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:

app'2(app'2(minus, x), 0) -> x
app'2(app'2(minus, app'2(s, x)), app'2(s, y)) -> app'2(app'2(minus, x), y)
app'2(app'2(minus, app'2(app'2(minus, x), y)), z) -> app'2(app'2(minus, x), app'2(app'2(plus, y), z))
app'2(app'2(quot, 0), app'2(s, y)) -> 0
app'2(app'2(quot, app'2(s, x)), app'2(s, y)) -> app'2(s, app'2(app'2(quot, app'2(app'2(minus, x), y)), app'2(s, y)))
app'2(app'2(plus, 0), y) -> y
app'2(app'2(plus, app'2(s, x)), y) -> app'2(s, app'2(app'2(plus, x), y))
app'2(app'2(app, nil), k) -> k
app'2(app'2(app, l), nil) -> l
app'2(app'2(app, app'2(app'2(cons, x), l)), k) -> app'2(app'2(cons, x), app'2(app'2(app, l), k))
app'2(sum, app'2(app'2(cons, x), nil)) -> app'2(app'2(cons, x), nil)
app'2(sum, app'2(app'2(cons, x), app'2(app'2(cons, y), l))) -> app'2(sum, app'2(app'2(cons, app'2(app'2(plus, x), y)), l))
app'2(sum, app'2(app'2(app, l), app'2(app'2(cons, x), app'2(app'2(cons, y), k)))) -> app'2(sum, app'2(app'2(app, l), app'2(sum, app'2(app'2(cons, x), app'2(app'2(cons, y), k)))))
app'2(app'2(map, f), nil) -> nil
app'2(app'2(map, f), app'2(app'2(cons, x), xs)) -> app'2(app'2(cons, app'2(f, x)), app'2(app'2(map, f), xs))
app'2(app'2(filter, f), nil) -> nil
app'2(app'2(filter, f), app'2(app'2(cons, x), xs)) -> app'2(app'2(app'2(app'2(filter2, app'2(f, x)), f), x), xs)
app'2(app'2(app'2(app'2(filter2, true), f), x), xs) -> app'2(app'2(cons, x), app'2(app'2(filter, f), xs))
app'2(app'2(app'2(app'2(filter2, false), f), x), xs) -> app'2(app'2(filter, f), xs)

Q is empty.


QTRS
  ↳ DependencyPairsProof

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

app'2(app'2(minus, x), 0) -> x
app'2(app'2(minus, app'2(s, x)), app'2(s, y)) -> app'2(app'2(minus, x), y)
app'2(app'2(minus, app'2(app'2(minus, x), y)), z) -> app'2(app'2(minus, x), app'2(app'2(plus, y), z))
app'2(app'2(quot, 0), app'2(s, y)) -> 0
app'2(app'2(quot, app'2(s, x)), app'2(s, y)) -> app'2(s, app'2(app'2(quot, app'2(app'2(minus, x), y)), app'2(s, y)))
app'2(app'2(plus, 0), y) -> y
app'2(app'2(plus, app'2(s, x)), y) -> app'2(s, app'2(app'2(plus, x), y))
app'2(app'2(app, nil), k) -> k
app'2(app'2(app, l), nil) -> l
app'2(app'2(app, app'2(app'2(cons, x), l)), k) -> app'2(app'2(cons, x), app'2(app'2(app, l), k))
app'2(sum, app'2(app'2(cons, x), nil)) -> app'2(app'2(cons, x), nil)
app'2(sum, app'2(app'2(cons, x), app'2(app'2(cons, y), l))) -> app'2(sum, app'2(app'2(cons, app'2(app'2(plus, x), y)), l))
app'2(sum, app'2(app'2(app, l), app'2(app'2(cons, x), app'2(app'2(cons, y), k)))) -> app'2(sum, app'2(app'2(app, l), app'2(sum, app'2(app'2(cons, x), app'2(app'2(cons, y), k)))))
app'2(app'2(map, f), nil) -> nil
app'2(app'2(map, f), app'2(app'2(cons, x), xs)) -> app'2(app'2(cons, app'2(f, x)), app'2(app'2(map, f), xs))
app'2(app'2(filter, f), nil) -> nil
app'2(app'2(filter, f), app'2(app'2(cons, x), xs)) -> app'2(app'2(app'2(app'2(filter2, app'2(f, x)), f), x), xs)
app'2(app'2(app'2(app'2(filter2, true), f), x), xs) -> app'2(app'2(cons, x), app'2(app'2(filter, f), xs))
app'2(app'2(app'2(app'2(filter2, false), f), x), xs) -> app'2(app'2(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:

APP'2(app'2(minus, app'2(s, x)), app'2(s, y)) -> APP'2(app'2(minus, x), y)
APP'2(app'2(app'2(app'2(filter2, false), f), x), xs) -> APP'2(filter, f)
APP'2(app'2(map, f), app'2(app'2(cons, x), xs)) -> APP'2(app'2(map, f), xs)
APP'2(sum, app'2(app'2(cons, x), app'2(app'2(cons, y), l))) -> APP'2(plus, x)
APP'2(app'2(quot, app'2(s, x)), app'2(s, y)) -> APP'2(quot, app'2(app'2(minus, x), y))
APP'2(app'2(filter, f), app'2(app'2(cons, x), xs)) -> APP'2(f, x)
APP'2(app'2(plus, app'2(s, x)), y) -> APP'2(app'2(plus, x), y)
APP'2(app'2(plus, app'2(s, x)), y) -> APP'2(s, app'2(app'2(plus, x), y))
APP'2(app'2(filter, f), app'2(app'2(cons, x), xs)) -> APP'2(app'2(app'2(filter2, app'2(f, x)), f), x)
APP'2(app'2(plus, app'2(s, x)), y) -> APP'2(plus, x)
APP'2(app'2(quot, app'2(s, x)), app'2(s, y)) -> APP'2(minus, x)
APP'2(app'2(minus, app'2(app'2(minus, x), y)), z) -> APP'2(app'2(plus, y), z)
APP'2(app'2(app'2(app'2(filter2, true), f), x), xs) -> APP'2(app'2(filter, f), xs)
APP'2(app'2(minus, app'2(app'2(minus, x), y)), z) -> APP'2(plus, y)
APP'2(app'2(quot, app'2(s, x)), app'2(s, y)) -> APP'2(s, app'2(app'2(quot, app'2(app'2(minus, x), y)), app'2(s, y)))
APP'2(app'2(map, f), app'2(app'2(cons, x), xs)) -> APP'2(app'2(cons, app'2(f, x)), app'2(app'2(map, f), xs))
APP'2(sum, app'2(app'2(cons, x), app'2(app'2(cons, y), l))) -> APP'2(app'2(cons, app'2(app'2(plus, x), y)), l)
APP'2(app'2(minus, app'2(s, x)), app'2(s, y)) -> APP'2(minus, x)
APP'2(sum, app'2(app'2(cons, x), app'2(app'2(cons, y), l))) -> APP'2(cons, app'2(app'2(plus, x), y))
APP'2(app'2(app'2(app'2(filter2, true), f), x), xs) -> APP'2(app'2(cons, x), app'2(app'2(filter, f), xs))
APP'2(app'2(app, app'2(app'2(cons, x), l)), k) -> APP'2(app, l)
APP'2(app'2(quot, app'2(s, x)), app'2(s, y)) -> APP'2(app'2(minus, x), y)
APP'2(app'2(filter, f), app'2(app'2(cons, x), xs)) -> APP'2(app'2(app'2(app'2(filter2, app'2(f, x)), f), x), xs)
APP'2(app'2(app'2(app'2(filter2, true), f), x), xs) -> APP'2(cons, x)
APP'2(app'2(map, f), app'2(app'2(cons, x), xs)) -> APP'2(f, x)
APP'2(app'2(quot, app'2(s, x)), app'2(s, y)) -> APP'2(app'2(quot, app'2(app'2(minus, x), y)), app'2(s, y))
APP'2(sum, app'2(app'2(cons, x), app'2(app'2(cons, y), l))) -> APP'2(app'2(plus, x), y)
APP'2(sum, app'2(app'2(app, l), app'2(app'2(cons, x), app'2(app'2(cons, y), k)))) -> APP'2(sum, app'2(app'2(app, l), app'2(sum, app'2(app'2(cons, x), app'2(app'2(cons, y), k)))))
APP'2(sum, app'2(app'2(cons, x), app'2(app'2(cons, y), l))) -> APP'2(sum, app'2(app'2(cons, app'2(app'2(plus, x), y)), l))
APP'2(app'2(app, app'2(app'2(cons, x), l)), k) -> APP'2(app'2(cons, x), app'2(app'2(app, l), k))
APP'2(sum, app'2(app'2(app, l), app'2(app'2(cons, x), app'2(app'2(cons, y), k)))) -> APP'2(app'2(app, l), app'2(sum, app'2(app'2(cons, x), app'2(app'2(cons, y), k))))
APP'2(app'2(app'2(app'2(filter2, false), f), x), xs) -> APP'2(app'2(filter, f), xs)
APP'2(app'2(app'2(app'2(filter2, true), f), x), xs) -> APP'2(filter, f)
APP'2(app'2(map, f), app'2(app'2(cons, x), xs)) -> APP'2(cons, app'2(f, x))
APP'2(app'2(filter, f), app'2(app'2(cons, x), xs)) -> APP'2(filter2, app'2(f, x))
APP'2(sum, app'2(app'2(app, l), app'2(app'2(cons, x), app'2(app'2(cons, y), k)))) -> APP'2(sum, app'2(app'2(cons, x), app'2(app'2(cons, y), k)))
APP'2(app'2(filter, f), app'2(app'2(cons, x), xs)) -> APP'2(app'2(filter2, app'2(f, x)), f)
APP'2(app'2(minus, app'2(app'2(minus, x), y)), z) -> APP'2(app'2(minus, x), app'2(app'2(plus, y), z))
APP'2(app'2(app, app'2(app'2(cons, x), l)), k) -> APP'2(app'2(app, l), k)

The TRS R consists of the following rules:

app'2(app'2(minus, x), 0) -> x
app'2(app'2(minus, app'2(s, x)), app'2(s, y)) -> app'2(app'2(minus, x), y)
app'2(app'2(minus, app'2(app'2(minus, x), y)), z) -> app'2(app'2(minus, x), app'2(app'2(plus, y), z))
app'2(app'2(quot, 0), app'2(s, y)) -> 0
app'2(app'2(quot, app'2(s, x)), app'2(s, y)) -> app'2(s, app'2(app'2(quot, app'2(app'2(minus, x), y)), app'2(s, y)))
app'2(app'2(plus, 0), y) -> y
app'2(app'2(plus, app'2(s, x)), y) -> app'2(s, app'2(app'2(plus, x), y))
app'2(app'2(app, nil), k) -> k
app'2(app'2(app, l), nil) -> l
app'2(app'2(app, app'2(app'2(cons, x), l)), k) -> app'2(app'2(cons, x), app'2(app'2(app, l), k))
app'2(sum, app'2(app'2(cons, x), nil)) -> app'2(app'2(cons, x), nil)
app'2(sum, app'2(app'2(cons, x), app'2(app'2(cons, y), l))) -> app'2(sum, app'2(app'2(cons, app'2(app'2(plus, x), y)), l))
app'2(sum, app'2(app'2(app, l), app'2(app'2(cons, x), app'2(app'2(cons, y), k)))) -> app'2(sum, app'2(app'2(app, l), app'2(sum, app'2(app'2(cons, x), app'2(app'2(cons, y), k)))))
app'2(app'2(map, f), nil) -> nil
app'2(app'2(map, f), app'2(app'2(cons, x), xs)) -> app'2(app'2(cons, app'2(f, x)), app'2(app'2(map, f), xs))
app'2(app'2(filter, f), nil) -> nil
app'2(app'2(filter, f), app'2(app'2(cons, x), xs)) -> app'2(app'2(app'2(app'2(filter2, app'2(f, x)), f), x), xs)
app'2(app'2(app'2(app'2(filter2, true), f), x), xs) -> app'2(app'2(cons, x), app'2(app'2(filter, f), xs))
app'2(app'2(app'2(app'2(filter2, false), f), x), xs) -> app'2(app'2(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:

APP'2(app'2(minus, app'2(s, x)), app'2(s, y)) -> APP'2(app'2(minus, x), y)
APP'2(app'2(app'2(app'2(filter2, false), f), x), xs) -> APP'2(filter, f)
APP'2(app'2(map, f), app'2(app'2(cons, x), xs)) -> APP'2(app'2(map, f), xs)
APP'2(sum, app'2(app'2(cons, x), app'2(app'2(cons, y), l))) -> APP'2(plus, x)
APP'2(app'2(quot, app'2(s, x)), app'2(s, y)) -> APP'2(quot, app'2(app'2(minus, x), y))
APP'2(app'2(filter, f), app'2(app'2(cons, x), xs)) -> APP'2(f, x)
APP'2(app'2(plus, app'2(s, x)), y) -> APP'2(app'2(plus, x), y)
APP'2(app'2(plus, app'2(s, x)), y) -> APP'2(s, app'2(app'2(plus, x), y))
APP'2(app'2(filter, f), app'2(app'2(cons, x), xs)) -> APP'2(app'2(app'2(filter2, app'2(f, x)), f), x)
APP'2(app'2(plus, app'2(s, x)), y) -> APP'2(plus, x)
APP'2(app'2(quot, app'2(s, x)), app'2(s, y)) -> APP'2(minus, x)
APP'2(app'2(minus, app'2(app'2(minus, x), y)), z) -> APP'2(app'2(plus, y), z)
APP'2(app'2(app'2(app'2(filter2, true), f), x), xs) -> APP'2(app'2(filter, f), xs)
APP'2(app'2(minus, app'2(app'2(minus, x), y)), z) -> APP'2(plus, y)
APP'2(app'2(quot, app'2(s, x)), app'2(s, y)) -> APP'2(s, app'2(app'2(quot, app'2(app'2(minus, x), y)), app'2(s, y)))
APP'2(app'2(map, f), app'2(app'2(cons, x), xs)) -> APP'2(app'2(cons, app'2(f, x)), app'2(app'2(map, f), xs))
APP'2(sum, app'2(app'2(cons, x), app'2(app'2(cons, y), l))) -> APP'2(app'2(cons, app'2(app'2(plus, x), y)), l)
APP'2(app'2(minus, app'2(s, x)), app'2(s, y)) -> APP'2(minus, x)
APP'2(sum, app'2(app'2(cons, x), app'2(app'2(cons, y), l))) -> APP'2(cons, app'2(app'2(plus, x), y))
APP'2(app'2(app'2(app'2(filter2, true), f), x), xs) -> APP'2(app'2(cons, x), app'2(app'2(filter, f), xs))
APP'2(app'2(app, app'2(app'2(cons, x), l)), k) -> APP'2(app, l)
APP'2(app'2(quot, app'2(s, x)), app'2(s, y)) -> APP'2(app'2(minus, x), y)
APP'2(app'2(filter, f), app'2(app'2(cons, x), xs)) -> APP'2(app'2(app'2(app'2(filter2, app'2(f, x)), f), x), xs)
APP'2(app'2(app'2(app'2(filter2, true), f), x), xs) -> APP'2(cons, x)
APP'2(app'2(map, f), app'2(app'2(cons, x), xs)) -> APP'2(f, x)
APP'2(app'2(quot, app'2(s, x)), app'2(s, y)) -> APP'2(app'2(quot, app'2(app'2(minus, x), y)), app'2(s, y))
APP'2(sum, app'2(app'2(cons, x), app'2(app'2(cons, y), l))) -> APP'2(app'2(plus, x), y)
APP'2(sum, app'2(app'2(app, l), app'2(app'2(cons, x), app'2(app'2(cons, y), k)))) -> APP'2(sum, app'2(app'2(app, l), app'2(sum, app'2(app'2(cons, x), app'2(app'2(cons, y), k)))))
APP'2(sum, app'2(app'2(cons, x), app'2(app'2(cons, y), l))) -> APP'2(sum, app'2(app'2(cons, app'2(app'2(plus, x), y)), l))
APP'2(app'2(app, app'2(app'2(cons, x), l)), k) -> APP'2(app'2(cons, x), app'2(app'2(app, l), k))
APP'2(sum, app'2(app'2(app, l), app'2(app'2(cons, x), app'2(app'2(cons, y), k)))) -> APP'2(app'2(app, l), app'2(sum, app'2(app'2(cons, x), app'2(app'2(cons, y), k))))
APP'2(app'2(app'2(app'2(filter2, false), f), x), xs) -> APP'2(app'2(filter, f), xs)
APP'2(app'2(app'2(app'2(filter2, true), f), x), xs) -> APP'2(filter, f)
APP'2(app'2(map, f), app'2(app'2(cons, x), xs)) -> APP'2(cons, app'2(f, x))
APP'2(app'2(filter, f), app'2(app'2(cons, x), xs)) -> APP'2(filter2, app'2(f, x))
APP'2(sum, app'2(app'2(app, l), app'2(app'2(cons, x), app'2(app'2(cons, y), k)))) -> APP'2(sum, app'2(app'2(cons, x), app'2(app'2(cons, y), k)))
APP'2(app'2(filter, f), app'2(app'2(cons, x), xs)) -> APP'2(app'2(filter2, app'2(f, x)), f)
APP'2(app'2(minus, app'2(app'2(minus, x), y)), z) -> APP'2(app'2(minus, x), app'2(app'2(plus, y), z))
APP'2(app'2(app, app'2(app'2(cons, x), l)), k) -> APP'2(app'2(app, l), k)

The TRS R consists of the following rules:

app'2(app'2(minus, x), 0) -> x
app'2(app'2(minus, app'2(s, x)), app'2(s, y)) -> app'2(app'2(minus, x), y)
app'2(app'2(minus, app'2(app'2(minus, x), y)), z) -> app'2(app'2(minus, x), app'2(app'2(plus, y), z))
app'2(app'2(quot, 0), app'2(s, y)) -> 0
app'2(app'2(quot, app'2(s, x)), app'2(s, y)) -> app'2(s, app'2(app'2(quot, app'2(app'2(minus, x), y)), app'2(s, y)))
app'2(app'2(plus, 0), y) -> y
app'2(app'2(plus, app'2(s, x)), y) -> app'2(s, app'2(app'2(plus, x), y))
app'2(app'2(app, nil), k) -> k
app'2(app'2(app, l), nil) -> l
app'2(app'2(app, app'2(app'2(cons, x), l)), k) -> app'2(app'2(cons, x), app'2(app'2(app, l), k))
app'2(sum, app'2(app'2(cons, x), nil)) -> app'2(app'2(cons, x), nil)
app'2(sum, app'2(app'2(cons, x), app'2(app'2(cons, y), l))) -> app'2(sum, app'2(app'2(cons, app'2(app'2(plus, x), y)), l))
app'2(sum, app'2(app'2(app, l), app'2(app'2(cons, x), app'2(app'2(cons, y), k)))) -> app'2(sum, app'2(app'2(app, l), app'2(sum, app'2(app'2(cons, x), app'2(app'2(cons, y), k)))))
app'2(app'2(map, f), nil) -> nil
app'2(app'2(map, f), app'2(app'2(cons, x), xs)) -> app'2(app'2(cons, app'2(f, x)), app'2(app'2(map, f), xs))
app'2(app'2(filter, f), nil) -> nil
app'2(app'2(filter, f), app'2(app'2(cons, x), xs)) -> app'2(app'2(app'2(app'2(filter2, app'2(f, x)), f), x), xs)
app'2(app'2(app'2(app'2(filter2, true), f), x), xs) -> app'2(app'2(cons, x), app'2(app'2(filter, f), xs))
app'2(app'2(app'2(app'2(filter2, false), f), x), xs) -> app'2(app'2(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 7 SCCs with 26 less nodes.

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

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

APP'2(app'2(app, app'2(app'2(cons, x), l)), k) -> APP'2(app'2(app, l), k)

The TRS R consists of the following rules:

app'2(app'2(minus, x), 0) -> x
app'2(app'2(minus, app'2(s, x)), app'2(s, y)) -> app'2(app'2(minus, x), y)
app'2(app'2(minus, app'2(app'2(minus, x), y)), z) -> app'2(app'2(minus, x), app'2(app'2(plus, y), z))
app'2(app'2(quot, 0), app'2(s, y)) -> 0
app'2(app'2(quot, app'2(s, x)), app'2(s, y)) -> app'2(s, app'2(app'2(quot, app'2(app'2(minus, x), y)), app'2(s, y)))
app'2(app'2(plus, 0), y) -> y
app'2(app'2(plus, app'2(s, x)), y) -> app'2(s, app'2(app'2(plus, x), y))
app'2(app'2(app, nil), k) -> k
app'2(app'2(app, l), nil) -> l
app'2(app'2(app, app'2(app'2(cons, x), l)), k) -> app'2(app'2(cons, x), app'2(app'2(app, l), k))
app'2(sum, app'2(app'2(cons, x), nil)) -> app'2(app'2(cons, x), nil)
app'2(sum, app'2(app'2(cons, x), app'2(app'2(cons, y), l))) -> app'2(sum, app'2(app'2(cons, app'2(app'2(plus, x), y)), l))
app'2(sum, app'2(app'2(app, l), app'2(app'2(cons, x), app'2(app'2(cons, y), k)))) -> app'2(sum, app'2(app'2(app, l), app'2(sum, app'2(app'2(cons, x), app'2(app'2(cons, y), k)))))
app'2(app'2(map, f), nil) -> nil
app'2(app'2(map, f), app'2(app'2(cons, x), xs)) -> app'2(app'2(cons, app'2(f, x)), app'2(app'2(map, f), xs))
app'2(app'2(filter, f), nil) -> nil
app'2(app'2(filter, f), app'2(app'2(cons, x), xs)) -> app'2(app'2(app'2(app'2(filter2, app'2(f, x)), f), x), xs)
app'2(app'2(app'2(app'2(filter2, true), f), x), xs) -> app'2(app'2(cons, x), app'2(app'2(filter, f), xs))
app'2(app'2(app'2(app'2(filter2, false), f), x), xs) -> app'2(app'2(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.


APP'2(app'2(app, app'2(app'2(cons, x), l)), k) -> APP'2(app'2(app, l), k)
The remaining pairs can at least be oriented weakly.
none
Used ordering: Polynomial interpretation [21]:

POL(APP'2(x1, x2)) = x1   
POL(app) = 0   
POL(app'2(x1, x2)) = 1 + x2   
POL(cons) = 0   

The following usable rules [14] were oriented: none



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

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

app'2(app'2(minus, x), 0) -> x
app'2(app'2(minus, app'2(s, x)), app'2(s, y)) -> app'2(app'2(minus, x), y)
app'2(app'2(minus, app'2(app'2(minus, x), y)), z) -> app'2(app'2(minus, x), app'2(app'2(plus, y), z))
app'2(app'2(quot, 0), app'2(s, y)) -> 0
app'2(app'2(quot, app'2(s, x)), app'2(s, y)) -> app'2(s, app'2(app'2(quot, app'2(app'2(minus, x), y)), app'2(s, y)))
app'2(app'2(plus, 0), y) -> y
app'2(app'2(plus, app'2(s, x)), y) -> app'2(s, app'2(app'2(plus, x), y))
app'2(app'2(app, nil), k) -> k
app'2(app'2(app, l), nil) -> l
app'2(app'2(app, app'2(app'2(cons, x), l)), k) -> app'2(app'2(cons, x), app'2(app'2(app, l), k))
app'2(sum, app'2(app'2(cons, x), nil)) -> app'2(app'2(cons, x), nil)
app'2(sum, app'2(app'2(cons, x), app'2(app'2(cons, y), l))) -> app'2(sum, app'2(app'2(cons, app'2(app'2(plus, x), y)), l))
app'2(sum, app'2(app'2(app, l), app'2(app'2(cons, x), app'2(app'2(cons, y), k)))) -> app'2(sum, app'2(app'2(app, l), app'2(sum, app'2(app'2(cons, x), app'2(app'2(cons, y), k)))))
app'2(app'2(map, f), nil) -> nil
app'2(app'2(map, f), app'2(app'2(cons, x), xs)) -> app'2(app'2(cons, app'2(f, x)), app'2(app'2(map, f), xs))
app'2(app'2(filter, f), nil) -> nil
app'2(app'2(filter, f), app'2(app'2(cons, x), xs)) -> app'2(app'2(app'2(app'2(filter2, app'2(f, x)), f), x), xs)
app'2(app'2(app'2(app'2(filter2, true), f), x), xs) -> app'2(app'2(cons, x), app'2(app'2(filter, f), xs))
app'2(app'2(app'2(app'2(filter2, false), f), x), xs) -> app'2(app'2(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.

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

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

APP'2(app'2(plus, app'2(s, x)), y) -> APP'2(app'2(plus, x), y)

The TRS R consists of the following rules:

app'2(app'2(minus, x), 0) -> x
app'2(app'2(minus, app'2(s, x)), app'2(s, y)) -> app'2(app'2(minus, x), y)
app'2(app'2(minus, app'2(app'2(minus, x), y)), z) -> app'2(app'2(minus, x), app'2(app'2(plus, y), z))
app'2(app'2(quot, 0), app'2(s, y)) -> 0
app'2(app'2(quot, app'2(s, x)), app'2(s, y)) -> app'2(s, app'2(app'2(quot, app'2(app'2(minus, x), y)), app'2(s, y)))
app'2(app'2(plus, 0), y) -> y
app'2(app'2(plus, app'2(s, x)), y) -> app'2(s, app'2(app'2(plus, x), y))
app'2(app'2(app, nil), k) -> k
app'2(app'2(app, l), nil) -> l
app'2(app'2(app, app'2(app'2(cons, x), l)), k) -> app'2(app'2(cons, x), app'2(app'2(app, l), k))
app'2(sum, app'2(app'2(cons, x), nil)) -> app'2(app'2(cons, x), nil)
app'2(sum, app'2(app'2(cons, x), app'2(app'2(cons, y), l))) -> app'2(sum, app'2(app'2(cons, app'2(app'2(plus, x), y)), l))
app'2(sum, app'2(app'2(app, l), app'2(app'2(cons, x), app'2(app'2(cons, y), k)))) -> app'2(sum, app'2(app'2(app, l), app'2(sum, app'2(app'2(cons, x), app'2(app'2(cons, y), k)))))
app'2(app'2(map, f), nil) -> nil
app'2(app'2(map, f), app'2(app'2(cons, x), xs)) -> app'2(app'2(cons, app'2(f, x)), app'2(app'2(map, f), xs))
app'2(app'2(filter, f), nil) -> nil
app'2(app'2(filter, f), app'2(app'2(cons, x), xs)) -> app'2(app'2(app'2(app'2(filter2, app'2(f, x)), f), x), xs)
app'2(app'2(app'2(app'2(filter2, true), f), x), xs) -> app'2(app'2(cons, x), app'2(app'2(filter, f), xs))
app'2(app'2(app'2(app'2(filter2, false), f), x), xs) -> app'2(app'2(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.


APP'2(app'2(plus, app'2(s, x)), y) -> APP'2(app'2(plus, x), y)
The remaining pairs can at least be oriented weakly.
none
Used ordering: Polynomial interpretation [21]:

POL(APP'2(x1, x2)) = x1   
POL(app'2(x1, x2)) = x1 + x2   
POL(plus) = 0   
POL(s) = 1   

The following usable rules [14] were oriented: none



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

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

app'2(app'2(minus, x), 0) -> x
app'2(app'2(minus, app'2(s, x)), app'2(s, y)) -> app'2(app'2(minus, x), y)
app'2(app'2(minus, app'2(app'2(minus, x), y)), z) -> app'2(app'2(minus, x), app'2(app'2(plus, y), z))
app'2(app'2(quot, 0), app'2(s, y)) -> 0
app'2(app'2(quot, app'2(s, x)), app'2(s, y)) -> app'2(s, app'2(app'2(quot, app'2(app'2(minus, x), y)), app'2(s, y)))
app'2(app'2(plus, 0), y) -> y
app'2(app'2(plus, app'2(s, x)), y) -> app'2(s, app'2(app'2(plus, x), y))
app'2(app'2(app, nil), k) -> k
app'2(app'2(app, l), nil) -> l
app'2(app'2(app, app'2(app'2(cons, x), l)), k) -> app'2(app'2(cons, x), app'2(app'2(app, l), k))
app'2(sum, app'2(app'2(cons, x), nil)) -> app'2(app'2(cons, x), nil)
app'2(sum, app'2(app'2(cons, x), app'2(app'2(cons, y), l))) -> app'2(sum, app'2(app'2(cons, app'2(app'2(plus, x), y)), l))
app'2(sum, app'2(app'2(app, l), app'2(app'2(cons, x), app'2(app'2(cons, y), k)))) -> app'2(sum, app'2(app'2(app, l), app'2(sum, app'2(app'2(cons, x), app'2(app'2(cons, y), k)))))
app'2(app'2(map, f), nil) -> nil
app'2(app'2(map, f), app'2(app'2(cons, x), xs)) -> app'2(app'2(cons, app'2(f, x)), app'2(app'2(map, f), xs))
app'2(app'2(filter, f), nil) -> nil
app'2(app'2(filter, f), app'2(app'2(cons, x), xs)) -> app'2(app'2(app'2(app'2(filter2, app'2(f, x)), f), x), xs)
app'2(app'2(app'2(app'2(filter2, true), f), x), xs) -> app'2(app'2(cons, x), app'2(app'2(filter, f), xs))
app'2(app'2(app'2(app'2(filter2, false), f), x), xs) -> app'2(app'2(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.

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

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

APP'2(sum, app'2(app'2(cons, x), app'2(app'2(cons, y), l))) -> APP'2(sum, app'2(app'2(cons, app'2(app'2(plus, x), y)), l))

The TRS R consists of the following rules:

app'2(app'2(minus, x), 0) -> x
app'2(app'2(minus, app'2(s, x)), app'2(s, y)) -> app'2(app'2(minus, x), y)
app'2(app'2(minus, app'2(app'2(minus, x), y)), z) -> app'2(app'2(minus, x), app'2(app'2(plus, y), z))
app'2(app'2(quot, 0), app'2(s, y)) -> 0
app'2(app'2(quot, app'2(s, x)), app'2(s, y)) -> app'2(s, app'2(app'2(quot, app'2(app'2(minus, x), y)), app'2(s, y)))
app'2(app'2(plus, 0), y) -> y
app'2(app'2(plus, app'2(s, x)), y) -> app'2(s, app'2(app'2(plus, x), y))
app'2(app'2(app, nil), k) -> k
app'2(app'2(app, l), nil) -> l
app'2(app'2(app, app'2(app'2(cons, x), l)), k) -> app'2(app'2(cons, x), app'2(app'2(app, l), k))
app'2(sum, app'2(app'2(cons, x), nil)) -> app'2(app'2(cons, x), nil)
app'2(sum, app'2(app'2(cons, x), app'2(app'2(cons, y), l))) -> app'2(sum, app'2(app'2(cons, app'2(app'2(plus, x), y)), l))
app'2(sum, app'2(app'2(app, l), app'2(app'2(cons, x), app'2(app'2(cons, y), k)))) -> app'2(sum, app'2(app'2(app, l), app'2(sum, app'2(app'2(cons, x), app'2(app'2(cons, y), k)))))
app'2(app'2(map, f), nil) -> nil
app'2(app'2(map, f), app'2(app'2(cons, x), xs)) -> app'2(app'2(cons, app'2(f, x)), app'2(app'2(map, f), xs))
app'2(app'2(filter, f), nil) -> nil
app'2(app'2(filter, f), app'2(app'2(cons, x), xs)) -> app'2(app'2(app'2(app'2(filter2, app'2(f, x)), f), x), xs)
app'2(app'2(app'2(app'2(filter2, true), f), x), xs) -> app'2(app'2(cons, x), app'2(app'2(filter, f), xs))
app'2(app'2(app'2(app'2(filter2, false), f), x), xs) -> app'2(app'2(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.


APP'2(sum, app'2(app'2(cons, x), app'2(app'2(cons, y), l))) -> APP'2(sum, app'2(app'2(cons, app'2(app'2(plus, x), y)), l))
The remaining pairs can at least be oriented weakly.
none
Used ordering: Polynomial interpretation [21]:

POL(0) = 0   
POL(APP'2(x1, x2)) = x2   
POL(app'2(x1, x2)) = 1 + x2   
POL(cons) = 0   
POL(plus) = 0   
POL(s) = 0   
POL(sum) = 0   

The following usable rules [14] were oriented: none



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

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

app'2(app'2(minus, x), 0) -> x
app'2(app'2(minus, app'2(s, x)), app'2(s, y)) -> app'2(app'2(minus, x), y)
app'2(app'2(minus, app'2(app'2(minus, x), y)), z) -> app'2(app'2(minus, x), app'2(app'2(plus, y), z))
app'2(app'2(quot, 0), app'2(s, y)) -> 0
app'2(app'2(quot, app'2(s, x)), app'2(s, y)) -> app'2(s, app'2(app'2(quot, app'2(app'2(minus, x), y)), app'2(s, y)))
app'2(app'2(plus, 0), y) -> y
app'2(app'2(plus, app'2(s, x)), y) -> app'2(s, app'2(app'2(plus, x), y))
app'2(app'2(app, nil), k) -> k
app'2(app'2(app, l), nil) -> l
app'2(app'2(app, app'2(app'2(cons, x), l)), k) -> app'2(app'2(cons, x), app'2(app'2(app, l), k))
app'2(sum, app'2(app'2(cons, x), nil)) -> app'2(app'2(cons, x), nil)
app'2(sum, app'2(app'2(cons, x), app'2(app'2(cons, y), l))) -> app'2(sum, app'2(app'2(cons, app'2(app'2(plus, x), y)), l))
app'2(sum, app'2(app'2(app, l), app'2(app'2(cons, x), app'2(app'2(cons, y), k)))) -> app'2(sum, app'2(app'2(app, l), app'2(sum, app'2(app'2(cons, x), app'2(app'2(cons, y), k)))))
app'2(app'2(map, f), nil) -> nil
app'2(app'2(map, f), app'2(app'2(cons, x), xs)) -> app'2(app'2(cons, app'2(f, x)), app'2(app'2(map, f), xs))
app'2(app'2(filter, f), nil) -> nil
app'2(app'2(filter, f), app'2(app'2(cons, x), xs)) -> app'2(app'2(app'2(app'2(filter2, app'2(f, x)), f), x), xs)
app'2(app'2(app'2(app'2(filter2, true), f), x), xs) -> app'2(app'2(cons, x), app'2(app'2(filter, f), xs))
app'2(app'2(app'2(app'2(filter2, false), f), x), xs) -> app'2(app'2(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.

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

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

APP'2(sum, app'2(app'2(app, l), app'2(app'2(cons, x), app'2(app'2(cons, y), k)))) -> APP'2(sum, app'2(app'2(app, l), app'2(sum, app'2(app'2(cons, x), app'2(app'2(cons, y), k)))))

The TRS R consists of the following rules:

app'2(app'2(minus, x), 0) -> x
app'2(app'2(minus, app'2(s, x)), app'2(s, y)) -> app'2(app'2(minus, x), y)
app'2(app'2(minus, app'2(app'2(minus, x), y)), z) -> app'2(app'2(minus, x), app'2(app'2(plus, y), z))
app'2(app'2(quot, 0), app'2(s, y)) -> 0
app'2(app'2(quot, app'2(s, x)), app'2(s, y)) -> app'2(s, app'2(app'2(quot, app'2(app'2(minus, x), y)), app'2(s, y)))
app'2(app'2(plus, 0), y) -> y
app'2(app'2(plus, app'2(s, x)), y) -> app'2(s, app'2(app'2(plus, x), y))
app'2(app'2(app, nil), k) -> k
app'2(app'2(app, l), nil) -> l
app'2(app'2(app, app'2(app'2(cons, x), l)), k) -> app'2(app'2(cons, x), app'2(app'2(app, l), k))
app'2(sum, app'2(app'2(cons, x), nil)) -> app'2(app'2(cons, x), nil)
app'2(sum, app'2(app'2(cons, x), app'2(app'2(cons, y), l))) -> app'2(sum, app'2(app'2(cons, app'2(app'2(plus, x), y)), l))
app'2(sum, app'2(app'2(app, l), app'2(app'2(cons, x), app'2(app'2(cons, y), k)))) -> app'2(sum, app'2(app'2(app, l), app'2(sum, app'2(app'2(cons, x), app'2(app'2(cons, y), k)))))
app'2(app'2(map, f), nil) -> nil
app'2(app'2(map, f), app'2(app'2(cons, x), xs)) -> app'2(app'2(cons, app'2(f, x)), app'2(app'2(map, f), xs))
app'2(app'2(filter, f), nil) -> nil
app'2(app'2(filter, f), app'2(app'2(cons, x), xs)) -> app'2(app'2(app'2(app'2(filter2, app'2(f, x)), f), x), xs)
app'2(app'2(app'2(app'2(filter2, true), f), x), xs) -> app'2(app'2(cons, x), app'2(app'2(filter, f), xs))
app'2(app'2(app'2(app'2(filter2, false), f), x), xs) -> app'2(app'2(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
QDP
            ↳ QDPOrderProof
          ↳ QDP
          ↳ QDP

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

APP'2(app'2(minus, app'2(s, x)), app'2(s, y)) -> APP'2(app'2(minus, x), y)
APP'2(app'2(minus, app'2(app'2(minus, x), y)), z) -> APP'2(app'2(minus, x), app'2(app'2(plus, y), z))

The TRS R consists of the following rules:

app'2(app'2(minus, x), 0) -> x
app'2(app'2(minus, app'2(s, x)), app'2(s, y)) -> app'2(app'2(minus, x), y)
app'2(app'2(minus, app'2(app'2(minus, x), y)), z) -> app'2(app'2(minus, x), app'2(app'2(plus, y), z))
app'2(app'2(quot, 0), app'2(s, y)) -> 0
app'2(app'2(quot, app'2(s, x)), app'2(s, y)) -> app'2(s, app'2(app'2(quot, app'2(app'2(minus, x), y)), app'2(s, y)))
app'2(app'2(plus, 0), y) -> y
app'2(app'2(plus, app'2(s, x)), y) -> app'2(s, app'2(app'2(plus, x), y))
app'2(app'2(app, nil), k) -> k
app'2(app'2(app, l), nil) -> l
app'2(app'2(app, app'2(app'2(cons, x), l)), k) -> app'2(app'2(cons, x), app'2(app'2(app, l), k))
app'2(sum, app'2(app'2(cons, x), nil)) -> app'2(app'2(cons, x), nil)
app'2(sum, app'2(app'2(cons, x), app'2(app'2(cons, y), l))) -> app'2(sum, app'2(app'2(cons, app'2(app'2(plus, x), y)), l))
app'2(sum, app'2(app'2(app, l), app'2(app'2(cons, x), app'2(app'2(cons, y), k)))) -> app'2(sum, app'2(app'2(app, l), app'2(sum, app'2(app'2(cons, x), app'2(app'2(cons, y), k)))))
app'2(app'2(map, f), nil) -> nil
app'2(app'2(map, f), app'2(app'2(cons, x), xs)) -> app'2(app'2(cons, app'2(f, x)), app'2(app'2(map, f), xs))
app'2(app'2(filter, f), nil) -> nil
app'2(app'2(filter, f), app'2(app'2(cons, x), xs)) -> app'2(app'2(app'2(app'2(filter2, app'2(f, x)), f), x), xs)
app'2(app'2(app'2(app'2(filter2, true), f), x), xs) -> app'2(app'2(cons, x), app'2(app'2(filter, f), xs))
app'2(app'2(app'2(app'2(filter2, false), f), x), xs) -> app'2(app'2(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.


APP'2(app'2(minus, app'2(s, x)), app'2(s, y)) -> APP'2(app'2(minus, x), y)
APP'2(app'2(minus, app'2(app'2(minus, x), y)), z) -> APP'2(app'2(minus, x), app'2(app'2(plus, y), z))
The remaining pairs can at least be oriented weakly.
none
Used ordering: Polynomial interpretation [21]:

POL(0) = 0   
POL(APP'2(x1, x2)) = x1   
POL(app'2(x1, x2)) = x1 + x2   
POL(minus) = 1   
POL(plus) = 0   
POL(s) = 1   

The following usable rules [14] were oriented: none



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

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

app'2(app'2(minus, x), 0) -> x
app'2(app'2(minus, app'2(s, x)), app'2(s, y)) -> app'2(app'2(minus, x), y)
app'2(app'2(minus, app'2(app'2(minus, x), y)), z) -> app'2(app'2(minus, x), app'2(app'2(plus, y), z))
app'2(app'2(quot, 0), app'2(s, y)) -> 0
app'2(app'2(quot, app'2(s, x)), app'2(s, y)) -> app'2(s, app'2(app'2(quot, app'2(app'2(minus, x), y)), app'2(s, y)))
app'2(app'2(plus, 0), y) -> y
app'2(app'2(plus, app'2(s, x)), y) -> app'2(s, app'2(app'2(plus, x), y))
app'2(app'2(app, nil), k) -> k
app'2(app'2(app, l), nil) -> l
app'2(app'2(app, app'2(app'2(cons, x), l)), k) -> app'2(app'2(cons, x), app'2(app'2(app, l), k))
app'2(sum, app'2(app'2(cons, x), nil)) -> app'2(app'2(cons, x), nil)
app'2(sum, app'2(app'2(cons, x), app'2(app'2(cons, y), l))) -> app'2(sum, app'2(app'2(cons, app'2(app'2(plus, x), y)), l))
app'2(sum, app'2(app'2(app, l), app'2(app'2(cons, x), app'2(app'2(cons, y), k)))) -> app'2(sum, app'2(app'2(app, l), app'2(sum, app'2(app'2(cons, x), app'2(app'2(cons, y), k)))))
app'2(app'2(map, f), nil) -> nil
app'2(app'2(map, f), app'2(app'2(cons, x), xs)) -> app'2(app'2(cons, app'2(f, x)), app'2(app'2(map, f), xs))
app'2(app'2(filter, f), nil) -> nil
app'2(app'2(filter, f), app'2(app'2(cons, x), xs)) -> app'2(app'2(app'2(app'2(filter2, app'2(f, x)), f), x), xs)
app'2(app'2(app'2(app'2(filter2, true), f), x), xs) -> app'2(app'2(cons, x), app'2(app'2(filter, f), xs))
app'2(app'2(app'2(app'2(filter2, false), f), x), xs) -> app'2(app'2(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.

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

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

APP'2(app'2(quot, app'2(s, x)), app'2(s, y)) -> APP'2(app'2(quot, app'2(app'2(minus, x), y)), app'2(s, y))

The TRS R consists of the following rules:

app'2(app'2(minus, x), 0) -> x
app'2(app'2(minus, app'2(s, x)), app'2(s, y)) -> app'2(app'2(minus, x), y)
app'2(app'2(minus, app'2(app'2(minus, x), y)), z) -> app'2(app'2(minus, x), app'2(app'2(plus, y), z))
app'2(app'2(quot, 0), app'2(s, y)) -> 0
app'2(app'2(quot, app'2(s, x)), app'2(s, y)) -> app'2(s, app'2(app'2(quot, app'2(app'2(minus, x), y)), app'2(s, y)))
app'2(app'2(plus, 0), y) -> y
app'2(app'2(plus, app'2(s, x)), y) -> app'2(s, app'2(app'2(plus, x), y))
app'2(app'2(app, nil), k) -> k
app'2(app'2(app, l), nil) -> l
app'2(app'2(app, app'2(app'2(cons, x), l)), k) -> app'2(app'2(cons, x), app'2(app'2(app, l), k))
app'2(sum, app'2(app'2(cons, x), nil)) -> app'2(app'2(cons, x), nil)
app'2(sum, app'2(app'2(cons, x), app'2(app'2(cons, y), l))) -> app'2(sum, app'2(app'2(cons, app'2(app'2(plus, x), y)), l))
app'2(sum, app'2(app'2(app, l), app'2(app'2(cons, x), app'2(app'2(cons, y), k)))) -> app'2(sum, app'2(app'2(app, l), app'2(sum, app'2(app'2(cons, x), app'2(app'2(cons, y), k)))))
app'2(app'2(map, f), nil) -> nil
app'2(app'2(map, f), app'2(app'2(cons, x), xs)) -> app'2(app'2(cons, app'2(f, x)), app'2(app'2(map, f), xs))
app'2(app'2(filter, f), nil) -> nil
app'2(app'2(filter, f), app'2(app'2(cons, x), xs)) -> app'2(app'2(app'2(app'2(filter2, app'2(f, x)), f), x), xs)
app'2(app'2(app'2(app'2(filter2, true), f), x), xs) -> app'2(app'2(cons, x), app'2(app'2(filter, f), xs))
app'2(app'2(app'2(app'2(filter2, false), f), x), xs) -> app'2(app'2(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
          ↳ QDP
          ↳ QDP
QDP
            ↳ QDPOrderProof

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

APP'2(app'2(app'2(app'2(filter2, true), f), x), xs) -> APP'2(app'2(filter, f), xs)
APP'2(app'2(map, f), app'2(app'2(cons, x), xs)) -> APP'2(f, x)
APP'2(app'2(map, f), app'2(app'2(cons, x), xs)) -> APP'2(app'2(map, f), xs)
APP'2(app'2(filter, f), app'2(app'2(cons, x), xs)) -> APP'2(f, x)
APP'2(app'2(filter, f), app'2(app'2(cons, x), xs)) -> APP'2(app'2(app'2(app'2(filter2, app'2(f, x)), f), x), xs)
APP'2(app'2(app'2(app'2(filter2, false), f), x), xs) -> APP'2(app'2(filter, f), xs)

The TRS R consists of the following rules:

app'2(app'2(minus, x), 0) -> x
app'2(app'2(minus, app'2(s, x)), app'2(s, y)) -> app'2(app'2(minus, x), y)
app'2(app'2(minus, app'2(app'2(minus, x), y)), z) -> app'2(app'2(minus, x), app'2(app'2(plus, y), z))
app'2(app'2(quot, 0), app'2(s, y)) -> 0
app'2(app'2(quot, app'2(s, x)), app'2(s, y)) -> app'2(s, app'2(app'2(quot, app'2(app'2(minus, x), y)), app'2(s, y)))
app'2(app'2(plus, 0), y) -> y
app'2(app'2(plus, app'2(s, x)), y) -> app'2(s, app'2(app'2(plus, x), y))
app'2(app'2(app, nil), k) -> k
app'2(app'2(app, l), nil) -> l
app'2(app'2(app, app'2(app'2(cons, x), l)), k) -> app'2(app'2(cons, x), app'2(app'2(app, l), k))
app'2(sum, app'2(app'2(cons, x), nil)) -> app'2(app'2(cons, x), nil)
app'2(sum, app'2(app'2(cons, x), app'2(app'2(cons, y), l))) -> app'2(sum, app'2(app'2(cons, app'2(app'2(plus, x), y)), l))
app'2(sum, app'2(app'2(app, l), app'2(app'2(cons, x), app'2(app'2(cons, y), k)))) -> app'2(sum, app'2(app'2(app, l), app'2(sum, app'2(app'2(cons, x), app'2(app'2(cons, y), k)))))
app'2(app'2(map, f), nil) -> nil
app'2(app'2(map, f), app'2(app'2(cons, x), xs)) -> app'2(app'2(cons, app'2(f, x)), app'2(app'2(map, f), xs))
app'2(app'2(filter, f), nil) -> nil
app'2(app'2(filter, f), app'2(app'2(cons, x), xs)) -> app'2(app'2(app'2(app'2(filter2, app'2(f, x)), f), x), xs)
app'2(app'2(app'2(app'2(filter2, true), f), x), xs) -> app'2(app'2(cons, x), app'2(app'2(filter, f), xs))
app'2(app'2(app'2(app'2(filter2, false), f), x), xs) -> app'2(app'2(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.


APP'2(app'2(map, f), app'2(app'2(cons, x), xs)) -> APP'2(f, x)
APP'2(app'2(map, f), app'2(app'2(cons, x), xs)) -> APP'2(app'2(map, f), xs)
APP'2(app'2(filter, f), app'2(app'2(cons, x), xs)) -> APP'2(f, x)
APP'2(app'2(filter, f), app'2(app'2(cons, x), xs)) -> APP'2(app'2(app'2(app'2(filter2, app'2(f, x)), f), x), xs)
The remaining pairs can at least be oriented weakly.

APP'2(app'2(app'2(app'2(filter2, true), f), x), xs) -> APP'2(app'2(filter, f), xs)
APP'2(app'2(app'2(app'2(filter2, false), f), x), xs) -> APP'2(app'2(filter, f), xs)
Used ordering: Polynomial interpretation [21]:

POL(0) = 0   
POL(APP'2(x1, x2)) = x2   
POL(app) = 0   
POL(app'2(x1, x2)) = 1 + x1 + x2   
POL(cons) = 0   
POL(false) = 0   
POL(filter) = 0   
POL(filter2) = 0   
POL(map) = 0   
POL(minus) = 0   
POL(nil) = 0   
POL(plus) = 0   
POL(quot) = 0   
POL(s) = 0   
POL(sum) = 0   
POL(true) = 0   

The following usable rules [14] were oriented: none



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

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

APP'2(app'2(app'2(app'2(filter2, true), f), x), xs) -> APP'2(app'2(filter, f), xs)
APP'2(app'2(app'2(app'2(filter2, false), f), x), xs) -> APP'2(app'2(filter, f), xs)

The TRS R consists of the following rules:

app'2(app'2(minus, x), 0) -> x
app'2(app'2(minus, app'2(s, x)), app'2(s, y)) -> app'2(app'2(minus, x), y)
app'2(app'2(minus, app'2(app'2(minus, x), y)), z) -> app'2(app'2(minus, x), app'2(app'2(plus, y), z))
app'2(app'2(quot, 0), app'2(s, y)) -> 0
app'2(app'2(quot, app'2(s, x)), app'2(s, y)) -> app'2(s, app'2(app'2(quot, app'2(app'2(minus, x), y)), app'2(s, y)))
app'2(app'2(plus, 0), y) -> y
app'2(app'2(plus, app'2(s, x)), y) -> app'2(s, app'2(app'2(plus, x), y))
app'2(app'2(app, nil), k) -> k
app'2(app'2(app, l), nil) -> l
app'2(app'2(app, app'2(app'2(cons, x), l)), k) -> app'2(app'2(cons, x), app'2(app'2(app, l), k))
app'2(sum, app'2(app'2(cons, x), nil)) -> app'2(app'2(cons, x), nil)
app'2(sum, app'2(app'2(cons, x), app'2(app'2(cons, y), l))) -> app'2(sum, app'2(app'2(cons, app'2(app'2(plus, x), y)), l))
app'2(sum, app'2(app'2(app, l), app'2(app'2(cons, x), app'2(app'2(cons, y), k)))) -> app'2(sum, app'2(app'2(app, l), app'2(sum, app'2(app'2(cons, x), app'2(app'2(cons, y), k)))))
app'2(app'2(map, f), nil) -> nil
app'2(app'2(map, f), app'2(app'2(cons, x), xs)) -> app'2(app'2(cons, app'2(f, x)), app'2(app'2(map, f), xs))
app'2(app'2(filter, f), nil) -> nil
app'2(app'2(filter, f), app'2(app'2(cons, x), xs)) -> app'2(app'2(app'2(app'2(filter2, app'2(f, x)), f), x), xs)
app'2(app'2(app'2(app'2(filter2, true), f), x), xs) -> app'2(app'2(cons, x), app'2(app'2(filter, f), xs))
app'2(app'2(app'2(app'2(filter2, false), f), x), xs) -> app'2(app'2(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 0 SCCs with 2 less nodes.