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(f, 0) → true
app(f, 1) → false
app(f, app(s, x)) → app(f, x)
app(app(app(if, true), app(s, x)), app(s, y)) → app(s, x)
app(app(app(if, false), app(s, x)), app(s, y)) → app(s, y)
app(app(g, x), app(c, y)) → app(c, app(app(g, x), y))
app(app(g, x), app(c, y)) → app(app(g, x), app(app(app(if, app(f, x)), app(c, app(app(g, app(s, x)), y))), app(c, y)))
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) DependencyPairsProof (EQUIVALENT transformation)

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

(2) Obligation:

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

APP(f, app(s, x)) → APP(f, x)
APP(app(g, x), app(c, y)) → APP(c, app(app(g, x), y))
APP(app(g, x), app(c, y)) → APP(app(g, x), y)
APP(app(g, x), app(c, y)) → APP(app(g, x), app(app(app(if, app(f, x)), app(c, app(app(g, app(s, x)), y))), app(c, y)))
APP(app(g, x), app(c, y)) → APP(app(app(if, app(f, x)), app(c, app(app(g, app(s, x)), y))), app(c, y))
APP(app(g, x), app(c, y)) → APP(app(if, app(f, x)), app(c, app(app(g, app(s, x)), y)))
APP(app(g, x), app(c, y)) → APP(if, app(f, x))
APP(app(g, x), app(c, y)) → APP(f, x)
APP(app(g, x), app(c, y)) → APP(c, app(app(g, app(s, x)), y))
APP(app(g, x), app(c, y)) → APP(app(g, app(s, x)), y)
APP(app(g, x), app(c, y)) → APP(g, app(s, x))
APP(app(g, x), app(c, y)) → APP(s, 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(f, 0) → true
app(f, 1) → false
app(f, app(s, x)) → app(f, x)
app(app(app(if, true), app(s, x)), app(s, y)) → app(s, x)
app(app(app(if, false), app(s, x)), app(s, y)) → app(s, y)
app(app(g, x), app(c, y)) → app(c, app(app(g, x), y))
app(app(g, x), app(c, y)) → app(app(g, x), app(app(app(if, app(f, x)), app(c, app(app(g, app(s, x)), y))), app(c, y)))
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.
We have to consider all minimal (P,Q,R)-chains.

(3) DependencyGraphProof (EQUIVALENT transformation)

The approximation of the Dependency Graph [LPAR04,FROCOS05,EDGSTAR] contains 3 SCCs with 18 less nodes.

(4) Complex Obligation (AND)

(5) Obligation:

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

APP(f, app(s, x)) → APP(f, x)

The TRS R consists of the following rules:

app(f, 0) → true
app(f, 1) → false
app(f, app(s, x)) → app(f, x)
app(app(app(if, true), app(s, x)), app(s, y)) → app(s, x)
app(app(app(if, false), app(s, x)), app(s, y)) → app(s, y)
app(app(g, x), app(c, y)) → app(c, app(app(g, x), y))
app(app(g, x), app(c, y)) → app(app(g, x), app(app(app(if, app(f, x)), app(c, app(app(g, app(s, x)), y))), app(c, y)))
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.
We have to consider all minimal (P,Q,R)-chains.

(6) Obligation:

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

APP(app(g, x), app(c, y)) → APP(app(g, app(s, x)), y)
APP(app(g, x), app(c, y)) → APP(app(g, x), y)

The TRS R consists of the following rules:

app(f, 0) → true
app(f, 1) → false
app(f, app(s, x)) → app(f, x)
app(app(app(if, true), app(s, x)), app(s, y)) → app(s, x)
app(app(app(if, false), app(s, x)), app(s, y)) → app(s, y)
app(app(g, x), app(c, y)) → app(c, app(app(g, x), y))
app(app(g, x), app(c, y)) → app(app(g, x), app(app(app(if, app(f, x)), app(c, app(app(g, app(s, x)), y))), app(c, y)))
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.
We have to consider all minimal (P,Q,R)-chains.

(7) 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(f, 0) → true
app(f, 1) → false
app(f, app(s, x)) → app(f, x)
app(app(app(if, true), app(s, x)), app(s, y)) → app(s, x)
app(app(app(if, false), app(s, x)), app(s, y)) → app(s, y)
app(app(g, x), app(c, y)) → app(c, app(app(g, x), y))
app(app(g, x), app(c, y)) → app(app(g, x), app(app(app(if, app(f, x)), app(c, app(app(g, app(s, x)), y))), app(c, y)))
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.
We have to consider all minimal (P,Q,R)-chains.