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, app2(s, x)) -> app2(f, x)
app2(g, app2(app2(cons, 0), y)) -> app2(g, y)
app2(g, app2(app2(cons, app2(s, x)), y)) -> app2(s, x)
app2(h, app2(app2(cons, x), y)) -> app2(h, app2(g, app2(app2(cons, x), y)))
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, app2(s, x)) -> app2(f, x)
app2(g, app2(app2(cons, 0), y)) -> app2(g, y)
app2(g, app2(app2(cons, app2(s, x)), y)) -> app2(s, x)
app2(h, app2(app2(cons, x), y)) -> app2(h, app2(g, app2(app2(cons, x), y)))
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(f, app2(s, x)) -> APP2(f, x)
APP2(g, app2(app2(cons, 0), y)) -> APP2(g, y)
APP2(app2(map, fun), app2(app2(cons, x), xs)) -> APP2(app2(cons, app2(fun, x)), app2(app2(map, fun), xs))
APP2(h, app2(app2(cons, x), y)) -> APP2(g, app2(app2(cons, x), y))
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(h, app2(app2(cons, x), y)) -> APP2(h, app2(g, app2(app2(cons, x), y)))
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(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(app2(map, fun), app2(app2(cons, x), xs)) -> APP2(cons, app2(fun, x))

The TRS R consists of the following rules:

app2(f, app2(s, x)) -> app2(f, x)
app2(g, app2(app2(cons, 0), y)) -> app2(g, y)
app2(g, app2(app2(cons, app2(s, x)), y)) -> app2(s, x)
app2(h, app2(app2(cons, x), y)) -> app2(h, app2(g, app2(app2(cons, x), y)))
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(f, app2(s, x)) -> APP2(f, x)
APP2(g, app2(app2(cons, 0), y)) -> APP2(g, y)
APP2(app2(map, fun), app2(app2(cons, x), xs)) -> APP2(app2(cons, app2(fun, x)), app2(app2(map, fun), xs))
APP2(h, app2(app2(cons, x), y)) -> APP2(g, app2(app2(cons, x), y))
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(h, app2(app2(cons, x), y)) -> APP2(h, app2(g, app2(app2(cons, x), y)))
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(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(app2(map, fun), app2(app2(cons, x), xs)) -> APP2(cons, app2(fun, x))

The TRS R consists of the following rules:

app2(f, app2(s, x)) -> app2(f, x)
app2(g, app2(app2(cons, 0), y)) -> app2(g, y)
app2(g, app2(app2(cons, app2(s, x)), y)) -> app2(s, x)
app2(h, app2(app2(cons, x), y)) -> app2(h, app2(g, app2(app2(cons, x), y)))
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 4 SCCs with 11 less nodes.

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

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

APP2(g, app2(app2(cons, 0), y)) -> APP2(g, y)

The TRS R consists of the following rules:

app2(f, app2(s, x)) -> app2(f, x)
app2(g, app2(app2(cons, 0), y)) -> app2(g, y)
app2(g, app2(app2(cons, app2(s, x)), y)) -> app2(s, x)
app2(h, app2(app2(cons, x), y)) -> app2(h, app2(g, app2(app2(cons, x), y)))
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(g, app2(app2(cons, 0), y)) -> APP2(g, y)
The remaining pairs can at least be oriented weakly.
none
Used ordering: Polynomial Order [17,21] with Interpretation:

POL( 0 ) = 2


POL( APP2(x1, x2) ) = 3x1 + x2 + 3


POL( g ) = 3


POL( app2(x1, x2) ) = 3x1 + 2x2 + 1


POL( cons ) = 2



The following usable rules [14] were oriented: none



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

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

app2(f, app2(s, x)) -> app2(f, x)
app2(g, app2(app2(cons, 0), y)) -> app2(g, y)
app2(g, app2(app2(cons, app2(s, x)), y)) -> app2(s, x)
app2(h, app2(app2(cons, x), y)) -> app2(h, app2(g, app2(app2(cons, x), y)))
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.

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

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

APP2(h, app2(app2(cons, x), y)) -> APP2(h, app2(g, app2(app2(cons, x), y)))

The TRS R consists of the following rules:

app2(f, app2(s, x)) -> app2(f, x)
app2(g, app2(app2(cons, 0), y)) -> app2(g, y)
app2(g, app2(app2(cons, app2(s, x)), y)) -> app2(s, x)
app2(h, app2(app2(cons, x), y)) -> app2(h, app2(g, app2(app2(cons, x), y)))
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(h, app2(app2(cons, x), y)) -> APP2(h, app2(g, app2(app2(cons, x), y)))
The remaining pairs can at least be oriented weakly.
none
Used ordering: Polynomial Order [17,21] with Interpretation:

POL( app2(x1, x2) ) = max{0, 2x1 - 1}


POL( 0 ) = 1


POL( s ) = 0


POL( APP2(x1, x2) ) = max{0, 2x2 - 3}


POL( g ) = 1


POL( h ) = 1


POL( cons ) = 3



The following usable rules [14] were oriented:

app2(g, app2(app2(cons, app2(s, x)), y)) -> app2(s, x)
app2(g, app2(app2(cons, 0), y)) -> app2(g, y)



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

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

app2(f, app2(s, x)) -> app2(f, x)
app2(g, app2(app2(cons, 0), y)) -> app2(g, y)
app2(g, app2(app2(cons, app2(s, x)), y)) -> app2(s, x)
app2(h, app2(app2(cons, x), y)) -> app2(h, app2(g, app2(app2(cons, x), y)))
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.

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

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

APP2(f, app2(s, x)) -> APP2(f, x)

The TRS R consists of the following rules:

app2(f, app2(s, x)) -> app2(f, x)
app2(g, app2(app2(cons, 0), y)) -> app2(g, y)
app2(g, app2(app2(cons, app2(s, x)), y)) -> app2(s, x)
app2(h, app2(app2(cons, x), y)) -> app2(h, app2(g, app2(app2(cons, x), y)))
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(f, app2(s, x)) -> APP2(f, x)
The remaining pairs can at least be oriented weakly.
none
Used ordering: Polynomial Order [17,21] with Interpretation:

POL( f ) = 1


POL( s ) = 3


POL( APP2(x1, x2) ) = x2


POL( app2(x1, x2) ) = max{0, x1 + x2 - 2}



The following usable rules [14] were oriented: none



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

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

app2(f, app2(s, x)) -> app2(f, x)
app2(g, app2(app2(cons, 0), y)) -> app2(g, y)
app2(g, app2(app2(cons, app2(s, x)), y)) -> app2(s, x)
app2(h, app2(app2(cons, x), y)) -> app2(h, app2(g, app2(app2(cons, x), y)))
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.

↳ QTRS
  ↳ DependencyPairsProof
    ↳ QDP
      ↳ DependencyGraphProof
        ↳ AND
          ↳ QDP
          ↳ QDP
          ↳ QDP
          ↳ 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, app2(s, x)) -> app2(f, x)
app2(g, app2(app2(cons, 0), y)) -> app2(g, y)
app2(g, app2(app2(cons, app2(s, x)), y)) -> app2(s, x)
app2(h, app2(app2(cons, x), y)) -> app2(h, app2(g, app2(app2(cons, x), y)))
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(map, fun), app2(app2(cons, x), xs)) -> APP2(fun, x)
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.

APP2(app2(app2(app2(filter2, true), fun), x), xs) -> APP2(app2(filter, fun), xs)
APP2(app2(app2(app2(filter2, false), fun), x), xs) -> APP2(app2(filter, fun), xs)
Used ordering: Polynomial Order [17,21] with Interpretation:

POL( filter ) = 2


POL( true ) = 0


POL( false ) = max{0, -3}


POL( map ) = 3


POL( app2(x1, x2) ) = x1 + 2x2 + 1


POL( filter2 ) = 2


POL( APP2(x1, x2) ) = 2x2 + 2


POL( cons ) = max{0, -1}



The following usable rules [14] were oriented: none



↳ QTRS
  ↳ DependencyPairsProof
    ↳ QDP
      ↳ DependencyGraphProof
        ↳ AND
          ↳ QDP
          ↳ QDP
          ↳ QDP
          ↳ QDP
            ↳ QDPOrderProof
              ↳ 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(app2(app2(filter2, false), fun), x), xs) -> APP2(app2(filter, fun), xs)

The TRS R consists of the following rules:

app2(f, app2(s, x)) -> app2(f, x)
app2(g, app2(app2(cons, 0), y)) -> app2(g, y)
app2(g, app2(app2(cons, app2(s, x)), y)) -> app2(s, x)
app2(h, app2(app2(cons, x), y)) -> app2(h, app2(g, app2(app2(cons, x), y)))
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 0 SCCs with 2 less nodes.