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(app2(app2(consif, true), x), ys) -> app2(app2(cons, x), ys)
app2(app2(app2(consif, false), x), ys) -> ys
app2(app2(filter, f), nil) -> nil
app2(app2(filter, f), app2(app2(cons, x), xs)) -> app2(app2(app2(consif, app2(f, x)), x), app2(app2(filter, f), xs))

Q is empty.


↳ QTRS
  ↳ DependencyPairsProof

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

app2(app2(app2(consif, true), x), ys) -> app2(app2(cons, x), ys)
app2(app2(app2(consif, false), x), ys) -> ys
app2(app2(filter, f), nil) -> nil
app2(app2(filter, f), app2(app2(cons, x), xs)) -> app2(app2(app2(consif, app2(f, x)), x), app2(app2(filter, f), 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(filter, f), app2(app2(cons, x), xs)) -> APP2(app2(app2(consif, app2(f, x)), x), app2(app2(filter, f), xs))
APP2(app2(app2(consif, true), x), ys) -> APP2(app2(cons, x), ys)
APP2(app2(filter, f), app2(app2(cons, x), xs)) -> APP2(consif, app2(f, x))
APP2(app2(filter, f), app2(app2(cons, x), xs)) -> APP2(app2(consif, app2(f, x)), x)
APP2(app2(app2(consif, true), x), ys) -> APP2(cons, x)
APP2(app2(filter, f), app2(app2(cons, x), xs)) -> APP2(f, x)
APP2(app2(filter, f), app2(app2(cons, x), xs)) -> APP2(app2(filter, f), xs)

The TRS R consists of the following rules:

app2(app2(app2(consif, true), x), ys) -> app2(app2(cons, x), ys)
app2(app2(app2(consif, false), x), ys) -> ys
app2(app2(filter, f), nil) -> nil
app2(app2(filter, f), app2(app2(cons, x), xs)) -> app2(app2(app2(consif, app2(f, x)), x), app2(app2(filter, f), 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(filter, f), app2(app2(cons, x), xs)) -> APP2(app2(app2(consif, app2(f, x)), x), app2(app2(filter, f), xs))
APP2(app2(app2(consif, true), x), ys) -> APP2(app2(cons, x), ys)
APP2(app2(filter, f), app2(app2(cons, x), xs)) -> APP2(consif, app2(f, x))
APP2(app2(filter, f), app2(app2(cons, x), xs)) -> APP2(app2(consif, app2(f, x)), x)
APP2(app2(app2(consif, true), x), ys) -> APP2(cons, x)
APP2(app2(filter, f), app2(app2(cons, x), xs)) -> APP2(f, x)
APP2(app2(filter, f), app2(app2(cons, x), xs)) -> APP2(app2(filter, f), xs)

The TRS R consists of the following rules:

app2(app2(app2(consif, true), x), ys) -> app2(app2(cons, x), ys)
app2(app2(app2(consif, false), x), ys) -> ys
app2(app2(filter, f), nil) -> nil
app2(app2(filter, f), app2(app2(cons, x), xs)) -> app2(app2(app2(consif, app2(f, x)), x), app2(app2(filter, f), 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 5 less nodes.

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

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

APP2(app2(filter, f), app2(app2(cons, x), xs)) -> APP2(f, x)
APP2(app2(filter, f), app2(app2(cons, x), xs)) -> APP2(app2(filter, f), xs)

The TRS R consists of the following rules:

app2(app2(app2(consif, true), x), ys) -> app2(app2(cons, x), ys)
app2(app2(app2(consif, false), x), ys) -> ys
app2(app2(filter, f), nil) -> nil
app2(app2(filter, f), app2(app2(cons, x), xs)) -> app2(app2(app2(consif, app2(f, x)), x), app2(app2(filter, f), 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(filter, f), app2(app2(cons, x), xs)) -> APP2(f, x)
The remaining pairs can at least be oriented weakly.

APP2(app2(filter, f), app2(app2(cons, x), xs)) -> APP2(app2(filter, f), xs)
Used ordering: Polynomial Order [17,21] with Interpretation:

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


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


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


POL( cons ) = 1



The following usable rules [14] were oriented: none



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

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

APP2(app2(filter, f), app2(app2(cons, x), xs)) -> APP2(app2(filter, f), xs)

The TRS R consists of the following rules:

app2(app2(app2(consif, true), x), ys) -> app2(app2(cons, x), ys)
app2(app2(app2(consif, false), x), ys) -> ys
app2(app2(filter, f), nil) -> nil
app2(app2(filter, f), app2(app2(cons, x), xs)) -> app2(app2(app2(consif, app2(f, x)), x), app2(app2(filter, f), 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(filter, f), app2(app2(cons, x), xs)) -> APP2(app2(filter, f), xs)
The remaining pairs can at least be oriented weakly.
none
Used ordering: Polynomial Order [17,21] with Interpretation:

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


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


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


POL( cons ) = 3



The following usable rules [14] were oriented: none



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

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

app2(app2(app2(consif, true), x), ys) -> app2(app2(cons, x), ys)
app2(app2(app2(consif, false), x), ys) -> ys
app2(app2(filter, f), nil) -> nil
app2(app2(filter, f), app2(app2(cons, x), xs)) -> app2(app2(app2(consif, app2(f, x)), x), app2(app2(filter, f), 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.