Termination w.r.t. Q of the given QTRS could be proven:

(0) Obligation:

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

app(f, app(app(cons, nil), y)) → y
app(f, app(app(cons, app(f, app(app(cons, nil), y))), z)) → app(app(app(copy, n), y), z)
app(app(app(copy, 0), y), z) → app(f, z)
app(app(app(copy, app(s, x)), y), z) → app(app(app(copy, x), y), app(app(cons, app(f, y)), z))
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(app(cons, app(f, app(app(cons, nil), y))), z)) → APP(app(app(copy, n), y), z)
APP(f, app(app(cons, app(f, app(app(cons, nil), y))), z)) → APP(app(copy, n), y)
APP(f, app(app(cons, app(f, app(app(cons, nil), y))), z)) → APP(copy, n)
APP(app(app(copy, 0), y), z) → APP(f, z)
APP(app(app(copy, app(s, x)), y), z) → APP(app(app(copy, x), y), app(app(cons, app(f, y)), z))
APP(app(app(copy, app(s, x)), y), z) → APP(app(copy, x), y)
APP(app(app(copy, app(s, x)), y), z) → APP(copy, x)
APP(app(app(copy, app(s, x)), y), z) → APP(app(cons, app(f, y)), z)
APP(app(app(copy, app(s, x)), y), z) → APP(cons, app(f, y))
APP(app(app(copy, app(s, x)), y), z) → APP(f, y)
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, app(app(cons, nil), y)) → y
app(f, app(app(cons, app(f, app(app(cons, nil), y))), z)) → app(app(app(copy, n), y), z)
app(app(app(copy, 0), y), z) → app(f, z)
app(app(app(copy, app(s, x)), y), z) → app(app(app(copy, x), y), app(app(cons, app(f, y)), z))
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 2 SCCs with 18 less nodes.

(4) Complex Obligation (AND)

(5) Obligation:

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

APP(app(app(copy, app(s, x)), y), z) → APP(app(app(copy, x), y), app(app(cons, app(f, y)), z))

The TRS R consists of the following rules:

app(f, app(app(cons, nil), y)) → y
app(f, app(app(cons, app(f, app(app(cons, nil), y))), z)) → app(app(app(copy, n), y), z)
app(app(app(copy, 0), y), z) → app(f, z)
app(app(app(copy, app(s, x)), y), z) → app(app(app(copy, x), y), app(app(cons, app(f, y)), z))
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) QDPOrderProof (EQUIVALENT transformation)

We use the reduction pair processor [LPAR04].

The following pairs can be oriented strictly and are deleted.

APP(app(app(copy, app(s, x)), y), z) → APP(app(app(copy, x), y), app(app(cons, app(f, y)), z))

The remaining pairs can at least be oriented weakly.
Used ordering: Lexicographic path order with status [LPO].
Quasi-Precedence:

APP₂ > [copy, f, n] > [app₂, cons]
s > [copy, f, n] > [app₂, cons]
nil > [app₂, cons]

Status:

APP₂: [1,2]
app₂: [2,1]
copy: []
s: []
cons: []
f: []
nil: []
n: []

The following usable rules [FROCOS05] were oriented:

app(f, app(app(cons, nil), y)) → y
app(f, app(app(cons, app(f, app(app(cons, nil), y))), z)) → app(app(app(copy, n), y), z)

(7) Obligation:

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

app(f, app(app(cons, nil), y)) → y
app(f, app(app(cons, app(f, app(app(cons, nil), y))), z)) → app(app(app(copy, n), y), z)
app(app(app(copy, 0), y), z) → app(f, z)
app(app(app(copy, app(s, x)), y), z) → app(app(app(copy, x), y), app(app(cons, app(f, y)), z))
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.

(8) PisEmptyProof (EQUIVALENT transformation)

The TRS P is empty. Hence, there is no (P,Q,R) chain.

(9) TRUE

(10) 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, app(app(cons, nil), y)) → y
app(f, app(app(cons, app(f, app(app(cons, nil), y))), z)) → app(app(app(copy, n), y), z)
app(app(app(copy, 0), y), z) → app(f, z)
app(app(app(copy, app(s, x)), y), z) → app(app(app(copy, x), y), app(app(cons, app(f, y)), z))
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.

(11) QDPOrderProof (EQUIVALENT transformation)

We use the reduction pair processor [LPAR04].

The following pairs can be oriented strictly and are deleted.

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

The remaining pairs can at least be oriented weakly.
Used ordering: Combined order from the following AFS and order.
APP(x1, x2)  =  APP(x2)
app(x1, x2)  =  app(x1, x2)
map  =  map
cons  =  cons
filter  =  filter
filter2  =  filter2
true  =  true
false  =  false
f  =  f
nil  =  nil
copy  =  copy
n  =  n
0  =  0
s  =  s

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

map > cons > APP₁ > filter > [app₂, filter2, true, s]
map > cons > copy > [app₂, filter2, true, s]
false > APP₁ > filter > [app₂, filter2, true, s]
f > copy > [app₂, filter2, true, s]
f > n > [app₂, filter2, true, s]
nil > copy > [app₂, filter2, true, s]
nil > n > [app₂, filter2, true, s]
0 > [app₂, filter2, true, s]

Status:

APP₁: [1]
app₂: [2,1]
map: []
cons: []
filter: []
filter2: []
true: []
false: []
f: []
nil: []
copy: []
n: []
0: []
s: []

The following usable rules [FROCOS05] were oriented: none

(12) Obligation:

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(f, app(app(cons, nil), y)) → y
app(f, app(app(cons, app(f, app(app(cons, nil), y))), z)) → app(app(app(copy, n), y), z)
app(app(app(copy, 0), y), z) → app(f, z)
app(app(app(copy, app(s, x)), y), z) → app(app(app(copy, x), y), app(app(cons, app(f, y)), z))
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.

(13) DependencyGraphProof (EQUIVALENT transformation)

The approximation of the Dependency Graph [LPAR04,FROCOS05,EDGSTAR] contains 0 SCCs with 2 less nodes.

(14) TRUE