Termination w.r.t. Q proof of TRS/currying/Ste92perfect2.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, 0), y) → 0
app(app(minus, app(s, x)), 0) → app(s, x)
app(app(minus, app(s, x)), app(s, y)) → app(app(minus, x), y)
app(app(le, 0), y) → true
app(app(le, app(s, x)), 0) → false
app(app(le, app(s, x)), app(s, y)) → app(app(le, x), y)
app(app(app(if, true), x), y) → x
app(app(app(if, false), x), y) → y
app(perfectp, 0) → false
app(perfectp, app(s, x)) → app(app(app(app(f, x), app(s, 0)), app(s, x)), app(s, x))
app(app(app(app(f, 0), y), 0), u) → true
app(app(app(app(f, 0), y), app(s, z)), u) → false
app(app(app(app(f, app(s, x)), 0), z), u) → app(app(app(app(f, x), u), app(app(minus, z), app(s, x))), u)
app(app(app(app(f, app(s, x)), app(s, y)), z), u) → app(app(app(if, app(app(le, x), y)), app(app(app(app(f, app(s, x)), app(app(minus, y), x)), z), u)), app(app(app(app(f, x), u), z), u))
app(app(map, fun), nil) → nil
app(app(map, fun), app(app(cons, x), xs)) → app(app(cons, app(fun, x)), app(app(map, fun), xs))
app(app(filter, fun), nil) → nil
app(app(filter, fun), app(app(cons, x), xs)) → app(app(app(app(filter2, app(fun, x)), fun), x), xs)
app(app(app(app(filter2, true), fun), x), xs) → app(app(cons, x), app(app(filter, fun), xs))
app(app(app(app(filter2, false), fun), x), xs) → app(app(filter, fun), xs)

Q is empty.

↳ QTRS

  ↳ Overlay + Local Confluence

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

app(app(minus, 0), y) → 0
app(app(minus, app(s, x)), 0) → app(s, x)
app(app(minus, app(s, x)), app(s, y)) → app(app(minus, x), y)
app(app(le, 0), y) → true
app(app(le, app(s, x)), 0) → false
app(app(le, app(s, x)), app(s, y)) → app(app(le, x), y)
app(app(app(if, true), x), y) → x
app(app(app(if, false), x), y) → y
app(perfectp, 0) → false
app(perfectp, app(s, x)) → app(app(app(app(f, x), app(s, 0)), app(s, x)), app(s, x))
app(app(app(app(f, 0), y), 0), u) → true
app(app(app(app(f, 0), y), app(s, z)), u) → false
app(app(app(app(f, app(s, x)), 0), z), u) → app(app(app(app(f, x), u), app(app(minus, z), app(s, x))), u)
app(app(app(app(f, app(s, x)), app(s, y)), z), u) → app(app(app(if, app(app(le, x), y)), app(app(app(app(f, app(s, x)), app(app(minus, y), x)), z), u)), app(app(app(app(f, x), u), z), u))
app(app(map, fun), nil) → nil
app(app(map, fun), app(app(cons, x), xs)) → app(app(cons, app(fun, x)), app(app(map, fun), xs))
app(app(filter, fun), nil) → nil
app(app(filter, fun), app(app(cons, x), xs)) → app(app(app(app(filter2, app(fun, x)), fun), x), xs)
app(app(app(app(filter2, true), fun), x), xs) → app(app(cons, x), app(app(filter, fun), xs))
app(app(app(app(filter2, false), fun), x), xs) → app(app(filter, fun), xs)

Q is empty.

The TRS is overlay and locally confluent. By [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(minus, 0), y) → 0
app(app(minus, app(s, x)), 0) → app(s, x)
app(app(minus, app(s, x)), app(s, y)) → app(app(minus, x), y)
app(app(le, 0), y) → true
app(app(le, app(s, x)), 0) → false
app(app(le, app(s, x)), app(s, y)) → app(app(le, x), y)
app(app(app(if, true), x), y) → x
app(app(app(if, false), x), y) → y
app(perfectp, 0) → false
app(perfectp, app(s, x)) → app(app(app(app(f, x), app(s, 0)), app(s, x)), app(s, x))
app(app(app(app(f, 0), y), 0), u) → true
app(app(app(app(f, 0), y), app(s, z)), u) → false
app(app(app(app(f, app(s, x)), 0), z), u) → app(app(app(app(f, x), u), app(app(minus, z), app(s, x))), u)
app(app(app(app(f, app(s, x)), app(s, y)), z), u) → app(app(app(if, app(app(le, x), y)), app(app(app(app(f, app(s, x)), app(app(minus, y), x)), z), u)), app(app(app(app(f, x), u), z), u))
app(app(map, fun), nil) → nil
app(app(map, fun), app(app(cons, x), xs)) → app(app(cons, app(fun, x)), app(app(map, fun), xs))
app(app(filter, fun), nil) → nil
app(app(filter, fun), app(app(cons, x), xs)) → app(app(app(app(filter2, app(fun, x)), fun), x), xs)
app(app(app(app(filter2, true), fun), x), xs) → app(app(cons, x), app(app(filter, fun), xs))
app(app(app(app(filter2, false), fun), x), xs) → app(app(filter, fun), xs)

The set Q consists of the following terms:

app(app(minus, 0), x0)
app(app(minus, app(s, x0)), 0)
app(app(minus, app(s, x0)), app(s, x1))
app(app(le, 0), x0)
app(app(le, app(s, x0)), 0)
app(app(le, app(s, x0)), app(s, x1))
app(app(app(if, true), x0), x1)
app(app(app(if, false), x0), x1)
app(perfectp, 0)
app(perfectp, app(s, x0))
app(app(app(app(f, 0), x0), 0), x1)
app(app(app(app(f, 0), x0), app(s, x1)), x2)
app(app(app(app(f, app(s, x0)), 0), x1), x2)
app(app(app(app(f, app(s, x0)), app(s, x1)), x2), x3)
app(app(map, x0), nil)
app(app(map, x0), app(app(cons, x1), x2))
app(app(filter, x0), nil)
app(app(filter, x0), app(app(cons, x1), x2))
app(app(app(app(filter2, true), x0), x1), x2)
app(app(app(app(filter2, false), x0), x1), x2)

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

APP(app(app(app(f, app(s, x)), app(s, y)), z), u) → APP(app(f, app(s, x)), app(app(minus, y), x))
APP(app(app(app(f, app(s, x)), app(s, y)), z), u) → APP(app(minus, y), x)
APP(app(minus, app(s, x)), app(s, y)) → APP(minus, x)
APP(perfectp, app(s, x)) → APP(app(app(f, x), app(s, 0)), app(s, x))
APP(app(map, fun), app(app(cons, x), xs)) → APP(app(cons, app(fun, x)), app(app(map, fun), xs))
APP(perfectp, app(s, x)) → APP(f, x)
APP(app(app(app(f, app(s, x)), app(s, y)), z), u) → APP(app(f, x), u)
APP(app(filter, fun), app(app(cons, x), xs)) → APP(app(app(filter2, app(fun, x)), fun), x)
APP(app(app(app(f, app(s, x)), app(s, y)), z), u) → APP(le, x)
APP(app(app(app(f, app(s, x)), app(s, y)), z), u) → APP(app(app(app(f, app(s, x)), app(app(minus, y), x)), z), u)
APP(app(filter, fun), app(app(cons, x), xs)) → APP(app(app(app(filter2, app(fun, x)), fun), x), xs)
APP(app(app(app(f, app(s, x)), app(s, y)), z), u) → APP(if, app(app(le, x), y))
APP(app(le, app(s, x)), app(s, y)) → APP(le, x)
APP(app(app(app(f, app(s, x)), 0), z), u) → APP(minus, z)
APP(app(app(app(filter2, true), fun), x), xs) → APP(app(cons, x), app(app(filter, fun), xs))
APP(perfectp, app(s, x)) → APP(app(f, x), app(s, 0))
APP(app(app(app(f, app(s, x)), app(s, y)), z), u) → APP(minus, y)
APP(app(app(app(f, app(s, x)), app(s, y)), z), u) → APP(app(app(f, x), u), z)
APP(app(filter, fun), app(app(cons, x), xs)) → APP(filter2, app(fun, x))
APP(app(map, fun), app(app(cons, x), xs)) → APP(fun, x)
APP(app(minus, app(s, x)), app(s, y)) → APP(app(minus, x), y)
APP(app(app(app(filter2, true), fun), x), xs) → APP(app(filter, fun), xs)
APP(app(app(app(f, app(s, x)), 0), z), u) → APP(app(f, x), u)
APP(app(filter, fun), app(app(cons, x), xs)) → APP(app(filter2, app(fun, x)), fun)
APP(app(app(app(filter2, false), fun), x), xs) → APP(filter, fun)
APP(perfectp, app(s, x)) → APP(app(app(app(f, x), app(s, 0)), app(s, x)), app(s, x))
APP(app(le, app(s, x)), app(s, y)) → APP(app(le, x), y)
APP(app(app(app(f, app(s, x)), app(s, y)), z), u) → APP(app(app(f, app(s, x)), app(app(minus, y), x)), z)
APP(app(filter, fun), app(app(cons, x), xs)) → APP(fun, x)
APP(app(map, fun), app(app(cons, x), xs)) → APP(app(map, fun), xs)
APP(app(app(app(f, app(s, x)), app(s, y)), z), u) → APP(app(app(app(f, x), u), z), u)
APP(app(app(app(f, app(s, x)), 0), z), u) → APP(f, x)
APP(perfectp, app(s, x)) → APP(s, 0)
APP(app(app(app(f, app(s, x)), app(s, y)), z), u) → APP(app(le, x), y)
APP(app(app(app(f, app(s, x)), 0), z), u) → APP(app(app(f, x), u), app(app(minus, z), app(s, x)))
APP(app(app(app(filter2, true), fun), x), xs) → APP(filter, fun)
APP(app(app(app(f, app(s, x)), 0), z), u) → APP(app(minus, z), app(s, x))
APP(app(app(app(f, app(s, x)), 0), z), u) → APP(app(app(app(f, x), u), app(app(minus, z), app(s, x))), u)
APP(app(app(app(f, app(s, x)), app(s, y)), z), u) → APP(f, x)
APP(app(app(app(f, app(s, x)), app(s, y)), z), u) → APP(app(app(if, app(app(le, x), y)), app(app(app(app(f, app(s, x)), app(app(minus, y), x)), z), u)), app(app(app(app(f, x), u), z), u))
APP(app(app(app(f, app(s, x)), app(s, y)), z), u) → APP(app(if, app(app(le, x), y)), app(app(app(app(f, app(s, x)), app(app(minus, y), x)), z), u))
APP(app(app(app(filter2, true), fun), x), xs) → APP(cons, x)
APP(app(app(app(filter2, false), fun), x), xs) → APP(app(filter, fun), xs)
APP(app(map, fun), app(app(cons, x), xs)) → APP(cons, app(fun, x))

The TRS R consists of the following rules:

app(app(minus, 0), y) → 0
app(app(minus, app(s, x)), 0) → app(s, x)
app(app(minus, app(s, x)), app(s, y)) → app(app(minus, x), y)
app(app(le, 0), y) → true
app(app(le, app(s, x)), 0) → false
app(app(le, app(s, x)), app(s, y)) → app(app(le, x), y)
app(app(app(if, true), x), y) → x
app(app(app(if, false), x), y) → y
app(perfectp, 0) → false
app(perfectp, app(s, x)) → app(app(app(app(f, x), app(s, 0)), app(s, x)), app(s, x))
app(app(app(app(f, 0), y), 0), u) → true
app(app(app(app(f, 0), y), app(s, z)), u) → false
app(app(app(app(f, app(s, x)), 0), z), u) → app(app(app(app(f, x), u), app(app(minus, z), app(s, x))), u)
app(app(app(app(f, app(s, x)), app(s, y)), z), u) → app(app(app(if, app(app(le, x), y)), app(app(app(app(f, app(s, x)), app(app(minus, y), x)), z), u)), app(app(app(app(f, x), u), z), u))
app(app(map, fun), nil) → nil
app(app(map, fun), app(app(cons, x), xs)) → app(app(cons, app(fun, x)), app(app(map, fun), xs))
app(app(filter, fun), nil) → nil
app(app(filter, fun), app(app(cons, x), xs)) → app(app(app(app(filter2, app(fun, x)), fun), x), xs)
app(app(app(app(filter2, true), fun), x), xs) → app(app(cons, x), app(app(filter, fun), xs))
app(app(app(app(filter2, false), fun), x), xs) → app(app(filter, fun), xs)

The set Q consists of the following terms:

app(app(minus, 0), x0)
app(app(minus, app(s, x0)), 0)
app(app(minus, app(s, x0)), app(s, x1))
app(app(le, 0), x0)
app(app(le, app(s, x0)), 0)
app(app(le, app(s, x0)), app(s, x1))
app(app(app(if, true), x0), x1)
app(app(app(if, false), x0), x1)
app(perfectp, 0)
app(perfectp, app(s, x0))
app(app(app(app(f, 0), x0), 0), x1)
app(app(app(app(f, 0), x0), app(s, x1)), x2)
app(app(app(app(f, app(s, x0)), 0), x1), x2)
app(app(app(app(f, app(s, x0)), app(s, x1)), x2), x3)
app(app(map, x0), nil)
app(app(map, x0), app(app(cons, x1), x2))
app(app(filter, x0), nil)
app(app(filter, x0), app(app(cons, x1), x2))
app(app(app(app(filter2, true), x0), x1), x2)
app(app(app(app(filter2, false), x0), x1), x2)

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

↳ QTRS

  ↳ Overlay + Local Confluence

    ↳ QTRS

      ↳ DependencyPairsProof

        ↳ QDP

          ↳ EdgeDeletionProof

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

APP(app(app(app(f, app(s, x)), app(s, y)), z), u) → APP(app(f, app(s, x)), app(app(minus, y), x))
APP(app(app(app(f, app(s, x)), app(s, y)), z), u) → APP(app(minus, y), x)
APP(app(minus, app(s, x)), app(s, y)) → APP(minus, x)
APP(perfectp, app(s, x)) → APP(app(app(f, x), app(s, 0)), app(s, x))
APP(app(map, fun), app(app(cons, x), xs)) → APP(app(cons, app(fun, x)), app(app(map, fun), xs))
APP(perfectp, app(s, x)) → APP(f, x)
APP(app(app(app(f, app(s, x)), app(s, y)), z), u) → APP(app(f, x), u)
APP(app(filter, fun), app(app(cons, x), xs)) → APP(app(app(filter2, app(fun, x)), fun), x)
APP(app(app(app(f, app(s, x)), app(s, y)), z), u) → APP(le, x)
APP(app(app(app(f, app(s, x)), app(s, y)), z), u) → APP(app(app(app(f, app(s, x)), app(app(minus, y), x)), z), u)
APP(app(filter, fun), app(app(cons, x), xs)) → APP(app(app(app(filter2, app(fun, x)), fun), x), xs)
APP(app(app(app(f, app(s, x)), app(s, y)), z), u) → APP(if, app(app(le, x), y))
APP(app(le, app(s, x)), app(s, y)) → APP(le, x)
APP(app(app(app(f, app(s, x)), 0), z), u) → APP(minus, z)
APP(app(app(app(filter2, true), fun), x), xs) → APP(app(cons, x), app(app(filter, fun), xs))
APP(perfectp, app(s, x)) → APP(app(f, x), app(s, 0))
APP(app(app(app(f, app(s, x)), app(s, y)), z), u) → APP(minus, y)
APP(app(app(app(f, app(s, x)), app(s, y)), z), u) → APP(app(app(f, x), u), z)
APP(app(filter, fun), app(app(cons, x), xs)) → APP(filter2, app(fun, x))
APP(app(map, fun), app(app(cons, x), xs)) → APP(fun, x)
APP(app(minus, app(s, x)), app(s, y)) → APP(app(minus, x), y)
APP(app(app(app(filter2, true), fun), x), xs) → APP(app(filter, fun), xs)
APP(app(app(app(f, app(s, x)), 0), z), u) → APP(app(f, x), u)
APP(app(filter, fun), app(app(cons, x), xs)) → APP(app(filter2, app(fun, x)), fun)
APP(app(app(app(filter2, false), fun), x), xs) → APP(filter, fun)
APP(perfectp, app(s, x)) → APP(app(app(app(f, x), app(s, 0)), app(s, x)), app(s, x))
APP(app(le, app(s, x)), app(s, y)) → APP(app(le, x), y)
APP(app(app(app(f, app(s, x)), app(s, y)), z), u) → APP(app(app(f, app(s, x)), app(app(minus, y), x)), z)
APP(app(filter, fun), app(app(cons, x), xs)) → APP(fun, x)
APP(app(map, fun), app(app(cons, x), xs)) → APP(app(map, fun), xs)
APP(app(app(app(f, app(s, x)), app(s, y)), z), u) → APP(app(app(app(f, x), u), z), u)
APP(app(app(app(f, app(s, x)), 0), z), u) → APP(f, x)
APP(perfectp, app(s, x)) → APP(s, 0)
APP(app(app(app(f, app(s, x)), app(s, y)), z), u) → APP(app(le, x), y)
APP(app(app(app(f, app(s, x)), 0), z), u) → APP(app(app(f, x), u), app(app(minus, z), app(s, x)))
APP(app(app(app(filter2, true), fun), x), xs) → APP(filter, fun)
APP(app(app(app(f, app(s, x)), 0), z), u) → APP(app(minus, z), app(s, x))
APP(app(app(app(f, app(s, x)), 0), z), u) → APP(app(app(app(f, x), u), app(app(minus, z), app(s, x))), u)
APP(app(app(app(f, app(s, x)), app(s, y)), z), u) → APP(f, x)
APP(app(app(app(f, app(s, x)), app(s, y)), z), u) → APP(app(app(if, app(app(le, x), y)), app(app(app(app(f, app(s, x)), app(app(minus, y), x)), z), u)), app(app(app(app(f, x), u), z), u))
APP(app(app(app(f, app(s, x)), app(s, y)), z), u) → APP(app(if, app(app(le, x), y)), app(app(app(app(f, app(s, x)), app(app(minus, y), x)), z), u))
APP(app(app(app(filter2, true), fun), x), xs) → APP(cons, x)
APP(app(app(app(filter2, false), fun), x), xs) → APP(app(filter, fun), xs)
APP(app(map, fun), app(app(cons, x), xs)) → APP(cons, app(fun, x))

The TRS R consists of the following rules:

app(app(minus, 0), y) → 0
app(app(minus, app(s, x)), 0) → app(s, x)
app(app(minus, app(s, x)), app(s, y)) → app(app(minus, x), y)
app(app(le, 0), y) → true
app(app(le, app(s, x)), 0) → false
app(app(le, app(s, x)), app(s, y)) → app(app(le, x), y)
app(app(app(if, true), x), y) → x
app(app(app(if, false), x), y) → y
app(perfectp, 0) → false
app(perfectp, app(s, x)) → app(app(app(app(f, x), app(s, 0)), app(s, x)), app(s, x))
app(app(app(app(f, 0), y), 0), u) → true
app(app(app(app(f, 0), y), app(s, z)), u) → false
app(app(app(app(f, app(s, x)), 0), z), u) → app(app(app(app(f, x), u), app(app(minus, z), app(s, x))), u)
app(app(app(app(f, app(s, x)), app(s, y)), z), u) → app(app(app(if, app(app(le, x), y)), app(app(app(app(f, app(s, x)), app(app(minus, y), x)), z), u)), app(app(app(app(f, x), u), z), u))
app(app(map, fun), nil) → nil
app(app(map, fun), app(app(cons, x), xs)) → app(app(cons, app(fun, x)), app(app(map, fun), xs))
app(app(filter, fun), nil) → nil
app(app(filter, fun), app(app(cons, x), xs)) → app(app(app(app(filter2, app(fun, x)), fun), x), xs)
app(app(app(app(filter2, true), fun), x), xs) → app(app(cons, x), app(app(filter, fun), xs))
app(app(app(app(filter2, false), fun), x), xs) → app(app(filter, fun), xs)

The set Q consists of the following terms:

app(app(minus, 0), x0)
app(app(minus, app(s, x0)), 0)
app(app(minus, app(s, x0)), app(s, x1))
app(app(le, 0), x0)
app(app(le, app(s, x0)), 0)
app(app(le, app(s, x0)), app(s, x1))
app(app(app(if, true), x0), x1)
app(app(app(if, false), x0), x1)
app(perfectp, 0)
app(perfectp, app(s, x0))
app(app(app(app(f, 0), x0), 0), x1)
app(app(app(app(f, 0), x0), app(s, x1)), x2)
app(app(app(app(f, app(s, x0)), 0), x1), x2)
app(app(app(app(f, app(s, x0)), app(s, x1)), x2), x3)
app(app(map, x0), nil)
app(app(map, x0), app(app(cons, x1), x2))
app(app(filter, x0), nil)
app(app(filter, x0), app(app(cons, x1), x2))
app(app(app(app(filter2, true), x0), x1), x2)
app(app(app(app(filter2, false), x0), x1), x2)

We have to consider all minimal (P,Q,R)-chains.
We 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(app(app(f, app(s, x)), app(s, y)), z), u) → APP(app(minus, y), x)
APP(app(app(app(f, app(s, x)), app(s, y)), z), u) → APP(app(f, app(s, x)), app(app(minus, y), x))
APP(app(minus, app(s, x)), app(s, y)) → APP(minus, x)
APP(perfectp, app(s, x)) → APP(app(app(f, x), app(s, 0)), app(s, x))
APP(app(map, fun), app(app(cons, x), xs)) → APP(app(cons, app(fun, x)), app(app(map, fun), xs))
APP(perfectp, app(s, x)) → APP(f, x)
APP(app(app(app(f, app(s, x)), app(s, y)), z), u) → APP(app(f, x), u)
APP(app(filter, fun), app(app(cons, x), xs)) → APP(app(app(filter2, app(fun, x)), fun), x)
APP(app(app(app(f, app(s, x)), app(s, y)), z), u) → APP(app(app(app(f, app(s, x)), app(app(minus, y), x)), z), u)
APP(app(app(app(f, app(s, x)), app(s, y)), z), u) → APP(le, x)
APP(app(app(app(f, app(s, x)), app(s, y)), z), u) → APP(if, app(app(le, x), y))
APP(app(filter, fun), app(app(cons, x), xs)) → APP(app(app(app(filter2, app(fun, x)), fun), x), xs)
APP(app(le, app(s, x)), app(s, y)) → APP(le, x)
APP(app(app(app(f, app(s, x)), 0), z), u) → APP(minus, z)
APP(app(app(app(filter2, true), fun), x), xs) → APP(app(cons, x), app(app(filter, fun), xs))
APP(perfectp, app(s, x)) → APP(app(f, x), app(s, 0))
APP(app(app(app(f, app(s, x)), app(s, y)), z), u) → APP(minus, y)
APP(app(app(app(f, app(s, x)), app(s, y)), z), u) → APP(app(app(f, x), u), z)
APP(app(map, fun), app(app(cons, x), xs)) → APP(fun, x)
APP(app(filter, fun), app(app(cons, x), xs)) → APP(filter2, app(fun, x))
APP(app(minus, app(s, x)), app(s, y)) → APP(app(minus, x), y)
APP(app(app(app(f, app(s, x)), 0), z), u) → APP(app(f, x), u)
APP(app(app(app(filter2, true), fun), x), xs) → APP(app(filter, fun), xs)
APP(app(filter, fun), app(app(cons, x), xs)) → APP(app(filter2, app(fun, x)), fun)
APP(perfectp, app(s, x)) → APP(app(app(app(f, x), app(s, 0)), app(s, x)), app(s, x))
APP(app(app(app(filter2, false), fun), x), xs) → APP(filter, fun)
APP(app(app(app(f, app(s, x)), app(s, y)), z), u) → APP(app(app(f, app(s, x)), app(app(minus, y), x)), z)
APP(app(le, app(s, x)), app(s, y)) → APP(app(le, x), y)
APP(app(map, fun), app(app(cons, x), xs)) → APP(app(map, fun), xs)
APP(app(filter, fun), app(app(cons, x), xs)) → APP(fun, x)
APP(perfectp, app(s, x)) → APP(s, 0)
APP(app(app(app(f, app(s, x)), 0), z), u) → APP(f, x)
APP(app(app(app(f, app(s, x)), app(s, y)), z), u) → APP(app(app(app(f, x), u), z), u)
APP(app(app(app(f, app(s, x)), app(s, y)), z), u) → APP(app(le, x), y)
APP(app(app(app(f, app(s, x)), 0), z), u) → APP(app(app(f, x), u), app(app(minus, z), app(s, x)))
APP(app(app(app(filter2, true), fun), x), xs) → APP(filter, fun)
APP(app(app(app(f, app(s, x)), 0), z), u) → APP(app(minus, z), app(s, x))
APP(app(app(app(f, app(s, x)), 0), z), u) → APP(app(app(app(f, x), u), app(app(minus, z), app(s, x))), u)
APP(app(app(app(f, app(s, x)), app(s, y)), z), u) → APP(app(if, app(app(le, x), y)), app(app(app(app(f, app(s, x)), app(app(minus, y), x)), z), u))
APP(app(app(app(f, app(s, x)), app(s, y)), z), u) → APP(app(app(if, app(app(le, x), y)), app(app(app(app(f, app(s, x)), app(app(minus, y), x)), z), u)), app(app(app(app(f, x), u), z), u))
APP(app(app(app(f, app(s, x)), app(s, y)), z), u) → APP(f, x)
APP(app(app(app(filter2, true), fun), x), xs) → APP(cons, x)
APP(app(app(app(filter2, false), fun), x), xs) → APP(app(filter, fun), xs)
APP(app(map, fun), app(app(cons, x), xs)) → APP(cons, app(fun, x))

The TRS R consists of the following rules:

app(app(minus, 0), y) → 0
app(app(minus, app(s, x)), 0) → app(s, x)
app(app(minus, app(s, x)), app(s, y)) → app(app(minus, x), y)
app(app(le, 0), y) → true
app(app(le, app(s, x)), 0) → false
app(app(le, app(s, x)), app(s, y)) → app(app(le, x), y)
app(app(app(if, true), x), y) → x
app(app(app(if, false), x), y) → y
app(perfectp, 0) → false
app(perfectp, app(s, x)) → app(app(app(app(f, x), app(s, 0)), app(s, x)), app(s, x))
app(app(app(app(f, 0), y), 0), u) → true
app(app(app(app(f, 0), y), app(s, z)), u) → false
app(app(app(app(f, app(s, x)), 0), z), u) → app(app(app(app(f, x), u), app(app(minus, z), app(s, x))), u)
app(app(app(app(f, app(s, x)), app(s, y)), z), u) → app(app(app(if, app(app(le, x), y)), app(app(app(app(f, app(s, x)), app(app(minus, y), x)), z), u)), app(app(app(app(f, x), u), z), u))
app(app(map, fun), nil) → nil
app(app(map, fun), app(app(cons, x), xs)) → app(app(cons, app(fun, x)), app(app(map, fun), xs))
app(app(filter, fun), nil) → nil
app(app(filter, fun), app(app(cons, x), xs)) → app(app(app(app(filter2, app(fun, x)), fun), x), xs)
app(app(app(app(filter2, true), fun), x), xs) → app(app(cons, x), app(app(filter, fun), xs))
app(app(app(app(filter2, false), fun), x), xs) → app(app(filter, fun), xs)

The set Q consists of the following terms:

app(app(minus, 0), x0)
app(app(minus, app(s, x0)), 0)
app(app(minus, app(s, x0)), app(s, x1))
app(app(le, 0), x0)
app(app(le, app(s, x0)), 0)
app(app(le, app(s, x0)), app(s, x1))
app(app(app(if, true), x0), x1)
app(app(app(if, false), x0), x1)
app(perfectp, 0)
app(perfectp, app(s, x0))
app(app(app(app(f, 0), x0), 0), x1)
app(app(app(app(f, 0), x0), app(s, x1)), x2)
app(app(app(app(f, app(s, x0)), 0), x1), x2)
app(app(app(app(f, app(s, x0)), app(s, x1)), x2), x3)
app(app(map, x0), nil)
app(app(map, x0), app(app(cons, x1), x2))
app(app(filter, x0), nil)
app(app(filter, x0), app(app(cons, x1), x2))
app(app(app(app(filter2, true), x0), x1), x2)
app(app(app(app(filter2, false), x0), x1), x2)

We have to consider all minimal (P,Q,R)-chains.
The approximation of the Dependency Graph [13,14,18] contains 4 SCCs with 33 less nodes.

↳ QTRS

  ↳ Overlay + Local Confluence

    ↳ QTRS

      ↳ DependencyPairsProof

        ↳ QDP

          ↳ EdgeDeletionProof

            ↳ QDP

              ↳ DependencyGraphProof

                ↳ AND

                  ↳ QDP

                    ↳ QDPOrderProof

                  ↳ QDP

                  ↳ QDP

                  ↳ QDP

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

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

The TRS R consists of the following rules:

app(app(minus, 0), y) → 0
app(app(minus, app(s, x)), 0) → app(s, x)
app(app(minus, app(s, x)), app(s, y)) → app(app(minus, x), y)
app(app(le, 0), y) → true
app(app(le, app(s, x)), 0) → false
app(app(le, app(s, x)), app(s, y)) → app(app(le, x), y)
app(app(app(if, true), x), y) → x
app(app(app(if, false), x), y) → y
app(perfectp, 0) → false
app(perfectp, app(s, x)) → app(app(app(app(f, x), app(s, 0)), app(s, x)), app(s, x))
app(app(app(app(f, 0), y), 0), u) → true
app(app(app(app(f, 0), y), app(s, z)), u) → false
app(app(app(app(f, app(s, x)), 0), z), u) → app(app(app(app(f, x), u), app(app(minus, z), app(s, x))), u)
app(app(app(app(f, app(s, x)), app(s, y)), z), u) → app(app(app(if, app(app(le, x), y)), app(app(app(app(f, app(s, x)), app(app(minus, y), x)), z), u)), app(app(app(app(f, x), u), z), u))
app(app(map, fun), nil) → nil
app(app(map, fun), app(app(cons, x), xs)) → app(app(cons, app(fun, x)), app(app(map, fun), xs))
app(app(filter, fun), nil) → nil
app(app(filter, fun), app(app(cons, x), xs)) → app(app(app(app(filter2, app(fun, x)), fun), x), xs)
app(app(app(app(filter2, true), fun), x), xs) → app(app(cons, x), app(app(filter, fun), xs))
app(app(app(app(filter2, false), fun), x), xs) → app(app(filter, fun), xs)

The set Q consists of the following terms:

app(app(minus, 0), x0)
app(app(minus, app(s, x0)), 0)
app(app(minus, app(s, x0)), app(s, x1))
app(app(le, 0), x0)
app(app(le, app(s, x0)), 0)
app(app(le, app(s, x0)), app(s, x1))
app(app(app(if, true), x0), x1)
app(app(app(if, false), x0), x1)
app(perfectp, 0)
app(perfectp, app(s, x0))
app(app(app(app(f, 0), x0), 0), x1)
app(app(app(app(f, 0), x0), app(s, x1)), x2)
app(app(app(app(f, app(s, x0)), 0), x1), x2)
app(app(app(app(f, app(s, x0)), app(s, x1)), x2), x3)
app(app(map, x0), nil)
app(app(map, x0), app(app(cons, x1), x2))
app(app(filter, x0), nil)
app(app(filter, x0), app(app(cons, x1), x2))
app(app(app(app(filter2, true), x0), x1), x2)
app(app(app(app(filter2, false), x0), x1), x2)

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

LE(s(x), s(y)) → LE(x, y)

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

minus(0, x0)
minus(s(x0), 0)
minus(s(x0), s(x1))
le(0, x0)
le(s(x0), 0)
le(s(x0), s(x1))
if(true, x0, x1)
if(false, x0, x1)
perfectp(0)
perfectp(s(x0))
f(0, x0, 0, x1)
f(0, x0, s(x1), x2)
f(s(x0), 0, x1, x2)
f(s(x0), s(x1), x2, x3)
map(x0, nil)
map(x0, cons(x1, x2))
filter(x0, nil)
filter(x0, cons(x1, x2))
filter2(true, x0, x1, x2)
filter2(false, x0, x1, x2)

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

The following pairs can be oriented strictly and are deleted.

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

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

Lexicographic path order with status [19].
Quasi-Precedence:

trivial

Status:

trivial

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

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

app(app(minus, 0), y) → 0
app(app(minus, app(s, x)), 0) → app(s, x)
app(app(minus, app(s, x)), app(s, y)) → app(app(minus, x), y)
app(app(le, 0), y) → true
app(app(le, app(s, x)), 0) → false
app(app(le, app(s, x)), app(s, y)) → app(app(le, x), y)
app(app(app(if, true), x), y) → x
app(app(app(if, false), x), y) → y
app(perfectp, 0) → false
app(perfectp, app(s, x)) → app(app(app(app(f, x), app(s, 0)), app(s, x)), app(s, x))
app(app(app(app(f, 0), y), 0), u) → true
app(app(app(app(f, 0), y), app(s, z)), u) → false
app(app(app(app(f, app(s, x)), 0), z), u) → app(app(app(app(f, x), u), app(app(minus, z), app(s, x))), u)
app(app(app(app(f, app(s, x)), app(s, y)), z), u) → app(app(app(if, app(app(le, x), y)), app(app(app(app(f, app(s, x)), app(app(minus, y), x)), z), u)), app(app(app(app(f, x), u), z), u))
app(app(map, fun), nil) → nil
app(app(map, fun), app(app(cons, x), xs)) → app(app(cons, app(fun, x)), app(app(map, fun), xs))
app(app(filter, fun), nil) → nil
app(app(filter, fun), app(app(cons, x), xs)) → app(app(app(app(filter2, app(fun, x)), fun), x), xs)
app(app(app(app(filter2, true), fun), x), xs) → app(app(cons, x), app(app(filter, fun), xs))
app(app(app(app(filter2, false), fun), x), xs) → app(app(filter, fun), xs)

The set Q consists of the following terms:

app(app(minus, 0), x0)
app(app(minus, app(s, x0)), 0)
app(app(minus, app(s, x0)), app(s, x1))
app(app(le, 0), x0)
app(app(le, app(s, x0)), 0)
app(app(le, app(s, x0)), app(s, x1))
app(app(app(if, true), x0), x1)
app(app(app(if, false), x0), x1)
app(perfectp, 0)
app(perfectp, app(s, x0))
app(app(app(app(f, 0), x0), 0), x1)
app(app(app(app(f, 0), x0), app(s, x1)), x2)
app(app(app(app(f, app(s, x0)), 0), x1), x2)
app(app(app(app(f, app(s, x0)), app(s, x1)), x2), x3)
app(app(map, x0), nil)
app(app(map, x0), app(app(cons, x1), x2))
app(app(filter, x0), nil)
app(app(filter, x0), app(app(cons, x1), x2))
app(app(app(app(filter2, true), x0), x1), x2)
app(app(app(app(filter2, false), x0), x1), x2)

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

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, 0), y) → 0
app(app(minus, app(s, x)), 0) → app(s, x)
app(app(minus, app(s, x)), app(s, y)) → app(app(minus, x), y)
app(app(le, 0), y) → true
app(app(le, app(s, x)), 0) → false
app(app(le, app(s, x)), app(s, y)) → app(app(le, x), y)
app(app(app(if, true), x), y) → x
app(app(app(if, false), x), y) → y
app(perfectp, 0) → false
app(perfectp, app(s, x)) → app(app(app(app(f, x), app(s, 0)), app(s, x)), app(s, x))
app(app(app(app(f, 0), y), 0), u) → true
app(app(app(app(f, 0), y), app(s, z)), u) → false
app(app(app(app(f, app(s, x)), 0), z), u) → app(app(app(app(f, x), u), app(app(minus, z), app(s, x))), u)
app(app(app(app(f, app(s, x)), app(s, y)), z), u) → app(app(app(if, app(app(le, x), y)), app(app(app(app(f, app(s, x)), app(app(minus, y), x)), z), u)), app(app(app(app(f, x), u), z), u))
app(app(map, fun), nil) → nil
app(app(map, fun), app(app(cons, x), xs)) → app(app(cons, app(fun, x)), app(app(map, fun), xs))
app(app(filter, fun), nil) → nil
app(app(filter, fun), app(app(cons, x), xs)) → app(app(app(app(filter2, app(fun, x)), fun), x), xs)
app(app(app(app(filter2, true), fun), x), xs) → app(app(cons, x), app(app(filter, fun), xs))
app(app(app(app(filter2, false), fun), x), xs) → app(app(filter, fun), xs)

The set Q consists of the following terms:

app(app(minus, 0), x0)
app(app(minus, app(s, x0)), 0)
app(app(minus, app(s, x0)), app(s, x1))
app(app(le, 0), x0)
app(app(le, app(s, x0)), 0)
app(app(le, app(s, x0)), app(s, x1))
app(app(app(if, true), x0), x1)
app(app(app(if, false), x0), x1)
app(perfectp, 0)
app(perfectp, app(s, x0))
app(app(app(app(f, 0), x0), 0), x1)
app(app(app(app(f, 0), x0), app(s, x1)), x2)
app(app(app(app(f, app(s, x0)), 0), x1), x2)
app(app(app(app(f, app(s, x0)), app(s, x1)), x2), x3)
app(app(map, x0), nil)
app(app(map, x0), app(app(cons, x1), x2))
app(app(filter, x0), nil)
app(app(filter, x0), app(app(cons, x1), x2))
app(app(app(app(filter2, true), x0), x1), x2)
app(app(app(app(filter2, false), x0), x1), x2)

MINUS(s(x), s(y)) → MINUS(x, y)

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

minus(0, x0)
minus(s(x0), 0)
minus(s(x0), s(x1))
le(0, x0)
le(s(x0), 0)
le(s(x0), s(x1))
if(true, x0, x1)
if(false, x0, x1)
perfectp(0)
perfectp(s(x0))
f(0, x0, 0, x1)
f(0, x0, s(x1), x2)
f(s(x0), 0, x1, x2)
f(s(x0), s(x1), x2, x3)
map(x0, nil)
map(x0, cons(x1, x2))
filter(x0, nil)
filter(x0, cons(x1, x2))
filter2(true, x0, x1, x2)
filter2(false, x0, x1, x2)

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

The following pairs can be oriented strictly and are deleted.

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

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

Lexicographic path order with status [19].
Quasi-Precedence:

trivial

Status:

trivial

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

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

app(app(minus, 0), y) → 0
app(app(minus, app(s, x)), 0) → app(s, x)
app(app(minus, app(s, x)), app(s, y)) → app(app(minus, x), y)
app(app(le, 0), y) → true
app(app(le, app(s, x)), 0) → false
app(app(le, app(s, x)), app(s, y)) → app(app(le, x), y)
app(app(app(if, true), x), y) → x
app(app(app(if, false), x), y) → y
app(perfectp, 0) → false
app(perfectp, app(s, x)) → app(app(app(app(f, x), app(s, 0)), app(s, x)), app(s, x))
app(app(app(app(f, 0), y), 0), u) → true
app(app(app(app(f, 0), y), app(s, z)), u) → false
app(app(app(app(f, app(s, x)), 0), z), u) → app(app(app(app(f, x), u), app(app(minus, z), app(s, x))), u)
app(app(app(app(f, app(s, x)), app(s, y)), z), u) → app(app(app(if, app(app(le, x), y)), app(app(app(app(f, app(s, x)), app(app(minus, y), x)), z), u)), app(app(app(app(f, x), u), z), u))
app(app(map, fun), nil) → nil
app(app(map, fun), app(app(cons, x), xs)) → app(app(cons, app(fun, x)), app(app(map, fun), xs))
app(app(filter, fun), nil) → nil
app(app(filter, fun), app(app(cons, x), xs)) → app(app(app(app(filter2, app(fun, x)), fun), x), xs)
app(app(app(app(filter2, true), fun), x), xs) → app(app(cons, x), app(app(filter, fun), xs))
app(app(app(app(filter2, false), fun), x), xs) → app(app(filter, fun), xs)

The set Q consists of the following terms:

app(app(minus, 0), x0)
app(app(minus, app(s, x0)), 0)
app(app(minus, app(s, x0)), app(s, x1))
app(app(le, 0), x0)
app(app(le, app(s, x0)), 0)
app(app(le, app(s, x0)), app(s, x1))
app(app(app(if, true), x0), x1)
app(app(app(if, false), x0), x1)
app(perfectp, 0)
app(perfectp, app(s, x0))
app(app(app(app(f, 0), x0), 0), x1)
app(app(app(app(f, 0), x0), app(s, x1)), x2)
app(app(app(app(f, app(s, x0)), 0), x1), x2)
app(app(app(app(f, app(s, x0)), app(s, x1)), x2), x3)
app(app(map, x0), nil)
app(app(map, x0), app(app(cons, x1), x2))
app(app(filter, x0), nil)
app(app(filter, x0), app(app(cons, x1), x2))
app(app(app(app(filter2, true), x0), x1), x2)
app(app(app(app(filter2, false), x0), x1), x2)

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

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

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

The TRS R consists of the following rules:

app(app(minus, 0), y) → 0
app(app(minus, app(s, x)), 0) → app(s, x)
app(app(minus, app(s, x)), app(s, y)) → app(app(minus, x), y)
app(app(le, 0), y) → true
app(app(le, app(s, x)), 0) → false
app(app(le, app(s, x)), app(s, y)) → app(app(le, x), y)
app(app(app(if, true), x), y) → x
app(app(app(if, false), x), y) → y
app(perfectp, 0) → false
app(perfectp, app(s, x)) → app(app(app(app(f, x), app(s, 0)), app(s, x)), app(s, x))
app(app(app(app(f, 0), y), 0), u) → true
app(app(app(app(f, 0), y), app(s, z)), u) → false
app(app(app(app(f, app(s, x)), 0), z), u) → app(app(app(app(f, x), u), app(app(minus, z), app(s, x))), u)
app(app(app(app(f, app(s, x)), app(s, y)), z), u) → app(app(app(if, app(app(le, x), y)), app(app(app(app(f, app(s, x)), app(app(minus, y), x)), z), u)), app(app(app(app(f, x), u), z), u))
app(app(map, fun), nil) → nil
app(app(map, fun), app(app(cons, x), xs)) → app(app(cons, app(fun, x)), app(app(map, fun), xs))
app(app(filter, fun), nil) → nil
app(app(filter, fun), app(app(cons, x), xs)) → app(app(app(app(filter2, app(fun, x)), fun), x), xs)
app(app(app(app(filter2, true), fun), x), xs) → app(app(cons, x), app(app(filter, fun), xs))
app(app(app(app(filter2, false), fun), x), xs) → app(app(filter, fun), xs)

The set Q consists of the following terms:

app(app(minus, 0), x0)
app(app(minus, app(s, x0)), 0)
app(app(minus, app(s, x0)), app(s, x1))
app(app(le, 0), x0)
app(app(le, app(s, x0)), 0)
app(app(le, app(s, x0)), app(s, x1))
app(app(app(if, true), x0), x1)
app(app(app(if, false), x0), x1)
app(perfectp, 0)
app(perfectp, app(s, x0))
app(app(app(app(f, 0), x0), 0), x1)
app(app(app(app(f, 0), x0), app(s, x1)), x2)
app(app(app(app(f, app(s, x0)), 0), x1), x2)
app(app(app(app(f, app(s, x0)), app(s, x1)), x2), x3)
app(app(map, x0), nil)
app(app(map, x0), app(app(cons, x1), x2))
app(app(filter, x0), nil)
app(app(filter, x0), app(app(cons, x1), x2))
app(app(app(app(filter2, true), x0), x1), x2)
app(app(app(app(filter2, false), x0), x1), x2)

F(s(x), s(y), z, u) → F(s(x), minus(y, x), z, u)
F(s(x), s(y), z, u) → F(x, u, z, u)
F(s(x), 0, z, u) → F(x, u, minus(z, s(x)), u)

The TRS R consists of the following rules:

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

The set Q consists of the following terms:

minus(0, x0)
minus(s(x0), 0)
minus(s(x0), s(x1))
le(0, x0)
le(s(x0), 0)
le(s(x0), s(x1))
if(true, x0, x1)
if(false, x0, x1)
perfectp(0)
perfectp(s(x0))
f(0, x0, 0, x1)
f(0, x0, s(x1), x2)
f(s(x0), 0, x1, x2)
f(s(x0), s(x1), x2, x3)
map(x0, nil)
map(x0, cons(x1, x2))
filter(x0, nil)
filter(x0, cons(x1, x2))
filter2(true, x0, x1, x2)
filter2(false, x0, x1, x2)

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

The following pairs can be oriented strictly and are deleted.

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

The remaining pairs can at least be oriented weakly.

APP(app(app(app(f, app(s, x)), app(s, y)), z), u) → APP(app(app(app(f, app(s, x)), app(app(minus, y), x)), z), u)

Used ordering: Combined order from the following AFS and order.
F(x1, x2, x3, x4) = x1
s(x1) = s(x1)

Lexicographic path order with status [19].
Quasi-Precedence:

trivial

Status:

trivial

The following usable rules [14] were oriented: none

↳ QTRS

  ↳ Overlay + Local Confluence

    ↳ QTRS

      ↳ DependencyPairsProof

        ↳ QDP

          ↳ EdgeDeletionProof

            ↳ QDP

              ↳ DependencyGraphProof

                ↳ AND

                  ↳ QDP

                  ↳ QDP

                  ↳ QDP

                    ↳ QDPOrderProof

                      ↳ QDP

                        ↳ QDPOrderProof

                  ↳ QDP

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

APP(app(app(app(f, app(s, x)), app(s, y)), z), u) → APP(app(app(app(f, app(s, x)), app(app(minus, y), x)), z), u)

The TRS R consists of the following rules:

app(app(minus, 0), y) → 0
app(app(minus, app(s, x)), 0) → app(s, x)
app(app(minus, app(s, x)), app(s, y)) → app(app(minus, x), y)
app(app(le, 0), y) → true
app(app(le, app(s, x)), 0) → false
app(app(le, app(s, x)), app(s, y)) → app(app(le, x), y)
app(app(app(if, true), x), y) → x
app(app(app(if, false), x), y) → y
app(perfectp, 0) → false
app(perfectp, app(s, x)) → app(app(app(app(f, x), app(s, 0)), app(s, x)), app(s, x))
app(app(app(app(f, 0), y), 0), u) → true
app(app(app(app(f, 0), y), app(s, z)), u) → false
app(app(app(app(f, app(s, x)), 0), z), u) → app(app(app(app(f, x), u), app(app(minus, z), app(s, x))), u)
app(app(app(app(f, app(s, x)), app(s, y)), z), u) → app(app(app(if, app(app(le, x), y)), app(app(app(app(f, app(s, x)), app(app(minus, y), x)), z), u)), app(app(app(app(f, x), u), z), u))
app(app(map, fun), nil) → nil
app(app(map, fun), app(app(cons, x), xs)) → app(app(cons, app(fun, x)), app(app(map, fun), xs))
app(app(filter, fun), nil) → nil
app(app(filter, fun), app(app(cons, x), xs)) → app(app(app(app(filter2, app(fun, x)), fun), x), xs)
app(app(app(app(filter2, true), fun), x), xs) → app(app(cons, x), app(app(filter, fun), xs))
app(app(app(app(filter2, false), fun), x), xs) → app(app(filter, fun), xs)

The set Q consists of the following terms:

app(app(minus, 0), x0)
app(app(minus, app(s, x0)), 0)
app(app(minus, app(s, x0)), app(s, x1))
app(app(le, 0), x0)
app(app(le, app(s, x0)), 0)
app(app(le, app(s, x0)), app(s, x1))
app(app(app(if, true), x0), x1)
app(app(app(if, false), x0), x1)
app(perfectp, 0)
app(perfectp, app(s, x0))
app(app(app(app(f, 0), x0), 0), x1)
app(app(app(app(f, 0), x0), app(s, x1)), x2)
app(app(app(app(f, app(s, x0)), 0), x1), x2)
app(app(app(app(f, app(s, x0)), app(s, x1)), x2), x3)
app(app(map, x0), nil)
app(app(map, x0), app(app(cons, x1), x2))
app(app(filter, x0), nil)
app(app(filter, x0), app(app(cons, x1), x2))
app(app(app(app(filter2, true), x0), x1), x2)
app(app(app(app(filter2, false), x0), x1), x2)

F(s(x), s(y), z, u) → F(s(x), minus(y, x), z, u)

The TRS R consists of the following rules:

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

The set Q consists of the following terms:

minus(0, x0)
minus(s(x0), 0)
minus(s(x0), s(x1))
le(0, x0)
le(s(x0), 0)
le(s(x0), s(x1))
if(true, x0, x1)
if(false, x0, x1)
perfectp(0)
perfectp(s(x0))
f(0, x0, 0, x1)
f(0, x0, s(x1), x2)
f(s(x0), 0, x1, x2)
f(s(x0), s(x1), x2, x3)
map(x0, nil)
map(x0, cons(x1, x2))
filter(x0, nil)
filter(x0, cons(x1, x2))
filter2(true, x0, x1, x2)
filter2(false, x0, x1, x2)

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

The following pairs can be oriented strictly and are deleted.

APP(app(app(app(f, app(s, x)), app(s, y)), z), u) → APP(app(app(app(f, app(s, x)), app(app(minus, y), x)), z), u)

The remaining pairs can at least be oriented weakly.
none
Used ordering: Combined order from the following AFS and order.
F(x1, x2, x3, x4) = x2
s(x1) = s(x1)
minus(x1, x2) = x1
0 = 0

Lexicographic path order with status [19].
Quasi-Precedence:

trivial

Status:

trivial

The following usable rules [14] were oriented:

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

↳ QTRS

  ↳ Overlay + Local Confluence

    ↳ QTRS

      ↳ DependencyPairsProof

        ↳ QDP

          ↳ EdgeDeletionProof

            ↳ QDP

              ↳ DependencyGraphProof

                ↳ AND

                  ↳ QDP

                  ↳ QDP

                  ↳ QDP

                    ↳ QDPOrderProof

                      ↳ QDP

                        ↳ QDPOrderProof

                          ↳ QDP

                            ↳ PisEmptyProof

                  ↳ QDP

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

app(app(minus, 0), y) → 0
app(app(minus, app(s, x)), 0) → app(s, x)
app(app(minus, app(s, x)), app(s, y)) → app(app(minus, x), y)
app(app(le, 0), y) → true
app(app(le, app(s, x)), 0) → false
app(app(le, app(s, x)), app(s, y)) → app(app(le, x), y)
app(app(app(if, true), x), y) → x
app(app(app(if, false), x), y) → y
app(perfectp, 0) → false
app(perfectp, app(s, x)) → app(app(app(app(f, x), app(s, 0)), app(s, x)), app(s, x))
app(app(app(app(f, 0), y), 0), u) → true
app(app(app(app(f, 0), y), app(s, z)), u) → false
app(app(app(app(f, app(s, x)), 0), z), u) → app(app(app(app(f, x), u), app(app(minus, z), app(s, x))), u)
app(app(app(app(f, app(s, x)), app(s, y)), z), u) → app(app(app(if, app(app(le, x), y)), app(app(app(app(f, app(s, x)), app(app(minus, y), x)), z), u)), app(app(app(app(f, x), u), z), u))
app(app(map, fun), nil) → nil
app(app(map, fun), app(app(cons, x), xs)) → app(app(cons, app(fun, x)), app(app(map, fun), xs))
app(app(filter, fun), nil) → nil
app(app(filter, fun), app(app(cons, x), xs)) → app(app(app(app(filter2, app(fun, x)), fun), x), xs)
app(app(app(app(filter2, true), fun), x), xs) → app(app(cons, x), app(app(filter, fun), xs))
app(app(app(app(filter2, false), fun), x), xs) → app(app(filter, fun), xs)

The set Q consists of the following terms:

app(app(minus, 0), x0)
app(app(minus, app(s, x0)), 0)
app(app(minus, app(s, x0)), app(s, x1))
app(app(le, 0), x0)
app(app(le, app(s, x0)), 0)
app(app(le, app(s, x0)), app(s, x1))
app(app(app(if, true), x0), x1)
app(app(app(if, false), x0), x1)
app(perfectp, 0)
app(perfectp, app(s, x0))
app(app(app(app(f, 0), x0), 0), x1)
app(app(app(app(f, 0), x0), app(s, x1)), x2)
app(app(app(app(f, app(s, x0)), 0), x1), x2)
app(app(app(app(f, app(s, x0)), app(s, x1)), x2), x3)
app(app(map, x0), nil)
app(app(map, x0), app(app(cons, x1), x2))
app(app(filter, x0), nil)
app(app(filter, x0), app(app(cons, x1), x2))
app(app(app(app(filter2, true), x0), x1), x2)
app(app(app(app(filter2, false), x0), x1), x2)

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

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

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

The TRS R consists of the following rules:

app(app(minus, 0), y) → 0
app(app(minus, app(s, x)), 0) → app(s, x)
app(app(minus, app(s, x)), app(s, y)) → app(app(minus, x), y)
app(app(le, 0), y) → true
app(app(le, app(s, x)), 0) → false
app(app(le, app(s, x)), app(s, y)) → app(app(le, x), y)
app(app(app(if, true), x), y) → x
app(app(app(if, false), x), y) → y
app(perfectp, 0) → false
app(perfectp, app(s, x)) → app(app(app(app(f, x), app(s, 0)), app(s, x)), app(s, x))
app(app(app(app(f, 0), y), 0), u) → true
app(app(app(app(f, 0), y), app(s, z)), u) → false
app(app(app(app(f, app(s, x)), 0), z), u) → app(app(app(app(f, x), u), app(app(minus, z), app(s, x))), u)
app(app(app(app(f, app(s, x)), app(s, y)), z), u) → app(app(app(if, app(app(le, x), y)), app(app(app(app(f, app(s, x)), app(app(minus, y), x)), z), u)), app(app(app(app(f, x), u), z), u))
app(app(map, fun), nil) → nil
app(app(map, fun), app(app(cons, x), xs)) → app(app(cons, app(fun, x)), app(app(map, fun), xs))
app(app(filter, fun), nil) → nil
app(app(filter, fun), app(app(cons, x), xs)) → app(app(app(app(filter2, app(fun, x)), fun), x), xs)
app(app(app(app(filter2, true), fun), x), xs) → app(app(cons, x), app(app(filter, fun), xs))
app(app(app(app(filter2, false), fun), x), xs) → app(app(filter, fun), xs)

The set Q consists of the following terms:

app(app(minus, 0), x0)
app(app(minus, app(s, x0)), 0)
app(app(minus, app(s, x0)), app(s, x1))
app(app(le, 0), x0)
app(app(le, app(s, x0)), 0)
app(app(le, app(s, x0)), app(s, x1))
app(app(app(if, true), x0), x1)
app(app(app(if, false), x0), x1)
app(perfectp, 0)
app(perfectp, app(s, x0))
app(app(app(app(f, 0), x0), 0), x1)
app(app(app(app(f, 0), x0), app(s, x1)), x2)
app(app(app(app(f, app(s, x0)), 0), x1), x2)
app(app(app(app(f, app(s, x0)), app(s, x1)), x2), x3)
app(app(map, x0), nil)
app(app(map, x0), app(app(cons, x1), x2))
app(app(filter, x0), nil)
app(app(filter, x0), app(app(cons, x1), x2))
app(app(app(app(filter2, true), x0), x1), x2)
app(app(app(app(filter2, false), x0), x1), x2)

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

The following pairs can be oriented strictly and are deleted.

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

The remaining pairs can at least be oriented weakly.

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

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

Lexicographic path order with status [19].
Quasi-Precedence:

trivial

Status:

trivial

The following usable rules [14] were oriented: none

↳ QTRS

  ↳ Overlay + Local Confluence

    ↳ QTRS

      ↳ DependencyPairsProof

        ↳ QDP

          ↳ EdgeDeletionProof

            ↳ QDP

              ↳ DependencyGraphProof

                ↳ AND

                  ↳ QDP

                  ↳ QDP

                  ↳ QDP

                  ↳ QDP

                    ↳ QDPOrderProof

                      ↳ QDP

                        ↳ DependencyGraphProof

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

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

The TRS R consists of the following rules:

app(app(minus, 0), y) → 0
app(app(minus, app(s, x)), 0) → app(s, x)
app(app(minus, app(s, x)), app(s, y)) → app(app(minus, x), y)
app(app(le, 0), y) → true
app(app(le, app(s, x)), 0) → false
app(app(le, app(s, x)), app(s, y)) → app(app(le, x), y)
app(app(app(if, true), x), y) → x
app(app(app(if, false), x), y) → y
app(perfectp, 0) → false
app(perfectp, app(s, x)) → app(app(app(app(f, x), app(s, 0)), app(s, x)), app(s, x))
app(app(app(app(f, 0), y), 0), u) → true
app(app(app(app(f, 0), y), app(s, z)), u) → false
app(app(app(app(f, app(s, x)), 0), z), u) → app(app(app(app(f, x), u), app(app(minus, z), app(s, x))), u)
app(app(app(app(f, app(s, x)), app(s, y)), z), u) → app(app(app(if, app(app(le, x), y)), app(app(app(app(f, app(s, x)), app(app(minus, y), x)), z), u)), app(app(app(app(f, x), u), z), u))
app(app(map, fun), nil) → nil
app(app(map, fun), app(app(cons, x), xs)) → app(app(cons, app(fun, x)), app(app(map, fun), xs))
app(app(filter, fun), nil) → nil
app(app(filter, fun), app(app(cons, x), xs)) → app(app(app(app(filter2, app(fun, x)), fun), x), xs)
app(app(app(app(filter2, true), fun), x), xs) → app(app(cons, x), app(app(filter, fun), xs))
app(app(app(app(filter2, false), fun), x), xs) → app(app(filter, fun), xs)

The set Q consists of the following terms:

app(app(minus, 0), x0)
app(app(minus, app(s, x0)), 0)
app(app(minus, app(s, x0)), app(s, x1))
app(app(le, 0), x0)
app(app(le, app(s, x0)), 0)
app(app(le, app(s, x0)), app(s, x1))
app(app(app(if, true), x0), x1)
app(app(app(if, false), x0), x1)
app(perfectp, 0)
app(perfectp, app(s, x0))
app(app(app(app(f, 0), x0), 0), x1)
app(app(app(app(f, 0), x0), app(s, x1)), x2)
app(app(app(app(f, app(s, x0)), 0), x1), x2)
app(app(app(app(f, app(s, x0)), app(s, x1)), x2), x3)
app(app(map, x0), nil)
app(app(map, x0), app(app(cons, x1), x2))
app(app(filter, x0), nil)
app(app(filter, x0), app(app(cons, x1), x2))
app(app(app(app(filter2, true), x0), x1), x2)
app(app(app(app(filter2, false), x0), x1), x2)

We have to consider all minimal (P,Q,R)-chains.
The approximation of the Dependency Graph [13,14,18] contains 0 SCCs with 2 less nodes.