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(minus, x), 0) -> x
app2(app2(minus, app2(s, x)), app2(s, y)) -> app2(app2(minus, x), y)
app2(app2(quot, 0), app2(s, y)) -> 0
app2(app2(quot, app2(s, x)), app2(s, y)) -> app2(s, app2(app2(quot, app2(app2(minus, x), y)), app2(s, y)))
app2(app2(plus, 0), y) -> y
app2(app2(plus, app2(s, x)), y) -> app2(s, app2(app2(plus, x), y))
app2(app2(plus, app2(app2(minus, x), app2(s, 0))), app2(app2(minus, y), app2(s, app2(s, z)))) -> app2(app2(plus, app2(app2(minus, y), app2(s, app2(s, z)))), app2(app2(minus, x), app2(s, 0)))
app2(app2(plus, app2(app2(plus, x), app2(s, 0))), app2(app2(plus, y), app2(s, app2(s, z)))) -> app2(app2(plus, app2(app2(plus, y), app2(s, app2(s, z)))), app2(app2(plus, x), app2(s, 0)))
app2(app2(map, f), nil) -> nil
app2(app2(map, f), app2(app2(cons, x), xs)) -> app2(app2(cons, app2(f, x)), app2(app2(map, f), xs))
app2(app2(filter, f), nil) -> nil
app2(app2(filter, f), app2(app2(cons, x), xs)) -> app2(app2(app2(app2(filter2, app2(f, x)), f), x), xs)
app2(app2(app2(app2(filter2, true), f), x), xs) -> app2(app2(cons, x), app2(app2(filter, f), xs))
app2(app2(app2(app2(filter2, false), f), x), 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(minus, x), 0) -> x
app2(app2(minus, app2(s, x)), app2(s, y)) -> app2(app2(minus, x), y)
app2(app2(quot, 0), app2(s, y)) -> 0
app2(app2(quot, app2(s, x)), app2(s, y)) -> app2(s, app2(app2(quot, app2(app2(minus, x), y)), app2(s, y)))
app2(app2(plus, 0), y) -> y
app2(app2(plus, app2(s, x)), y) -> app2(s, app2(app2(plus, x), y))
app2(app2(plus, app2(app2(minus, x), app2(s, 0))), app2(app2(minus, y), app2(s, app2(s, z)))) -> app2(app2(plus, app2(app2(minus, y), app2(s, app2(s, z)))), app2(app2(minus, x), app2(s, 0)))
app2(app2(plus, app2(app2(plus, x), app2(s, 0))), app2(app2(plus, y), app2(s, app2(s, z)))) -> app2(app2(plus, app2(app2(plus, y), app2(s, app2(s, z)))), app2(app2(plus, x), app2(s, 0)))
app2(app2(map, f), nil) -> nil
app2(app2(map, f), app2(app2(cons, x), xs)) -> app2(app2(cons, app2(f, x)), app2(app2(map, f), xs))
app2(app2(filter, f), nil) -> nil
app2(app2(filter, f), app2(app2(cons, x), xs)) -> app2(app2(app2(app2(filter2, app2(f, x)), f), x), xs)
app2(app2(app2(app2(filter2, true), f), x), xs) -> app2(app2(cons, x), app2(app2(filter, f), xs))
app2(app2(app2(app2(filter2, false), f), x), 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(minus, app2(s, x)), app2(s, y)) -> APP2(minus, x)
APP2(app2(plus, app2(s, x)), y) -> APP2(s, app2(app2(plus, x), y))
APP2(app2(map, f), app2(app2(cons, x), xs)) -> APP2(app2(cons, app2(f, x)), app2(app2(map, f), xs))
APP2(app2(quot, app2(s, x)), app2(s, y)) -> APP2(minus, x)
APP2(app2(quot, app2(s, x)), app2(s, y)) -> APP2(app2(minus, x), y)
APP2(app2(filter, f), app2(app2(cons, x), xs)) -> APP2(app2(app2(filter2, app2(f, x)), f), x)
APP2(app2(filter, f), app2(app2(cons, x), xs)) -> APP2(app2(app2(app2(filter2, app2(f, x)), f), x), xs)
APP2(app2(plus, app2(app2(plus, x), app2(s, 0))), app2(app2(plus, y), app2(s, app2(s, z)))) -> APP2(app2(plus, app2(app2(plus, y), app2(s, app2(s, z)))), app2(app2(plus, x), app2(s, 0)))
APP2(app2(app2(app2(filter2, true), f), x), xs) -> APP2(app2(cons, x), app2(app2(filter, f), xs))
APP2(app2(quot, app2(s, x)), app2(s, y)) -> APP2(quot, app2(app2(minus, x), y))
APP2(app2(map, f), app2(app2(cons, x), xs)) -> APP2(f, x)
APP2(app2(filter, f), app2(app2(cons, x), xs)) -> APP2(filter2, app2(f, x))
APP2(app2(plus, app2(s, x)), y) -> APP2(plus, x)
APP2(app2(minus, app2(s, x)), app2(s, y)) -> APP2(app2(minus, x), y)
APP2(app2(quot, app2(s, x)), app2(s, y)) -> APP2(app2(quot, app2(app2(minus, x), y)), app2(s, y))
APP2(app2(plus, app2(app2(minus, x), app2(s, 0))), app2(app2(minus, y), app2(s, app2(s, z)))) -> APP2(app2(plus, app2(app2(minus, y), app2(s, app2(s, z)))), app2(app2(minus, x), app2(s, 0)))
APP2(app2(quot, app2(s, x)), app2(s, y)) -> APP2(s, app2(app2(quot, app2(app2(minus, x), y)), app2(s, y)))
APP2(app2(app2(app2(filter2, true), f), x), xs) -> APP2(app2(filter, f), xs)
APP2(app2(filter, f), app2(app2(cons, x), xs)) -> APP2(app2(filter2, app2(f, x)), f)
APP2(app2(app2(app2(filter2, false), f), x), xs) -> APP2(filter, f)
APP2(app2(map, f), app2(app2(cons, x), xs)) -> APP2(app2(map, f), xs)
APP2(app2(filter, f), app2(app2(cons, x), xs)) -> APP2(f, x)
APP2(app2(app2(app2(filter2, true), f), x), xs) -> APP2(filter, f)
APP2(app2(plus, app2(s, x)), y) -> APP2(app2(plus, x), y)
APP2(app2(plus, app2(app2(minus, x), app2(s, 0))), app2(app2(minus, y), app2(s, app2(s, z)))) -> APP2(plus, app2(app2(minus, y), app2(s, app2(s, z))))
APP2(app2(app2(app2(filter2, true), f), x), xs) -> APP2(cons, x)
APP2(app2(plus, app2(app2(plus, x), app2(s, 0))), app2(app2(plus, y), app2(s, app2(s, z)))) -> APP2(plus, app2(app2(plus, y), app2(s, app2(s, z))))
APP2(app2(app2(app2(filter2, false), f), x), xs) -> APP2(app2(filter, f), xs)
APP2(app2(map, f), app2(app2(cons, x), xs)) -> APP2(cons, app2(f, x))

The TRS R consists of the following rules:

app2(app2(minus, x), 0) -> x
app2(app2(minus, app2(s, x)), app2(s, y)) -> app2(app2(minus, x), y)
app2(app2(quot, 0), app2(s, y)) -> 0
app2(app2(quot, app2(s, x)), app2(s, y)) -> app2(s, app2(app2(quot, app2(app2(minus, x), y)), app2(s, y)))
app2(app2(plus, 0), y) -> y
app2(app2(plus, app2(s, x)), y) -> app2(s, app2(app2(plus, x), y))
app2(app2(plus, app2(app2(minus, x), app2(s, 0))), app2(app2(minus, y), app2(s, app2(s, z)))) -> app2(app2(plus, app2(app2(minus, y), app2(s, app2(s, z)))), app2(app2(minus, x), app2(s, 0)))
app2(app2(plus, app2(app2(plus, x), app2(s, 0))), app2(app2(plus, y), app2(s, app2(s, z)))) -> app2(app2(plus, app2(app2(plus, y), app2(s, app2(s, z)))), app2(app2(plus, x), app2(s, 0)))
app2(app2(map, f), nil) -> nil
app2(app2(map, f), app2(app2(cons, x), xs)) -> app2(app2(cons, app2(f, x)), app2(app2(map, f), xs))
app2(app2(filter, f), nil) -> nil
app2(app2(filter, f), app2(app2(cons, x), xs)) -> app2(app2(app2(app2(filter2, app2(f, x)), f), x), xs)
app2(app2(app2(app2(filter2, true), f), x), xs) -> app2(app2(cons, x), app2(app2(filter, f), xs))
app2(app2(app2(app2(filter2, false), f), x), 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(minus, app2(s, x)), app2(s, y)) -> APP2(minus, x)
APP2(app2(plus, app2(s, x)), y) -> APP2(s, app2(app2(plus, x), y))
APP2(app2(map, f), app2(app2(cons, x), xs)) -> APP2(app2(cons, app2(f, x)), app2(app2(map, f), xs))
APP2(app2(quot, app2(s, x)), app2(s, y)) -> APP2(minus, x)
APP2(app2(quot, app2(s, x)), app2(s, y)) -> APP2(app2(minus, x), y)
APP2(app2(filter, f), app2(app2(cons, x), xs)) -> APP2(app2(app2(filter2, app2(f, x)), f), x)
APP2(app2(filter, f), app2(app2(cons, x), xs)) -> APP2(app2(app2(app2(filter2, app2(f, x)), f), x), xs)
APP2(app2(plus, app2(app2(plus, x), app2(s, 0))), app2(app2(plus, y), app2(s, app2(s, z)))) -> APP2(app2(plus, app2(app2(plus, y), app2(s, app2(s, z)))), app2(app2(plus, x), app2(s, 0)))
APP2(app2(app2(app2(filter2, true), f), x), xs) -> APP2(app2(cons, x), app2(app2(filter, f), xs))
APP2(app2(quot, app2(s, x)), app2(s, y)) -> APP2(quot, app2(app2(minus, x), y))
APP2(app2(map, f), app2(app2(cons, x), xs)) -> APP2(f, x)
APP2(app2(filter, f), app2(app2(cons, x), xs)) -> APP2(filter2, app2(f, x))
APP2(app2(plus, app2(s, x)), y) -> APP2(plus, x)
APP2(app2(minus, app2(s, x)), app2(s, y)) -> APP2(app2(minus, x), y)
APP2(app2(quot, app2(s, x)), app2(s, y)) -> APP2(app2(quot, app2(app2(minus, x), y)), app2(s, y))
APP2(app2(plus, app2(app2(minus, x), app2(s, 0))), app2(app2(minus, y), app2(s, app2(s, z)))) -> APP2(app2(plus, app2(app2(minus, y), app2(s, app2(s, z)))), app2(app2(minus, x), app2(s, 0)))
APP2(app2(quot, app2(s, x)), app2(s, y)) -> APP2(s, app2(app2(quot, app2(app2(minus, x), y)), app2(s, y)))
APP2(app2(app2(app2(filter2, true), f), x), xs) -> APP2(app2(filter, f), xs)
APP2(app2(filter, f), app2(app2(cons, x), xs)) -> APP2(app2(filter2, app2(f, x)), f)
APP2(app2(app2(app2(filter2, false), f), x), xs) -> APP2(filter, f)
APP2(app2(map, f), app2(app2(cons, x), xs)) -> APP2(app2(map, f), xs)
APP2(app2(filter, f), app2(app2(cons, x), xs)) -> APP2(f, x)
APP2(app2(app2(app2(filter2, true), f), x), xs) -> APP2(filter, f)
APP2(app2(plus, app2(s, x)), y) -> APP2(app2(plus, x), y)
APP2(app2(plus, app2(app2(minus, x), app2(s, 0))), app2(app2(minus, y), app2(s, app2(s, z)))) -> APP2(plus, app2(app2(minus, y), app2(s, app2(s, z))))
APP2(app2(app2(app2(filter2, true), f), x), xs) -> APP2(cons, x)
APP2(app2(plus, app2(app2(plus, x), app2(s, 0))), app2(app2(plus, y), app2(s, app2(s, z)))) -> APP2(plus, app2(app2(plus, y), app2(s, app2(s, z))))
APP2(app2(app2(app2(filter2, false), f), x), xs) -> APP2(app2(filter, f), xs)
APP2(app2(map, f), app2(app2(cons, x), xs)) -> APP2(cons, app2(f, x))

The TRS R consists of the following rules:

app2(app2(minus, x), 0) -> x
app2(app2(minus, app2(s, x)), app2(s, y)) -> app2(app2(minus, x), y)
app2(app2(quot, 0), app2(s, y)) -> 0
app2(app2(quot, app2(s, x)), app2(s, y)) -> app2(s, app2(app2(quot, app2(app2(minus, x), y)), app2(s, y)))
app2(app2(plus, 0), y) -> y
app2(app2(plus, app2(s, x)), y) -> app2(s, app2(app2(plus, x), y))
app2(app2(plus, app2(app2(minus, x), app2(s, 0))), app2(app2(minus, y), app2(s, app2(s, z)))) -> app2(app2(plus, app2(app2(minus, y), app2(s, app2(s, z)))), app2(app2(minus, x), app2(s, 0)))
app2(app2(plus, app2(app2(plus, x), app2(s, 0))), app2(app2(plus, y), app2(s, app2(s, z)))) -> app2(app2(plus, app2(app2(plus, y), app2(s, app2(s, z)))), app2(app2(plus, x), app2(s, 0)))
app2(app2(map, f), nil) -> nil
app2(app2(map, f), app2(app2(cons, x), xs)) -> app2(app2(cons, app2(f, x)), app2(app2(map, f), xs))
app2(app2(filter, f), nil) -> nil
app2(app2(filter, f), app2(app2(cons, x), xs)) -> app2(app2(app2(app2(filter2, app2(f, x)), f), x), xs)
app2(app2(app2(app2(filter2, true), f), x), xs) -> app2(app2(cons, x), app2(app2(filter, f), xs))
app2(app2(app2(app2(filter2, false), f), x), 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 4 SCCs with 18 less nodes.

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

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

APP2(app2(plus, app2(s, x)), y) -> APP2(app2(plus, x), y)
APP2(app2(plus, app2(app2(plus, x), app2(s, 0))), app2(app2(plus, y), app2(s, app2(s, z)))) -> APP2(app2(plus, app2(app2(plus, y), app2(s, app2(s, z)))), app2(app2(plus, x), app2(s, 0)))
APP2(app2(plus, app2(app2(minus, x), app2(s, 0))), app2(app2(minus, y), app2(s, app2(s, z)))) -> APP2(app2(plus, app2(app2(minus, y), app2(s, app2(s, z)))), app2(app2(minus, x), app2(s, 0)))

The TRS R consists of the following rules:

app2(app2(minus, x), 0) -> x
app2(app2(minus, app2(s, x)), app2(s, y)) -> app2(app2(minus, x), y)
app2(app2(quot, 0), app2(s, y)) -> 0
app2(app2(quot, app2(s, x)), app2(s, y)) -> app2(s, app2(app2(quot, app2(app2(minus, x), y)), app2(s, y)))
app2(app2(plus, 0), y) -> y
app2(app2(plus, app2(s, x)), y) -> app2(s, app2(app2(plus, x), y))
app2(app2(plus, app2(app2(minus, x), app2(s, 0))), app2(app2(minus, y), app2(s, app2(s, z)))) -> app2(app2(plus, app2(app2(minus, y), app2(s, app2(s, z)))), app2(app2(minus, x), app2(s, 0)))
app2(app2(plus, app2(app2(plus, x), app2(s, 0))), app2(app2(plus, y), app2(s, app2(s, z)))) -> app2(app2(plus, app2(app2(plus, y), app2(s, app2(s, z)))), app2(app2(plus, x), app2(s, 0)))
app2(app2(map, f), nil) -> nil
app2(app2(map, f), app2(app2(cons, x), xs)) -> app2(app2(cons, app2(f, x)), app2(app2(map, f), xs))
app2(app2(filter, f), nil) -> nil
app2(app2(filter, f), app2(app2(cons, x), xs)) -> app2(app2(app2(app2(filter2, app2(f, x)), f), x), xs)
app2(app2(app2(app2(filter2, true), f), x), xs) -> app2(app2(cons, x), app2(app2(filter, f), xs))
app2(app2(app2(app2(filter2, false), f), x), xs) -> app2(app2(filter, f), xs)

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

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

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

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

The TRS R consists of the following rules:

app2(app2(minus, x), 0) -> x
app2(app2(minus, app2(s, x)), app2(s, y)) -> app2(app2(minus, x), y)
app2(app2(quot, 0), app2(s, y)) -> 0
app2(app2(quot, app2(s, x)), app2(s, y)) -> app2(s, app2(app2(quot, app2(app2(minus, x), y)), app2(s, y)))
app2(app2(plus, 0), y) -> y
app2(app2(plus, app2(s, x)), y) -> app2(s, app2(app2(plus, x), y))
app2(app2(plus, app2(app2(minus, x), app2(s, 0))), app2(app2(minus, y), app2(s, app2(s, z)))) -> app2(app2(plus, app2(app2(minus, y), app2(s, app2(s, z)))), app2(app2(minus, x), app2(s, 0)))
app2(app2(plus, app2(app2(plus, x), app2(s, 0))), app2(app2(plus, y), app2(s, app2(s, z)))) -> app2(app2(plus, app2(app2(plus, y), app2(s, app2(s, z)))), app2(app2(plus, x), app2(s, 0)))
app2(app2(map, f), nil) -> nil
app2(app2(map, f), app2(app2(cons, x), xs)) -> app2(app2(cons, app2(f, x)), app2(app2(map, f), xs))
app2(app2(filter, f), nil) -> nil
app2(app2(filter, f), app2(app2(cons, x), xs)) -> app2(app2(app2(app2(filter2, app2(f, x)), f), x), xs)
app2(app2(app2(app2(filter2, true), f), x), xs) -> app2(app2(cons, x), app2(app2(filter, f), xs))
app2(app2(app2(app2(filter2, false), f), x), xs) -> app2(app2(filter, f), xs)

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

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

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

APP2(app2(quot, app2(s, x)), app2(s, y)) -> APP2(app2(quot, app2(app2(minus, x), y)), app2(s, y))

The TRS R consists of the following rules:

app2(app2(minus, x), 0) -> x
app2(app2(minus, app2(s, x)), app2(s, y)) -> app2(app2(minus, x), y)
app2(app2(quot, 0), app2(s, y)) -> 0
app2(app2(quot, app2(s, x)), app2(s, y)) -> app2(s, app2(app2(quot, app2(app2(minus, x), y)), app2(s, y)))
app2(app2(plus, 0), y) -> y
app2(app2(plus, app2(s, x)), y) -> app2(s, app2(app2(plus, x), y))
app2(app2(plus, app2(app2(minus, x), app2(s, 0))), app2(app2(minus, y), app2(s, app2(s, z)))) -> app2(app2(plus, app2(app2(minus, y), app2(s, app2(s, z)))), app2(app2(minus, x), app2(s, 0)))
app2(app2(plus, app2(app2(plus, x), app2(s, 0))), app2(app2(plus, y), app2(s, app2(s, z)))) -> app2(app2(plus, app2(app2(plus, y), app2(s, app2(s, z)))), app2(app2(plus, x), app2(s, 0)))
app2(app2(map, f), nil) -> nil
app2(app2(map, f), app2(app2(cons, x), xs)) -> app2(app2(cons, app2(f, x)), app2(app2(map, f), xs))
app2(app2(filter, f), nil) -> nil
app2(app2(filter, f), app2(app2(cons, x), xs)) -> app2(app2(app2(app2(filter2, app2(f, x)), f), x), xs)
app2(app2(app2(app2(filter2, true), f), x), xs) -> app2(app2(cons, x), app2(app2(filter, f), xs))
app2(app2(app2(app2(filter2, false), f), x), xs) -> app2(app2(filter, f), xs)

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

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

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

APP2(app2(app2(app2(filter2, true), f), x), xs) -> APP2(app2(filter, f), xs)
APP2(app2(map, f), app2(app2(cons, x), xs)) -> APP2(f, x)
APP2(app2(filter, f), app2(app2(cons, x), xs)) -> APP2(app2(app2(app2(filter2, app2(f, x)), f), x), xs)
APP2(app2(app2(app2(filter2, false), f), x), xs) -> APP2(app2(filter, f), xs)
APP2(app2(filter, f), app2(app2(cons, x), xs)) -> APP2(f, x)
APP2(app2(map, f), app2(app2(cons, x), xs)) -> APP2(app2(map, f), xs)

The TRS R consists of the following rules:

app2(app2(minus, x), 0) -> x
app2(app2(minus, app2(s, x)), app2(s, y)) -> app2(app2(minus, x), y)
app2(app2(quot, 0), app2(s, y)) -> 0
app2(app2(quot, app2(s, x)), app2(s, y)) -> app2(s, app2(app2(quot, app2(app2(minus, x), y)), app2(s, y)))
app2(app2(plus, 0), y) -> y
app2(app2(plus, app2(s, x)), y) -> app2(s, app2(app2(plus, x), y))
app2(app2(plus, app2(app2(minus, x), app2(s, 0))), app2(app2(minus, y), app2(s, app2(s, z)))) -> app2(app2(plus, app2(app2(minus, y), app2(s, app2(s, z)))), app2(app2(minus, x), app2(s, 0)))
app2(app2(plus, app2(app2(plus, x), app2(s, 0))), app2(app2(plus, y), app2(s, app2(s, z)))) -> app2(app2(plus, app2(app2(plus, y), app2(s, app2(s, z)))), app2(app2(plus, x), app2(s, 0)))
app2(app2(map, f), nil) -> nil
app2(app2(map, f), app2(app2(cons, x), xs)) -> app2(app2(cons, app2(f, x)), app2(app2(map, f), xs))
app2(app2(filter, f), nil) -> nil
app2(app2(filter, f), app2(app2(cons, x), xs)) -> app2(app2(app2(app2(filter2, app2(f, x)), f), x), xs)
app2(app2(app2(app2(filter2, true), f), x), xs) -> app2(app2(cons, x), app2(app2(filter, f), xs))
app2(app2(app2(app2(filter2, false), f), x), xs) -> app2(app2(filter, f), xs)

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