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(rev, nil) → nil
app(rev, app(app(cons, x), l)) → app(app(cons, app(app(rev1, x), l)), app(app(rev2, x), l))
app(app(rev1, 0), nil) → 0
app(app(rev1, app(s, x)), nil) → app(s, x)
app(app(rev1, x), app(app(cons, y), l)) → app(app(rev1, y), l)
app(app(rev2, x), nil) → nil
app(app(rev2, x), app(app(cons, y), l)) → app(rev, app(app(cons, x), app(app(rev2, y), l)))
app(app(map, f), nil) → nil
app(app(map, f), app(app(cons, x), xs)) → app(app(cons, app(f, x)), app(app(map, f), xs))
app(app(filter, f), nil) → nil
app(app(filter, f), app(app(cons, x), xs)) → app(app(app(app(filter2, app(f, x)), f), x), xs)
app(app(app(app(filter2, true), f), x), xs) → app(app(cons, x), app(app(filter, f), xs))
app(app(app(app(filter2, false), f), x), xs) → app(app(filter, f), xs)

Q is empty.

(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(rev, nil) → nil
app(rev, app(app(cons, x), l)) → app(app(cons, app(app(rev1, x), l)), app(app(rev2, x), l))
app(app(rev1, 0), nil) → 0
app(app(rev1, app(s, x)), nil) → app(s, x)
app(app(rev1, x), app(app(cons, y), l)) → app(app(rev1, y), l)
app(app(rev2, x), nil) → nil
app(app(rev2, x), app(app(cons, y), l)) → app(rev, app(app(cons, x), app(app(rev2, y), l)))
app(app(map, f), nil) → nil
app(app(map, f), app(app(cons, x), xs)) → app(app(cons, app(f, x)), app(app(map, f), xs))
app(app(filter, f), nil) → nil
app(app(filter, f), app(app(cons, x), xs)) → app(app(app(app(filter2, app(f, x)), f), x), xs)
app(app(app(app(filter2, true), f), x), xs) → app(app(cons, x), app(app(filter, f), xs))
app(app(app(app(filter2, false), f), x), xs) → app(app(filter, f), xs)

The set Q consists of the following terms:

app(rev, nil)
app(rev, app(app(cons, x0), x1))
app(app(rev1, 0), nil)
app(app(rev1, app(s, x0)), nil)
app(app(rev1, x0), app(app(cons, x1), x2))
app(app(rev2, x0), nil)
app(app(rev2, x0), app(app(cons, x1), x2))
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(rev, app(app(cons, x), l)) → APP(app(cons, app(app(rev1, x), l)), app(app(rev2, x), l))
APP(rev, app(app(cons, x), l)) → APP(cons, app(app(rev1, x), l))
APP(rev, app(app(cons, x), l)) → APP(app(rev1, x), l)
APP(rev, app(app(cons, x), l)) → APP(rev1, x)
APP(rev, app(app(cons, x), l)) → APP(app(rev2, x), l)
APP(rev, app(app(cons, x), l)) → APP(rev2, x)
APP(app(rev1, x), app(app(cons, y), l)) → APP(app(rev1, y), l)
APP(app(rev1, x), app(app(cons, y), l)) → APP(rev1, y)
APP(app(rev2, x), app(app(cons, y), l)) → APP(rev, app(app(cons, x), app(app(rev2, y), l)))
APP(app(rev2, x), app(app(cons, y), l)) → APP(app(cons, x), app(app(rev2, y), l))
APP(app(rev2, x), app(app(cons, y), l)) → APP(cons, x)
APP(app(rev2, x), app(app(cons, y), l)) → APP(app(rev2, y), l)
APP(app(rev2, x), app(app(cons, y), l)) → APP(rev2, y)
APP(app(map, f), app(app(cons, x), xs)) → APP(app(cons, app(f, x)), app(app(map, f), xs))
APP(app(map, f), app(app(cons, x), xs)) → APP(cons, app(f, x))
APP(app(map, f), app(app(cons, x), xs)) → APP(f, x)
APP(app(map, f), app(app(cons, x), xs)) → APP(app(map, f), xs)
APP(app(filter, f), app(app(cons, x), xs)) → APP(app(app(app(filter2, app(f, x)), f), x), xs)
APP(app(filter, f), app(app(cons, x), xs)) → APP(app(app(filter2, app(f, x)), f), x)
APP(app(filter, f), app(app(cons, x), xs)) → APP(app(filter2, app(f, x)), f)
APP(app(filter, f), app(app(cons, x), xs)) → APP(filter2, app(f, x))
APP(app(filter, f), app(app(cons, x), xs)) → APP(f, x)
APP(app(app(app(filter2, true), f), x), xs) → APP(app(cons, x), app(app(filter, f), xs))
APP(app(app(app(filter2, true), f), x), xs) → APP(cons, x)
APP(app(app(app(filter2, true), f), x), xs) → APP(app(filter, f), xs)
APP(app(app(app(filter2, true), f), x), xs) → APP(filter, f)
APP(app(app(app(filter2, false), f), x), xs) → APP(app(filter, f), xs)
APP(app(app(app(filter2, false), f), x), xs) → APP(filter, f)

The TRS R consists of the following rules:

app(rev, nil) → nil
app(rev, app(app(cons, x), l)) → app(app(cons, app(app(rev1, x), l)), app(app(rev2, x), l))
app(app(rev1, 0), nil) → 0
app(app(rev1, app(s, x)), nil) → app(s, x)
app(app(rev1, x), app(app(cons, y), l)) → app(app(rev1, y), l)
app(app(rev2, x), nil) → nil
app(app(rev2, x), app(app(cons, y), l)) → app(rev, app(app(cons, x), app(app(rev2, y), l)))
app(app(map, f), nil) → nil
app(app(map, f), app(app(cons, x), xs)) → app(app(cons, app(f, x)), app(app(map, f), xs))
app(app(filter, f), nil) → nil
app(app(filter, f), app(app(cons, x), xs)) → app(app(app(app(filter2, app(f, x)), f), x), xs)
app(app(app(app(filter2, true), f), x), xs) → app(app(cons, x), app(app(filter, f), xs))
app(app(app(app(filter2, false), f), x), xs) → app(app(filter, f), xs)

The set Q consists of the following terms:

app(rev, nil)
app(rev, app(app(cons, x0), x1))
app(app(rev1, 0), nil)
app(app(rev1, app(s, x0)), nil)
app(app(rev1, x0), app(app(cons, x1), x2))
app(app(rev2, x0), nil)
app(app(rev2, x0), app(app(cons, x1), x2))
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 3 SCCs with 19 less nodes.

(6) Complex Obligation (AND)

(7) Obligation:

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

APP(app(rev1, x), app(app(cons, y), l)) → APP(app(rev1, y), l)

The TRS R consists of the following rules:

app(rev, nil) → nil
app(rev, app(app(cons, x), l)) → app(app(cons, app(app(rev1, x), l)), app(app(rev2, x), l))
app(app(rev1, 0), nil) → 0
app(app(rev1, app(s, x)), nil) → app(s, x)
app(app(rev1, x), app(app(cons, y), l)) → app(app(rev1, y), l)
app(app(rev2, x), nil) → nil
app(app(rev2, x), app(app(cons, y), l)) → app(rev, app(app(cons, x), app(app(rev2, y), l)))
app(app(map, f), nil) → nil
app(app(map, f), app(app(cons, x), xs)) → app(app(cons, app(f, x)), app(app(map, f), xs))
app(app(filter, f), nil) → nil
app(app(filter, f), app(app(cons, x), xs)) → app(app(app(app(filter2, app(f, x)), f), x), xs)
app(app(app(app(filter2, true), f), x), xs) → app(app(cons, x), app(app(filter, f), xs))
app(app(app(app(filter2, false), f), x), xs) → app(app(filter, f), xs)

The set Q consists of the following terms:

app(rev, nil)
app(rev, app(app(cons, x0), x1))
app(app(rev1, 0), nil)
app(app(rev1, app(s, x0)), nil)
app(app(rev1, x0), app(app(cons, x1), x2))
app(app(rev2, x0), nil)
app(app(rev2, x0), app(app(cons, x1), x2))
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(rev, app(app(cons, x), l)) → APP(app(rev2, x), l)
APP(app(rev2, x), app(app(cons, y), l)) → APP(rev, app(app(cons, x), app(app(rev2, y), l)))
APP(app(rev2, x), app(app(cons, y), l)) → APP(app(rev2, y), l)

The TRS R consists of the following rules:

app(rev, nil) → nil
app(rev, app(app(cons, x), l)) → app(app(cons, app(app(rev1, x), l)), app(app(rev2, x), l))
app(app(rev1, 0), nil) → 0
app(app(rev1, app(s, x)), nil) → app(s, x)
app(app(rev1, x), app(app(cons, y), l)) → app(app(rev1, y), l)
app(app(rev2, x), nil) → nil
app(app(rev2, x), app(app(cons, y), l)) → app(rev, app(app(cons, x), app(app(rev2, y), l)))
app(app(map, f), nil) → nil
app(app(map, f), app(app(cons, x), xs)) → app(app(cons, app(f, x)), app(app(map, f), xs))
app(app(filter, f), nil) → nil
app(app(filter, f), app(app(cons, x), xs)) → app(app(app(app(filter2, app(f, x)), f), x), xs)
app(app(app(app(filter2, true), f), x), xs) → app(app(cons, x), app(app(filter, f), xs))
app(app(app(app(filter2, false), f), x), xs) → app(app(filter, f), xs)

The set Q consists of the following terms:

app(rev, nil)
app(rev, app(app(cons, x0), x1))
app(app(rev1, 0), nil)
app(app(rev1, app(s, x0)), nil)
app(app(rev1, x0), app(app(cons, x1), x2))
app(app(rev2, x0), nil)
app(app(rev2, x0), app(app(cons, x1), x2))
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.

(9) Obligation:

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

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

The TRS R consists of the following rules:

app(rev, nil) → nil
app(rev, app(app(cons, x), l)) → app(app(cons, app(app(rev1, x), l)), app(app(rev2, x), l))
app(app(rev1, 0), nil) → 0
app(app(rev1, app(s, x)), nil) → app(s, x)
app(app(rev1, x), app(app(cons, y), l)) → app(app(rev1, y), l)
app(app(rev2, x), nil) → nil
app(app(rev2, x), app(app(cons, y), l)) → app(rev, app(app(cons, x), app(app(rev2, y), l)))
app(app(map, f), nil) → nil
app(app(map, f), app(app(cons, x), xs)) → app(app(cons, app(f, x)), app(app(map, f), xs))
app(app(filter, f), nil) → nil
app(app(filter, f), app(app(cons, x), xs)) → app(app(app(app(filter2, app(f, x)), f), x), xs)
app(app(app(app(filter2, true), f), x), xs) → app(app(cons, x), app(app(filter, f), xs))
app(app(app(app(filter2, false), f), x), xs) → app(app(filter, f), xs)

The set Q consists of the following terms:

app(rev, nil)
app(rev, app(app(cons, x0), x1))
app(app(rev1, 0), nil)
app(app(rev1, app(s, x0)), nil)
app(app(rev1, x0), app(app(cons, x1), x2))
app(app(rev2, x0), nil)
app(app(rev2, x0), app(app(cons, x1), x2))
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.