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(app(app(f, 0), 1), x) → app(app(app(f, app(s, x)), x), x)
app(app(app(f, x), y), app(s, z)) → app(s, app(app(app(f, 0), 1), 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
  ↳ Overlay + Local Confluence

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

app(app(app(f, 0), 1), x) → app(app(app(f, app(s, x)), x), x)
app(app(app(f, x), y), app(s, z)) → app(s, app(app(app(f, 0), 1), 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.

The TRS is overlay and locally confluent. By [15] we can switch to innermost.

↳ QTRS
  ↳ Overlay + Local Confluence
    ↳ QTRS
      ↳ DependencyPairsProof

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

app(app(app(f, 0), 1), x) → app(app(app(f, app(s, x)), x), x)
app(app(app(f, x), y), app(s, z)) → app(s, app(app(app(f, 0), 1), 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)

The set Q consists of the following terms:

app(app(app(f, 0), 1), x0)
app(app(app(f, x0), x1), app(s, 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)


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(app(filter2, true), fun), x), xs) → APP(app(filter, fun), xs)
APP(app(map, fun), app(app(cons, x), xs)) → APP(app(cons, app(fun, x)), app(app(map, fun), xs))
APP(app(filter, fun), app(app(cons, x), xs)) → APP(app(filter2, app(fun, x)), fun)
APP(app(app(app(filter2, false), fun), x), xs) → APP(filter, fun)
APP(app(app(f, 0), 1), x) → APP(f, app(s, x))
APP(app(filter, fun), app(app(cons, x), xs)) → APP(app(app(filter2, app(fun, x)), fun), x)
APP(app(app(f, x), y), app(s, z)) → APP(app(f, 0), 1)
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(app(f, x), y), app(s, z)) → APP(app(app(f, 0), 1), z)
APP(app(map, fun), app(app(cons, x), xs)) → APP(app(map, fun), xs)
APP(app(app(app(filter2, true), fun), x), xs) → APP(app(cons, x), app(app(filter, fun), xs))
APP(app(app(f, x), y), app(s, z)) → APP(s, app(app(app(f, 0), 1), z))
APP(app(app(app(filter2, true), fun), x), xs) → APP(filter, fun)
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(app(filter2, true), fun), x), xs) → APP(cons, x)
APP(app(app(f, 0), 1), x) → APP(app(app(f, app(s, x)), x), x)
APP(app(app(f, 0), 1), x) → APP(app(f, app(s, x)), x)
APP(app(app(app(filter2, false), fun), x), xs) → APP(app(filter, fun), xs)
APP(app(app(f, x), y), app(s, z)) → APP(f, 0)
APP(app(map, fun), app(app(cons, x), xs)) → APP(cons, app(fun, x))
APP(app(app(f, 0), 1), x) → APP(s, x)

The TRS R consists of the following rules:

app(app(app(f, 0), 1), x) → app(app(app(f, app(s, x)), x), x)
app(app(app(f, x), y), app(s, z)) → app(s, app(app(app(f, 0), 1), 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)

The set Q consists of the following terms:

app(app(app(f, 0), 1), x0)
app(app(app(f, x0), x1), app(s, 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.

↳ QTRS
  ↳ Overlay + Local Confluence
    ↳ QTRS
      ↳ DependencyPairsProof
        ↳ QDP
          ↳ EdgeDeletionProof

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(app(cons, app(fun, x)), app(app(map, fun), xs))
APP(app(filter, fun), app(app(cons, x), xs)) → APP(app(filter2, app(fun, x)), fun)
APP(app(app(app(filter2, false), fun), x), xs) → APP(filter, fun)
APP(app(app(f, 0), 1), x) → APP(f, app(s, x))
APP(app(filter, fun), app(app(cons, x), xs)) → APP(app(app(filter2, app(fun, x)), fun), x)
APP(app(app(f, x), y), app(s, z)) → APP(app(f, 0), 1)
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(app(f, x), y), app(s, z)) → APP(app(app(f, 0), 1), z)
APP(app(map, fun), app(app(cons, x), xs)) → APP(app(map, fun), xs)
APP(app(app(app(filter2, true), fun), x), xs) → APP(app(cons, x), app(app(filter, fun), xs))
APP(app(app(f, x), y), app(s, z)) → APP(s, app(app(app(f, 0), 1), z))
APP(app(app(app(filter2, true), fun), x), xs) → APP(filter, fun)
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(app(filter2, true), fun), x), xs) → APP(cons, x)
APP(app(app(f, 0), 1), x) → APP(app(app(f, app(s, x)), x), x)
APP(app(app(f, 0), 1), x) → APP(app(f, app(s, x)), x)
APP(app(app(app(filter2, false), fun), x), xs) → APP(app(filter, fun), xs)
APP(app(app(f, x), y), app(s, z)) → APP(f, 0)
APP(app(map, fun), app(app(cons, x), xs)) → APP(cons, app(fun, x))
APP(app(app(f, 0), 1), x) → APP(s, x)

The TRS R consists of the following rules:

app(app(app(f, 0), 1), x) → app(app(app(f, app(s, x)), x), x)
app(app(app(f, x), y), app(s, z)) → app(s, app(app(app(f, 0), 1), 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)

The set Q consists of the following terms:

app(app(app(f, 0), 1), x0)
app(app(app(f, x0), x1), app(s, 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.
We deleted some edges using various graph approximations

↳ QTRS
  ↳ Overlay + Local Confluence
    ↳ QTRS
      ↳ DependencyPairsProof
        ↳ QDP
          ↳ EdgeDeletionProof
            ↳ 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(map, fun), app(app(cons, x), xs)) → APP(app(cons, app(fun, x)), app(app(map, fun), xs))
APP(app(filter, fun), app(app(cons, x), xs)) → APP(app(filter2, app(fun, x)), fun)
APP(app(app(app(filter2, false), fun), x), xs) → APP(filter, fun)
APP(app(app(f, 0), 1), x) → APP(f, app(s, x))
APP(app(filter, fun), app(app(cons, x), xs)) → APP(app(app(filter2, app(fun, x)), fun), x)
APP(app(app(f, x), y), app(s, z)) → APP(app(f, 0), 1)
APP(app(filter, fun), app(app(cons, x), xs)) → APP(app(app(app(filter2, app(fun, x)), fun), x), xs)
APP(app(map, fun), app(app(cons, x), xs)) → APP(app(map, fun), xs)
APP(app(app(f, x), y), app(s, z)) → APP(app(app(f, 0), 1), z)
APP(app(filter, fun), app(app(cons, x), xs)) → APP(fun, x)
APP(app(app(f, x), y), app(s, z)) → APP(s, app(app(app(f, 0), 1), z))
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(filter, fun)
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(app(filter2, true), fun), x), xs) → APP(cons, x)
APP(app(app(f, 0), 1), x) → APP(app(app(f, app(s, x)), x), x)
APP(app(app(f, 0), 1), x) → APP(app(f, app(s, x)), x)
APP(app(app(f, x), y), app(s, z)) → APP(f, 0)
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))
APP(app(app(f, 0), 1), x) → APP(s, x)

The TRS R consists of the following rules:

app(app(app(f, 0), 1), x) → app(app(app(f, app(s, x)), x), x)
app(app(app(f, x), y), app(s, z)) → app(s, app(app(app(f, 0), 1), 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)

The set Q consists of the following terms:

app(app(app(f, 0), 1), x0)
app(app(app(f, x0), x1), app(s, 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.
The approximation of the Dependency Graph [13,14,18] contains 2 SCCs with 16 less nodes.

↳ QTRS
  ↳ Overlay + Local Confluence
    ↳ QTRS
      ↳ DependencyPairsProof
        ↳ QDP
          ↳ EdgeDeletionProof
            ↳ QDP
              ↳ DependencyGraphProof
                ↳ AND
                  ↳ QDP
                    ↳ QDPOrderProof
                  ↳ QDP

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

APP(app(app(f, 0), 1), x) → APP(app(app(f, app(s, x)), x), x)
APP(app(app(f, x), y), app(s, z)) → APP(app(app(f, 0), 1), z)

The TRS R consists of the following rules:

app(app(app(f, 0), 1), x) → app(app(app(f, app(s, x)), x), x)
app(app(app(f, x), y), app(s, z)) → app(s, app(app(app(f, 0), 1), 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)

The set Q consists of the following terms:

app(app(app(f, 0), 1), x0)
app(app(app(f, x0), x1), app(s, 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.
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:

F(x, y, s(z)) → F(0, 1, z)
F(0, 1, x) → F(s(x), x, x)

R is empty.
The set Q consists of the following terms:

f(0, 1, x0)
f(x0, x1, s(x2))
map(x0, nil)
map(x0, cons(x1, x2))
filter(x0, nil)
filter(x0, cons(x1, x2))
filter2(true, x0, x1, x2)
filter2(false, x0, x1, x2)

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


The following pairs can be oriented strictly and are deleted.


APP(app(app(f, x), y), app(s, z)) → APP(app(app(f, 0), 1), z)
The remaining pairs can at least be oriented weakly.

APP(app(app(f, 0), 1), x) → APP(app(app(f, app(s, x)), x), x)
Used ordering: Combined order from the following AFS and order.
F(x1, x2, x3)  =  F(x3)
s(x1)  =  s(x1)
0  =  0
1  =  1

Lexicographic path order with status [19].
Quasi-Precedence:
F1 > 0 > [s1, 1]

Status:
1: multiset
0: multiset
s1: [1]
F1: [1]


The following usable rules [14] were oriented: none



↳ QTRS
  ↳ Overlay + Local Confluence
    ↳ QTRS
      ↳ DependencyPairsProof
        ↳ QDP
          ↳ EdgeDeletionProof
            ↳ QDP
              ↳ DependencyGraphProof
                ↳ AND
                  ↳ QDP
                    ↳ QDPOrderProof
                      ↳ QDP
                        ↳ DependencyGraphProof
                  ↳ QDP

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

APP(app(app(f, 0), 1), x) → APP(app(app(f, app(s, x)), x), x)

The TRS R consists of the following rules:

app(app(app(f, 0), 1), x) → app(app(app(f, app(s, x)), x), x)
app(app(app(f, x), y), app(s, z)) → app(s, app(app(app(f, 0), 1), 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)

The set Q consists of the following terms:

app(app(app(f, 0), 1), x0)
app(app(app(f, x0), x1), app(s, 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.
The approximation of the Dependency Graph [13,14,18] contains 0 SCCs with 1 less node.

↳ QTRS
  ↳ Overlay + Local Confluence
    ↳ 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(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(app(app(f, 0), 1), x) → app(app(app(f, app(s, x)), x), x)
app(app(app(f, x), y), app(s, z)) → app(s, app(app(app(f, 0), 1), 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)

The set Q consists of the following terms:

app(app(app(f, 0), 1), x0)
app(app(app(f, x0), x1), app(s, 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.
We use the reduction pair processor [13].


The following pairs can be oriented strictly and are deleted.


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(app(app(filter2, false), fun), x), xs) → APP(app(filter, fun), xs)
APP(app(filter, fun), app(app(cons, x), xs)) → APP(fun, x)
The remaining pairs can at least be oriented weakly.

APP(app(map, fun), app(app(cons, x), xs)) → APP(app(map, fun), xs)
Used ordering: Combined order from the following AFS and order.
APP(x1, x2)  =  APP(x1)
app(x1, x2)  =  app(x1, x2)
filter2  =  filter2
true  =  true
filter  =  filter
map  =  map
cons  =  cons
false  =  false

Lexicographic path order with status [19].
Quasi-Precedence:
true > filter > [APP1, app2, filter2, cons]
map > [APP1, app2, filter2, cons]
false > filter > [APP1, app2, filter2, cons]

Status:
filter: multiset
true: multiset
APP1: [1]
map: multiset
false: multiset
app2: [2,1]
filter2: multiset
cons: multiset


The following usable rules [14] were oriented: none



↳ QTRS
  ↳ Overlay + Local Confluence
    ↳ QTRS
      ↳ DependencyPairsProof
        ↳ QDP
          ↳ EdgeDeletionProof
            ↳ QDP
              ↳ DependencyGraphProof
                ↳ AND
                  ↳ QDP
                  ↳ QDP
                    ↳ QDPOrderProof
                      ↳ QDP
                        ↳ QDPOrderProof

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)

The TRS R consists of the following rules:

app(app(app(f, 0), 1), x) → app(app(app(f, app(s, x)), x), x)
app(app(app(f, x), y), app(s, z)) → app(s, app(app(app(f, 0), 1), 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)

The set Q consists of the following terms:

app(app(app(f, 0), 1), x0)
app(app(app(f, x0), x1), app(s, 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.
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:

MAP(fun, cons(x, xs)) → MAP(fun, xs)

R is empty.
The set Q consists of the following terms:

f(0, 1, x0)
f(x0, x1, s(x2))
map(x0, nil)
map(x0, cons(x1, x2))
filter(x0, nil)
filter(x0, cons(x1, x2))
filter2(true, x0, x1, x2)
filter2(false, x0, x1, x2)

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


The following pairs can be oriented strictly and are deleted.


APP(app(map, fun), app(app(cons, x), xs)) → APP(app(map, fun), xs)
The remaining pairs can at least be oriented weakly.
none
Used ordering: Combined order from the following AFS and order.
MAP(x1, x2)  =  MAP(x2)
cons(x1, x2)  =  cons(x1, x2)

Lexicographic path order with status [19].
Quasi-Precedence:
cons2 > MAP1

Status:
MAP1: [1]
cons2: [1,2]


The following usable rules [14] were oriented: none



↳ QTRS
  ↳ Overlay + Local Confluence
    ↳ QTRS
      ↳ DependencyPairsProof
        ↳ QDP
          ↳ EdgeDeletionProof
            ↳ QDP
              ↳ DependencyGraphProof
                ↳ AND
                  ↳ QDP
                  ↳ QDP
                    ↳ QDPOrderProof
                      ↳ QDP
                        ↳ QDPOrderProof
                          ↳ QDP
                            ↳ PisEmptyProof

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

app(app(app(f, 0), 1), x) → app(app(app(f, app(s, x)), x), x)
app(app(app(f, x), y), app(s, z)) → app(s, app(app(app(f, 0), 1), 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)

The set Q consists of the following terms:

app(app(app(f, 0), 1), x0)
app(app(app(f, x0), x1), app(s, 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.
The TRS P is empty. Hence, there is no (P,Q,R) chain.