Termination w.r.t. Q of the given QTRS could not be shown:

(0) Obligation:

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

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.

(1) Overlay + Local Confluence (EQUIVALENT transformation)

The TRS is overlay and locally confluent. By [NOC] we can switch to innermost.

(2) Obligation:

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

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(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)

(3) DependencyPairsProof (EQUIVALENT transformation)

Using Dependency Pairs [AG00,LPAR04] we result in the following initial DP problem.

(4) Obligation:

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

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

The TRS R consists of the following rules:

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(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.

(5) DependencyGraphProof (EQUIVALENT transformation)

The approximation of the Dependency Graph [LPAR04,FROCOS05,EDGSTAR] contains 2 SCCs with 32 less nodes.

(6) Complex Obligation (AND)

(7) Obligation:

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, x), u), z), u)
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)

The TRS R consists of the following rules:

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(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.

(8) Obligation:

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

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

The TRS R consists of the following rules:

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(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.