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
  ↳ Non-Overlap Check

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.

The TRS is non-overlapping. Hence, we can switch to innermost.

↳ QTRS
  ↳ Non-Overlap Check
    ↳ 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))

The set Q consists of the following terms:

app2(app2(app2(if, true), x0), x1)
app2(app2(app2(if, false), x0), x1)
app2(app2(sub, x0), 0)
app2(app2(sub, app2(s, x0)), app2(s, x1))
app2(app2(gtr, 0), x0)
app2(app2(gtr, app2(s, x0)), 0)
app2(app2(gtr, app2(s, x0)), app2(s, x1))
app2(app2(d, x0), 0)
app2(app2(d, app2(s, x0)), app2(s, x1))
app2(len, nil)
app2(len, app2(app2(cons, x0), x1))
app2(app2(filter, x0), nil)
app2(app2(filter, x0), app2(app2(cons, x1), x2))


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))

The set Q consists of the following terms:

app2(app2(app2(if, true), x0), x1)
app2(app2(app2(if, false), x0), x1)
app2(app2(sub, x0), 0)
app2(app2(sub, app2(s, x0)), app2(s, x1))
app2(app2(gtr, 0), x0)
app2(app2(gtr, app2(s, x0)), 0)
app2(app2(gtr, app2(s, x0)), app2(s, x1))
app2(app2(d, x0), 0)
app2(app2(d, app2(s, x0)), app2(s, x1))
app2(len, nil)
app2(len, app2(app2(cons, x0), x1))
app2(app2(filter, x0), nil)
app2(app2(filter, x0), app2(app2(cons, x1), x2))

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

↳ QTRS
  ↳ Non-Overlap Check
    ↳ 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))

The set Q consists of the following terms:

app2(app2(app2(if, true), x0), x1)
app2(app2(app2(if, false), x0), x1)
app2(app2(sub, x0), 0)
app2(app2(sub, app2(s, x0)), app2(s, x1))
app2(app2(gtr, 0), x0)
app2(app2(gtr, app2(s, x0)), 0)
app2(app2(gtr, app2(s, x0)), app2(s, x1))
app2(app2(d, x0), 0)
app2(app2(d, app2(s, x0)), app2(s, x1))
app2(len, nil)
app2(len, app2(app2(cons, x0), x1))
app2(app2(filter, x0), nil)
app2(app2(filter, x0), app2(app2(cons, x1), x2))

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
  ↳ Non-Overlap Check
    ↳ 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))

The set Q consists of the following terms:

app2(app2(app2(if, true), x0), x1)
app2(app2(app2(if, false), x0), x1)
app2(app2(sub, x0), 0)
app2(app2(sub, app2(s, x0)), app2(s, x1))
app2(app2(gtr, 0), x0)
app2(app2(gtr, app2(s, x0)), 0)
app2(app2(gtr, app2(s, x0)), app2(s, x1))
app2(app2(d, x0), 0)
app2(app2(d, app2(s, x0)), app2(s, x1))
app2(len, nil)
app2(len, app2(app2(cons, x0), x1))
app2(app2(filter, x0), nil)
app2(app2(filter, x0), app2(app2(cons, x1), x2))

We have to consider all minimal (P,Q,R)-chains.
We use the reduction pair processor [13].


The following pairs can be strictly oriented and are deleted.


APP2(len, app2(app2(cons, x), xs)) -> APP2(len, xs)
The remaining pairs can at least by weakly be oriented.
none
Used ordering: Combined order from the following AFS and order.
APP2(x1, x2)  =  APP1(x2)
len  =  len
app2(x1, x2)  =  app1(x2)
cons  =  cons

Lexicographic Path Order [19].
Precedence:
[APP1, app1] > len
cons > len


The following usable rules [14] were oriented: none



↳ QTRS
  ↳ Non-Overlap Check
    ↳ 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))

The set Q consists of the following terms:

app2(app2(app2(if, true), x0), x1)
app2(app2(app2(if, false), x0), x1)
app2(app2(sub, x0), 0)
app2(app2(sub, app2(s, x0)), app2(s, x1))
app2(app2(gtr, 0), x0)
app2(app2(gtr, app2(s, x0)), 0)
app2(app2(gtr, app2(s, x0)), app2(s, x1))
app2(app2(d, x0), 0)
app2(app2(d, app2(s, x0)), app2(s, x1))
app2(len, nil)
app2(len, app2(app2(cons, x0), x1))
app2(app2(filter, x0), nil)
app2(app2(filter, x0), app2(app2(cons, x1), x2))

We have to consider all minimal (P,Q,R)-chains.
The TRS P is empty. Hence, there is no (P,Q,R) chain.

↳ QTRS
  ↳ Non-Overlap Check
    ↳ 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))

The set Q consists of the following terms:

app2(app2(app2(if, true), x0), x1)
app2(app2(app2(if, false), x0), x1)
app2(app2(sub, x0), 0)
app2(app2(sub, app2(s, x0)), app2(s, x1))
app2(app2(gtr, 0), x0)
app2(app2(gtr, app2(s, x0)), 0)
app2(app2(gtr, app2(s, x0)), app2(s, x1))
app2(app2(d, x0), 0)
app2(app2(d, app2(s, x0)), app2(s, x1))
app2(len, nil)
app2(len, app2(app2(cons, x0), x1))
app2(app2(filter, x0), nil)
app2(app2(filter, x0), app2(app2(cons, x1), x2))

We have to consider all minimal (P,Q,R)-chains.
We use the reduction pair processor [13].


The following pairs can be strictly oriented and are deleted.


APP2(app2(gtr, app2(s, x)), app2(s, y)) -> APP2(app2(gtr, x), y)
The remaining pairs can at least by weakly be oriented.
none
Used ordering: Combined order from the following AFS and order.
APP2(x1, x2)  =  x1
app2(x1, x2)  =  app2(x1, x2)
gtr  =  gtr
s  =  s

Lexicographic Path Order [19].
Precedence:
trivial


The following usable rules [14] were oriented: none



↳ QTRS
  ↳ Non-Overlap Check
    ↳ 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))

The set Q consists of the following terms:

app2(app2(app2(if, true), x0), x1)
app2(app2(app2(if, false), x0), x1)
app2(app2(sub, x0), 0)
app2(app2(sub, app2(s, x0)), app2(s, x1))
app2(app2(gtr, 0), x0)
app2(app2(gtr, app2(s, x0)), 0)
app2(app2(gtr, app2(s, x0)), app2(s, x1))
app2(app2(d, x0), 0)
app2(app2(d, app2(s, x0)), app2(s, x1))
app2(len, nil)
app2(len, app2(app2(cons, x0), x1))
app2(app2(filter, x0), nil)
app2(app2(filter, x0), app2(app2(cons, x1), x2))

We have to consider all minimal (P,Q,R)-chains.
The TRS P is empty. Hence, there is no (P,Q,R) chain.

↳ QTRS
  ↳ Non-Overlap Check
    ↳ 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))

The set Q consists of the following terms:

app2(app2(app2(if, true), x0), x1)
app2(app2(app2(if, false), x0), x1)
app2(app2(sub, x0), 0)
app2(app2(sub, app2(s, x0)), app2(s, x1))
app2(app2(gtr, 0), x0)
app2(app2(gtr, app2(s, x0)), 0)
app2(app2(gtr, app2(s, x0)), app2(s, x1))
app2(app2(d, x0), 0)
app2(app2(d, app2(s, x0)), app2(s, x1))
app2(len, nil)
app2(len, app2(app2(cons, x0), x1))
app2(app2(filter, x0), nil)
app2(app2(filter, x0), app2(app2(cons, x1), x2))

We have to consider all minimal (P,Q,R)-chains.
We use the reduction pair processor [13].


The following pairs can be strictly oriented and are deleted.


APP2(app2(sub, app2(s, x)), app2(s, y)) -> APP2(app2(sub, x), y)
The remaining pairs can at least by weakly be oriented.
none
Used ordering: Combined order from the following AFS and order.
APP2(x1, x2)  =  x1
app2(x1, x2)  =  app2(x1, x2)
sub  =  sub
s  =  s

Lexicographic Path Order [19].
Precedence:
trivial


The following usable rules [14] were oriented: none



↳ QTRS
  ↳ Non-Overlap Check
    ↳ 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))

The set Q consists of the following terms:

app2(app2(app2(if, true), x0), x1)
app2(app2(app2(if, false), x0), x1)
app2(app2(sub, x0), 0)
app2(app2(sub, app2(s, x0)), app2(s, x1))
app2(app2(gtr, 0), x0)
app2(app2(gtr, app2(s, x0)), 0)
app2(app2(gtr, app2(s, x0)), app2(s, x1))
app2(app2(d, x0), 0)
app2(app2(d, app2(s, x0)), app2(s, x1))
app2(len, nil)
app2(len, app2(app2(cons, x0), x1))
app2(app2(filter, x0), nil)
app2(app2(filter, x0), app2(app2(cons, x1), x2))

We have to consider all minimal (P,Q,R)-chains.
The TRS P is empty. Hence, there is no (P,Q,R) chain.

↳ QTRS
  ↳ Non-Overlap Check
    ↳ 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))

The set Q consists of the following terms:

app2(app2(app2(if, true), x0), x1)
app2(app2(app2(if, false), x0), x1)
app2(app2(sub, x0), 0)
app2(app2(sub, app2(s, x0)), app2(s, x1))
app2(app2(gtr, 0), x0)
app2(app2(gtr, app2(s, x0)), 0)
app2(app2(gtr, app2(s, x0)), app2(s, x1))
app2(app2(d, x0), 0)
app2(app2(d, app2(s, x0)), app2(s, x1))
app2(len, nil)
app2(len, app2(app2(cons, x0), x1))
app2(app2(filter, x0), nil)
app2(app2(filter, x0), app2(app2(cons, x1), x2))

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

↳ QTRS
  ↳ Non-Overlap Check
    ↳ 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))

The set Q consists of the following terms:

app2(app2(app2(if, true), x0), x1)
app2(app2(app2(if, false), x0), x1)
app2(app2(sub, x0), 0)
app2(app2(sub, app2(s, x0)), app2(s, x1))
app2(app2(gtr, 0), x0)
app2(app2(gtr, app2(s, x0)), 0)
app2(app2(gtr, app2(s, x0)), app2(s, x1))
app2(app2(d, x0), 0)
app2(app2(d, app2(s, x0)), app2(s, x1))
app2(len, nil)
app2(len, app2(app2(cons, x0), x1))
app2(app2(filter, x0), nil)
app2(app2(filter, x0), app2(app2(cons, x1), x2))

We have to consider all minimal (P,Q,R)-chains.
We use the reduction pair processor [13].


The following pairs can be strictly oriented and are deleted.


APP2(app2(filter, p), app2(app2(cons, x), xs)) -> APP2(p, x)
The remaining pairs can at least by weakly be oriented.

APP2(app2(filter, p), app2(app2(cons, x), xs)) -> APP2(app2(filter, p), xs)
Used ordering: Combined order from the following AFS and order.
APP2(x1, x2)  =  APP1(x1)
app2(x1, x2)  =  app1(x2)
filter  =  filter
cons  =  cons

Lexicographic Path Order [19].
Precedence:
APP1 > filter
app1 > filter
cons > filter


The following usable rules [14] were oriented: none



↳ QTRS
  ↳ Non-Overlap Check
    ↳ QTRS
      ↳ DependencyPairsProof
        ↳ QDP
          ↳ DependencyGraphProof
            ↳ AND
              ↳ QDP
              ↳ QDP
              ↳ QDP
              ↳ QDP
              ↳ QDP
                ↳ QDPOrderProof
                  ↳ QDP
                    ↳ QDPOrderProof

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

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))

The set Q consists of the following terms:

app2(app2(app2(if, true), x0), x1)
app2(app2(app2(if, false), x0), x1)
app2(app2(sub, x0), 0)
app2(app2(sub, app2(s, x0)), app2(s, x1))
app2(app2(gtr, 0), x0)
app2(app2(gtr, app2(s, x0)), 0)
app2(app2(gtr, app2(s, x0)), app2(s, x1))
app2(app2(d, x0), 0)
app2(app2(d, app2(s, x0)), app2(s, x1))
app2(len, nil)
app2(len, app2(app2(cons, x0), x1))
app2(app2(filter, x0), nil)
app2(app2(filter, x0), app2(app2(cons, x1), x2))

We have to consider all minimal (P,Q,R)-chains.
We use the reduction pair processor [13].


The following pairs can be strictly oriented and are deleted.


APP2(app2(filter, p), app2(app2(cons, x), xs)) -> APP2(app2(filter, p), xs)
The remaining pairs can at least by weakly be oriented.
none
Used ordering: Combined order from the following AFS and order.
APP2(x1, x2)  =  APP1(x2)
app2(x1, x2)  =  app1(x2)
filter  =  filter
cons  =  cons

Lexicographic Path Order [19].
Precedence:
[APP1, app1] > [filter, cons]


The following usable rules [14] were oriented: none



↳ QTRS
  ↳ Non-Overlap Check
    ↳ QTRS
      ↳ DependencyPairsProof
        ↳ QDP
          ↳ DependencyGraphProof
            ↳ AND
              ↳ QDP
              ↳ QDP
              ↳ QDP
              ↳ QDP
              ↳ QDP
                ↳ QDPOrderProof
                  ↳ 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))

The set Q consists of the following terms:

app2(app2(app2(if, true), x0), x1)
app2(app2(app2(if, false), x0), x1)
app2(app2(sub, x0), 0)
app2(app2(sub, app2(s, x0)), app2(s, x1))
app2(app2(gtr, 0), x0)
app2(app2(gtr, app2(s, x0)), 0)
app2(app2(gtr, app2(s, x0)), app2(s, x1))
app2(app2(d, x0), 0)
app2(app2(d, app2(s, x0)), app2(s, x1))
app2(len, nil)
app2(len, app2(app2(cons, x0), x1))
app2(app2(filter, x0), nil)
app2(app2(filter, x0), app2(app2(cons, x1), x2))

We have to consider all minimal (P,Q,R)-chains.
The TRS P is empty. Hence, there is no (P,Q,R) chain.