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(if, true), x), y) -> x
app2(app2(app2(if, false), x), y) -> y
app2(app2(filter, f), nil) -> nil
app2(app2(filter, f), app2(app2(cons, x), xs)) -> app2(app2(app2(if, app2(f, x)), app2(app2(cons, x), app2(app2(filter, f), xs))), 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(if, true), x), y) -> x
app2(app2(app2(if, false), x), y) -> y
app2(app2(filter, f), nil) -> nil
app2(app2(filter, f), app2(app2(cons, x), xs)) -> app2(app2(app2(if, app2(f, x)), app2(app2(cons, x), app2(app2(filter, f), xs))), 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(if, app2(f, x)), app2(app2(cons, x), app2(app2(filter, f), xs))), app2(app2(filter, f), xs))
APP2(app2(filter, f), app2(app2(cons, x), xs)) -> APP2(if, app2(f, x))
APP2(app2(filter, f), app2(app2(cons, x), xs)) -> APP2(f, x)
APP2(app2(filter, f), app2(app2(cons, x), xs)) -> APP2(app2(if, app2(f, x)), app2(app2(cons, x), app2(app2(filter, f), xs)))
APP2(app2(filter, f), app2(app2(cons, x), xs)) -> APP2(app2(filter, f), xs)
APP2(app2(filter, f), app2(app2(cons, x), xs)) -> APP2(app2(cons, x), app2(app2(filter, f), xs))

The TRS R consists of the following rules:

app2(app2(app2(if, true), x), y) -> x
app2(app2(app2(if, false), x), y) -> y
app2(app2(filter, f), nil) -> nil
app2(app2(filter, f), app2(app2(cons, x), xs)) -> app2(app2(app2(if, app2(f, x)), app2(app2(cons, x), app2(app2(filter, f), xs))), 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(if, app2(f, x)), app2(app2(cons, x), app2(app2(filter, f), xs))), app2(app2(filter, f), xs))
APP2(app2(filter, f), app2(app2(cons, x), xs)) -> APP2(if, app2(f, x))
APP2(app2(filter, f), app2(app2(cons, x), xs)) -> APP2(f, x)
APP2(app2(filter, f), app2(app2(cons, x), xs)) -> APP2(app2(if, app2(f, x)), app2(app2(cons, x), app2(app2(filter, f), xs)))
APP2(app2(filter, f), app2(app2(cons, x), xs)) -> APP2(app2(filter, f), xs)
APP2(app2(filter, f), app2(app2(cons, x), xs)) -> APP2(app2(cons, x), app2(app2(filter, f), xs))

The TRS R consists of the following rules:

app2(app2(app2(if, true), x), y) -> x
app2(app2(app2(if, false), x), y) -> y
app2(app2(filter, f), nil) -> nil
app2(app2(filter, f), app2(app2(cons, x), xs)) -> app2(app2(app2(if, app2(f, x)), app2(app2(cons, x), app2(app2(filter, f), xs))), 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 4 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(if, true), x), y) -> x
app2(app2(app2(if, false), x), y) -> y
app2(app2(filter, f), nil) -> nil
app2(app2(filter, f), app2(app2(cons, x), xs)) -> app2(app2(app2(if, app2(f, x)), app2(app2(cons, x), app2(app2(filter, f), xs))), 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)
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 ) = 1


POL( APP2(x1, x2) ) = x2


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


POL( cons ) = 0



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(app2(app2(if, true), x), y) -> x
app2(app2(app2(if, false), x), y) -> y
app2(app2(filter, f), nil) -> nil
app2(app2(filter, f), app2(app2(cons, x), xs)) -> app2(app2(app2(if, app2(f, x)), app2(app2(cons, x), app2(app2(filter, f), xs))), 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.