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

Q is empty.


↳ QTRS
  ↳ DependencyPairsProof

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

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

The TRS R consists of the following rules:

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

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

↳ QTRS
  ↳ DependencyPairsProof
    ↳ QDP
      ↳ EdgeDeletionProof

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

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

The TRS R consists of the following rules:

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

Q is empty.
We have to consider all minimal (P,Q,R)-chains.
We deleted some edges using various graph approximations

↳ QTRS
  ↳ DependencyPairsProof
    ↳ QDP
      ↳ EdgeDeletionProof
        ↳ QDP
          ↳ DependencyGraphProof

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

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

The TRS R consists of the following rules:

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

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

APP'(app'(app, app'(app'(cons, x), l)), k) → APP'(app'(app, l), k)

The TRS R consists of the following rules:

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

Q is empty.
We have to consider all minimal (P,Q,R)-chains.
We use the reduction pair processor [13]. Here, we combined the reduction pair processor with the A-transformation [14] which results in the following intermediate Q-DP Problem.
Q DP problem:
The TRS P consists of the following rules:

APP(cons(x, l), k) → APP(l, k)

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


The following pairs can be oriented strictly and are deleted.


APP'(app'(app, app'(app'(cons, x), l)), k) → APP'(app'(app, l), k)
The remaining pairs can at least be oriented weakly.
none
Used ordering: Combined order from the following AFS and order.
APP(x1, x2)  =  x1
cons(x1, x2)  =  cons(x1, x2)

Lexicographic Path Order [19].
Precedence:
trivial

The following usable rules [14] were oriented: none



↳ QTRS
  ↳ DependencyPairsProof
    ↳ QDP
      ↳ EdgeDeletionProof
        ↳ 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'(app'(minus, x), 0) → x
app'(app'(minus, app'(s, x)), app'(s, y)) → app'(app'(minus, x), y)
app'(app'(minus, app'(app'(minus, x), y)), z) → app'(app'(minus, x), app'(app'(plus, y), z))
app'(app'(quot, 0), app'(s, y)) → 0
app'(app'(quot, app'(s, x)), app'(s, y)) → app'(s, app'(app'(quot, app'(app'(minus, x), y)), app'(s, y)))
app'(app'(plus, 0), y) → y
app'(app'(plus, app'(s, x)), y) → app'(s, app'(app'(plus, x), y))
app'(app'(app, nil), k) → k
app'(app'(app, l), nil) → l
app'(app'(app, app'(app'(cons, x), l)), k) → app'(app'(cons, x), app'(app'(app, l), k))
app'(sum, app'(app'(cons, x), nil)) → app'(app'(cons, x), nil)
app'(sum, app'(app'(cons, x), app'(app'(cons, y), l))) → app'(sum, app'(app'(cons, app'(app'(plus, x), y)), l))
app'(sum, app'(app'(app, l), app'(app'(cons, x), app'(app'(cons, y), k)))) → app'(sum, app'(app'(app, l), app'(sum, app'(app'(cons, x), app'(app'(cons, y), k)))))
app'(app'(map, f), nil) → nil
app'(app'(map, f), app'(app'(cons, x), xs)) → app'(app'(cons, app'(f, x)), app'(app'(map, f), xs))
app'(app'(filter, f), nil) → nil
app'(app'(filter, f), app'(app'(cons, x), xs)) → app'(app'(app'(app'(filter2, app'(f, x)), f), x), xs)
app'(app'(app'(app'(filter2, true), f), x), xs) → app'(app'(cons, x), app'(app'(filter, f), xs))
app'(app'(app'(app'(filter2, false), f), x), xs) → app'(app'(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
      ↳ EdgeDeletionProof
        ↳ QDP
          ↳ DependencyGraphProof
            ↳ AND
              ↳ QDP
              ↳ QDP
                ↳ QDPOrderProof
              ↳ QDP
              ↳ QDP
              ↳ QDP
              ↳ QDP
              ↳ QDP

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

APP'(app'(plus, app'(s, x)), y) → APP'(app'(plus, x), y)

The TRS R consists of the following rules:

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

Q is empty.
We have to consider all minimal (P,Q,R)-chains.
We use the reduction pair processor [13]. Here, we combined the reduction pair processor with the A-transformation [14] which results in the following intermediate Q-DP Problem.
Q DP problem:
The TRS P consists of the following rules:

PLUS(s(x), y) → PLUS(x, y)

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


The following pairs can be oriented strictly and are deleted.


APP'(app'(plus, app'(s, x)), y) → APP'(app'(plus, x), y)
The remaining pairs can at least be oriented weakly.
none
Used ordering: Combined order from the following AFS and order.
PLUS(x1, x2)  =  PLUS(x1)
s(x1)  =  s(x1)

Lexicographic Path Order [19].
Precedence:
s1 > PLUS1

The following usable rules [14] were oriented: none



↳ QTRS
  ↳ DependencyPairsProof
    ↳ QDP
      ↳ EdgeDeletionProof
        ↳ 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'(app'(minus, x), 0) → x
app'(app'(minus, app'(s, x)), app'(s, y)) → app'(app'(minus, x), y)
app'(app'(minus, app'(app'(minus, x), y)), z) → app'(app'(minus, x), app'(app'(plus, y), z))
app'(app'(quot, 0), app'(s, y)) → 0
app'(app'(quot, app'(s, x)), app'(s, y)) → app'(s, app'(app'(quot, app'(app'(minus, x), y)), app'(s, y)))
app'(app'(plus, 0), y) → y
app'(app'(plus, app'(s, x)), y) → app'(s, app'(app'(plus, x), y))
app'(app'(app, nil), k) → k
app'(app'(app, l), nil) → l
app'(app'(app, app'(app'(cons, x), l)), k) → app'(app'(cons, x), app'(app'(app, l), k))
app'(sum, app'(app'(cons, x), nil)) → app'(app'(cons, x), nil)
app'(sum, app'(app'(cons, x), app'(app'(cons, y), l))) → app'(sum, app'(app'(cons, app'(app'(plus, x), y)), l))
app'(sum, app'(app'(app, l), app'(app'(cons, x), app'(app'(cons, y), k)))) → app'(sum, app'(app'(app, l), app'(sum, app'(app'(cons, x), app'(app'(cons, y), k)))))
app'(app'(map, f), nil) → nil
app'(app'(map, f), app'(app'(cons, x), xs)) → app'(app'(cons, app'(f, x)), app'(app'(map, f), xs))
app'(app'(filter, f), nil) → nil
app'(app'(filter, f), app'(app'(cons, x), xs)) → app'(app'(app'(app'(filter2, app'(f, x)), f), x), xs)
app'(app'(app'(app'(filter2, true), f), x), xs) → app'(app'(cons, x), app'(app'(filter, f), xs))
app'(app'(app'(app'(filter2, false), f), x), xs) → app'(app'(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
      ↳ EdgeDeletionProof
        ↳ QDP
          ↳ DependencyGraphProof
            ↳ AND
              ↳ QDP
              ↳ QDP
              ↳ QDP
                ↳ QDPOrderProof
              ↳ QDP
              ↳ QDP
              ↳ QDP
              ↳ QDP

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

APP'(sum, app'(app'(cons, x), app'(app'(cons, y), l))) → APP'(sum, app'(app'(cons, app'(app'(plus, x), y)), l))

The TRS R consists of the following rules:

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

Q is empty.
We have to consider all minimal (P,Q,R)-chains.
We use the reduction pair processor [13]. Here, we combined the reduction pair processor with the A-transformation [14] which results in the following intermediate Q-DP Problem.
Q DP problem:
The TRS P consists of the following rules:

SUM(cons(x, cons(y, l))) → SUM(cons(plus(x, y), l))

The TRS R consists of the following rules:

plus(0, y) → y
plus(s(x), y) → s(plus(x, y))

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


The following pairs can be oriented strictly and are deleted.


APP'(sum, app'(app'(cons, x), app'(app'(cons, y), l))) → APP'(sum, app'(app'(cons, app'(app'(plus, x), y)), l))
The remaining pairs can at least be oriented weakly.
none
Used ordering: Combined order from the following AFS and order.
SUM(x1)  =  SUM(x1)
cons(x1, x2)  =  cons(x1, x2)
plus(x1, x2)  =  x2
0  =  0
s(x1)  =  x1

Lexicographic Path Order [19].
Precedence:
SUM1 > cons2
0 > cons2

The following usable rules [14] were oriented:

app'(app'(plus, 0), y) → y
app'(app'(plus, app'(s, x)), y) → app'(s, app'(app'(plus, x), y))



↳ QTRS
  ↳ DependencyPairsProof
    ↳ QDP
      ↳ EdgeDeletionProof
        ↳ 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'(app'(minus, x), 0) → x
app'(app'(minus, app'(s, x)), app'(s, y)) → app'(app'(minus, x), y)
app'(app'(minus, app'(app'(minus, x), y)), z) → app'(app'(minus, x), app'(app'(plus, y), z))
app'(app'(quot, 0), app'(s, y)) → 0
app'(app'(quot, app'(s, x)), app'(s, y)) → app'(s, app'(app'(quot, app'(app'(minus, x), y)), app'(s, y)))
app'(app'(plus, 0), y) → y
app'(app'(plus, app'(s, x)), y) → app'(s, app'(app'(plus, x), y))
app'(app'(app, nil), k) → k
app'(app'(app, l), nil) → l
app'(app'(app, app'(app'(cons, x), l)), k) → app'(app'(cons, x), app'(app'(app, l), k))
app'(sum, app'(app'(cons, x), nil)) → app'(app'(cons, x), nil)
app'(sum, app'(app'(cons, x), app'(app'(cons, y), l))) → app'(sum, app'(app'(cons, app'(app'(plus, x), y)), l))
app'(sum, app'(app'(app, l), app'(app'(cons, x), app'(app'(cons, y), k)))) → app'(sum, app'(app'(app, l), app'(sum, app'(app'(cons, x), app'(app'(cons, y), k)))))
app'(app'(map, f), nil) → nil
app'(app'(map, f), app'(app'(cons, x), xs)) → app'(app'(cons, app'(f, x)), app'(app'(map, f), xs))
app'(app'(filter, f), nil) → nil
app'(app'(filter, f), app'(app'(cons, x), xs)) → app'(app'(app'(app'(filter2, app'(f, x)), f), x), xs)
app'(app'(app'(app'(filter2, true), f), x), xs) → app'(app'(cons, x), app'(app'(filter, f), xs))
app'(app'(app'(app'(filter2, false), f), x), xs) → app'(app'(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
      ↳ EdgeDeletionProof
        ↳ QDP
          ↳ DependencyGraphProof
            ↳ AND
              ↳ QDP
              ↳ QDP
              ↳ QDP
              ↳ QDP
              ↳ QDP
              ↳ QDP
              ↳ QDP

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

APP'(sum, app'(app'(app, l), app'(app'(cons, x), app'(app'(cons, y), k)))) → APP'(sum, app'(app'(app, l), app'(sum, app'(app'(cons, x), app'(app'(cons, y), k)))))

The TRS R consists of the following rules:

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

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

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

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

APP'(app'(minus, app'(s, x)), app'(s, y)) → APP'(app'(minus, x), y)
APP'(app'(minus, app'(app'(minus, x), y)), z) → APP'(app'(minus, x), app'(app'(plus, y), z))

The TRS R consists of the following rules:

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

Q is empty.
We have to consider all minimal (P,Q,R)-chains.
We use the reduction pair processor [13]. Here, we combined the reduction pair processor with the A-transformation [14] which results in the following intermediate Q-DP Problem.
Q DP problem:
The TRS P consists of the following rules:

MINUS(minus(x, y), z) → MINUS(x, plus(y, z))
MINUS(s(x), s(y)) → MINUS(x, y)

The TRS R consists of the following rules:

plus(0, y) → y
plus(s(x), y) → s(plus(x, y))

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


The following pairs can be oriented strictly and are deleted.


APP'(app'(minus, app'(app'(minus, x), y)), z) → APP'(app'(minus, x), app'(app'(plus, y), z))
The remaining pairs can at least be oriented weakly.

APP'(app'(minus, app'(s, x)), app'(s, y)) → APP'(app'(minus, x), y)
Used ordering: Combined order from the following AFS and order.
MINUS(x1, x2)  =  MINUS(x1, x2)
minus(x1, x2)  =  minus(x1)
plus(x1, x2)  =  x2
s(x1)  =  x1
0  =  0

Lexicographic Path Order [19].
Precedence:
minus1 > MINUS2

The following usable rules [14] were oriented:

app'(app'(plus, app'(s, x)), y) → app'(s, app'(app'(plus, x), y))
app'(app'(plus, 0), y) → y



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

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

APP'(app'(minus, app'(s, x)), app'(s, y)) → APP'(app'(minus, x), y)

The TRS R consists of the following rules:

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

Q is empty.
We have to consider all minimal (P,Q,R)-chains.
We use the reduction pair processor [13]. Here, we combined the reduction pair processor with the A-transformation [14] which results in the following intermediate Q-DP Problem.
Q DP problem:
The TRS P consists of the following rules:

MINUS(s(x), s(y)) → MINUS(x, y)

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


The following pairs can be oriented strictly and are deleted.


APP'(app'(minus, app'(s, x)), app'(s, y)) → APP'(app'(minus, x), y)
The remaining pairs can at least be oriented weakly.
none
Used ordering: Combined order from the following AFS and order.
MINUS(x1, x2)  =  MINUS(x2)
s(x1)  =  s(x1)

Lexicographic Path Order [19].
Precedence:
s1 > MINUS1

The following usable rules [14] were oriented: none



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

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

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

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

APP'(app'(quot, app'(s, x)), app'(s, y)) → APP'(app'(quot, app'(app'(minus, x), y)), app'(s, y))

The TRS R consists of the following rules:

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

Q is empty.
We have to consider all minimal (P,Q,R)-chains.
We use the reduction pair processor [13]. Here, we combined the reduction pair processor with the A-transformation [14] which results in the following intermediate Q-DP Problem.
Q DP problem:
The TRS P consists of the following rules:

QUOT(s(x), s(y)) → QUOT(minus(x, y), s(y))

The TRS R consists of the following rules:

minus(x, 0) → x
minus(minus(x, y), z) → minus(x, plus(y, z))
minus(s(x), s(y)) → minus(x, y)
plus(0, y) → y
plus(s(x), y) → s(plus(x, y))

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


The following pairs can be oriented strictly and are deleted.


APP'(app'(quot, app'(s, x)), app'(s, y)) → APP'(app'(quot, app'(app'(minus, x), y)), app'(s, y))
The remaining pairs can at least be oriented weakly.
none
Used ordering: Combined order from the following AFS and order.
QUOT(x1, x2)  =  QUOT(x1)
s(x1)  =  s(x1)
minus(x1, x2)  =  x1
plus(x1, x2)  =  plus(x1, x2)
0  =  0

Lexicographic Path Order [19].
Precedence:
plus2 > s1 > QUOT1
0 > QUOT1

The following usable rules [14] were oriented:

app'(app'(minus, x), 0) → x
app'(app'(minus, app'(app'(minus, x), y)), z) → app'(app'(minus, x), app'(app'(plus, y), z))
app'(app'(minus, app'(s, x)), app'(s, y)) → app'(app'(minus, x), y)



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

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

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

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

APP'(app'(app'(app'(filter2, true), f), x), xs) → APP'(app'(filter, f), xs)
APP'(app'(map, f), app'(app'(cons, x), xs)) → APP'(f, x)
APP'(app'(map, f), app'(app'(cons, x), xs)) → APP'(app'(map, f), xs)
APP'(app'(filter, f), app'(app'(cons, x), xs)) → APP'(f, x)
APP'(app'(filter, f), app'(app'(cons, x), xs)) → APP'(app'(app'(app'(filter2, app'(f, x)), f), x), xs)
APP'(app'(app'(app'(filter2, false), f), x), xs) → APP'(app'(filter, f), xs)

The TRS R consists of the following rules:

app'(app'(minus, x), 0) → x
app'(app'(minus, app'(s, x)), app'(s, y)) → app'(app'(minus, x), y)
app'(app'(minus, app'(app'(minus, x), y)), z) → app'(app'(minus, x), app'(app'(plus, y), z))
app'(app'(quot, 0), app'(s, y)) → 0
app'(app'(quot, app'(s, x)), app'(s, y)) → app'(s, app'(app'(quot, app'(app'(minus, x), y)), app'(s, y)))
app'(app'(plus, 0), y) → y
app'(app'(plus, app'(s, x)), y) → app'(s, app'(app'(plus, x), y))
app'(app'(app, nil), k) → k
app'(app'(app, l), nil) → l
app'(app'(app, app'(app'(cons, x), l)), k) → app'(app'(cons, x), app'(app'(app, l), k))
app'(sum, app'(app'(cons, x), nil)) → app'(app'(cons, x), nil)
app'(sum, app'(app'(cons, x), app'(app'(cons, y), l))) → app'(sum, app'(app'(cons, app'(app'(plus, x), y)), l))
app'(sum, app'(app'(app, l), app'(app'(cons, x), app'(app'(cons, y), k)))) → app'(sum, app'(app'(app, l), app'(sum, app'(app'(cons, x), app'(app'(cons, y), k)))))
app'(app'(map, f), nil) → nil
app'(app'(map, f), app'(app'(cons, x), xs)) → app'(app'(cons, app'(f, x)), app'(app'(map, f), xs))
app'(app'(filter, f), nil) → nil
app'(app'(filter, f), app'(app'(cons, x), xs)) → app'(app'(app'(app'(filter2, app'(f, x)), f), x), xs)
app'(app'(app'(app'(filter2, true), f), x), xs) → app'(app'(cons, x), app'(app'(filter, f), xs))
app'(app'(app'(app'(filter2, false), f), x), xs) → app'(app'(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'(app'(map, f), app'(app'(cons, x), xs)) → APP'(f, x)
APP'(app'(map, f), app'(app'(cons, x), xs)) → APP'(app'(map, f), xs)
APP'(app'(filter, f), app'(app'(cons, x), xs)) → APP'(f, x)
APP'(app'(filter, f), app'(app'(cons, x), xs)) → APP'(app'(app'(app'(filter2, app'(f, x)), f), x), xs)
The remaining pairs can at least be oriented weakly.

APP'(app'(app'(app'(filter2, true), f), x), xs) → APP'(app'(filter, f), xs)
APP'(app'(app'(app'(filter2, false), f), x), xs) → APP'(app'(filter, f), xs)
Used ordering: Combined order from the following AFS and order.
APP'(x1, x2)  =  x2
app'(x1, x2)  =  app'(x1, x2)
filter2  =  filter2
true  =  true
filter  =  filter
map  =  map
cons  =  cons
false  =  false
sum  =  sum
nil  =  nil
minus  =  minus
plus  =  plus
s  =  s
quot  =  quot
app  =  app
0  =  0

Lexicographic Path Order [19].
Precedence:
filter2 > app'2 > sum
true > app'2 > sum
filter > app'2 > sum
map > app'2 > sum
cons > app'2 > sum
false > app'2 > sum
nil > app'2 > sum
minus > plus > app'2 > sum
minus > plus > s > sum
quot > sum
app > app'2 > sum
0 > sum

The following usable rules [14] were oriented: none



↳ QTRS
  ↳ DependencyPairsProof
    ↳ QDP
      ↳ EdgeDeletionProof
        ↳ 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'(app'(app'(app'(filter2, true), f), x), xs) → APP'(app'(filter, f), xs)
APP'(app'(app'(app'(filter2, false), f), x), xs) → APP'(app'(filter, f), xs)

The TRS R consists of the following rules:

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