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(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.


↳ QTRS
  ↳ Overlay + Local Confluence

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.

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(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)


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

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.

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

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

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

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.
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(map, f), app(app(cons, x), xs)) → APP(app(cons, app(f, x)), app(app(map, f), xs))
APP(rev, app(app(cons, x), l)) → APP(app(rev1, x), l)
APP(rev, app(app(cons, x), l)) → APP(rev2, x)
APP(rev, app(app(cons, x), l)) → APP(app(rev2, x), l)
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(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(filter, f), app(app(cons, x), xs)) → APP(filter2, app(f, x))
APP(app(map, f), app(app(cons, x), xs)) → APP(f, x)
APP(app(rev2, x), app(app(cons, y), l)) → APP(cons, x)
APP(app(rev2, x), app(app(cons, y), l)) → APP(rev, app(app(cons, x), app(app(rev2, y), l)))
APP(app(app(app(filter2, true), f), x), xs) → APP(app(filter, f), xs)
APP(rev, app(app(cons, x), l)) → APP(app(cons, app(app(rev1, x), l)), app(app(rev2, x), l))
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(filter, f), app(app(cons, x), xs)) → APP(app(filter2, app(f, x)), f)
APP(app(app(app(filter2, false), f), x), xs) → APP(filter, f)
APP(app(filter, 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(rev2, x), app(app(cons, y), l)) → APP(app(rev2, y), l)
APP(app(app(app(filter2, true), f), x), xs) → APP(filter, f)
APP(rev, app(app(cons, x), l)) → APP(rev1, x)
APP(app(rev2, x), app(app(cons, y), l)) → APP(rev2, y)
APP(app(app(app(filter2, true), f), x), xs) → APP(cons, x)
APP(app(rev2, x), app(app(cons, y), l)) → APP(app(cons, x), app(app(rev2, y), l))
APP(app(app(app(filter2, false), f), x), xs) → APP(app(filter, f), xs)
APP(app(map, f), app(app(cons, x), xs)) → APP(cons, app(f, x))
APP(rev, app(app(cons, x), l)) → APP(cons, app(app(rev1, x), 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.
The approximation of the Dependency Graph [13,14,18] contains 3 SCCs with 19 less nodes.

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

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.
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:

REV1(x, cons(y, l)) → REV1(y, l)

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

REV(nil)
REV(cons(x0, x1))
rev1(0, nil)
rev1(s(x0), nil)
rev1(x0, cons(x1, x2))
rev2(x0, nil)
rev2(x0, cons(x1, 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(rev1, x), app(app(cons, y), l)) → APP(app(rev1, y), l)
The remaining pairs can at least be oriented weakly.
none
Used ordering: Combined order from the following AFS and order.
REV1(x1, x2)  =  x2
cons(x1, x2)  =  cons(x2)

Lexicographic Path Order [19].
Precedence:
trivial


The following usable rules [14] were oriented: none



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

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

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

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(app(rev2, y), l)
APP(app(rev2, x), app(app(cons, y), l)) → APP(rev, app(app(cons, x), 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.
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:

REV2(x, cons(y, l)) → REV2(y, l)
REV(cons(x, l)) → REV2(x, l)
REV2(x, cons(y, l)) → REV(cons(x, rev2(y, l)))

The TRS R consists of the following rules:

rev2(x, nil) → nil
rev2(x, cons(y, l)) → rev(cons(x, rev2(y, l)))
rev(cons(x, l)) → cons(rev1(x, l), rev2(x, l))
rev1(0, nil) → 0
rev1(s(x), nil) → s(x)
rev1(x, cons(y, l)) → rev1(y, l)

The set Q consists of the following terms:

rev(nil)
rev(cons(x0, x1))
rev1(0, nil)
rev1(s(x0), nil)
rev1(x0, cons(x1, x2))
rev2(x0, nil)
rev2(x0, cons(x1, 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(rev, app(app(cons, x), l)) → APP(app(rev2, x), l)
APP(app(rev2, x), app(app(cons, y), l)) → APP(app(rev2, y), l)
The remaining pairs can at least be oriented weakly.

APP(app(rev2, x), app(app(cons, y), l)) → APP(rev, app(app(cons, x), app(app(rev2, y), l)))
Used ordering: Combined order from the following AFS and order.
REV2(x1, x2)  =  x2
cons(x1, x2)  =  cons(x2)
REV(x1)  =  x1
rev2(x1, x2)  =  x2
nil  =  nil
rev(x1)  =  x1

Lexicographic Path Order [19].
Precedence:
trivial


The following usable rules [14] were oriented:

app(app(rev2, x), nil) → nil
app(rev, app(app(cons, x), l)) → app(app(cons, app(app(rev1, 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)))



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

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

APP(app(rev2, x), app(app(cons, y), l)) → APP(rev, app(app(cons, x), 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.
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
                  ↳ QDP
                    ↳ QDPOrderProof

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

APP(app(app(app(filter2, true), f), x), xs) → APP(app(filter, f), xs)
APP(app(map, f), app(app(cons, x), xs)) → APP(f, x)
APP(app(app(app(filter2, false), f), x), xs) → APP(app(filter, f), xs)
APP(app(filter, f), app(app(cons, x), xs)) → APP(f, x)
APP(app(map, f), app(app(cons, x), xs)) → APP(app(map, 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.
We use the reduction pair processor [13].


The following pairs can be oriented strictly and are deleted.


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

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)
Used ordering: Combined order from the following AFS and order.
APP(x1, x2)  =  x2
app(x1, x2)  =  app(x1, x2)
cons  =  cons

Lexicographic Path Order [19].
Precedence:
trivial


The following usable rules [14] were oriented: none



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

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

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