Termination w.r.t. Q proof of TRS/higher-order/KusakariEx7

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(app(if, true), xs), ys) → xs
app(app(app(if, false), xs), ys) → ys
app(app(sub, x), 0) → x
app(app(sub, app(s, x)), app(s, y)) → app(app(sub, x), y)
app(app(gtr, 0), y) → false
app(app(gtr, app(s, x)), 0) → true
app(app(gtr, app(s, x)), app(s, y)) → app(app(gtr, x), y)
app(app(d, x), 0) → true
app(app(d, app(s, x)), app(s, y)) → app(app(app(if, app(app(gtr, x), y)), false), app(app(d, app(s, x)), app(app(sub, y), x)))
app(len, nil) → 0
app(len, app(app(cons, x), xs)) → app(s, app(len, xs))
app(app(filter, p), nil) → nil
app(app(filter, p), app(app(cons, x), xs)) → app(app(app(if, app(p, x)), app(app(cons, x), app(app(filter, p), xs))), app(app(filter, p), xs))

Q is empty.

↳ QTRS

  ↳ Overlay + Local Confluence

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

app(app(app(if, true), xs), ys) → xs
app(app(app(if, false), xs), ys) → ys
app(app(sub, x), 0) → x
app(app(sub, app(s, x)), app(s, y)) → app(app(sub, x), y)
app(app(gtr, 0), y) → false
app(app(gtr, app(s, x)), 0) → true
app(app(gtr, app(s, x)), app(s, y)) → app(app(gtr, x), y)
app(app(d, x), 0) → true
app(app(d, app(s, x)), app(s, y)) → app(app(app(if, app(app(gtr, x), y)), false), app(app(d, app(s, x)), app(app(sub, y), x)))
app(len, nil) → 0
app(len, app(app(cons, x), xs)) → app(s, app(len, xs))
app(app(filter, p), nil) → nil
app(app(filter, p), app(app(cons, x), xs)) → app(app(app(if, app(p, x)), app(app(cons, x), app(app(filter, p), xs))), app(app(filter, p), xs))

Q is empty.

The TRS is overlay and locally confluent. By [15] 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(app(if, true), xs), ys) → xs
app(app(app(if, false), xs), ys) → ys
app(app(sub, x), 0) → x
app(app(sub, app(s, x)), app(s, y)) → app(app(sub, x), y)
app(app(gtr, 0), y) → false
app(app(gtr, app(s, x)), 0) → true
app(app(gtr, app(s, x)), app(s, y)) → app(app(gtr, x), y)
app(app(d, x), 0) → true
app(app(d, app(s, x)), app(s, y)) → app(app(app(if, app(app(gtr, x), y)), false), app(app(d, app(s, x)), app(app(sub, y), x)))
app(len, nil) → 0
app(len, app(app(cons, x), xs)) → app(s, app(len, xs))
app(app(filter, p), nil) → nil
app(app(filter, p), app(app(cons, x), xs)) → app(app(app(if, app(p, x)), app(app(cons, x), app(app(filter, p), xs))), app(app(filter, p), xs))

The set Q consists of the following terms:

app(app(app(if, true), x0), x1)
app(app(app(if, false), x0), x1)
app(app(sub, x0), 0)
app(app(sub, app(s, x0)), app(s, x1))
app(app(gtr, 0), x0)
app(app(gtr, app(s, x0)), 0)
app(app(gtr, app(s, x0)), app(s, x1))
app(app(d, x0), 0)
app(app(d, app(s, x0)), app(s, x1))
app(len, nil)
app(len, app(app(cons, x0), x1))
app(app(filter, x0), nil)
app(app(filter, x0), app(app(cons, x1), x2))

Using Dependency Pairs [1,13] we result in the following initial DP problem:
Q DP problem:
The TRS P consists of the following rules:

APP(app(filter, p), app(app(cons, x), xs)) → APP(app(app(if, app(p, x)), app(app(cons, x), app(app(filter, p), xs))), app(app(filter, p), xs))
APP(app(gtr, app(s, x)), app(s, y)) → APP(app(gtr, x), y)
APP(app(gtr, app(s, x)), app(s, y)) → APP(gtr, x)
APP(app(d, app(s, x)), app(s, y)) → APP(app(app(if, app(app(gtr, x), y)), false), app(app(d, app(s, x)), app(app(sub, y), x)))
APP(len, app(app(cons, x), xs)) → APP(len, xs)
APP(app(filter, p), app(app(cons, x), xs)) → APP(if, app(p, x))
APP(app(sub, app(s, x)), app(s, y)) → APP(app(sub, x), y)
APP(app(d, app(s, x)), app(s, y)) → APP(app(d, app(s, x)), app(app(sub, y), x))
APP(app(sub, app(s, x)), app(s, y)) → APP(sub, x)
APP(app(d, app(s, x)), app(s, y)) → APP(gtr, x)
APP(app(d, app(s, x)), app(s, y)) → APP(sub, y)
APP(app(filter, p), app(app(cons, x), xs)) → APP(p, x)
APP(app(d, app(s, x)), app(s, y)) → APP(app(if, app(app(gtr, x), y)), false)
APP(app(d, app(s, x)), app(s, y)) → APP(if, app(app(gtr, x), y))
APP(app(filter, p), app(app(cons, x), xs)) → APP(app(filter, p), xs)
APP(app(filter, p), app(app(cons, x), xs)) → APP(app(cons, x), app(app(filter, p), xs))
APP(app(d, app(s, x)), app(s, y)) → APP(app(gtr, x), y)
APP(app(d, app(s, x)), app(s, y)) → APP(app(sub, y), x)
APP(len, app(app(cons, x), xs)) → APP(s, app(len, xs))
APP(app(filter, p), app(app(cons, x), xs)) → APP(app(if, app(p, x)), app(app(cons, x), app(app(filter, p), xs)))

The TRS R consists of the following rules:

app(app(app(if, true), xs), ys) → xs
app(app(app(if, false), xs), ys) → ys
app(app(sub, x), 0) → x
app(app(sub, app(s, x)), app(s, y)) → app(app(sub, x), y)
app(app(gtr, 0), y) → false
app(app(gtr, app(s, x)), 0) → true
app(app(gtr, app(s, x)), app(s, y)) → app(app(gtr, x), y)
app(app(d, x), 0) → true
app(app(d, app(s, x)), app(s, y)) → app(app(app(if, app(app(gtr, x), y)), false), app(app(d, app(s, x)), app(app(sub, y), x)))
app(len, nil) → 0
app(len, app(app(cons, x), xs)) → app(s, app(len, xs))
app(app(filter, p), nil) → nil
app(app(filter, p), app(app(cons, x), xs)) → app(app(app(if, app(p, x)), app(app(cons, x), app(app(filter, p), xs))), app(app(filter, p), xs))

The set Q consists of the following terms:

app(app(app(if, true), x0), x1)
app(app(app(if, false), x0), x1)
app(app(sub, x0), 0)
app(app(sub, app(s, x0)), app(s, x1))
app(app(gtr, 0), x0)
app(app(gtr, app(s, x0)), 0)
app(app(gtr, app(s, x0)), app(s, x1))
app(app(d, x0), 0)
app(app(d, app(s, x0)), app(s, x1))
app(len, nil)
app(len, app(app(cons, x0), x1))
app(app(filter, x0), nil)
app(app(filter, x0), app(app(cons, x1), x2))

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

↳ QTRS

  ↳ Overlay + Local Confluence

    ↳ QTRS

      ↳ DependencyPairsProof

        ↳ QDP

          ↳ EdgeDeletionProof

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

APP(app(filter, p), app(app(cons, x), xs)) → APP(app(app(if, app(p, x)), app(app(cons, x), app(app(filter, p), xs))), app(app(filter, p), xs))
APP(app(gtr, app(s, x)), app(s, y)) → APP(app(gtr, x), y)
APP(app(gtr, app(s, x)), app(s, y)) → APP(gtr, x)
APP(app(d, app(s, x)), app(s, y)) → APP(app(app(if, app(app(gtr, x), y)), false), app(app(d, app(s, x)), app(app(sub, y), x)))
APP(len, app(app(cons, x), xs)) → APP(len, xs)
APP(app(filter, p), app(app(cons, x), xs)) → APP(if, app(p, x))
APP(app(sub, app(s, x)), app(s, y)) → APP(app(sub, x), y)
APP(app(d, app(s, x)), app(s, y)) → APP(app(d, app(s, x)), app(app(sub, y), x))
APP(app(sub, app(s, x)), app(s, y)) → APP(sub, x)
APP(app(d, app(s, x)), app(s, y)) → APP(gtr, x)
APP(app(d, app(s, x)), app(s, y)) → APP(sub, y)
APP(app(filter, p), app(app(cons, x), xs)) → APP(p, x)
APP(app(d, app(s, x)), app(s, y)) → APP(app(if, app(app(gtr, x), y)), false)
APP(app(d, app(s, x)), app(s, y)) → APP(if, app(app(gtr, x), y))
APP(app(filter, p), app(app(cons, x), xs)) → APP(app(filter, p), xs)
APP(app(filter, p), app(app(cons, x), xs)) → APP(app(cons, x), app(app(filter, p), xs))
APP(app(d, app(s, x)), app(s, y)) → APP(app(gtr, x), y)
APP(app(d, app(s, x)), app(s, y)) → APP(app(sub, y), x)
APP(len, app(app(cons, x), xs)) → APP(s, app(len, xs))
APP(app(filter, p), app(app(cons, x), xs)) → APP(app(if, app(p, x)), app(app(cons, x), app(app(filter, p), xs)))

The TRS R consists of the following rules:

app(app(app(if, true), xs), ys) → xs
app(app(app(if, false), xs), ys) → ys
app(app(sub, x), 0) → x
app(app(sub, app(s, x)), app(s, y)) → app(app(sub, x), y)
app(app(gtr, 0), y) → false
app(app(gtr, app(s, x)), 0) → true
app(app(gtr, app(s, x)), app(s, y)) → app(app(gtr, x), y)
app(app(d, x), 0) → true
app(app(d, app(s, x)), app(s, y)) → app(app(app(if, app(app(gtr, x), y)), false), app(app(d, app(s, x)), app(app(sub, y), x)))
app(len, nil) → 0
app(len, app(app(cons, x), xs)) → app(s, app(len, xs))
app(app(filter, p), nil) → nil
app(app(filter, p), app(app(cons, x), xs)) → app(app(app(if, app(p, x)), app(app(cons, x), app(app(filter, p), xs))), app(app(filter, p), xs))

The set Q consists of the following terms:

app(app(app(if, true), x0), x1)
app(app(app(if, false), x0), x1)
app(app(sub, x0), 0)
app(app(sub, app(s, x0)), app(s, x1))
app(app(gtr, 0), x0)
app(app(gtr, app(s, x0)), 0)
app(app(gtr, app(s, x0)), app(s, x1))
app(app(d, x0), 0)
app(app(d, app(s, x0)), app(s, x1))
app(len, nil)
app(len, app(app(cons, x0), x1))
app(app(filter, x0), nil)
app(app(filter, x0), app(app(cons, x1), x2))

We have to consider all minimal (P,Q,R)-chains.
We deleted some edges using various graph approximations

↳ QTRS

  ↳ Overlay + Local Confluence

    ↳ QTRS

      ↳ DependencyPairsProof

        ↳ QDP

          ↳ EdgeDeletionProof

            ↳ QDP

              ↳ DependencyGraphProof

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

APP(app(gtr, app(s, x)), app(s, y)) → APP(app(gtr, x), y)
APP(app(filter, p), app(app(cons, x), xs)) → APP(app(app(if, app(p, x)), app(app(cons, x), app(app(filter, p), xs))), app(app(filter, p), xs))
APP(app(gtr, app(s, x)), app(s, y)) → APP(gtr, x)
APP(app(d, app(s, x)), app(s, y)) → APP(app(app(if, app(app(gtr, x), y)), false), app(app(d, app(s, x)), app(app(sub, y), x)))
APP(len, app(app(cons, x), xs)) → APP(len, xs)
APP(app(filter, p), app(app(cons, x), xs)) → APP(if, app(p, x))
APP(app(d, app(s, x)), app(s, y)) → APP(app(d, app(s, x)), app(app(sub, y), x))
APP(app(sub, app(s, x)), app(s, y)) → APP(app(sub, x), y)
APP(app(sub, app(s, x)), app(s, y)) → APP(sub, x)
APP(app(d, app(s, x)), app(s, y)) → APP(gtr, x)
APP(app(d, app(s, x)), app(s, y)) → APP(sub, y)
APP(app(d, app(s, x)), app(s, y)) → APP(app(if, app(app(gtr, x), y)), false)
APP(app(filter, p), app(app(cons, x), xs)) → APP(p, x)
APP(app(d, app(s, x)), app(s, y)) → APP(if, app(app(gtr, x), y))
APP(app(filter, p), app(app(cons, x), xs)) → APP(app(filter, p), xs)
APP(app(filter, p), app(app(cons, x), xs)) → APP(app(cons, x), app(app(filter, p), xs))
APP(app(d, app(s, x)), app(s, y)) → APP(app(gtr, x), y)
APP(app(d, app(s, x)), app(s, y)) → APP(app(sub, y), x)
APP(len, app(app(cons, x), xs)) → APP(s, app(len, xs))
APP(app(filter, p), app(app(cons, x), xs)) → APP(app(if, app(p, x)), app(app(cons, x), app(app(filter, p), xs)))

The TRS R consists of the following rules:

app(app(app(if, true), xs), ys) → xs
app(app(app(if, false), xs), ys) → ys
app(app(sub, x), 0) → x
app(app(sub, app(s, x)), app(s, y)) → app(app(sub, x), y)
app(app(gtr, 0), y) → false
app(app(gtr, app(s, x)), 0) → true
app(app(gtr, app(s, x)), app(s, y)) → app(app(gtr, x), y)
app(app(d, x), 0) → true
app(app(d, app(s, x)), app(s, y)) → app(app(app(if, app(app(gtr, x), y)), false), app(app(d, app(s, x)), app(app(sub, y), x)))
app(len, nil) → 0
app(len, app(app(cons, x), xs)) → app(s, app(len, xs))
app(app(filter, p), nil) → nil
app(app(filter, p), app(app(cons, x), xs)) → app(app(app(if, app(p, x)), app(app(cons, x), app(app(filter, p), xs))), app(app(filter, p), xs))

The set Q consists of the following terms:

app(app(app(if, true), x0), x1)
app(app(app(if, false), x0), x1)
app(app(sub, x0), 0)
app(app(sub, app(s, x0)), app(s, x1))
app(app(gtr, 0), x0)
app(app(gtr, app(s, x0)), 0)
app(app(gtr, app(s, x0)), app(s, x1))
app(app(d, x0), 0)
app(app(d, app(s, x0)), app(s, x1))
app(len, nil)
app(len, app(app(cons, x0), x1))
app(app(filter, x0), nil)
app(app(filter, x0), app(app(cons, x1), x2))

We have to consider all minimal (P,Q,R)-chains.
The approximation of the Dependency Graph [13,14,18] contains 5 SCCs with 14 less nodes.

↳ QTRS

  ↳ Overlay + Local Confluence

    ↳ QTRS

      ↳ DependencyPairsProof

        ↳ QDP

          ↳ EdgeDeletionProof

            ↳ QDP

              ↳ DependencyGraphProof

                ↳ AND

                  ↳ QDP

                    ↳ QDPOrderProof

                  ↳ QDP

                  ↳ QDP

                  ↳ QDP

                  ↳ QDP

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

APP(len, app(app(cons, x), xs)) → APP(len, xs)

The TRS R consists of the following rules:

app(app(app(if, true), xs), ys) → xs
app(app(app(if, false), xs), ys) → ys
app(app(sub, x), 0) → x
app(app(sub, app(s, x)), app(s, y)) → app(app(sub, x), y)
app(app(gtr, 0), y) → false
app(app(gtr, app(s, x)), 0) → true
app(app(gtr, app(s, x)), app(s, y)) → app(app(gtr, x), y)
app(app(d, x), 0) → true
app(app(d, app(s, x)), app(s, y)) → app(app(app(if, app(app(gtr, x), y)), false), app(app(d, app(s, x)), app(app(sub, y), x)))
app(len, nil) → 0
app(len, app(app(cons, x), xs)) → app(s, app(len, xs))
app(app(filter, p), nil) → nil
app(app(filter, p), app(app(cons, x), xs)) → app(app(app(if, app(p, x)), app(app(cons, x), app(app(filter, p), xs))), app(app(filter, p), xs))

The set Q consists of the following terms:

app(app(app(if, true), x0), x1)
app(app(app(if, false), x0), x1)
app(app(sub, x0), 0)
app(app(sub, app(s, x0)), app(s, x1))
app(app(gtr, 0), x0)
app(app(gtr, app(s, x0)), 0)
app(app(gtr, app(s, x0)), app(s, x1))
app(app(d, x0), 0)
app(app(d, app(s, x0)), app(s, x1))
app(len, nil)
app(len, app(app(cons, x0), x1))
app(app(filter, x0), nil)
app(app(filter, x0), app(app(cons, x1), x2))

We have to consider all minimal (P,Q,R)-chains.
We use the reduction pair processor [13]. Here, we combined the reduction pair processor with the A-transformation [14] which results in the following intermediate Q-DP Problem.
Q DP problem:
The TRS P consists of the following rules:

LEN(cons(x, xs)) → LEN(xs)

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

if(true, x0, x1)
if(false, x0, x1)
sub(x0, 0)
sub(s(x0), s(x1))
gtr(0, x0)
gtr(s(x0), 0)
gtr(s(x0), s(x1))
d(x0, 0)
d(s(x0), s(x1))
len(nil)
len(cons(x0, x1))
filter(x0, nil)
filter(x0, cons(x1, x2))

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

The following pairs can be oriented strictly and are deleted.

APP(len, app(app(cons, x), xs)) → APP(len, xs)

The remaining pairs can at least be oriented weakly.
none
Used ordering: Combined order from the following AFS and order.
LEN(x1) = LEN(x1)
cons(x1, x2) = cons(x1, x2)

Lexicographic path order with status [19].
Precedence:

cons₂ > LEN₁

Status:

LEN₁: [1]
cons₂: multiset

The following usable rules [14] were oriented: none

↳ QTRS

  ↳ Overlay + Local Confluence

    ↳ QTRS

      ↳ DependencyPairsProof

        ↳ QDP

          ↳ EdgeDeletionProof

            ↳ QDP

              ↳ DependencyGraphProof

                ↳ AND

                  ↳ QDP

                    ↳ QDPOrderProof

                      ↳ QDP

                        ↳ PisEmptyProof

                  ↳ QDP

                  ↳ QDP

                  ↳ QDP

                  ↳ QDP

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

app(app(app(if, true), xs), ys) → xs
app(app(app(if, false), xs), ys) → ys
app(app(sub, x), 0) → x
app(app(sub, app(s, x)), app(s, y)) → app(app(sub, x), y)
app(app(gtr, 0), y) → false
app(app(gtr, app(s, x)), 0) → true
app(app(gtr, app(s, x)), app(s, y)) → app(app(gtr, x), y)
app(app(d, x), 0) → true
app(app(d, app(s, x)), app(s, y)) → app(app(app(if, app(app(gtr, x), y)), false), app(app(d, app(s, x)), app(app(sub, y), x)))
app(len, nil) → 0
app(len, app(app(cons, x), xs)) → app(s, app(len, xs))
app(app(filter, p), nil) → nil
app(app(filter, p), app(app(cons, x), xs)) → app(app(app(if, app(p, x)), app(app(cons, x), app(app(filter, p), xs))), app(app(filter, p), xs))

The set Q consists of the following terms:

app(app(app(if, true), x0), x1)
app(app(app(if, false), x0), x1)
app(app(sub, x0), 0)
app(app(sub, app(s, x0)), app(s, x1))
app(app(gtr, 0), x0)
app(app(gtr, app(s, x0)), 0)
app(app(gtr, app(s, x0)), app(s, x1))
app(app(d, x0), 0)
app(app(d, app(s, x0)), app(s, x1))
app(len, nil)
app(len, app(app(cons, x0), x1))
app(app(filter, x0), nil)
app(app(filter, x0), app(app(cons, x1), x2))

We have to consider all minimal (P,Q,R)-chains.
The TRS P is empty. Hence, there is no (P,Q,R) chain.

↳ QTRS

  ↳ Overlay + Local Confluence

    ↳ QTRS

      ↳ DependencyPairsProof

        ↳ QDP

          ↳ EdgeDeletionProof

            ↳ QDP

              ↳ DependencyGraphProof

                ↳ AND

                  ↳ QDP

                  ↳ QDP

                    ↳ QDPOrderProof

                  ↳ QDP

                  ↳ QDP

                  ↳ QDP

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

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

The TRS R consists of the following rules:

app(app(app(if, true), xs), ys) → xs
app(app(app(if, false), xs), ys) → ys
app(app(sub, x), 0) → x
app(app(sub, app(s, x)), app(s, y)) → app(app(sub, x), y)
app(app(gtr, 0), y) → false
app(app(gtr, app(s, x)), 0) → true
app(app(gtr, app(s, x)), app(s, y)) → app(app(gtr, x), y)
app(app(d, x), 0) → true
app(app(d, app(s, x)), app(s, y)) → app(app(app(if, app(app(gtr, x), y)), false), app(app(d, app(s, x)), app(app(sub, y), x)))
app(len, nil) → 0
app(len, app(app(cons, x), xs)) → app(s, app(len, xs))
app(app(filter, p), nil) → nil
app(app(filter, p), app(app(cons, x), xs)) → app(app(app(if, app(p, x)), app(app(cons, x), app(app(filter, p), xs))), app(app(filter, p), xs))

The set Q consists of the following terms:

app(app(app(if, true), x0), x1)
app(app(app(if, false), x0), x1)
app(app(sub, x0), 0)
app(app(sub, app(s, x0)), app(s, x1))
app(app(gtr, 0), x0)
app(app(gtr, app(s, x0)), 0)
app(app(gtr, app(s, x0)), app(s, x1))
app(app(d, x0), 0)
app(app(d, app(s, x0)), app(s, x1))
app(len, nil)
app(len, app(app(cons, x0), x1))
app(app(filter, x0), nil)
app(app(filter, x0), app(app(cons, x1), x2))

GTR(s(x), s(y)) → GTR(x, y)

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

if(true, x0, x1)
if(false, x0, x1)
sub(x0, 0)
sub(s(x0), s(x1))
gtr(0, x0)
gtr(s(x0), 0)
gtr(s(x0), s(x1))
d(x0, 0)
d(s(x0), s(x1))
len(nil)
len(cons(x0, x1))
filter(x0, nil)
filter(x0, cons(x1, x2))

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

The following pairs can be oriented strictly and are deleted.

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

The remaining pairs can at least be oriented weakly.
none
Used ordering: Combined order from the following AFS and order.
GTR(x1, x2) = GTR(x1)
s(x1) = s(x1)

Lexicographic path order with status [19].
Precedence:

s₁ > GTR₁

Status:

GTR₁: [1]
s₁: multiset

The following usable rules [14] were oriented: none

↳ QTRS

  ↳ Overlay + Local Confluence

    ↳ QTRS

      ↳ DependencyPairsProof

        ↳ QDP

          ↳ EdgeDeletionProof

            ↳ QDP

              ↳ DependencyGraphProof

                ↳ AND

                  ↳ QDP

                  ↳ QDP

                    ↳ QDPOrderProof

                      ↳ QDP

                        ↳ PisEmptyProof

                  ↳ QDP

                  ↳ QDP

                  ↳ QDP

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

app(app(app(if, true), xs), ys) → xs
app(app(app(if, false), xs), ys) → ys
app(app(sub, x), 0) → x
app(app(sub, app(s, x)), app(s, y)) → app(app(sub, x), y)
app(app(gtr, 0), y) → false
app(app(gtr, app(s, x)), 0) → true
app(app(gtr, app(s, x)), app(s, y)) → app(app(gtr, x), y)
app(app(d, x), 0) → true
app(app(d, app(s, x)), app(s, y)) → app(app(app(if, app(app(gtr, x), y)), false), app(app(d, app(s, x)), app(app(sub, y), x)))
app(len, nil) → 0
app(len, app(app(cons, x), xs)) → app(s, app(len, xs))
app(app(filter, p), nil) → nil
app(app(filter, p), app(app(cons, x), xs)) → app(app(app(if, app(p, x)), app(app(cons, x), app(app(filter, p), xs))), app(app(filter, p), xs))

The set Q consists of the following terms:

app(app(app(if, true), x0), x1)
app(app(app(if, false), x0), x1)
app(app(sub, x0), 0)
app(app(sub, app(s, x0)), app(s, x1))
app(app(gtr, 0), x0)
app(app(gtr, app(s, x0)), 0)
app(app(gtr, app(s, x0)), app(s, x1))
app(app(d, x0), 0)
app(app(d, app(s, x0)), app(s, x1))
app(len, nil)
app(len, app(app(cons, x0), x1))
app(app(filter, x0), nil)
app(app(filter, x0), app(app(cons, x1), x2))

We have to consider all minimal (P,Q,R)-chains.
The TRS P is empty. Hence, there is no (P,Q,R) chain.

↳ QTRS

  ↳ Overlay + Local Confluence

    ↳ QTRS

      ↳ DependencyPairsProof

        ↳ QDP

          ↳ EdgeDeletionProof

            ↳ QDP

              ↳ DependencyGraphProof

                ↳ AND

                  ↳ QDP

                  ↳ QDP

                  ↳ QDP

                    ↳ QDPOrderProof

                  ↳ QDP

                  ↳ QDP

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

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

The TRS R consists of the following rules:

app(app(app(if, true), xs), ys) → xs
app(app(app(if, false), xs), ys) → ys
app(app(sub, x), 0) → x
app(app(sub, app(s, x)), app(s, y)) → app(app(sub, x), y)
app(app(gtr, 0), y) → false
app(app(gtr, app(s, x)), 0) → true
app(app(gtr, app(s, x)), app(s, y)) → app(app(gtr, x), y)
app(app(d, x), 0) → true
app(app(d, app(s, x)), app(s, y)) → app(app(app(if, app(app(gtr, x), y)), false), app(app(d, app(s, x)), app(app(sub, y), x)))
app(len, nil) → 0
app(len, app(app(cons, x), xs)) → app(s, app(len, xs))
app(app(filter, p), nil) → nil
app(app(filter, p), app(app(cons, x), xs)) → app(app(app(if, app(p, x)), app(app(cons, x), app(app(filter, p), xs))), app(app(filter, p), xs))

The set Q consists of the following terms:

app(app(app(if, true), x0), x1)
app(app(app(if, false), x0), x1)
app(app(sub, x0), 0)
app(app(sub, app(s, x0)), app(s, x1))
app(app(gtr, 0), x0)
app(app(gtr, app(s, x0)), 0)
app(app(gtr, app(s, x0)), app(s, x1))
app(app(d, x0), 0)
app(app(d, app(s, x0)), app(s, x1))
app(len, nil)
app(len, app(app(cons, x0), x1))
app(app(filter, x0), nil)
app(app(filter, x0), app(app(cons, x1), x2))

SUB(s(x), s(y)) → SUB(x, y)

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

if(true, x0, x1)
if(false, x0, x1)
sub(x0, 0)
sub(s(x0), s(x1))
gtr(0, x0)
gtr(s(x0), 0)
gtr(s(x0), s(x1))
d(x0, 0)
d(s(x0), s(x1))
len(nil)
len(cons(x0, x1))
filter(x0, nil)
filter(x0, cons(x1, x2))

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

The following pairs can be oriented strictly and are deleted.

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

The remaining pairs can at least be oriented weakly.
none
Used ordering: Combined order from the following AFS and order.
SUB(x1, x2) = SUB(x1)
s(x1) = s(x1)

Lexicographic path order with status [19].
Precedence:

s₁ > SUB₁

Status:

SUB₁: [1]
s₁: multiset

The following usable rules [14] were oriented: none

↳ QTRS

  ↳ Overlay + Local Confluence

    ↳ QTRS

      ↳ DependencyPairsProof

        ↳ QDP

          ↳ EdgeDeletionProof

            ↳ QDP

              ↳ DependencyGraphProof

                ↳ AND

                  ↳ QDP

                  ↳ QDP

                  ↳ QDP

                    ↳ QDPOrderProof

                      ↳ QDP

                        ↳ PisEmptyProof

                  ↳ QDP

                  ↳ QDP

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

app(app(app(if, true), xs), ys) → xs
app(app(app(if, false), xs), ys) → ys
app(app(sub, x), 0) → x
app(app(sub, app(s, x)), app(s, y)) → app(app(sub, x), y)
app(app(gtr, 0), y) → false
app(app(gtr, app(s, x)), 0) → true
app(app(gtr, app(s, x)), app(s, y)) → app(app(gtr, x), y)
app(app(d, x), 0) → true
app(app(d, app(s, x)), app(s, y)) → app(app(app(if, app(app(gtr, x), y)), false), app(app(d, app(s, x)), app(app(sub, y), x)))
app(len, nil) → 0
app(len, app(app(cons, x), xs)) → app(s, app(len, xs))
app(app(filter, p), nil) → nil
app(app(filter, p), app(app(cons, x), xs)) → app(app(app(if, app(p, x)), app(app(cons, x), app(app(filter, p), xs))), app(app(filter, p), xs))

The set Q consists of the following terms:

app(app(app(if, true), x0), x1)
app(app(app(if, false), x0), x1)
app(app(sub, x0), 0)
app(app(sub, app(s, x0)), app(s, x1))
app(app(gtr, 0), x0)
app(app(gtr, app(s, x0)), 0)
app(app(gtr, app(s, x0)), app(s, x1))
app(app(d, x0), 0)
app(app(d, app(s, x0)), app(s, x1))
app(len, nil)
app(len, app(app(cons, x0), x1))
app(app(filter, x0), nil)
app(app(filter, x0), app(app(cons, x1), x2))

We have to consider all minimal (P,Q,R)-chains.
The TRS P is empty. Hence, there is no (P,Q,R) chain.

↳ QTRS

  ↳ Overlay + Local Confluence

    ↳ QTRS

      ↳ DependencyPairsProof

        ↳ QDP

          ↳ EdgeDeletionProof

            ↳ QDP

              ↳ DependencyGraphProof

                ↳ AND

                  ↳ QDP

                  ↳ QDP

                  ↳ QDP

                  ↳ QDP

                    ↳ QDPOrderProof

                  ↳ QDP

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

APP(app(d, app(s, x)), app(s, y)) → APP(app(d, app(s, x)), app(app(sub, y), x))

The TRS R consists of the following rules:

app(app(app(if, true), xs), ys) → xs
app(app(app(if, false), xs), ys) → ys
app(app(sub, x), 0) → x
app(app(sub, app(s, x)), app(s, y)) → app(app(sub, x), y)
app(app(gtr, 0), y) → false
app(app(gtr, app(s, x)), 0) → true
app(app(gtr, app(s, x)), app(s, y)) → app(app(gtr, x), y)
app(app(d, x), 0) → true
app(app(d, app(s, x)), app(s, y)) → app(app(app(if, app(app(gtr, x), y)), false), app(app(d, app(s, x)), app(app(sub, y), x)))
app(len, nil) → 0
app(len, app(app(cons, x), xs)) → app(s, app(len, xs))
app(app(filter, p), nil) → nil
app(app(filter, p), app(app(cons, x), xs)) → app(app(app(if, app(p, x)), app(app(cons, x), app(app(filter, p), xs))), app(app(filter, p), xs))

The set Q consists of the following terms:

app(app(app(if, true), x0), x1)
app(app(app(if, false), x0), x1)
app(app(sub, x0), 0)
app(app(sub, app(s, x0)), app(s, x1))
app(app(gtr, 0), x0)
app(app(gtr, app(s, x0)), 0)
app(app(gtr, app(s, x0)), app(s, x1))
app(app(d, x0), 0)
app(app(d, app(s, x0)), app(s, x1))
app(len, nil)
app(len, app(app(cons, x0), x1))
app(app(filter, x0), nil)
app(app(filter, x0), app(app(cons, x1), x2))

D(s(x), s(y)) → D(s(x), sub(y, x))

The TRS R consists of the following rules:

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

The set Q consists of the following terms:

if(true, x0, x1)
if(false, x0, x1)
sub(x0, 0)
sub(s(x0), s(x1))
gtr(0, x0)
gtr(s(x0), 0)
gtr(s(x0), s(x1))
d(x0, 0)
d(s(x0), s(x1))
len(nil)
len(cons(x0, x1))
filter(x0, nil)
filter(x0, cons(x1, x2))

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

The following pairs can be oriented strictly and are deleted.

APP(app(d, app(s, x)), app(s, y)) → APP(app(d, app(s, x)), app(app(sub, y), x))

The remaining pairs can at least be oriented weakly.
none
Used ordering: Combined order from the following AFS and order.
D(x1, x2) = x2
s(x1) = s(x1)
sub(x1, x2) = x1
0 = 0

Lexicographic path order with status [19].
Precedence:

trivial

Status:

0: multiset
s₁: [1]

The following usable rules [14] were oriented:

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

↳ QTRS

  ↳ Overlay + Local Confluence

    ↳ QTRS

      ↳ DependencyPairsProof

        ↳ QDP

          ↳ EdgeDeletionProof

            ↳ QDP

              ↳ DependencyGraphProof

                ↳ AND

                  ↳ QDP

                  ↳ QDP

                  ↳ QDP

                  ↳ QDP

                    ↳ QDPOrderProof

                      ↳ QDP

                        ↳ PisEmptyProof

                  ↳ QDP

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

app(app(app(if, true), xs), ys) → xs
app(app(app(if, false), xs), ys) → ys
app(app(sub, x), 0) → x
app(app(sub, app(s, x)), app(s, y)) → app(app(sub, x), y)
app(app(gtr, 0), y) → false
app(app(gtr, app(s, x)), 0) → true
app(app(gtr, app(s, x)), app(s, y)) → app(app(gtr, x), y)
app(app(d, x), 0) → true
app(app(d, app(s, x)), app(s, y)) → app(app(app(if, app(app(gtr, x), y)), false), app(app(d, app(s, x)), app(app(sub, y), x)))
app(len, nil) → 0
app(len, app(app(cons, x), xs)) → app(s, app(len, xs))
app(app(filter, p), nil) → nil
app(app(filter, p), app(app(cons, x), xs)) → app(app(app(if, app(p, x)), app(app(cons, x), app(app(filter, p), xs))), app(app(filter, p), xs))

The set Q consists of the following terms:

app(app(app(if, true), x0), x1)
app(app(app(if, false), x0), x1)
app(app(sub, x0), 0)
app(app(sub, app(s, x0)), app(s, x1))
app(app(gtr, 0), x0)
app(app(gtr, app(s, x0)), 0)
app(app(gtr, app(s, x0)), app(s, x1))
app(app(d, x0), 0)
app(app(d, app(s, x0)), app(s, x1))
app(len, nil)
app(len, app(app(cons, x0), x1))
app(app(filter, x0), nil)
app(app(filter, x0), app(app(cons, x1), x2))

We have to consider all minimal (P,Q,R)-chains.
The TRS P is empty. Hence, there is no (P,Q,R) chain.

↳ QTRS

  ↳ Overlay + Local Confluence

    ↳ QTRS

      ↳ DependencyPairsProof

        ↳ QDP

          ↳ EdgeDeletionProof

            ↳ QDP

              ↳ DependencyGraphProof

                ↳ AND

                  ↳ QDP

                  ↳ QDP

                  ↳ QDP

                  ↳ QDP

                  ↳ QDP

                    ↳ QDPOrderProof

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

APP(app(filter, p), app(app(cons, x), xs)) → APP(p, x)
APP(app(filter, p), app(app(cons, x), xs)) → APP(app(filter, p), xs)

The TRS R consists of the following rules:

app(app(app(if, true), xs), ys) → xs
app(app(app(if, false), xs), ys) → ys
app(app(sub, x), 0) → x
app(app(sub, app(s, x)), app(s, y)) → app(app(sub, x), y)
app(app(gtr, 0), y) → false
app(app(gtr, app(s, x)), 0) → true
app(app(gtr, app(s, x)), app(s, y)) → app(app(gtr, x), y)
app(app(d, x), 0) → true
app(app(d, app(s, x)), app(s, y)) → app(app(app(if, app(app(gtr, x), y)), false), app(app(d, app(s, x)), app(app(sub, y), x)))
app(len, nil) → 0
app(len, app(app(cons, x), xs)) → app(s, app(len, xs))
app(app(filter, p), nil) → nil
app(app(filter, p), app(app(cons, x), xs)) → app(app(app(if, app(p, x)), app(app(cons, x), app(app(filter, p), xs))), app(app(filter, p), xs))

The set Q consists of the following terms:

app(app(app(if, true), x0), x1)
app(app(app(if, false), x0), x1)
app(app(sub, x0), 0)
app(app(sub, app(s, x0)), app(s, x1))
app(app(gtr, 0), x0)
app(app(gtr, app(s, x0)), 0)
app(app(gtr, app(s, x0)), app(s, x1))
app(app(d, x0), 0)
app(app(d, app(s, x0)), app(s, x1))
app(len, nil)
app(len, app(app(cons, x0), x1))
app(app(filter, x0), nil)
app(app(filter, x0), app(app(cons, x1), x2))

We have to consider all minimal (P,Q,R)-chains.
We use the reduction pair processor [13].

The following pairs can be oriented strictly and are deleted.

APP(app(filter, p), app(app(cons, x), xs)) → APP(p, x)

The remaining pairs can at least be oriented weakly.

APP(app(filter, p), app(app(cons, x), xs)) → APP(app(filter, p), xs)

Used ordering: Combined order from the following AFS and order.
APP(x1, x2) = APP(x1)
app(x1, x2) = app(x1, x2)
filter = filter
cons = cons

Lexicographic path order with status [19].
Precedence:

app₂ > APP₁ > filter
cons > APP₁ > filter

Status:

filter: multiset
APP₁: [1]
app₂: [1,2]
cons: multiset

The following usable rules [14] were oriented: none

↳ QTRS

  ↳ Overlay + Local Confluence

    ↳ QTRS

      ↳ DependencyPairsProof

        ↳ QDP

          ↳ EdgeDeletionProof

            ↳ QDP

              ↳ DependencyGraphProof

                ↳ AND

                  ↳ QDP

                  ↳ QDP

                  ↳ QDP

                  ↳ QDP

                  ↳ QDP

                    ↳ QDPOrderProof

                      ↳ QDP

                        ↳ QDPOrderProof

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

APP(app(filter, p), app(app(cons, x), xs)) → APP(app(filter, p), xs)

The TRS R consists of the following rules:

app(app(app(if, true), xs), ys) → xs
app(app(app(if, false), xs), ys) → ys
app(app(sub, x), 0) → x
app(app(sub, app(s, x)), app(s, y)) → app(app(sub, x), y)
app(app(gtr, 0), y) → false
app(app(gtr, app(s, x)), 0) → true
app(app(gtr, app(s, x)), app(s, y)) → app(app(gtr, x), y)
app(app(d, x), 0) → true
app(app(d, app(s, x)), app(s, y)) → app(app(app(if, app(app(gtr, x), y)), false), app(app(d, app(s, x)), app(app(sub, y), x)))
app(len, nil) → 0
app(len, app(app(cons, x), xs)) → app(s, app(len, xs))
app(app(filter, p), nil) → nil
app(app(filter, p), app(app(cons, x), xs)) → app(app(app(if, app(p, x)), app(app(cons, x), app(app(filter, p), xs))), app(app(filter, p), xs))

The set Q consists of the following terms:

app(app(app(if, true), x0), x1)
app(app(app(if, false), x0), x1)
app(app(sub, x0), 0)
app(app(sub, app(s, x0)), app(s, x1))
app(app(gtr, 0), x0)
app(app(gtr, app(s, x0)), 0)
app(app(gtr, app(s, x0)), app(s, x1))
app(app(d, x0), 0)
app(app(d, app(s, x0)), app(s, x1))
app(len, nil)
app(len, app(app(cons, x0), x1))
app(app(filter, x0), nil)
app(app(filter, x0), app(app(cons, x1), x2))

FILTER(p, cons(x, xs)) → FILTER(p, xs)

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

if(true, x0, x1)
if(false, x0, x1)
sub(x0, 0)
sub(s(x0), s(x1))
gtr(0, x0)
gtr(s(x0), 0)
gtr(s(x0), s(x1))
d(x0, 0)
d(s(x0), s(x1))
len(nil)
len(cons(x0, x1))
filter(x0, nil)
filter(x0, cons(x1, x2))

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

The following pairs can be oriented strictly and are deleted.

APP(app(filter, p), app(app(cons, x), xs)) → APP(app(filter, p), xs)

The remaining pairs can at least be oriented weakly.
none
Used ordering: Combined order from the following AFS and order.
FILTER(x1, x2) = FILTER(x1, x2)
cons(x1, x2) = cons(x1, x2)

Lexicographic path order with status [19].
Precedence:

cons₂ > FILTER₂

Status:

FILTER₂: [1,2]
cons₂: multiset

The following usable rules [14] were oriented: none

↳ QTRS

  ↳ Overlay + Local Confluence

    ↳ QTRS

      ↳ DependencyPairsProof

        ↳ QDP

          ↳ EdgeDeletionProof

            ↳ QDP

              ↳ DependencyGraphProof

                ↳ AND

                  ↳ QDP

                  ↳ QDP

                  ↳ QDP

                  ↳ QDP

                  ↳ QDP

                    ↳ QDPOrderProof

                      ↳ QDP

                        ↳ QDPOrderProof

                          ↳ QDP

                            ↳ PisEmptyProof

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

app(app(app(if, true), xs), ys) → xs
app(app(app(if, false), xs), ys) → ys
app(app(sub, x), 0) → x
app(app(sub, app(s, x)), app(s, y)) → app(app(sub, x), y)
app(app(gtr, 0), y) → false
app(app(gtr, app(s, x)), 0) → true
app(app(gtr, app(s, x)), app(s, y)) → app(app(gtr, x), y)
app(app(d, x), 0) → true
app(app(d, app(s, x)), app(s, y)) → app(app(app(if, app(app(gtr, x), y)), false), app(app(d, app(s, x)), app(app(sub, y), x)))
app(len, nil) → 0
app(len, app(app(cons, x), xs)) → app(s, app(len, xs))
app(app(filter, p), nil) → nil
app(app(filter, p), app(app(cons, x), xs)) → app(app(app(if, app(p, x)), app(app(cons, x), app(app(filter, p), xs))), app(app(filter, p), xs))

The set Q consists of the following terms:

app(app(app(if, true), x0), x1)
app(app(app(if, false), x0), x1)
app(app(sub, x0), 0)
app(app(sub, app(s, x0)), app(s, x1))
app(app(gtr, 0), x0)
app(app(gtr, app(s, x0)), 0)
app(app(gtr, app(s, x0)), app(s, x1))
app(app(d, x0), 0)
app(app(d, app(s, x0)), app(s, x1))
app(len, nil)
app(len, app(app(cons, x0), x1))
app(app(filter, x0), nil)
app(app(filter, x0), app(app(cons, x1), x2))

We have to consider all minimal (P,Q,R)-chains.
The TRS P is empty. Hence, there is no (P,Q,R) chain.