Termination w.r.t. Q of the following Term Rewriting System could be proven:

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.


QTRS
  ↳ DependencyPairsProof

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.

Using Dependency Pairs [1,13] we result in the following initial DP problem:
Q DP problem:
The TRS P consists of the following rules:

APP(app(app(copy, app(s, x)), y), z) → APP(app(cons, app(f, y)), z)
APP(app(map, fun), app(app(cons, x), xs)) → APP(app(cons, app(fun, x)), app(app(map, fun), xs))
APP(f, app(app(cons, app(f, app(app(cons, nil), y))), z)) → APP(app(app(copy, n), 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(copy, x)
APP(f, app(app(cons, app(f, app(app(cons, nil), y))), z)) → APP(app(copy, n), y)
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(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(f, app(app(cons, app(f, app(app(cons, nil), y))), z)) → APP(copy, n)
APP(app(map, fun), app(app(cons, x), xs)) → APP(fun, x)
APP(app(filter, fun), app(app(cons, x), xs)) → APP(filter2, app(fun, x))
APP(app(app(copy, app(s, x)), y), z) → APP(f, y)
APP(app(app(app(filter2, true), fun), x), xs) → APP(app(filter, fun), xs)
APP(app(filter, fun), app(app(cons, x), xs)) → APP(app(filter2, app(fun, x)), fun)
APP(app(app(copy, app(s, x)), y), z) → APP(app(copy, x), y)
APP(app(app(app(filter2, false), fun), x), xs) → APP(filter, fun)
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), app(app(cons, x), xs)) → APP(app(map, fun), xs)
APP(app(filter, fun), app(app(cons, x), xs)) → APP(fun, x)
APP(app(app(app(filter2, true), fun), x), xs) → APP(filter, fun)
APP(app(app(copy, 0), y), z) → APP(f, z)
APP(app(app(app(filter2, true), fun), x), xs) → APP(cons, x)
APP(app(app(app(filter2, false), fun), x), xs) → APP(app(filter, fun), xs)
APP(app(map, fun), app(app(cons, x), xs)) → APP(cons, app(fun, x))

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.

↳ QTRS
  ↳ DependencyPairsProof
QDP
      ↳ EdgeDeletionProof

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

APP(app(app(copy, app(s, x)), y), z) → APP(app(cons, app(f, y)), z)
APP(app(map, fun), app(app(cons, x), xs)) → APP(app(cons, app(fun, x)), app(app(map, fun), xs))
APP(f, app(app(cons, app(f, app(app(cons, nil), y))), z)) → APP(app(app(copy, n), 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(copy, x)
APP(f, app(app(cons, app(f, app(app(cons, nil), y))), z)) → APP(app(copy, n), y)
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(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(f, app(app(cons, app(f, app(app(cons, nil), y))), z)) → APP(copy, n)
APP(app(map, fun), app(app(cons, x), xs)) → APP(fun, x)
APP(app(filter, fun), app(app(cons, x), xs)) → APP(filter2, app(fun, x))
APP(app(app(copy, app(s, x)), y), z) → APP(f, y)
APP(app(app(app(filter2, true), fun), x), xs) → APP(app(filter, fun), xs)
APP(app(filter, fun), app(app(cons, x), xs)) → APP(app(filter2, app(fun, x)), fun)
APP(app(app(copy, app(s, x)), y), z) → APP(app(copy, x), y)
APP(app(app(app(filter2, false), fun), x), xs) → APP(filter, fun)
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), app(app(cons, x), xs)) → APP(app(map, fun), xs)
APP(app(filter, fun), app(app(cons, x), xs)) → APP(fun, x)
APP(app(app(app(filter2, true), fun), x), xs) → APP(filter, fun)
APP(app(app(copy, 0), y), z) → APP(f, z)
APP(app(app(app(filter2, true), fun), x), xs) → APP(cons, x)
APP(app(app(app(filter2, false), fun), x), xs) → APP(app(filter, fun), xs)
APP(app(map, fun), app(app(cons, x), xs)) → APP(cons, app(fun, x))

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.
We deleted some edges using various graph approximations

↳ QTRS
  ↳ DependencyPairsProof
    ↳ QDP
      ↳ EdgeDeletionProof
QDP
          ↳ DependencyGraphProof

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

APP(app(app(copy, app(s, x)), y), z) → APP(app(cons, app(f, y)), z)
APP(app(map, fun), app(app(cons, x), xs)) → APP(app(cons, app(fun, x)), app(app(map, fun), xs))
APP(f, app(app(cons, app(f, app(app(cons, nil), y))), z)) → APP(app(app(copy, n), y), z)
APP(app(app(copy, app(s, x)), y), z) → APP(cons, app(f, y))
APP(app(filter, fun), app(app(cons, x), xs)) → APP(app(app(filter2, app(fun, x)), fun), x)
APP(f, app(app(cons, app(f, app(app(cons, nil), y))), z)) → APP(app(copy, n), y)
APP(app(app(copy, app(s, x)), y), z) → APP(copy, 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(cons, x), app(app(filter, fun), xs))
APP(f, app(app(cons, app(f, app(app(cons, nil), y))), z)) → APP(copy, n)
APP(app(filter, fun), app(app(cons, x), xs)) → APP(filter2, app(fun, x))
APP(app(map, fun), app(app(cons, x), xs)) → APP(fun, x)
APP(app(app(copy, app(s, x)), y), z) → APP(f, y)
APP(app(app(app(filter2, true), fun), x), xs) → APP(app(filter, fun), xs)
APP(app(filter, fun), app(app(cons, x), xs)) → APP(app(filter2, app(fun, x)), fun)
APP(app(app(copy, app(s, x)), y), z) → APP(app(copy, x), y)
APP(app(app(app(filter2, false), fun), x), xs) → APP(filter, fun)
APP(app(filter, 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(app(copy, app(s, x)), y), z) → APP(app(app(copy, x), y), app(app(cons, app(f, y)), z))
APP(app(app(app(filter2, true), fun), x), xs) → APP(filter, fun)
APP(app(app(copy, 0), y), z) → APP(f, z)
APP(app(app(app(filter2, true), fun), x), xs) → APP(cons, x)
APP(app(app(app(filter2, false), fun), x), xs) → APP(app(filter, fun), xs)
APP(app(map, fun), app(app(cons, x), xs)) → APP(cons, app(fun, x))

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.
The approximation of the Dependency Graph [13,14,18] contains 2 SCCs with 18 less nodes.

↳ QTRS
  ↳ DependencyPairsProof
    ↳ QDP
      ↳ EdgeDeletionProof
        ↳ QDP
          ↳ DependencyGraphProof
            ↳ AND
QDP
                ↳ QDPOrderProof
              ↳ QDP

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.
We use the reduction pair processor [13]. Here, we combined the reduction pair processor with the A-transformation [14] which results in the following intermediate Q-DP Problem.
Q DP problem:
The TRS P consists of the following rules:

COPY(s(x), y, z) → COPY(x, y, cons(F(y), z))

The TRS R consists of the following rules:

F(cons(nil, y)) → y
F(cons(F(cons(nil, y)), z)) → copy(n, y, z)

Q is empty.
We have to consider all minimal (P,Q,R)-chains.


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.
none
Used ordering: Combined order from the following AFS and order.
COPY(x1, x2, x3)  =  COPY(x1)
s(x1)  =  s(x1)
cons(x1, x2)  =  cons(x2)
F(x1)  =  x1
nil  =  nil
copy(x1, x2, x3)  =  copy(x1, x2)
n  =  n

Lexicographic path order with status [19].
Quasi-Precedence:
s1 > [COPY1, cons1]
[nil, copy2] > [COPY1, cons1]
n > [COPY1, cons1]

Status:
copy2: [1,2]
COPY1: [1]
s1: [1]
cons1: [1]
nil: multiset
n: multiset


The following usable rules [14] were oriented: none



↳ QTRS
  ↳ DependencyPairsProof
    ↳ QDP
      ↳ EdgeDeletionProof
        ↳ QDP
          ↳ DependencyGraphProof
            ↳ AND
              ↳ QDP
                ↳ QDPOrderProof
QDP
                    ↳ PisEmptyProof
              ↳ QDP

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

↳ QTRS
  ↳ DependencyPairsProof
    ↳ QDP
      ↳ EdgeDeletionProof
        ↳ QDP
          ↳ DependencyGraphProof
            ↳ AND
              ↳ QDP
QDP
                ↳ QDPOrderProof

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(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, false), fun), x), xs) → APP(app(filter, fun), xs)
APP(app(filter, fun), app(app(cons, x), xs)) → APP(fun, x)
APP(app(map, fun), app(app(cons, x), xs)) → APP(app(map, 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.
We use the reduction pair processor [13].


The following pairs can be oriented strictly and are deleted.


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)
APP(app(map, fun), app(app(cons, x), xs)) → APP(app(map, fun), xs)
The remaining pairs can at least be oriented weakly.

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)
Used ordering: Combined order from the following AFS and order.
APP(x1, x2)  =  APP(x2)
app(x1, x2)  =  app(x1, x2)
filter2  =  filter2
true  =  true
filter  =  filter
map  =  map
cons  =  cons
false  =  false
copy  =  copy
s  =  s
f  =  f
nil  =  nil
0  =  0
n  =  n

Lexicographic path order with status [19].
Quasi-Precedence:
map > cons > [filter2, filter] > [APP1, true] > [app2, false, s] > copy > [f, 0] > [nil, n]

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


The following usable rules [14] were oriented: none



↳ QTRS
  ↳ DependencyPairsProof
    ↳ QDP
      ↳ EdgeDeletionProof
        ↳ QDP
          ↳ DependencyGraphProof
            ↳ AND
              ↳ QDP
              ↳ QDP
                ↳ QDPOrderProof
QDP
                    ↳ DependencyGraphProof

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.
The approximation of the Dependency Graph [13,14,18] contains 0 SCCs with 2 less nodes.