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:

app2(f, a) -> app2(f, b)
app2(g, b) -> app2(g, a)
app2(app2(map, fun), nil) -> nil
app2(app2(map, fun), app2(app2(cons, x), xs)) -> app2(app2(cons, app2(fun, x)), app2(app2(map, fun), xs))
app2(app2(filter, fun), nil) -> nil
app2(app2(filter, fun), app2(app2(cons, x), xs)) -> app2(app2(app2(app2(filter2, app2(fun, x)), fun), x), xs)
app2(app2(app2(app2(filter2, true), fun), x), xs) -> app2(app2(cons, x), app2(app2(filter, fun), xs))
app2(app2(app2(app2(filter2, false), fun), x), xs) -> app2(app2(filter, fun), xs)

Q is empty.


↳ QTRS
  ↳ DependencyPairsProof

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

app2(f, a) -> app2(f, b)
app2(g, b) -> app2(g, a)
app2(app2(map, fun), nil) -> nil
app2(app2(map, fun), app2(app2(cons, x), xs)) -> app2(app2(cons, app2(fun, x)), app2(app2(map, fun), xs))
app2(app2(filter, fun), nil) -> nil
app2(app2(filter, fun), app2(app2(cons, x), xs)) -> app2(app2(app2(app2(filter2, app2(fun, x)), fun), x), xs)
app2(app2(app2(app2(filter2, true), fun), x), xs) -> app2(app2(cons, x), app2(app2(filter, fun), xs))
app2(app2(app2(app2(filter2, false), fun), x), xs) -> app2(app2(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:

APP2(app2(app2(app2(filter2, true), fun), x), xs) -> APP2(app2(filter, fun), xs)
APP2(app2(map, fun), app2(app2(cons, x), xs)) -> APP2(app2(cons, app2(fun, x)), app2(app2(map, fun), xs))
APP2(app2(filter, fun), app2(app2(cons, x), xs)) -> APP2(app2(filter2, app2(fun, x)), fun)
APP2(app2(app2(app2(filter2, false), fun), x), xs) -> APP2(filter, fun)
APP2(app2(filter, fun), app2(app2(cons, x), xs)) -> APP2(app2(app2(filter2, app2(fun, x)), fun), x)
APP2(app2(filter, fun), app2(app2(cons, x), xs)) -> APP2(app2(app2(app2(filter2, app2(fun, x)), fun), x), xs)
APP2(app2(map, fun), app2(app2(cons, x), xs)) -> APP2(app2(map, fun), xs)
APP2(app2(filter, fun), app2(app2(cons, x), xs)) -> APP2(fun, x)
APP2(app2(app2(app2(filter2, true), fun), x), xs) -> APP2(app2(cons, x), app2(app2(filter, fun), xs))
APP2(app2(app2(app2(filter2, true), fun), x), xs) -> APP2(filter, fun)
APP2(g, b) -> APP2(g, a)
APP2(app2(map, fun), app2(app2(cons, x), xs)) -> APP2(fun, x)
APP2(app2(filter, fun), app2(app2(cons, x), xs)) -> APP2(filter2, app2(fun, x))
APP2(app2(app2(app2(filter2, true), fun), x), xs) -> APP2(cons, x)
APP2(app2(app2(app2(filter2, false), fun), x), xs) -> APP2(app2(filter, fun), xs)
APP2(f, a) -> APP2(f, b)
APP2(app2(map, fun), app2(app2(cons, x), xs)) -> APP2(cons, app2(fun, x))

The TRS R consists of the following rules:

app2(f, a) -> app2(f, b)
app2(g, b) -> app2(g, a)
app2(app2(map, fun), nil) -> nil
app2(app2(map, fun), app2(app2(cons, x), xs)) -> app2(app2(cons, app2(fun, x)), app2(app2(map, fun), xs))
app2(app2(filter, fun), nil) -> nil
app2(app2(filter, fun), app2(app2(cons, x), xs)) -> app2(app2(app2(app2(filter2, app2(fun, x)), fun), x), xs)
app2(app2(app2(app2(filter2, true), fun), x), xs) -> app2(app2(cons, x), app2(app2(filter, fun), xs))
app2(app2(app2(app2(filter2, false), fun), x), xs) -> app2(app2(filter, fun), xs)

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

↳ QTRS
  ↳ DependencyPairsProof
    ↳ QDP
      ↳ DependencyGraphProof

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

APP2(app2(app2(app2(filter2, true), fun), x), xs) -> APP2(app2(filter, fun), xs)
APP2(app2(map, fun), app2(app2(cons, x), xs)) -> APP2(app2(cons, app2(fun, x)), app2(app2(map, fun), xs))
APP2(app2(filter, fun), app2(app2(cons, x), xs)) -> APP2(app2(filter2, app2(fun, x)), fun)
APP2(app2(app2(app2(filter2, false), fun), x), xs) -> APP2(filter, fun)
APP2(app2(filter, fun), app2(app2(cons, x), xs)) -> APP2(app2(app2(filter2, app2(fun, x)), fun), x)
APP2(app2(filter, fun), app2(app2(cons, x), xs)) -> APP2(app2(app2(app2(filter2, app2(fun, x)), fun), x), xs)
APP2(app2(map, fun), app2(app2(cons, x), xs)) -> APP2(app2(map, fun), xs)
APP2(app2(filter, fun), app2(app2(cons, x), xs)) -> APP2(fun, x)
APP2(app2(app2(app2(filter2, true), fun), x), xs) -> APP2(app2(cons, x), app2(app2(filter, fun), xs))
APP2(app2(app2(app2(filter2, true), fun), x), xs) -> APP2(filter, fun)
APP2(g, b) -> APP2(g, a)
APP2(app2(map, fun), app2(app2(cons, x), xs)) -> APP2(fun, x)
APP2(app2(filter, fun), app2(app2(cons, x), xs)) -> APP2(filter2, app2(fun, x))
APP2(app2(app2(app2(filter2, true), fun), x), xs) -> APP2(cons, x)
APP2(app2(app2(app2(filter2, false), fun), x), xs) -> APP2(app2(filter, fun), xs)
APP2(f, a) -> APP2(f, b)
APP2(app2(map, fun), app2(app2(cons, x), xs)) -> APP2(cons, app2(fun, x))

The TRS R consists of the following rules:

app2(f, a) -> app2(f, b)
app2(g, b) -> app2(g, a)
app2(app2(map, fun), nil) -> nil
app2(app2(map, fun), app2(app2(cons, x), xs)) -> app2(app2(cons, app2(fun, x)), app2(app2(map, fun), xs))
app2(app2(filter, fun), nil) -> nil
app2(app2(filter, fun), app2(app2(cons, x), xs)) -> app2(app2(app2(app2(filter2, app2(fun, x)), fun), x), xs)
app2(app2(app2(app2(filter2, true), fun), x), xs) -> app2(app2(cons, x), app2(app2(filter, fun), xs))
app2(app2(app2(app2(filter2, false), fun), x), xs) -> app2(app2(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 1 SCC with 12 less nodes.

↳ QTRS
  ↳ DependencyPairsProof
    ↳ QDP
      ↳ DependencyGraphProof
        ↳ QDP
          ↳ QDPOrderProof

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

APP2(app2(app2(app2(filter2, true), fun), x), xs) -> APP2(app2(filter, fun), xs)
APP2(app2(map, fun), app2(app2(cons, x), xs)) -> APP2(fun, x)
APP2(app2(app2(app2(filter2, false), fun), x), xs) -> APP2(app2(filter, fun), xs)
APP2(app2(filter, fun), app2(app2(cons, x), xs)) -> APP2(fun, x)
APP2(app2(map, fun), app2(app2(cons, x), xs)) -> APP2(app2(map, fun), xs)

The TRS R consists of the following rules:

app2(f, a) -> app2(f, b)
app2(g, b) -> app2(g, a)
app2(app2(map, fun), nil) -> nil
app2(app2(map, fun), app2(app2(cons, x), xs)) -> app2(app2(cons, app2(fun, x)), app2(app2(map, fun), xs))
app2(app2(filter, fun), nil) -> nil
app2(app2(filter, fun), app2(app2(cons, x), xs)) -> app2(app2(app2(app2(filter2, app2(fun, x)), fun), x), xs)
app2(app2(app2(app2(filter2, true), fun), x), xs) -> app2(app2(cons, x), app2(app2(filter, fun), xs))
app2(app2(app2(app2(filter2, false), fun), x), xs) -> app2(app2(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.


APP2(app2(app2(app2(filter2, true), fun), x), xs) -> APP2(app2(filter, fun), xs)
APP2(app2(map, fun), app2(app2(cons, x), xs)) -> APP2(fun, x)
APP2(app2(app2(app2(filter2, false), fun), x), xs) -> APP2(app2(filter, fun), xs)
APP2(app2(filter, fun), app2(app2(cons, x), xs)) -> APP2(fun, x)
APP2(app2(map, fun), app2(app2(cons, x), xs)) -> APP2(app2(map, fun), xs)
The remaining pairs can at least be oriented weakly.
none
Used ordering: Polynomial interpretation [21]:

POL(APP2(x1, x2)) = x1 + x2   
POL(app2(x1, x2)) = 1 + x1 + x2   
POL(cons) = 1   
POL(false) = 0   
POL(filter) = 1   
POL(filter2) = 1   
POL(map) = 0   
POL(true) = 1   

The following usable rules [14] were oriented: none



↳ QTRS
  ↳ DependencyPairsProof
    ↳ QDP
      ↳ DependencyGraphProof
        ↳ QDP
          ↳ QDPOrderProof
            ↳ QDP
              ↳ PisEmptyProof

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

app2(f, a) -> app2(f, b)
app2(g, b) -> app2(g, a)
app2(app2(map, fun), nil) -> nil
app2(app2(map, fun), app2(app2(cons, x), xs)) -> app2(app2(cons, app2(fun, x)), app2(app2(map, fun), xs))
app2(app2(filter, fun), nil) -> nil
app2(app2(filter, fun), app2(app2(cons, x), xs)) -> app2(app2(app2(app2(filter2, app2(fun, x)), fun), x), xs)
app2(app2(app2(app2(filter2, true), fun), x), xs) -> app2(app2(cons, x), app2(app2(filter, fun), xs))
app2(app2(app2(app2(filter2, false), fun), x), xs) -> app2(app2(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.