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

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

app2(app2(app2(if, true), xs), ys) -> xs
app2(app2(app2(if, false), xs), ys) -> ys
app2(app2(lt, app2(s, x)), app2(s, y)) -> app2(app2(lt, x), y)
app2(app2(lt, 0), app2(s, y)) -> true
app2(app2(lt, y), 0) -> false
app2(app2(eq, x), x) -> true
app2(app2(eq, app2(s, x)), 0) -> false
app2(app2(eq, 0), app2(s, x)) -> false
app2(app2(merge, xs), nil) -> xs
app2(app2(merge, nil), ys) -> ys
app2(app2(merge, app2(app2(cons, x), xs)), app2(app2(cons, y), ys)) -> app2(app2(app2(if, app2(app2(lt, x), y)), app2(app2(cons, x), app2(app2(merge, xs), app2(app2(cons, y), ys)))), app2(app2(app2(if, app2(app2(eq, x), y)), app2(app2(cons, x), app2(app2(merge, xs), ys))), app2(app2(cons, y), app2(app2(merge, app2(app2(cons, x), xs)), ys))))
app2(app2(map, f), nil) -> nil
app2(app2(map, f), app2(app2(cons, x), xs)) -> app2(app2(cons, app2(f, x)), app2(app2(map, f), xs))
app2(app2(mult, 0), x) -> 0
app2(app2(mult, app2(s, x)), y) -> app2(app2(plus, y), app2(app2(mult, x), y))
app2(app2(plus, 0), x) -> 0
app2(app2(plus, app2(s, x)), y) -> app2(s, app2(app2(plus, x), y))
list1 -> app2(app2(map, app2(mult, app2(s, app2(s, 0)))), hamming)
list2 -> app2(app2(map, app2(mult, app2(s, app2(s, app2(s, 0))))), hamming)
list3 -> app2(app2(map, app2(mult, app2(s, app2(s, app2(s, app2(s, app2(s, 0))))))), hamming)
hamming -> app2(app2(cons, app2(s, 0)), app2(app2(merge, list1), app2(app2(merge, list2), list3)))

Q is empty.


↳ QTRS
  ↳ DependencyPairsProof

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

app2(app2(app2(if, true), xs), ys) -> xs
app2(app2(app2(if, false), xs), ys) -> ys
app2(app2(lt, app2(s, x)), app2(s, y)) -> app2(app2(lt, x), y)
app2(app2(lt, 0), app2(s, y)) -> true
app2(app2(lt, y), 0) -> false
app2(app2(eq, x), x) -> true
app2(app2(eq, app2(s, x)), 0) -> false
app2(app2(eq, 0), app2(s, x)) -> false
app2(app2(merge, xs), nil) -> xs
app2(app2(merge, nil), ys) -> ys
app2(app2(merge, app2(app2(cons, x), xs)), app2(app2(cons, y), ys)) -> app2(app2(app2(if, app2(app2(lt, x), y)), app2(app2(cons, x), app2(app2(merge, xs), app2(app2(cons, y), ys)))), app2(app2(app2(if, app2(app2(eq, x), y)), app2(app2(cons, x), app2(app2(merge, xs), ys))), app2(app2(cons, y), app2(app2(merge, app2(app2(cons, x), xs)), ys))))
app2(app2(map, f), nil) -> nil
app2(app2(map, f), app2(app2(cons, x), xs)) -> app2(app2(cons, app2(f, x)), app2(app2(map, f), xs))
app2(app2(mult, 0), x) -> 0
app2(app2(mult, app2(s, x)), y) -> app2(app2(plus, y), app2(app2(mult, x), y))
app2(app2(plus, 0), x) -> 0
app2(app2(plus, app2(s, x)), y) -> app2(s, app2(app2(plus, x), y))
list1 -> app2(app2(map, app2(mult, app2(s, app2(s, 0)))), hamming)
list2 -> app2(app2(map, app2(mult, app2(s, app2(s, app2(s, 0))))), hamming)
list3 -> app2(app2(map, app2(mult, app2(s, app2(s, app2(s, app2(s, app2(s, 0))))))), hamming)
hamming -> app2(app2(cons, app2(s, 0)), app2(app2(merge, list1), app2(app2(merge, list2), list3)))

Q is empty.

Using Dependency Pairs [1,13] we result in the following initial DP problem:
Q DP problem:
The TRS P consists of the following rules:

APP2(app2(merge, app2(app2(cons, x), xs)), app2(app2(cons, y), ys)) -> APP2(app2(merge, xs), app2(app2(cons, y), ys))
LIST2 -> APP2(s, 0)
APP2(app2(plus, app2(s, x)), y) -> APP2(s, app2(app2(plus, x), y))
APP2(app2(map, f), app2(app2(cons, x), xs)) -> APP2(app2(cons, app2(f, x)), app2(app2(map, f), xs))
APP2(app2(merge, app2(app2(cons, x), xs)), app2(app2(cons, y), ys)) -> APP2(app2(if, app2(app2(lt, x), y)), app2(app2(cons, x), app2(app2(merge, xs), app2(app2(cons, y), ys))))
LIST3 -> HAMMING
APP2(app2(merge, app2(app2(cons, x), xs)), app2(app2(cons, y), ys)) -> APP2(lt, x)
HAMMING -> APP2(app2(cons, app2(s, 0)), app2(app2(merge, list1), app2(app2(merge, list2), list3)))
HAMMING -> LIST3
LIST3 -> APP2(s, 0)
LIST1 -> HAMMING
APP2(app2(mult, app2(s, x)), y) -> APP2(app2(plus, y), app2(app2(mult, x), y))
APP2(app2(merge, app2(app2(cons, x), xs)), app2(app2(cons, y), ys)) -> APP2(app2(app2(if, app2(app2(eq, x), y)), app2(app2(cons, x), app2(app2(merge, xs), ys))), app2(app2(cons, y), app2(app2(merge, app2(app2(cons, x), xs)), ys)))
APP2(app2(merge, app2(app2(cons, x), xs)), app2(app2(cons, y), ys)) -> APP2(eq, x)
APP2(app2(merge, app2(app2(cons, x), xs)), app2(app2(cons, y), ys)) -> APP2(app2(cons, x), app2(app2(merge, xs), app2(app2(cons, y), ys)))
APP2(app2(merge, app2(app2(cons, x), xs)), app2(app2(cons, y), ys)) -> APP2(app2(cons, y), app2(app2(merge, app2(app2(cons, x), xs)), ys))
LIST1 -> APP2(map, app2(mult, app2(s, app2(s, 0))))
APP2(app2(plus, app2(s, x)), y) -> APP2(app2(plus, x), y)
APP2(app2(mult, app2(s, x)), y) -> APP2(plus, y)
APP2(app2(merge, app2(app2(cons, x), xs)), app2(app2(cons, y), ys)) -> APP2(app2(lt, x), y)
APP2(app2(merge, app2(app2(cons, x), xs)), app2(app2(cons, y), ys)) -> APP2(app2(merge, app2(app2(cons, x), xs)), ys)
HAMMING -> APP2(merge, list2)
APP2(app2(mult, app2(s, x)), y) -> APP2(mult, x)
LIST3 -> APP2(s, app2(s, app2(s, app2(s, app2(s, 0)))))
APP2(app2(map, f), app2(app2(cons, x), xs)) -> APP2(cons, app2(f, x))
APP2(app2(merge, app2(app2(cons, x), xs)), app2(app2(cons, y), ys)) -> APP2(merge, xs)
LIST2 -> APP2(s, app2(s, app2(s, 0)))
HAMMING -> APP2(s, 0)
APP2(app2(merge, app2(app2(cons, x), xs)), app2(app2(cons, y), ys)) -> APP2(app2(eq, x), y)
LIST2 -> APP2(s, app2(s, 0))
LIST1 -> APP2(mult, app2(s, app2(s, 0)))
LIST2 -> APP2(mult, app2(s, app2(s, app2(s, 0))))
HAMMING -> APP2(app2(merge, list1), app2(app2(merge, list2), list3))
HAMMING -> LIST2
APP2(app2(merge, app2(app2(cons, x), xs)), app2(app2(cons, y), ys)) -> APP2(app2(if, app2(app2(eq, x), y)), app2(app2(cons, x), app2(app2(merge, xs), ys)))
LIST1 -> APP2(app2(map, app2(mult, app2(s, app2(s, 0)))), hamming)
APP2(app2(merge, app2(app2(cons, x), xs)), app2(app2(cons, y), ys)) -> APP2(app2(merge, xs), ys)
APP2(app2(lt, app2(s, x)), app2(s, y)) -> APP2(lt, x)
HAMMING -> APP2(cons, app2(s, 0))
APP2(app2(map, f), app2(app2(cons, x), xs)) -> APP2(f, x)
APP2(app2(mult, app2(s, x)), y) -> APP2(app2(mult, x), y)
LIST2 -> HAMMING
APP2(app2(merge, app2(app2(cons, x), xs)), app2(app2(cons, y), ys)) -> APP2(if, app2(app2(eq, x), y))
APP2(app2(plus, app2(s, x)), y) -> APP2(plus, x)
LIST3 -> APP2(s, app2(s, 0))
APP2(app2(lt, app2(s, x)), app2(s, y)) -> APP2(app2(lt, x), y)
LIST1 -> APP2(s, app2(s, 0))
LIST3 -> APP2(s, app2(s, app2(s, app2(s, 0))))
APP2(app2(merge, app2(app2(cons, x), xs)), app2(app2(cons, y), ys)) -> APP2(if, app2(app2(lt, x), y))
HAMMING -> APP2(app2(merge, list2), list3)
HAMMING -> LIST1
LIST2 -> APP2(app2(map, app2(mult, app2(s, app2(s, app2(s, 0))))), hamming)
LIST3 -> APP2(map, app2(mult, app2(s, app2(s, app2(s, app2(s, app2(s, 0)))))))
APP2(app2(map, f), app2(app2(cons, x), xs)) -> APP2(app2(map, f), xs)
LIST3 -> APP2(mult, app2(s, app2(s, app2(s, app2(s, app2(s, 0))))))
HAMMING -> APP2(merge, list1)
LIST3 -> APP2(s, app2(s, app2(s, 0)))
APP2(app2(merge, app2(app2(cons, x), xs)), app2(app2(cons, y), ys)) -> APP2(app2(app2(if, app2(app2(lt, x), y)), app2(app2(cons, x), app2(app2(merge, xs), app2(app2(cons, y), ys)))), app2(app2(app2(if, app2(app2(eq, x), y)), app2(app2(cons, x), app2(app2(merge, xs), ys))), app2(app2(cons, y), app2(app2(merge, app2(app2(cons, x), xs)), ys))))
LIST3 -> APP2(app2(map, app2(mult, app2(s, app2(s, app2(s, app2(s, app2(s, 0))))))), hamming)
LIST2 -> APP2(map, app2(mult, app2(s, app2(s, app2(s, 0)))))
LIST1 -> APP2(s, 0)
APP2(app2(merge, app2(app2(cons, x), xs)), app2(app2(cons, y), ys)) -> APP2(app2(cons, x), app2(app2(merge, xs), ys))

The TRS R consists of the following rules:

app2(app2(app2(if, true), xs), ys) -> xs
app2(app2(app2(if, false), xs), ys) -> ys
app2(app2(lt, app2(s, x)), app2(s, y)) -> app2(app2(lt, x), y)
app2(app2(lt, 0), app2(s, y)) -> true
app2(app2(lt, y), 0) -> false
app2(app2(eq, x), x) -> true
app2(app2(eq, app2(s, x)), 0) -> false
app2(app2(eq, 0), app2(s, x)) -> false
app2(app2(merge, xs), nil) -> xs
app2(app2(merge, nil), ys) -> ys
app2(app2(merge, app2(app2(cons, x), xs)), app2(app2(cons, y), ys)) -> app2(app2(app2(if, app2(app2(lt, x), y)), app2(app2(cons, x), app2(app2(merge, xs), app2(app2(cons, y), ys)))), app2(app2(app2(if, app2(app2(eq, x), y)), app2(app2(cons, x), app2(app2(merge, xs), ys))), app2(app2(cons, y), app2(app2(merge, app2(app2(cons, x), xs)), ys))))
app2(app2(map, f), nil) -> nil
app2(app2(map, f), app2(app2(cons, x), xs)) -> app2(app2(cons, app2(f, x)), app2(app2(map, f), xs))
app2(app2(mult, 0), x) -> 0
app2(app2(mult, app2(s, x)), y) -> app2(app2(plus, y), app2(app2(mult, x), y))
app2(app2(plus, 0), x) -> 0
app2(app2(plus, app2(s, x)), y) -> app2(s, app2(app2(plus, x), y))
list1 -> app2(app2(map, app2(mult, app2(s, app2(s, 0)))), hamming)
list2 -> app2(app2(map, app2(mult, app2(s, app2(s, app2(s, 0))))), hamming)
list3 -> app2(app2(map, app2(mult, app2(s, app2(s, app2(s, app2(s, app2(s, 0))))))), hamming)
hamming -> app2(app2(cons, app2(s, 0)), app2(app2(merge, list1), app2(app2(merge, list2), list3)))

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

↳ QTRS
  ↳ DependencyPairsProof
    ↳ QDP
      ↳ DependencyGraphProof

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

APP2(app2(merge, app2(app2(cons, x), xs)), app2(app2(cons, y), ys)) -> APP2(app2(merge, xs), app2(app2(cons, y), ys))
LIST2 -> APP2(s, 0)
APP2(app2(plus, app2(s, x)), y) -> APP2(s, app2(app2(plus, x), y))
APP2(app2(map, f), app2(app2(cons, x), xs)) -> APP2(app2(cons, app2(f, x)), app2(app2(map, f), xs))
APP2(app2(merge, app2(app2(cons, x), xs)), app2(app2(cons, y), ys)) -> APP2(app2(if, app2(app2(lt, x), y)), app2(app2(cons, x), app2(app2(merge, xs), app2(app2(cons, y), ys))))
LIST3 -> HAMMING
APP2(app2(merge, app2(app2(cons, x), xs)), app2(app2(cons, y), ys)) -> APP2(lt, x)
HAMMING -> APP2(app2(cons, app2(s, 0)), app2(app2(merge, list1), app2(app2(merge, list2), list3)))
HAMMING -> LIST3
LIST3 -> APP2(s, 0)
LIST1 -> HAMMING
APP2(app2(mult, app2(s, x)), y) -> APP2(app2(plus, y), app2(app2(mult, x), y))
APP2(app2(merge, app2(app2(cons, x), xs)), app2(app2(cons, y), ys)) -> APP2(app2(app2(if, app2(app2(eq, x), y)), app2(app2(cons, x), app2(app2(merge, xs), ys))), app2(app2(cons, y), app2(app2(merge, app2(app2(cons, x), xs)), ys)))
APP2(app2(merge, app2(app2(cons, x), xs)), app2(app2(cons, y), ys)) -> APP2(eq, x)
APP2(app2(merge, app2(app2(cons, x), xs)), app2(app2(cons, y), ys)) -> APP2(app2(cons, x), app2(app2(merge, xs), app2(app2(cons, y), ys)))
APP2(app2(merge, app2(app2(cons, x), xs)), app2(app2(cons, y), ys)) -> APP2(app2(cons, y), app2(app2(merge, app2(app2(cons, x), xs)), ys))
LIST1 -> APP2(map, app2(mult, app2(s, app2(s, 0))))
APP2(app2(plus, app2(s, x)), y) -> APP2(app2(plus, x), y)
APP2(app2(mult, app2(s, x)), y) -> APP2(plus, y)
APP2(app2(merge, app2(app2(cons, x), xs)), app2(app2(cons, y), ys)) -> APP2(app2(lt, x), y)
APP2(app2(merge, app2(app2(cons, x), xs)), app2(app2(cons, y), ys)) -> APP2(app2(merge, app2(app2(cons, x), xs)), ys)
HAMMING -> APP2(merge, list2)
APP2(app2(mult, app2(s, x)), y) -> APP2(mult, x)
LIST3 -> APP2(s, app2(s, app2(s, app2(s, app2(s, 0)))))
APP2(app2(map, f), app2(app2(cons, x), xs)) -> APP2(cons, app2(f, x))
APP2(app2(merge, app2(app2(cons, x), xs)), app2(app2(cons, y), ys)) -> APP2(merge, xs)
LIST2 -> APP2(s, app2(s, app2(s, 0)))
HAMMING -> APP2(s, 0)
APP2(app2(merge, app2(app2(cons, x), xs)), app2(app2(cons, y), ys)) -> APP2(app2(eq, x), y)
LIST2 -> APP2(s, app2(s, 0))
LIST1 -> APP2(mult, app2(s, app2(s, 0)))
LIST2 -> APP2(mult, app2(s, app2(s, app2(s, 0))))
HAMMING -> APP2(app2(merge, list1), app2(app2(merge, list2), list3))
HAMMING -> LIST2
APP2(app2(merge, app2(app2(cons, x), xs)), app2(app2(cons, y), ys)) -> APP2(app2(if, app2(app2(eq, x), y)), app2(app2(cons, x), app2(app2(merge, xs), ys)))
LIST1 -> APP2(app2(map, app2(mult, app2(s, app2(s, 0)))), hamming)
APP2(app2(merge, app2(app2(cons, x), xs)), app2(app2(cons, y), ys)) -> APP2(app2(merge, xs), ys)
APP2(app2(lt, app2(s, x)), app2(s, y)) -> APP2(lt, x)
HAMMING -> APP2(cons, app2(s, 0))
APP2(app2(map, f), app2(app2(cons, x), xs)) -> APP2(f, x)
APP2(app2(mult, app2(s, x)), y) -> APP2(app2(mult, x), y)
LIST2 -> HAMMING
APP2(app2(merge, app2(app2(cons, x), xs)), app2(app2(cons, y), ys)) -> APP2(if, app2(app2(eq, x), y))
APP2(app2(plus, app2(s, x)), y) -> APP2(plus, x)
LIST3 -> APP2(s, app2(s, 0))
APP2(app2(lt, app2(s, x)), app2(s, y)) -> APP2(app2(lt, x), y)
LIST1 -> APP2(s, app2(s, 0))
LIST3 -> APP2(s, app2(s, app2(s, app2(s, 0))))
APP2(app2(merge, app2(app2(cons, x), xs)), app2(app2(cons, y), ys)) -> APP2(if, app2(app2(lt, x), y))
HAMMING -> APP2(app2(merge, list2), list3)
HAMMING -> LIST1
LIST2 -> APP2(app2(map, app2(mult, app2(s, app2(s, app2(s, 0))))), hamming)
LIST3 -> APP2(map, app2(mult, app2(s, app2(s, app2(s, app2(s, app2(s, 0)))))))
APP2(app2(map, f), app2(app2(cons, x), xs)) -> APP2(app2(map, f), xs)
LIST3 -> APP2(mult, app2(s, app2(s, app2(s, app2(s, app2(s, 0))))))
HAMMING -> APP2(merge, list1)
LIST3 -> APP2(s, app2(s, app2(s, 0)))
APP2(app2(merge, app2(app2(cons, x), xs)), app2(app2(cons, y), ys)) -> APP2(app2(app2(if, app2(app2(lt, x), y)), app2(app2(cons, x), app2(app2(merge, xs), app2(app2(cons, y), ys)))), app2(app2(app2(if, app2(app2(eq, x), y)), app2(app2(cons, x), app2(app2(merge, xs), ys))), app2(app2(cons, y), app2(app2(merge, app2(app2(cons, x), xs)), ys))))
LIST3 -> APP2(app2(map, app2(mult, app2(s, app2(s, app2(s, app2(s, app2(s, 0))))))), hamming)
LIST2 -> APP2(map, app2(mult, app2(s, app2(s, app2(s, 0)))))
LIST1 -> APP2(s, 0)
APP2(app2(merge, app2(app2(cons, x), xs)), app2(app2(cons, y), ys)) -> APP2(app2(cons, x), app2(app2(merge, xs), ys))

The TRS R consists of the following rules:

app2(app2(app2(if, true), xs), ys) -> xs
app2(app2(app2(if, false), xs), ys) -> ys
app2(app2(lt, app2(s, x)), app2(s, y)) -> app2(app2(lt, x), y)
app2(app2(lt, 0), app2(s, y)) -> true
app2(app2(lt, y), 0) -> false
app2(app2(eq, x), x) -> true
app2(app2(eq, app2(s, x)), 0) -> false
app2(app2(eq, 0), app2(s, x)) -> false
app2(app2(merge, xs), nil) -> xs
app2(app2(merge, nil), ys) -> ys
app2(app2(merge, app2(app2(cons, x), xs)), app2(app2(cons, y), ys)) -> app2(app2(app2(if, app2(app2(lt, x), y)), app2(app2(cons, x), app2(app2(merge, xs), app2(app2(cons, y), ys)))), app2(app2(app2(if, app2(app2(eq, x), y)), app2(app2(cons, x), app2(app2(merge, xs), ys))), app2(app2(cons, y), app2(app2(merge, app2(app2(cons, x), xs)), ys))))
app2(app2(map, f), nil) -> nil
app2(app2(map, f), app2(app2(cons, x), xs)) -> app2(app2(cons, app2(f, x)), app2(app2(map, f), xs))
app2(app2(mult, 0), x) -> 0
app2(app2(mult, app2(s, x)), y) -> app2(app2(plus, y), app2(app2(mult, x), y))
app2(app2(plus, 0), x) -> 0
app2(app2(plus, app2(s, x)), y) -> app2(s, app2(app2(plus, x), y))
list1 -> app2(app2(map, app2(mult, app2(s, app2(s, 0)))), hamming)
list2 -> app2(app2(map, app2(mult, app2(s, app2(s, app2(s, 0))))), hamming)
list3 -> app2(app2(map, app2(mult, app2(s, app2(s, app2(s, app2(s, app2(s, 0))))))), hamming)
hamming -> app2(app2(cons, app2(s, 0)), app2(app2(merge, list1), app2(app2(merge, list2), list3)))

Q is empty.
We have to consider all minimal (P,Q,R)-chains.
The approximation of the Dependency Graph [13,14,18] contains 6 SCCs with 48 less nodes.

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

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

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

The TRS R consists of the following rules:

app2(app2(app2(if, true), xs), ys) -> xs
app2(app2(app2(if, false), xs), ys) -> ys
app2(app2(lt, app2(s, x)), app2(s, y)) -> app2(app2(lt, x), y)
app2(app2(lt, 0), app2(s, y)) -> true
app2(app2(lt, y), 0) -> false
app2(app2(eq, x), x) -> true
app2(app2(eq, app2(s, x)), 0) -> false
app2(app2(eq, 0), app2(s, x)) -> false
app2(app2(merge, xs), nil) -> xs
app2(app2(merge, nil), ys) -> ys
app2(app2(merge, app2(app2(cons, x), xs)), app2(app2(cons, y), ys)) -> app2(app2(app2(if, app2(app2(lt, x), y)), app2(app2(cons, x), app2(app2(merge, xs), app2(app2(cons, y), ys)))), app2(app2(app2(if, app2(app2(eq, x), y)), app2(app2(cons, x), app2(app2(merge, xs), ys))), app2(app2(cons, y), app2(app2(merge, app2(app2(cons, x), xs)), ys))))
app2(app2(map, f), nil) -> nil
app2(app2(map, f), app2(app2(cons, x), xs)) -> app2(app2(cons, app2(f, x)), app2(app2(map, f), xs))
app2(app2(mult, 0), x) -> 0
app2(app2(mult, app2(s, x)), y) -> app2(app2(plus, y), app2(app2(mult, x), y))
app2(app2(plus, 0), x) -> 0
app2(app2(plus, app2(s, x)), y) -> app2(s, app2(app2(plus, x), y))
list1 -> app2(app2(map, app2(mult, app2(s, app2(s, 0)))), hamming)
list2 -> app2(app2(map, app2(mult, app2(s, app2(s, app2(s, 0))))), hamming)
list3 -> app2(app2(map, app2(mult, app2(s, app2(s, app2(s, app2(s, app2(s, 0))))))), hamming)
hamming -> app2(app2(cons, app2(s, 0)), app2(app2(merge, list1), app2(app2(merge, list2), list3)))

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


The following pairs can be oriented strictly and are deleted.


APP2(app2(plus, app2(s, x)), y) -> APP2(app2(plus, x), y)
The remaining pairs can at least be oriented weakly.
none
Used ordering: Polynomial interpretation [21]:

POL(APP2(x1, x2)) = x1   
POL(app2(x1, x2)) = 1 + x2   
POL(plus) = 0   
POL(s) = 0   

The following usable rules [14] were oriented: none



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

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

app2(app2(app2(if, true), xs), ys) -> xs
app2(app2(app2(if, false), xs), ys) -> ys
app2(app2(lt, app2(s, x)), app2(s, y)) -> app2(app2(lt, x), y)
app2(app2(lt, 0), app2(s, y)) -> true
app2(app2(lt, y), 0) -> false
app2(app2(eq, x), x) -> true
app2(app2(eq, app2(s, x)), 0) -> false
app2(app2(eq, 0), app2(s, x)) -> false
app2(app2(merge, xs), nil) -> xs
app2(app2(merge, nil), ys) -> ys
app2(app2(merge, app2(app2(cons, x), xs)), app2(app2(cons, y), ys)) -> app2(app2(app2(if, app2(app2(lt, x), y)), app2(app2(cons, x), app2(app2(merge, xs), app2(app2(cons, y), ys)))), app2(app2(app2(if, app2(app2(eq, x), y)), app2(app2(cons, x), app2(app2(merge, xs), ys))), app2(app2(cons, y), app2(app2(merge, app2(app2(cons, x), xs)), ys))))
app2(app2(map, f), nil) -> nil
app2(app2(map, f), app2(app2(cons, x), xs)) -> app2(app2(cons, app2(f, x)), app2(app2(map, f), xs))
app2(app2(mult, 0), x) -> 0
app2(app2(mult, app2(s, x)), y) -> app2(app2(plus, y), app2(app2(mult, x), y))
app2(app2(plus, 0), x) -> 0
app2(app2(plus, app2(s, x)), y) -> app2(s, app2(app2(plus, x), y))
list1 -> app2(app2(map, app2(mult, app2(s, app2(s, 0)))), hamming)
list2 -> app2(app2(map, app2(mult, app2(s, app2(s, app2(s, 0))))), hamming)
list3 -> app2(app2(map, app2(mult, app2(s, app2(s, app2(s, app2(s, app2(s, 0))))))), hamming)
hamming -> app2(app2(cons, app2(s, 0)), app2(app2(merge, list1), app2(app2(merge, list2), list3)))

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

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

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

The TRS R consists of the following rules:

app2(app2(app2(if, true), xs), ys) -> xs
app2(app2(app2(if, false), xs), ys) -> ys
app2(app2(lt, app2(s, x)), app2(s, y)) -> app2(app2(lt, x), y)
app2(app2(lt, 0), app2(s, y)) -> true
app2(app2(lt, y), 0) -> false
app2(app2(eq, x), x) -> true
app2(app2(eq, app2(s, x)), 0) -> false
app2(app2(eq, 0), app2(s, x)) -> false
app2(app2(merge, xs), nil) -> xs
app2(app2(merge, nil), ys) -> ys
app2(app2(merge, app2(app2(cons, x), xs)), app2(app2(cons, y), ys)) -> app2(app2(app2(if, app2(app2(lt, x), y)), app2(app2(cons, x), app2(app2(merge, xs), app2(app2(cons, y), ys)))), app2(app2(app2(if, app2(app2(eq, x), y)), app2(app2(cons, x), app2(app2(merge, xs), ys))), app2(app2(cons, y), app2(app2(merge, app2(app2(cons, x), xs)), ys))))
app2(app2(map, f), nil) -> nil
app2(app2(map, f), app2(app2(cons, x), xs)) -> app2(app2(cons, app2(f, x)), app2(app2(map, f), xs))
app2(app2(mult, 0), x) -> 0
app2(app2(mult, app2(s, x)), y) -> app2(app2(plus, y), app2(app2(mult, x), y))
app2(app2(plus, 0), x) -> 0
app2(app2(plus, app2(s, x)), y) -> app2(s, app2(app2(plus, x), y))
list1 -> app2(app2(map, app2(mult, app2(s, app2(s, 0)))), hamming)
list2 -> app2(app2(map, app2(mult, app2(s, app2(s, app2(s, 0))))), hamming)
list3 -> app2(app2(map, app2(mult, app2(s, app2(s, app2(s, app2(s, app2(s, 0))))))), hamming)
hamming -> app2(app2(cons, app2(s, 0)), app2(app2(merge, list1), app2(app2(merge, list2), list3)))

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


The following pairs can be oriented strictly and are deleted.


APP2(app2(mult, app2(s, x)), y) -> APP2(app2(mult, x), y)
The remaining pairs can at least be oriented weakly.
none
Used ordering: Polynomial interpretation [21]:

POL(APP2(x1, x2)) = x1   
POL(app2(x1, x2)) = 1 + x2   
POL(mult) = 0   
POL(s) = 0   

The following usable rules [14] were oriented: none



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

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

app2(app2(app2(if, true), xs), ys) -> xs
app2(app2(app2(if, false), xs), ys) -> ys
app2(app2(lt, app2(s, x)), app2(s, y)) -> app2(app2(lt, x), y)
app2(app2(lt, 0), app2(s, y)) -> true
app2(app2(lt, y), 0) -> false
app2(app2(eq, x), x) -> true
app2(app2(eq, app2(s, x)), 0) -> false
app2(app2(eq, 0), app2(s, x)) -> false
app2(app2(merge, xs), nil) -> xs
app2(app2(merge, nil), ys) -> ys
app2(app2(merge, app2(app2(cons, x), xs)), app2(app2(cons, y), ys)) -> app2(app2(app2(if, app2(app2(lt, x), y)), app2(app2(cons, x), app2(app2(merge, xs), app2(app2(cons, y), ys)))), app2(app2(app2(if, app2(app2(eq, x), y)), app2(app2(cons, x), app2(app2(merge, xs), ys))), app2(app2(cons, y), app2(app2(merge, app2(app2(cons, x), xs)), ys))))
app2(app2(map, f), nil) -> nil
app2(app2(map, f), app2(app2(cons, x), xs)) -> app2(app2(cons, app2(f, x)), app2(app2(map, f), xs))
app2(app2(mult, 0), x) -> 0
app2(app2(mult, app2(s, x)), y) -> app2(app2(plus, y), app2(app2(mult, x), y))
app2(app2(plus, 0), x) -> 0
app2(app2(plus, app2(s, x)), y) -> app2(s, app2(app2(plus, x), y))
list1 -> app2(app2(map, app2(mult, app2(s, app2(s, 0)))), hamming)
list2 -> app2(app2(map, app2(mult, app2(s, app2(s, app2(s, 0))))), hamming)
list3 -> app2(app2(map, app2(mult, app2(s, app2(s, app2(s, app2(s, app2(s, 0))))))), hamming)
hamming -> app2(app2(cons, app2(s, 0)), app2(app2(merge, list1), app2(app2(merge, list2), list3)))

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

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

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

The TRS R consists of the following rules:

app2(app2(app2(if, true), xs), ys) -> xs
app2(app2(app2(if, false), xs), ys) -> ys
app2(app2(lt, app2(s, x)), app2(s, y)) -> app2(app2(lt, x), y)
app2(app2(lt, 0), app2(s, y)) -> true
app2(app2(lt, y), 0) -> false
app2(app2(eq, x), x) -> true
app2(app2(eq, app2(s, x)), 0) -> false
app2(app2(eq, 0), app2(s, x)) -> false
app2(app2(merge, xs), nil) -> xs
app2(app2(merge, nil), ys) -> ys
app2(app2(merge, app2(app2(cons, x), xs)), app2(app2(cons, y), ys)) -> app2(app2(app2(if, app2(app2(lt, x), y)), app2(app2(cons, x), app2(app2(merge, xs), app2(app2(cons, y), ys)))), app2(app2(app2(if, app2(app2(eq, x), y)), app2(app2(cons, x), app2(app2(merge, xs), ys))), app2(app2(cons, y), app2(app2(merge, app2(app2(cons, x), xs)), ys))))
app2(app2(map, f), nil) -> nil
app2(app2(map, f), app2(app2(cons, x), xs)) -> app2(app2(cons, app2(f, x)), app2(app2(map, f), xs))
app2(app2(mult, 0), x) -> 0
app2(app2(mult, app2(s, x)), y) -> app2(app2(plus, y), app2(app2(mult, x), y))
app2(app2(plus, 0), x) -> 0
app2(app2(plus, app2(s, x)), y) -> app2(s, app2(app2(plus, x), y))
list1 -> app2(app2(map, app2(mult, app2(s, app2(s, 0)))), hamming)
list2 -> app2(app2(map, app2(mult, app2(s, app2(s, app2(s, 0))))), hamming)
list3 -> app2(app2(map, app2(mult, app2(s, app2(s, app2(s, app2(s, app2(s, 0))))))), hamming)
hamming -> app2(app2(cons, app2(s, 0)), app2(app2(merge, list1), app2(app2(merge, list2), list3)))

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


The following pairs can be oriented strictly and are deleted.


APP2(app2(lt, app2(s, x)), app2(s, y)) -> APP2(app2(lt, x), y)
The remaining pairs can at least be oriented weakly.
none
Used ordering: Polynomial interpretation [21]:

POL(APP2(x1, x2)) = x1·x2   
POL(app2(x1, x2)) = 1 + x2   
POL(lt) = 0   
POL(s) = 0   

The following usable rules [14] were oriented: none



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

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

app2(app2(app2(if, true), xs), ys) -> xs
app2(app2(app2(if, false), xs), ys) -> ys
app2(app2(lt, app2(s, x)), app2(s, y)) -> app2(app2(lt, x), y)
app2(app2(lt, 0), app2(s, y)) -> true
app2(app2(lt, y), 0) -> false
app2(app2(eq, x), x) -> true
app2(app2(eq, app2(s, x)), 0) -> false
app2(app2(eq, 0), app2(s, x)) -> false
app2(app2(merge, xs), nil) -> xs
app2(app2(merge, nil), ys) -> ys
app2(app2(merge, app2(app2(cons, x), xs)), app2(app2(cons, y), ys)) -> app2(app2(app2(if, app2(app2(lt, x), y)), app2(app2(cons, x), app2(app2(merge, xs), app2(app2(cons, y), ys)))), app2(app2(app2(if, app2(app2(eq, x), y)), app2(app2(cons, x), app2(app2(merge, xs), ys))), app2(app2(cons, y), app2(app2(merge, app2(app2(cons, x), xs)), ys))))
app2(app2(map, f), nil) -> nil
app2(app2(map, f), app2(app2(cons, x), xs)) -> app2(app2(cons, app2(f, x)), app2(app2(map, f), xs))
app2(app2(mult, 0), x) -> 0
app2(app2(mult, app2(s, x)), y) -> app2(app2(plus, y), app2(app2(mult, x), y))
app2(app2(plus, 0), x) -> 0
app2(app2(plus, app2(s, x)), y) -> app2(s, app2(app2(plus, x), y))
list1 -> app2(app2(map, app2(mult, app2(s, app2(s, 0)))), hamming)
list2 -> app2(app2(map, app2(mult, app2(s, app2(s, app2(s, 0))))), hamming)
list3 -> app2(app2(map, app2(mult, app2(s, app2(s, app2(s, app2(s, app2(s, 0))))))), hamming)
hamming -> app2(app2(cons, app2(s, 0)), app2(app2(merge, list1), app2(app2(merge, list2), list3)))

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
            ↳ QDPOrderProof
          ↳ QDP
          ↳ QDP

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

APP2(app2(merge, app2(app2(cons, x), xs)), app2(app2(cons, y), ys)) -> APP2(app2(merge, xs), ys)
APP2(app2(merge, app2(app2(cons, x), xs)), app2(app2(cons, y), ys)) -> APP2(app2(merge, xs), app2(app2(cons, y), ys))
APP2(app2(merge, app2(app2(cons, x), xs)), app2(app2(cons, y), ys)) -> APP2(app2(merge, app2(app2(cons, x), xs)), ys)

The TRS R consists of the following rules:

app2(app2(app2(if, true), xs), ys) -> xs
app2(app2(app2(if, false), xs), ys) -> ys
app2(app2(lt, app2(s, x)), app2(s, y)) -> app2(app2(lt, x), y)
app2(app2(lt, 0), app2(s, y)) -> true
app2(app2(lt, y), 0) -> false
app2(app2(eq, x), x) -> true
app2(app2(eq, app2(s, x)), 0) -> false
app2(app2(eq, 0), app2(s, x)) -> false
app2(app2(merge, xs), nil) -> xs
app2(app2(merge, nil), ys) -> ys
app2(app2(merge, app2(app2(cons, x), xs)), app2(app2(cons, y), ys)) -> app2(app2(app2(if, app2(app2(lt, x), y)), app2(app2(cons, x), app2(app2(merge, xs), app2(app2(cons, y), ys)))), app2(app2(app2(if, app2(app2(eq, x), y)), app2(app2(cons, x), app2(app2(merge, xs), ys))), app2(app2(cons, y), app2(app2(merge, app2(app2(cons, x), xs)), ys))))
app2(app2(map, f), nil) -> nil
app2(app2(map, f), app2(app2(cons, x), xs)) -> app2(app2(cons, app2(f, x)), app2(app2(map, f), xs))
app2(app2(mult, 0), x) -> 0
app2(app2(mult, app2(s, x)), y) -> app2(app2(plus, y), app2(app2(mult, x), y))
app2(app2(plus, 0), x) -> 0
app2(app2(plus, app2(s, x)), y) -> app2(s, app2(app2(plus, x), y))
list1 -> app2(app2(map, app2(mult, app2(s, app2(s, 0)))), hamming)
list2 -> app2(app2(map, app2(mult, app2(s, app2(s, app2(s, 0))))), hamming)
list3 -> app2(app2(map, app2(mult, app2(s, app2(s, app2(s, app2(s, app2(s, 0))))))), hamming)
hamming -> app2(app2(cons, app2(s, 0)), app2(app2(merge, list1), app2(app2(merge, list2), list3)))

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


The following pairs can be oriented strictly and are deleted.


APP2(app2(merge, app2(app2(cons, x), xs)), app2(app2(cons, y), ys)) -> APP2(app2(merge, xs), ys)
APP2(app2(merge, app2(app2(cons, x), xs)), app2(app2(cons, y), ys)) -> APP2(app2(merge, app2(app2(cons, x), xs)), ys)
The remaining pairs can at least be oriented weakly.

APP2(app2(merge, app2(app2(cons, x), xs)), app2(app2(cons, y), ys)) -> APP2(app2(merge, xs), app2(app2(cons, y), ys))
Used ordering: Polynomial interpretation [21]:

POL(APP2(x1, x2)) = x1·x2   
POL(app2(x1, x2)) = 1 + x1·x2   
POL(cons) = 0   
POL(merge) = 0   

The following usable rules [14] were oriented: none



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

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

APP2(app2(merge, app2(app2(cons, x), xs)), app2(app2(cons, y), ys)) -> APP2(app2(merge, xs), app2(app2(cons, y), ys))

The TRS R consists of the following rules:

app2(app2(app2(if, true), xs), ys) -> xs
app2(app2(app2(if, false), xs), ys) -> ys
app2(app2(lt, app2(s, x)), app2(s, y)) -> app2(app2(lt, x), y)
app2(app2(lt, 0), app2(s, y)) -> true
app2(app2(lt, y), 0) -> false
app2(app2(eq, x), x) -> true
app2(app2(eq, app2(s, x)), 0) -> false
app2(app2(eq, 0), app2(s, x)) -> false
app2(app2(merge, xs), nil) -> xs
app2(app2(merge, nil), ys) -> ys
app2(app2(merge, app2(app2(cons, x), xs)), app2(app2(cons, y), ys)) -> app2(app2(app2(if, app2(app2(lt, x), y)), app2(app2(cons, x), app2(app2(merge, xs), app2(app2(cons, y), ys)))), app2(app2(app2(if, app2(app2(eq, x), y)), app2(app2(cons, x), app2(app2(merge, xs), ys))), app2(app2(cons, y), app2(app2(merge, app2(app2(cons, x), xs)), ys))))
app2(app2(map, f), nil) -> nil
app2(app2(map, f), app2(app2(cons, x), xs)) -> app2(app2(cons, app2(f, x)), app2(app2(map, f), xs))
app2(app2(mult, 0), x) -> 0
app2(app2(mult, app2(s, x)), y) -> app2(app2(plus, y), app2(app2(mult, x), y))
app2(app2(plus, 0), x) -> 0
app2(app2(plus, app2(s, x)), y) -> app2(s, app2(app2(plus, x), y))
list1 -> app2(app2(map, app2(mult, app2(s, app2(s, 0)))), hamming)
list2 -> app2(app2(map, app2(mult, app2(s, app2(s, app2(s, 0))))), hamming)
list3 -> app2(app2(map, app2(mult, app2(s, app2(s, app2(s, app2(s, app2(s, 0))))))), hamming)
hamming -> app2(app2(cons, app2(s, 0)), app2(app2(merge, list1), app2(app2(merge, list2), list3)))

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


The following pairs can be oriented strictly and are deleted.


APP2(app2(merge, app2(app2(cons, x), xs)), app2(app2(cons, y), ys)) -> APP2(app2(merge, xs), app2(app2(cons, y), ys))
The remaining pairs can at least be oriented weakly.
none
Used ordering: Polynomial interpretation [21]:

POL(APP2(x1, x2)) = x1·x2   
POL(app2(x1, x2)) = 1 + x1·x2   
POL(cons) = 0   
POL(merge) = 1   

The following usable rules [14] were oriented: none



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

app2(app2(app2(if, true), xs), ys) -> xs
app2(app2(app2(if, false), xs), ys) -> ys
app2(app2(lt, app2(s, x)), app2(s, y)) -> app2(app2(lt, x), y)
app2(app2(lt, 0), app2(s, y)) -> true
app2(app2(lt, y), 0) -> false
app2(app2(eq, x), x) -> true
app2(app2(eq, app2(s, x)), 0) -> false
app2(app2(eq, 0), app2(s, x)) -> false
app2(app2(merge, xs), nil) -> xs
app2(app2(merge, nil), ys) -> ys
app2(app2(merge, app2(app2(cons, x), xs)), app2(app2(cons, y), ys)) -> app2(app2(app2(if, app2(app2(lt, x), y)), app2(app2(cons, x), app2(app2(merge, xs), app2(app2(cons, y), ys)))), app2(app2(app2(if, app2(app2(eq, x), y)), app2(app2(cons, x), app2(app2(merge, xs), ys))), app2(app2(cons, y), app2(app2(merge, app2(app2(cons, x), xs)), ys))))
app2(app2(map, f), nil) -> nil
app2(app2(map, f), app2(app2(cons, x), xs)) -> app2(app2(cons, app2(f, x)), app2(app2(map, f), xs))
app2(app2(mult, 0), x) -> 0
app2(app2(mult, app2(s, x)), y) -> app2(app2(plus, y), app2(app2(mult, x), y))
app2(app2(plus, 0), x) -> 0
app2(app2(plus, app2(s, x)), y) -> app2(s, app2(app2(plus, x), y))
list1 -> app2(app2(map, app2(mult, app2(s, app2(s, 0)))), hamming)
list2 -> app2(app2(map, app2(mult, app2(s, app2(s, app2(s, 0))))), hamming)
list3 -> app2(app2(map, app2(mult, app2(s, app2(s, app2(s, app2(s, app2(s, 0))))))), hamming)
hamming -> app2(app2(cons, app2(s, 0)), app2(app2(merge, list1), app2(app2(merge, list2), list3)))

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
            ↳ QDPOrderProof
          ↳ QDP

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

APP2(app2(map, f), app2(app2(cons, x), xs)) -> APP2(f, x)
APP2(app2(map, f), app2(app2(cons, x), xs)) -> APP2(app2(map, f), xs)

The TRS R consists of the following rules:

app2(app2(app2(if, true), xs), ys) -> xs
app2(app2(app2(if, false), xs), ys) -> ys
app2(app2(lt, app2(s, x)), app2(s, y)) -> app2(app2(lt, x), y)
app2(app2(lt, 0), app2(s, y)) -> true
app2(app2(lt, y), 0) -> false
app2(app2(eq, x), x) -> true
app2(app2(eq, app2(s, x)), 0) -> false
app2(app2(eq, 0), app2(s, x)) -> false
app2(app2(merge, xs), nil) -> xs
app2(app2(merge, nil), ys) -> ys
app2(app2(merge, app2(app2(cons, x), xs)), app2(app2(cons, y), ys)) -> app2(app2(app2(if, app2(app2(lt, x), y)), app2(app2(cons, x), app2(app2(merge, xs), app2(app2(cons, y), ys)))), app2(app2(app2(if, app2(app2(eq, x), y)), app2(app2(cons, x), app2(app2(merge, xs), ys))), app2(app2(cons, y), app2(app2(merge, app2(app2(cons, x), xs)), ys))))
app2(app2(map, f), nil) -> nil
app2(app2(map, f), app2(app2(cons, x), xs)) -> app2(app2(cons, app2(f, x)), app2(app2(map, f), xs))
app2(app2(mult, 0), x) -> 0
app2(app2(mult, app2(s, x)), y) -> app2(app2(plus, y), app2(app2(mult, x), y))
app2(app2(plus, 0), x) -> 0
app2(app2(plus, app2(s, x)), y) -> app2(s, app2(app2(plus, x), y))
list1 -> app2(app2(map, app2(mult, app2(s, app2(s, 0)))), hamming)
list2 -> app2(app2(map, app2(mult, app2(s, app2(s, app2(s, 0))))), hamming)
list3 -> app2(app2(map, app2(mult, app2(s, app2(s, app2(s, app2(s, app2(s, 0))))))), hamming)
hamming -> app2(app2(cons, app2(s, 0)), app2(app2(merge, list1), app2(app2(merge, list2), list3)))

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


The following pairs can be oriented strictly and are deleted.


APP2(app2(map, f), app2(app2(cons, x), xs)) -> APP2(f, x)
APP2(app2(map, f), app2(app2(cons, x), xs)) -> APP2(app2(map, f), xs)
The remaining pairs can at least be oriented weakly.
none
Used ordering: Polynomial interpretation [21]:

POL(APP2(x1, x2)) = x1·x2   
POL(app2(x1, x2)) = 1 + x1 + x2   
POL(cons) = 0   
POL(map) = 0   

The following usable rules [14] were oriented: none



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

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

app2(app2(app2(if, true), xs), ys) -> xs
app2(app2(app2(if, false), xs), ys) -> ys
app2(app2(lt, app2(s, x)), app2(s, y)) -> app2(app2(lt, x), y)
app2(app2(lt, 0), app2(s, y)) -> true
app2(app2(lt, y), 0) -> false
app2(app2(eq, x), x) -> true
app2(app2(eq, app2(s, x)), 0) -> false
app2(app2(eq, 0), app2(s, x)) -> false
app2(app2(merge, xs), nil) -> xs
app2(app2(merge, nil), ys) -> ys
app2(app2(merge, app2(app2(cons, x), xs)), app2(app2(cons, y), ys)) -> app2(app2(app2(if, app2(app2(lt, x), y)), app2(app2(cons, x), app2(app2(merge, xs), app2(app2(cons, y), ys)))), app2(app2(app2(if, app2(app2(eq, x), y)), app2(app2(cons, x), app2(app2(merge, xs), ys))), app2(app2(cons, y), app2(app2(merge, app2(app2(cons, x), xs)), ys))))
app2(app2(map, f), nil) -> nil
app2(app2(map, f), app2(app2(cons, x), xs)) -> app2(app2(cons, app2(f, x)), app2(app2(map, f), xs))
app2(app2(mult, 0), x) -> 0
app2(app2(mult, app2(s, x)), y) -> app2(app2(plus, y), app2(app2(mult, x), y))
app2(app2(plus, 0), x) -> 0
app2(app2(plus, app2(s, x)), y) -> app2(s, app2(app2(plus, x), y))
list1 -> app2(app2(map, app2(mult, app2(s, app2(s, 0)))), hamming)
list2 -> app2(app2(map, app2(mult, app2(s, app2(s, app2(s, 0))))), hamming)
list3 -> app2(app2(map, app2(mult, app2(s, app2(s, app2(s, app2(s, app2(s, 0))))))), hamming)
hamming -> app2(app2(cons, app2(s, 0)), app2(app2(merge, list1), app2(app2(merge, list2), list3)))

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

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

HAMMING -> LIST3
LIST2 -> HAMMING
HAMMING -> LIST1
HAMMING -> LIST2
LIST1 -> HAMMING
LIST3 -> HAMMING

The TRS R consists of the following rules:

app2(app2(app2(if, true), xs), ys) -> xs
app2(app2(app2(if, false), xs), ys) -> ys
app2(app2(lt, app2(s, x)), app2(s, y)) -> app2(app2(lt, x), y)
app2(app2(lt, 0), app2(s, y)) -> true
app2(app2(lt, y), 0) -> false
app2(app2(eq, x), x) -> true
app2(app2(eq, app2(s, x)), 0) -> false
app2(app2(eq, 0), app2(s, x)) -> false
app2(app2(merge, xs), nil) -> xs
app2(app2(merge, nil), ys) -> ys
app2(app2(merge, app2(app2(cons, x), xs)), app2(app2(cons, y), ys)) -> app2(app2(app2(if, app2(app2(lt, x), y)), app2(app2(cons, x), app2(app2(merge, xs), app2(app2(cons, y), ys)))), app2(app2(app2(if, app2(app2(eq, x), y)), app2(app2(cons, x), app2(app2(merge, xs), ys))), app2(app2(cons, y), app2(app2(merge, app2(app2(cons, x), xs)), ys))))
app2(app2(map, f), nil) -> nil
app2(app2(map, f), app2(app2(cons, x), xs)) -> app2(app2(cons, app2(f, x)), app2(app2(map, f), xs))
app2(app2(mult, 0), x) -> 0
app2(app2(mult, app2(s, x)), y) -> app2(app2(plus, y), app2(app2(mult, x), y))
app2(app2(plus, 0), x) -> 0
app2(app2(plus, app2(s, x)), y) -> app2(s, app2(app2(plus, x), y))
list1 -> app2(app2(map, app2(mult, app2(s, app2(s, 0)))), hamming)
list2 -> app2(app2(map, app2(mult, app2(s, app2(s, app2(s, 0))))), hamming)
list3 -> app2(app2(map, app2(mult, app2(s, app2(s, app2(s, app2(s, app2(s, 0))))))), hamming)
hamming -> app2(app2(cons, app2(s, 0)), app2(app2(merge, list1), app2(app2(merge, list2), list3)))

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