(app'(app, l), nil) → l
app'(app'(app, app'(app'(cons, x), l)), k) → app'(app'(cons, x), app'(app'(app, l), k))
app'(sum, app'(app'(cons, x), nil)) → app'(app'(cons, x), nil)
app'(sum, app'(app'(cons, x), app'(app'(cons, y), l))) → app'(sum, app'(app'(cons, app'(app'(plus, x), y)), l))
app'(sum, app'(app'(app, l), app'(app'(cons, x), app'(app'(cons, y), k)))) → app'(sum, app'(app'(app, l), app'(sum, app'(app'(cons, x), app'(app'(cons, y), k)))))
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

          ↳ QDP

          ↳ QDP

            ↳ UsableRulesProof

              ↳ QDP

                ↳ UsableRulesReductionPairsProof

          ↳ QDP

          ↳ QDP

          ↳ QDP

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

APP'(sum, app'(app'(app, l), app'(app'(cons, x), app'(app'(cons, y), k)))) → APP'(sum, app'(app'(app, l), app'(sum, app'(app'(cons, x), app'(app'(cons, y), k)))))

The TRS R consists of the following rules:

app'(sum, app'(app'(cons, x), app'(app'(cons, y), l))) → app'(sum, app'(app'(cons, app'(app'(plus, x), y)), l))
app'(app'(app, nil), k) → k
app'(app'(app, l), nil) → l
app'(app'(app, app'(app'(cons, x), l)), k) → app'(app'(cons, x), app'(app'(app, l), k))
app'(app'(plus, 0), y) → y
app'(app'(plus, app'(s, x)), y) → app'(s, app'(app'(plus, x), y))
app'(sum, app'(app'(cons, x), nil)) → app'(app'(cons, x), nil)

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:

sum1(app(l, cons(x, cons(y, k)))) → sum1(app(l, sum(cons(x, cons(y, k)))))

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

sum(cons(x, cons(y, l))) → sum(cons(plus(x, y), l))
app(nil, k) → k
app(l, nil) → l
app(cons(x, l), k) → cons(x, app(l, k))
plus(0, y) → y
plus(s(x), y) → s(plus(x, y))
sum(cons(x, nil)) → cons(x, nil)

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.

The following rules are removed from R:

app(nil, k) → k
app(l, nil) → l
plus(0, y) → y

Used ordering: POLO with Polynomial interpretation [25]:

POL(0) = 0
POL(app(x₁, x₂)) = 2 + 2·x₁ + x₂
POL(cons(x₁, x₂)) = 2·x₁ + x₂
POL(nil) = 0
POL(plus(x₁, x₂)) = x₁ + x₂
POL(s(x₁)) = x₁
POL(sum(x₁)) = x₁
POL(sum1(x₁)) = x₁

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ DependencyGraphProof

        ↳ AND

          ↳ QDP

          ↳ QDP

          ↳ QDP

          ↳ QDP

            ↳ UsableRulesProof

              ↳ QDP

                ↳ UsableRulesReductionPairsProof

                  ↳ QDP

                    ↳ RuleRemovalProof

          ↳ QDP

          ↳ QDP

          ↳ QDP

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

sum1(app(l, cons(x, cons(y, k)))) → sum1(app(l, sum(cons(x, cons(y, k)))))

The TRS R consists of the following rules:

sum(cons(x, cons(y, l))) → sum(cons(plus(x, y), l))
app(cons(x, l), k) → cons(x, app(l, k))
plus(s(x), y) → s(plus(x, y))
sum(cons(x, nil)) → cons(x, nil)

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:

sum(cons(x, cons(y, l))) → sum(cons(plus(x, y), l))
app(cons(x, l), k) → cons(x, app(l, k))

Used ordering: POLO with Polynomial interpretation [25]:

POL(app(x₁, x₂)) = 1 + 2·x₁ + 2·x₂
POL(cons(x₁, x₂)) = 1 + x₁ + x₂
POL(nil) = 0
POL(plus(x₁, x₂)) = x₁ + x₂
POL(s(x₁)) = x₁
POL(sum(x₁)) = x₁
POL(sum1(x₁)) = x₁

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ DependencyGraphProof

        ↳ AND

          ↳ QDP

          ↳ QDP

          ↳ QDP

          ↳ QDP

            ↳ UsableRulesProof

              ↳ QDP

                ↳ UsableRulesReductionPairsProof

                  ↳ QDP

                    ↳ RuleRemovalProof

                      ↳ QDP

                        ↳ DependencyGraphProof

          ↳ QDP

          ↳ QDP

          ↳ QDP

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

sum1(app(l, cons(x, cons(y, k)))) → sum1(app(l, sum(cons(x, cons(y, k)))))

The TRS R consists of the following rules:

plus(s(x), y) → s(plus(x, y))
sum(cons(x, nil)) → cons(x, nil)

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 1 less node.

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ DependencyGraphProof

        ↳ AND

          ↳ QDP

          ↳ QDP

          ↳ QDP

          ↳ 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)
APP'(app'(minus, app'(app'(minus, x), y)), z) → APP'(app'(minus, x), app'(app'(plus, y), z))

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'(minus, app'(app'(minus, x), y)), z) → app'(app'(minus, x), app'(app'(plus, y), z))
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'(app, nil), k) → k
app'(app'(app, l), nil) → l
app'(app'(app, app'(app'(cons, x), l)), k) → app'(app'(cons, x), app'(app'(app, l), k))
app'(sum, app'(app'(cons, x), nil)) → app'(app'(cons, x), nil)
app'(sum, app'(app'(cons, x), app'(app'(cons, y), l))) → app'(sum, app'(app'(cons, app'(app'(plus, x), y)), l))
app'(sum, app'(app'(app, l), app'(app'(cons, x), app'(app'(cons, y), k)))) → app'(sum, app'(app'(app, l), app'(sum, app'(app'(cons, x), app'(app'(cons, y), k)))))
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

          ↳ QDP

          ↳ 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, x), y)
APP'(app'(minus, app'(app'(minus, x), y)), z) → APP'(app'(minus, x), app'(app'(plus, y), z))

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))

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

          ↳ QDP

          ↳ 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, x), y)
APP'(app'(minus, app'(app'(minus, x), y)), z) → APP'(app'(minus, x), app'(app'(plus, y), z))

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))

The set Q consists of the following terms:

app'(app'(plus, 0), x0)
app'(app'(plus, app'(s, x0)), 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

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ DependencyGraphProof

        ↳ AND

          ↳ QDP

          ↳ QDP

          ↳ QDP

          ↳ QDP

          ↳ QDP

            ↳ UsableRulesProof

              ↳ QDP

                ↳ MNOCProof

                  ↳ 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)
minus(minus1(x, y), z) → minus(x, plus(y, z))

The TRS R consists of the following rules:

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

The set Q consists of the following terms:

plus(0, x0)
plus(s(x0), 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:

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

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ DependencyGraphProof

        ↳ AND

          ↳ QDP

          ↳ QDP

          ↳ QDP

          ↳ QDP

          ↳ QDP

          ↳ QDP

            ↳ QDPOrderProof

          ↳ 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'(minus, app'(app'(minus, x), y)), z) → app'(app'(minus, x), app'(app'(plus, y), z))
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'(app, nil), k) → k
app'(app'(app, l), nil) → l
app'(app'(app, app'(app'(cons, x), l)), k) → app'(app'(cons, x), app'(app'(app, l), k))
app'(sum, app'(app'(cons, x), nil)) → app'(app'(cons, x), nil)
app'(sum, app'(app'(cons, x), app'(app'(cons, y), l))) → app'(sum, app'(app'(cons, app'(app'(plus, x), y)), l))
app'(sum, app'(app'(app, l), app'(app'(cons, x), app'(app'(cons, y), k)))) → app'(sum, app'(app'(app, l), app'(sum, app'(app'(cons, x), app'(app'(cons, y), k)))))
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(x, 0) → x
minus(s(x), s(y)) → minus(x, y)
minus(minus(x, y), z) → minus(x, plus(y, z))
plus(0, y) → y
plus(s(x), y) → s(plus(x, y))

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: Matrix interpretation [3]:
Non-tuple symbols:

M( plus(x₁, x₂) ) =

/	0	\
\	0	/

/	0	0	\
\	0	1	/

x₁

/	1	1	\
\	0	1	/

x₂

M( minus(x₁, x₂) ) =

/	1	\
\	0	/

/	1	0	\
\	0	1	/

x₁

/	0	0	\
\	0	0	/

x₂

M( s(x₁) ) =

/	0	\
\	1	/

/	1	0	\
\	0	1	/

x₁

M( 0 ) =

/	0	\
\	0	/

Tuple symbols:

M( quot1(x₁, x₂) ) =

[

]

x₁

[

]

x₂

Matrix type:
We used a basic matrix type which is not further parametrizeable.

The following usable rules [17] were oriented:

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

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ DependencyGraphProof

        ↳ AND

          ↳ QDP

          ↳ QDP

          ↳ QDP

          ↳ QDP

          ↳ QDP

          ↳ QDP

            ↳ QDPOrderProof

              ↳ 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'(minus, app'(app'(minus, x), y)), z) → app'(app'(minus, x), app'(app'(plus, y), z))
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'(app, nil), k) → k
app'(app'(app, l), nil) → l
app'(app'(app, app'(app'(cons, x), l)), k) → app'(app'(cons, x), app'(app'(app, l), k))
app'(sum, app'(app'(cons, x), nil)) → app'(app'(cons, x), nil)
app'(sum, app'(app'(cons, x), app'(app'(cons, y), l))) → app'(sum, app'(app'(cons, app'(app'(plus, x), y)), l))
app'(sum, app'(app'(app, l), app'(app'(cons, x), app'(app'(cons, y), k)))) → app'(sum, app'(app'(app, l), app'(sum, app'(app'(cons, x), app'(app'(cons, y), k)))))
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

          ↳ QDP

          ↳ QDP

          ↳ 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'(filter, 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'(app'(app'(filter2, true), f), x), xs) → APP'(app'(filter, f), xs)
APP'(app'(app'(app'(filter2, false), 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)

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'(minus, app'(app'(minus, x), y)), z) → app'(app'(minus, x), app'(app'(plus, y), z))
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'(app, nil), k) → k
app'(app'(app, l), nil) → l
app'(app'(app, app'(app'(cons, x), l)), k) → app'(app'(cons, x), app'(app'(app, l), k))
app'(sum, app'(app'(cons, x), nil)) → app'(app'(cons, x), nil)
app'(sum, app'(app'(cons, x), app'(app'(cons, y), l))) → app'(sum, app'(app'(cons, app'(app'(plus, x), y)), l))
app'(sum, app'(app'(app, l), app'(app'(cons, x), app'(app'(cons, y), k)))) → app'(sum, app'(app'(app, l), app'(sum, app'(app'(cons, x), app'(app'(cons, y), k)))))
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