Termination w.r.t. Q of the following Term Rewriting System could not be shown:

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

app2(app2(app2(if, true), xs), ys) -> xs
app2(app2(app2(if, false), xs), ys) -> ys
app2(app2(sub, x), 0) -> x
app2(app2(sub, app2(s, x)), app2(s, y)) -> app2(app2(sub, x), y)
app2(app2(gtr, 0), y) -> false
app2(app2(gtr, app2(s, x)), 0) -> true
app2(app2(gtr, app2(s, x)), app2(s, y)) -> app2(app2(gtr, x), y)
app2(app2(d, x), 0) -> true
app2(app2(d, app2(s, x)), app2(s, y)) -> app2(app2(app2(if, app2(app2(gtr, x), y)), false), app2(app2(d, app2(s, x)), app2(app2(sub, y), x)))
app2(len, nil) -> 0
app2(len, app2(app2(cons, x), xs)) -> app2(s, app2(len, xs))
app2(app2(filter, p), nil) -> nil
app2(app2(filter, p), app2(app2(cons, x), xs)) -> app2(app2(app2(if, app2(p, x)), app2(app2(cons, x), app2(app2(filter, p), xs))), app2(app2(filter, p), xs))

Q is empty.


↳ QTRS
  ↳ DependencyPairsProof

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

app2(app2(app2(if, true), xs), ys) -> xs
app2(app2(app2(if, false), xs), ys) -> ys
app2(app2(sub, x), 0) -> x
app2(app2(sub, app2(s, x)), app2(s, y)) -> app2(app2(sub, x), y)
app2(app2(gtr, 0), y) -> false
app2(app2(gtr, app2(s, x)), 0) -> true
app2(app2(gtr, app2(s, x)), app2(s, y)) -> app2(app2(gtr, x), y)
app2(app2(d, x), 0) -> true
app2(app2(d, app2(s, x)), app2(s, y)) -> app2(app2(app2(if, app2(app2(gtr, x), y)), false), app2(app2(d, app2(s, x)), app2(app2(sub, y), x)))
app2(len, nil) -> 0
app2(len, app2(app2(cons, x), xs)) -> app2(s, app2(len, xs))
app2(app2(filter, p), nil) -> nil
app2(app2(filter, p), app2(app2(cons, x), xs)) -> app2(app2(app2(if, app2(p, x)), app2(app2(cons, x), app2(app2(filter, p), xs))), app2(app2(filter, p), 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, p), app2(app2(cons, x), xs)) -> APP2(app2(app2(if, app2(p, x)), app2(app2(cons, x), app2(app2(filter, p), xs))), app2(app2(filter, p), xs))
APP2(app2(gtr, app2(s, x)), app2(s, y)) -> APP2(app2(gtr, x), y)
APP2(app2(gtr, app2(s, x)), app2(s, y)) -> APP2(gtr, x)
APP2(app2(d, app2(s, x)), app2(s, y)) -> APP2(app2(app2(if, app2(app2(gtr, x), y)), false), app2(app2(d, app2(s, x)), app2(app2(sub, y), x)))
APP2(len, app2(app2(cons, x), xs)) -> APP2(len, xs)
APP2(app2(filter, p), app2(app2(cons, x), xs)) -> APP2(if, app2(p, x))
APP2(app2(sub, app2(s, x)), app2(s, y)) -> APP2(app2(sub, x), y)
APP2(app2(d, app2(s, x)), app2(s, y)) -> APP2(app2(d, app2(s, x)), app2(app2(sub, y), x))
APP2(app2(sub, app2(s, x)), app2(s, y)) -> APP2(sub, x)
APP2(app2(d, app2(s, x)), app2(s, y)) -> APP2(gtr, x)
APP2(app2(d, app2(s, x)), app2(s, y)) -> APP2(sub, y)
APP2(app2(filter, p), app2(app2(cons, x), xs)) -> APP2(p, x)
APP2(app2(d, app2(s, x)), app2(s, y)) -> APP2(app2(if, app2(app2(gtr, x), y)), false)
APP2(app2(d, app2(s, x)), app2(s, y)) -> APP2(if, app2(app2(gtr, x), y))
APP2(app2(filter, p), app2(app2(cons, x), xs)) -> APP2(app2(filter, p), xs)
APP2(app2(filter, p), app2(app2(cons, x), xs)) -> APP2(app2(cons, x), app2(app2(filter, p), xs))
APP2(app2(d, app2(s, x)), app2(s, y)) -> APP2(app2(gtr, x), y)
APP2(app2(d, app2(s, x)), app2(s, y)) -> APP2(app2(sub, y), x)
APP2(len, app2(app2(cons, x), xs)) -> APP2(s, app2(len, xs))
APP2(app2(filter, p), app2(app2(cons, x), xs)) -> APP2(app2(if, app2(p, x)), app2(app2(cons, x), app2(app2(filter, p), xs)))

The TRS R consists of the following rules:

app2(app2(app2(if, true), xs), ys) -> xs
app2(app2(app2(if, false), xs), ys) -> ys
app2(app2(sub, x), 0) -> x
app2(app2(sub, app2(s, x)), app2(s, y)) -> app2(app2(sub, x), y)
app2(app2(gtr, 0), y) -> false
app2(app2(gtr, app2(s, x)), 0) -> true
app2(app2(gtr, app2(s, x)), app2(s, y)) -> app2(app2(gtr, x), y)
app2(app2(d, x), 0) -> true
app2(app2(d, app2(s, x)), app2(s, y)) -> app2(app2(app2(if, app2(app2(gtr, x), y)), false), app2(app2(d, app2(s, x)), app2(app2(sub, y), x)))
app2(len, nil) -> 0
app2(len, app2(app2(cons, x), xs)) -> app2(s, app2(len, xs))
app2(app2(filter, p), nil) -> nil
app2(app2(filter, p), app2(app2(cons, x), xs)) -> app2(app2(app2(if, app2(p, x)), app2(app2(cons, x), app2(app2(filter, p), xs))), app2(app2(filter, p), 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, p), app2(app2(cons, x), xs)) -> APP2(app2(app2(if, app2(p, x)), app2(app2(cons, x), app2(app2(filter, p), xs))), app2(app2(filter, p), xs))
APP2(app2(gtr, app2(s, x)), app2(s, y)) -> APP2(app2(gtr, x), y)
APP2(app2(gtr, app2(s, x)), app2(s, y)) -> APP2(gtr, x)
APP2(app2(d, app2(s, x)), app2(s, y)) -> APP2(app2(app2(if, app2(app2(gtr, x), y)), false), app2(app2(d, app2(s, x)), app2(app2(sub, y), x)))
APP2(len, app2(app2(cons, x), xs)) -> APP2(len, xs)
APP2(app2(filter, p), app2(app2(cons, x), xs)) -> APP2(if, app2(p, x))
APP2(app2(sub, app2(s, x)), app2(s, y)) -> APP2(app2(sub, x), y)
APP2(app2(d, app2(s, x)), app2(s, y)) -> APP2(app2(d, app2(s, x)), app2(app2(sub, y), x))
APP2(app2(sub, app2(s, x)), app2(s, y)) -> APP2(sub, x)
APP2(app2(d, app2(s, x)), app2(s, y)) -> APP2(gtr, x)
APP2(app2(d, app2(s, x)), app2(s, y)) -> APP2(sub, y)
APP2(app2(filter, p), app2(app2(cons, x), xs)) -> APP2(p, x)
APP2(app2(d, app2(s, x)), app2(s, y)) -> APP2(app2(if, app2(app2(gtr, x), y)), false)
APP2(app2(d, app2(s, x)), app2(s, y)) -> APP2(if, app2(app2(gtr, x), y))
APP2(app2(filter, p), app2(app2(cons, x), xs)) -> APP2(app2(filter, p), xs)
APP2(app2(filter, p), app2(app2(cons, x), xs)) -> APP2(app2(cons, x), app2(app2(filter, p), xs))
APP2(app2(d, app2(s, x)), app2(s, y)) -> APP2(app2(gtr, x), y)
APP2(app2(d, app2(s, x)), app2(s, y)) -> APP2(app2(sub, y), x)
APP2(len, app2(app2(cons, x), xs)) -> APP2(s, app2(len, xs))
APP2(app2(filter, p), app2(app2(cons, x), xs)) -> APP2(app2(if, app2(p, x)), app2(app2(cons, x), app2(app2(filter, p), xs)))

The TRS R consists of the following rules:

app2(app2(app2(if, true), xs), ys) -> xs
app2(app2(app2(if, false), xs), ys) -> ys
app2(app2(sub, x), 0) -> x
app2(app2(sub, app2(s, x)), app2(s, y)) -> app2(app2(sub, x), y)
app2(app2(gtr, 0), y) -> false
app2(app2(gtr, app2(s, x)), 0) -> true
app2(app2(gtr, app2(s, x)), app2(s, y)) -> app2(app2(gtr, x), y)
app2(app2(d, x), 0) -> true
app2(app2(d, app2(s, x)), app2(s, y)) -> app2(app2(app2(if, app2(app2(gtr, x), y)), false), app2(app2(d, app2(s, x)), app2(app2(sub, y), x)))
app2(len, nil) -> 0
app2(len, app2(app2(cons, x), xs)) -> app2(s, app2(len, xs))
app2(app2(filter, p), nil) -> nil
app2(app2(filter, p), app2(app2(cons, x), xs)) -> app2(app2(app2(if, app2(p, x)), app2(app2(cons, x), app2(app2(filter, p), xs))), app2(app2(filter, p), xs))

Q is empty.
We have to consider all minimal (P,Q,R)-chains.
The approximation of the Dependency Graph [13,14,18] contains 5 SCCs with 14 less nodes.

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

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

APP2(len, app2(app2(cons, x), xs)) -> APP2(len, xs)

The TRS R consists of the following rules:

app2(app2(app2(if, true), xs), ys) -> xs
app2(app2(app2(if, false), xs), ys) -> ys
app2(app2(sub, x), 0) -> x
app2(app2(sub, app2(s, x)), app2(s, y)) -> app2(app2(sub, x), y)
app2(app2(gtr, 0), y) -> false
app2(app2(gtr, app2(s, x)), 0) -> true
app2(app2(gtr, app2(s, x)), app2(s, y)) -> app2(app2(gtr, x), y)
app2(app2(d, x), 0) -> true
app2(app2(d, app2(s, x)), app2(s, y)) -> app2(app2(app2(if, app2(app2(gtr, x), y)), false), app2(app2(d, app2(s, x)), app2(app2(sub, y), x)))
app2(len, nil) -> 0
app2(len, app2(app2(cons, x), xs)) -> app2(s, app2(len, xs))
app2(app2(filter, p), nil) -> nil
app2(app2(filter, p), app2(app2(cons, x), xs)) -> app2(app2(app2(if, app2(p, x)), app2(app2(cons, x), app2(app2(filter, p), xs))), app2(app2(filter, p), 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(len, app2(app2(cons, x), xs)) -> APP2(len, xs)
The remaining pairs can at least be oriented weakly.
none
Used ordering: Polynomial Order [17,21] with Interpretation:

POL( APP2(x1, x2) ) = max{0, x2 - 1}


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


POL( cons ) = 0



The following usable rules [14] were oriented: none



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

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

app2(app2(app2(if, true), xs), ys) -> xs
app2(app2(app2(if, false), xs), ys) -> ys
app2(app2(sub, x), 0) -> x
app2(app2(sub, app2(s, x)), app2(s, y)) -> app2(app2(sub, x), y)
app2(app2(gtr, 0), y) -> false
app2(app2(gtr, app2(s, x)), 0) -> true
app2(app2(gtr, app2(s, x)), app2(s, y)) -> app2(app2(gtr, x), y)
app2(app2(d, x), 0) -> true
app2(app2(d, app2(s, x)), app2(s, y)) -> app2(app2(app2(if, app2(app2(gtr, x), y)), false), app2(app2(d, app2(s, x)), app2(app2(sub, y), x)))
app2(len, nil) -> 0
app2(len, app2(app2(cons, x), xs)) -> app2(s, app2(len, xs))
app2(app2(filter, p), nil) -> nil
app2(app2(filter, p), app2(app2(cons, x), xs)) -> app2(app2(app2(if, app2(p, x)), app2(app2(cons, x), app2(app2(filter, p), xs))), app2(app2(filter, p), 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
          ↳ QDP

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

APP2(app2(gtr, app2(s, x)), app2(s, y)) -> APP2(app2(gtr, x), y)

The TRS R consists of the following rules:

app2(app2(app2(if, true), xs), ys) -> xs
app2(app2(app2(if, false), xs), ys) -> ys
app2(app2(sub, x), 0) -> x
app2(app2(sub, app2(s, x)), app2(s, y)) -> app2(app2(sub, x), y)
app2(app2(gtr, 0), y) -> false
app2(app2(gtr, app2(s, x)), 0) -> true
app2(app2(gtr, app2(s, x)), app2(s, y)) -> app2(app2(gtr, x), y)
app2(app2(d, x), 0) -> true
app2(app2(d, app2(s, x)), app2(s, y)) -> app2(app2(app2(if, app2(app2(gtr, x), y)), false), app2(app2(d, app2(s, x)), app2(app2(sub, y), x)))
app2(len, nil) -> 0
app2(len, app2(app2(cons, x), xs)) -> app2(s, app2(len, xs))
app2(app2(filter, p), nil) -> nil
app2(app2(filter, p), app2(app2(cons, x), xs)) -> app2(app2(app2(if, app2(p, x)), app2(app2(cons, x), app2(app2(filter, p), xs))), app2(app2(filter, p), 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(gtr, app2(s, x)), app2(s, y)) -> APP2(app2(gtr, 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, x2 - 1}


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


POL( s ) = 1



The following usable rules [14] were oriented: none



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

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

app2(app2(app2(if, true), xs), ys) -> xs
app2(app2(app2(if, false), xs), ys) -> ys
app2(app2(sub, x), 0) -> x
app2(app2(sub, app2(s, x)), app2(s, y)) -> app2(app2(sub, x), y)
app2(app2(gtr, 0), y) -> false
app2(app2(gtr, app2(s, x)), 0) -> true
app2(app2(gtr, app2(s, x)), app2(s, y)) -> app2(app2(gtr, x), y)
app2(app2(d, x), 0) -> true
app2(app2(d, app2(s, x)), app2(s, y)) -> app2(app2(app2(if, app2(app2(gtr, x), y)), false), app2(app2(d, app2(s, x)), app2(app2(sub, y), x)))
app2(len, nil) -> 0
app2(len, app2(app2(cons, x), xs)) -> app2(s, app2(len, xs))
app2(app2(filter, p), nil) -> nil
app2(app2(filter, p), app2(app2(cons, x), xs)) -> app2(app2(app2(if, app2(p, x)), app2(app2(cons, x), app2(app2(filter, p), xs))), app2(app2(filter, p), 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
          ↳ QDP

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

APP2(app2(sub, app2(s, x)), app2(s, y)) -> APP2(app2(sub, x), y)

The TRS R consists of the following rules:

app2(app2(app2(if, true), xs), ys) -> xs
app2(app2(app2(if, false), xs), ys) -> ys
app2(app2(sub, x), 0) -> x
app2(app2(sub, app2(s, x)), app2(s, y)) -> app2(app2(sub, x), y)
app2(app2(gtr, 0), y) -> false
app2(app2(gtr, app2(s, x)), 0) -> true
app2(app2(gtr, app2(s, x)), app2(s, y)) -> app2(app2(gtr, x), y)
app2(app2(d, x), 0) -> true
app2(app2(d, app2(s, x)), app2(s, y)) -> app2(app2(app2(if, app2(app2(gtr, x), y)), false), app2(app2(d, app2(s, x)), app2(app2(sub, y), x)))
app2(len, nil) -> 0
app2(len, app2(app2(cons, x), xs)) -> app2(s, app2(len, xs))
app2(app2(filter, p), nil) -> nil
app2(app2(filter, p), app2(app2(cons, x), xs)) -> app2(app2(app2(if, app2(p, x)), app2(app2(cons, x), app2(app2(filter, p), xs))), app2(app2(filter, p), 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(sub, app2(s, x)), app2(s, y)) -> APP2(app2(sub, 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, x2 - 1}


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


POL( s ) = 1



The following usable rules [14] were oriented: none



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

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

app2(app2(app2(if, true), xs), ys) -> xs
app2(app2(app2(if, false), xs), ys) -> ys
app2(app2(sub, x), 0) -> x
app2(app2(sub, app2(s, x)), app2(s, y)) -> app2(app2(sub, x), y)
app2(app2(gtr, 0), y) -> false
app2(app2(gtr, app2(s, x)), 0) -> true
app2(app2(gtr, app2(s, x)), app2(s, y)) -> app2(app2(gtr, x), y)
app2(app2(d, x), 0) -> true
app2(app2(d, app2(s, x)), app2(s, y)) -> app2(app2(app2(if, app2(app2(gtr, x), y)), false), app2(app2(d, app2(s, x)), app2(app2(sub, y), x)))
app2(len, nil) -> 0
app2(len, app2(app2(cons, x), xs)) -> app2(s, app2(len, xs))
app2(app2(filter, p), nil) -> nil
app2(app2(filter, p), app2(app2(cons, x), xs)) -> app2(app2(app2(if, app2(p, x)), app2(app2(cons, x), app2(app2(filter, p), xs))), app2(app2(filter, p), 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
          ↳ QDP

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

APP2(app2(d, app2(s, x)), app2(s, y)) -> APP2(app2(d, app2(s, x)), app2(app2(sub, y), x))

The TRS R consists of the following rules:

app2(app2(app2(if, true), xs), ys) -> xs
app2(app2(app2(if, false), xs), ys) -> ys
app2(app2(sub, x), 0) -> x
app2(app2(sub, app2(s, x)), app2(s, y)) -> app2(app2(sub, x), y)
app2(app2(gtr, 0), y) -> false
app2(app2(gtr, app2(s, x)), 0) -> true
app2(app2(gtr, app2(s, x)), app2(s, y)) -> app2(app2(gtr, x), y)
app2(app2(d, x), 0) -> true
app2(app2(d, app2(s, x)), app2(s, y)) -> app2(app2(app2(if, app2(app2(gtr, x), y)), false), app2(app2(d, app2(s, x)), app2(app2(sub, y), x)))
app2(len, nil) -> 0
app2(len, app2(app2(cons, x), xs)) -> app2(s, app2(len, xs))
app2(app2(filter, p), nil) -> nil
app2(app2(filter, p), app2(app2(cons, x), xs)) -> app2(app2(app2(if, app2(p, x)), app2(app2(cons, x), app2(app2(filter, p), xs))), app2(app2(filter, p), xs))

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

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

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

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

The TRS R consists of the following rules:

app2(app2(app2(if, true), xs), ys) -> xs
app2(app2(app2(if, false), xs), ys) -> ys
app2(app2(sub, x), 0) -> x
app2(app2(sub, app2(s, x)), app2(s, y)) -> app2(app2(sub, x), y)
app2(app2(gtr, 0), y) -> false
app2(app2(gtr, app2(s, x)), 0) -> true
app2(app2(gtr, app2(s, x)), app2(s, y)) -> app2(app2(gtr, x), y)
app2(app2(d, x), 0) -> true
app2(app2(d, app2(s, x)), app2(s, y)) -> app2(app2(app2(if, app2(app2(gtr, x), y)), false), app2(app2(d, app2(s, x)), app2(app2(sub, y), x)))
app2(len, nil) -> 0
app2(len, app2(app2(cons, x), xs)) -> app2(s, app2(len, xs))
app2(app2(filter, p), nil) -> nil
app2(app2(filter, p), app2(app2(cons, x), xs)) -> app2(app2(app2(if, app2(p, x)), app2(app2(cons, x), app2(app2(filter, p), xs))), app2(app2(filter, p), 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, p), app2(app2(cons, x), xs)) -> APP2(p, x)
APP2(app2(filter, p), app2(app2(cons, x), xs)) -> APP2(app2(filter, p), xs)
The remaining pairs can at least be oriented weakly.
none
Used ordering: Polynomial Order [17,21] with Interpretation:

POL( APP2(x1, x2) ) = max{0, x2 - 1}


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


POL( cons ) = 0



The following usable rules [14] were oriented: none



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

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

app2(app2(app2(if, true), xs), ys) -> xs
app2(app2(app2(if, false), xs), ys) -> ys
app2(app2(sub, x), 0) -> x
app2(app2(sub, app2(s, x)), app2(s, y)) -> app2(app2(sub, x), y)
app2(app2(gtr, 0), y) -> false
app2(app2(gtr, app2(s, x)), 0) -> true
app2(app2(gtr, app2(s, x)), app2(s, y)) -> app2(app2(gtr, x), y)
app2(app2(d, x), 0) -> true
app2(app2(d, app2(s, x)), app2(s, y)) -> app2(app2(app2(if, app2(app2(gtr, x), y)), false), app2(app2(d, app2(s, x)), app2(app2(sub, y), x)))
app2(len, nil) -> 0
app2(len, app2(app2(cons, x), xs)) -> app2(s, app2(len, xs))
app2(app2(filter, p), nil) -> nil
app2(app2(filter, p), app2(app2(cons, x), xs)) -> app2(app2(app2(if, app2(p, x)), app2(app2(cons, x), app2(app2(filter, p), xs))), app2(app2(filter, p), 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.