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(app(., 1), x) → x
app(app(., x), 1) → x
app(app(., app(i, x)), x) → 1
app(app(., x), app(i, x)) → 1
app(app(., app(i, y)), app(app(., y), z)) → z
app(app(., y), app(app(., app(i, y)), z)) → z
app(app(., app(app(., x), y)), z) → app(app(., x), app(app(., y), z))
app(i, 1) → 1
app(i, app(i, x)) → x
app(i, app(app(., x), y)) → app(app(., app(i, y)), app(i, x))
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) 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(app(., app(app(., x), y)), z) → APP(app(., x), app(app(., y), z))
APP(app(., app(app(., x), y)), z) → APP(app(., y), z)
APP(app(., app(app(., x), y)), z) → APP(., y)
APP(i, app(app(., x), y)) → APP(app(., app(i, y)), app(i, x))
APP(i, app(app(., x), y)) → APP(., app(i, y))
APP(i, app(app(., x), y)) → APP(i, y)
APP(i, app(app(., x), y)) → APP(i, x)
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(app(., 1), x) → x
app(app(., x), 1) → x
app(app(., app(i, x)), x) → 1
app(app(., x), app(i, x)) → 1
app(app(., app(i, y)), app(app(., y), z)) → z
app(app(., y), app(app(., app(i, y)), z)) → z
app(app(., app(app(., x), y)), z) → app(app(., x), app(app(., y), z))
app(i, 1) → 1
app(i, app(i, x)) → x
app(i, app(app(., x), y)) → app(app(., app(i, y)), app(i, x))
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.
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 12 less nodes.

(4) Complex Obligation (AND)

(5) Obligation:

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

APP(app(., app(app(., x), y)), z) → APP(app(., y), z)
APP(app(., app(app(., x), y)), z) → APP(app(., x), app(app(., y), z))

The TRS R consists of the following rules:

app(app(., 1), x) → x
app(app(., x), 1) → x
app(app(., app(i, x)), x) → 1
app(app(., x), app(i, x)) → 1
app(app(., app(i, y)), app(app(., y), z)) → z
app(app(., y), app(app(., app(i, y)), z)) → z
app(app(., app(app(., x), y)), z) → app(app(., x), app(app(., y), z))
app(i, 1) → 1
app(i, app(i, x)) → x
app(i, app(app(., x), y)) → app(app(., app(i, y)), app(i, x))
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.
We have to consider all minimal (P,Q,R)-chains.

(6) Obligation:

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

APP(i, app(app(., x), y)) → APP(i, x)
APP(i, app(app(., x), y)) → APP(i, y)

The TRS R consists of the following rules:

app(app(., 1), x) → x
app(app(., x), 1) → x
app(app(., app(i, x)), x) → 1
app(app(., x), app(i, x)) → 1
app(app(., app(i, y)), app(app(., y), z)) → z
app(app(., y), app(app(., app(i, y)), z)) → z
app(app(., app(app(., x), y)), z) → app(app(., x), app(app(., y), z))
app(i, 1) → 1
app(i, app(i, x)) → x
app(i, app(app(., x), y)) → app(app(., app(i, y)), app(i, x))
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.
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, 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(app(app(app(filter2, app(f, x)), f), x), xs)
APP(app(app(app(filter2, true), f), x), xs) → APP(app(filter, f), xs)
APP(app(filter, f), app(app(cons, x), xs)) → APP(f, x)
APP(app(app(app(filter2, false), f), x), xs) → APP(app(filter, f), xs)

The TRS R consists of the following rules:

app(app(., 1), x) → x
app(app(., x), 1) → x
app(app(., app(i, x)), x) → 1
app(app(., x), app(i, x)) → 1
app(app(., app(i, y)), app(app(., y), z)) → z
app(app(., y), app(app(., app(i, y)), z)) → z
app(app(., app(app(., x), y)), z) → app(app(., x), app(app(., y), z))
app(i, 1) → 1
app(i, app(i, x)) → x
app(i, app(app(., x), y)) → app(app(., app(i, y)), app(i, x))
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.
We have to consider all minimal (P,Q,R)-chains.