Termination w.r.t. Q proof of /home/nowonder/forschung/aprove/satrung/aprove/lpar16/AProVESLASHAAECC.trs

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:

fstsplit₂(0, x) -> nil
fstsplit₂(s₁(n), nil) -> nil
fstsplit₂(s₁(n), cons₂(h, t)) -> cons₂(h, fstsplit₂(n, t))
sndsplit₂(0, x) -> x
sndsplit₂(s₁(n), nil) -> nil
sndsplit₂(s₁(n), cons₂(h, t)) -> sndsplit₂(n, t)
empty₁(nil) -> true
empty₁(cons₂(h, t)) -> false
leq₂(0, m) -> true
leq₂(s₁(n), 0) -> false
leq₂(s₁(n), s₁(m)) -> leq₂(n, m)
length₁(nil) -> 0
length₁(cons₂(h, t)) -> s₁(length₁(t))
app₂(nil, x) -> x
app₂(cons₂(h, t), x) -> cons₂(h, app₂(t, x))
map_f₂(pid, nil) -> nil
map_f₂(pid, cons₂(h, t)) -> app₂(f₂(pid, h), map_f₂(pid, t))
process₂(store, m) -> if1₃(store, m, leq₂(m, length₁(store)))
if1₃(store, m, true) -> if2₃(store, m, empty₁(fstsplit₂(m, store)))
if1₃(store, m, false) -> if3₃(store, m, empty₁(fstsplit₂(m, app₂(map_f₂(self, nil), store))))
if2₃(store, m, false) -> process₂(app₂(map_f₂(self, nil), sndsplit₂(m, store)), m)
if3₃(store, m, false) -> process₂(sndsplit₂(m, app₂(map_f₂(self, nil), store)), m)

Q is empty.

↳ QTRS

  ↳ Non-Overlap Check

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

fstsplit₂(0, x) -> nil
fstsplit₂(s₁(n), nil) -> nil
fstsplit₂(s₁(n), cons₂(h, t)) -> cons₂(h, fstsplit₂(n, t))
sndsplit₂(0, x) -> x
sndsplit₂(s₁(n), nil) -> nil
sndsplit₂(s₁(n), cons₂(h, t)) -> sndsplit₂(n, t)
empty₁(nil) -> true
empty₁(cons₂(h, t)) -> false
leq₂(0, m) -> true
leq₂(s₁(n), 0) -> false
leq₂(s₁(n), s₁(m)) -> leq₂(n, m)
length₁(nil) -> 0
length₁(cons₂(h, t)) -> s₁(length₁(t))
app₂(nil, x) -> x
app₂(cons₂(h, t), x) -> cons₂(h, app₂(t, x))
map_f₂(pid, nil) -> nil
map_f₂(pid, cons₂(h, t)) -> app₂(f₂(pid, h), map_f₂(pid, t))
process₂(store, m) -> if1₃(store, m, leq₂(m, length₁(store)))
if1₃(store, m, true) -> if2₃(store, m, empty₁(fstsplit₂(m, store)))
if1₃(store, m, false) -> if3₃(store, m, empty₁(fstsplit₂(m, app₂(map_f₂(self, nil), store))))
if2₃(store, m, false) -> process₂(app₂(map_f₂(self, nil), sndsplit₂(m, store)), m)
if3₃(store, m, false) -> process₂(sndsplit₂(m, app₂(map_f₂(self, nil), store)), m)

Q is empty.

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

↳ QTRS

  ↳ Non-Overlap Check

    ↳ QTRS

      ↳ DependencyPairsProof

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

fstsplit₂(0, x) -> nil
fstsplit₂(s₁(n), nil) -> nil
fstsplit₂(s₁(n), cons₂(h, t)) -> cons₂(h, fstsplit₂(n, t))
sndsplit₂(0, x) -> x
sndsplit₂(s₁(n), nil) -> nil
sndsplit₂(s₁(n), cons₂(h, t)) -> sndsplit₂(n, t)
empty₁(nil) -> true
empty₁(cons₂(h, t)) -> false
leq₂(0, m) -> true
leq₂(s₁(n), 0) -> false
leq₂(s₁(n), s₁(m)) -> leq₂(n, m)
length₁(nil) -> 0
length₁(cons₂(h, t)) -> s₁(length₁(t))
app₂(nil, x) -> x
app₂(cons₂(h, t), x) -> cons₂(h, app₂(t, x))
map_f₂(pid, nil) -> nil
map_f₂(pid, cons₂(h, t)) -> app₂(f₂(pid, h), map_f₂(pid, t))
process₂(store, m) -> if1₃(store, m, leq₂(m, length₁(store)))
if1₃(store, m, true) -> if2₃(store, m, empty₁(fstsplit₂(m, store)))
if1₃(store, m, false) -> if3₃(store, m, empty₁(fstsplit₂(m, app₂(map_f₂(self, nil), store))))
if2₃(store, m, false) -> process₂(app₂(map_f₂(self, nil), sndsplit₂(m, store)), m)
if3₃(store, m, false) -> process₂(sndsplit₂(m, app₂(map_f₂(self, nil), store)), m)

The set Q consists of the following terms:

fstsplit₂(0, x0)
fstsplit₂(s₁(x0), nil)
fstsplit₂(s₁(x0), cons₂(x1, x2))
sndsplit₂(0, x0)
sndsplit₂(s₁(x0), nil)
sndsplit₂(s₁(x0), cons₂(x1, x2))
empty₁(nil)
empty₁(cons₂(x0, x1))
leq₂(0, x0)
leq₂(s₁(x0), 0)
leq₂(s₁(x0), s₁(x1))
length₁(nil)
length₁(cons₂(x0, x1))
app₂(nil, x0)
app₂(cons₂(x0, x1), x2)
map_f₂(x0, nil)
map_f₂(x0, cons₂(x1, x2))
process₂(x0, x1)
if1₃(x0, x1, true)
if1₃(x0, x1, false)
if2₃(x0, x1, false)
if3₃(x0, x1, false)

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

IF3₃(store, m, false) -> SNDSPLIT₂(m, app₂(map_f₂(self, nil), store))
PROCESS₂(store, m) -> LENGTH₁(store)
SNDSPLIT₂(s₁(n), cons₂(h, t)) -> SNDSPLIT₂(n, t)
IF2₃(store, m, false) -> PROCESS₂(app₂(map_f₂(self, nil), sndsplit₂(m, store)), m)
IF3₃(store, m, false) -> PROCESS₂(sndsplit₂(m, app₂(map_f₂(self, nil), store)), m)
IF2₃(store, m, false) -> APP₂(map_f₂(self, nil), sndsplit₂(m, store))
IF1₃(store, m, true) -> FSTSPLIT₂(m, store)
IF2₃(store, m, false) -> MAP_F₂(self, nil)
MAP_F₂(pid, cons₂(h, t)) -> APP₂(f₂(pid, h), map_f₂(pid, t))
IF1₃(store, m, true) -> EMPTY₁(fstsplit₂(m, store))
PROCESS₂(store, m) -> IF1₃(store, m, leq₂(m, length₁(store)))
IF3₃(store, m, false) -> APP₂(map_f₂(self, nil), store)
FSTSPLIT₂(s₁(n), cons₂(h, t)) -> FSTSPLIT₂(n, t)
IF1₃(store, m, true) -> IF2₃(store, m, empty₁(fstsplit₂(m, store)))
IF1₃(store, m, false) -> MAP_F₂(self, nil)
IF1₃(store, m, false) -> FSTSPLIT₂(m, app₂(map_f₂(self, nil), store))
IF1₃(store, m, false) -> EMPTY₁(fstsplit₂(m, app₂(map_f₂(self, nil), store)))
APP₂(cons₂(h, t), x) -> APP₂(t, x)
IF3₃(store, m, false) -> MAP_F₂(self, nil)
LENGTH₁(cons₂(h, t)) -> LENGTH₁(t)
IF2₃(store, m, false) -> SNDSPLIT₂(m, store)
PROCESS₂(store, m) -> LEQ₂(m, length₁(store))
IF1₃(store, m, false) -> IF3₃(store, m, empty₁(fstsplit₂(m, app₂(map_f₂(self, nil), store))))
LEQ₂(s₁(n), s₁(m)) -> LEQ₂(n, m)
MAP_F₂(pid, cons₂(h, t)) -> MAP_F₂(pid, t)
IF1₃(store, m, false) -> APP₂(map_f₂(self, nil), store)

The TRS R consists of the following rules:

fstsplit₂(0, x) -> nil
fstsplit₂(s₁(n), nil) -> nil
fstsplit₂(s₁(n), cons₂(h, t)) -> cons₂(h, fstsplit₂(n, t))
sndsplit₂(0, x) -> x
sndsplit₂(s₁(n), nil) -> nil
sndsplit₂(s₁(n), cons₂(h, t)) -> sndsplit₂(n, t)
empty₁(nil) -> true
empty₁(cons₂(h, t)) -> false
leq₂(0, m) -> true
leq₂(s₁(n), 0) -> false
leq₂(s₁(n), s₁(m)) -> leq₂(n, m)
length₁(nil) -> 0
length₁(cons₂(h, t)) -> s₁(length₁(t))
app₂(nil, x) -> x
app₂(cons₂(h, t), x) -> cons₂(h, app₂(t, x))
map_f₂(pid, nil) -> nil
map_f₂(pid, cons₂(h, t)) -> app₂(f₂(pid, h), map_f₂(pid, t))
process₂(store, m) -> if1₃(store, m, leq₂(m, length₁(store)))
if1₃(store, m, true) -> if2₃(store, m, empty₁(fstsplit₂(m, store)))
if1₃(store, m, false) -> if3₃(store, m, empty₁(fstsplit₂(m, app₂(map_f₂(self, nil), store))))
if2₃(store, m, false) -> process₂(app₂(map_f₂(self, nil), sndsplit₂(m, store)), m)
if3₃(store, m, false) -> process₂(sndsplit₂(m, app₂(map_f₂(self, nil), store)), m)

The set Q consists of the following terms:

fstsplit₂(0, x0)
fstsplit₂(s₁(x0), nil)
fstsplit₂(s₁(x0), cons₂(x1, x2))
sndsplit₂(0, x0)
sndsplit₂(s₁(x0), nil)
sndsplit₂(s₁(x0), cons₂(x1, x2))
empty₁(nil)
empty₁(cons₂(x0, x1))
leq₂(0, x0)
leq₂(s₁(x0), 0)
leq₂(s₁(x0), s₁(x1))
length₁(nil)
length₁(cons₂(x0, x1))
app₂(nil, x0)
app₂(cons₂(x0, x1), x2)
map_f₂(x0, nil)
map_f₂(x0, cons₂(x1, x2))
process₂(x0, x1)
if1₃(x0, x1, true)
if1₃(x0, x1, false)
if2₃(x0, x1, false)
if3₃(x0, x1, false)

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

↳ QTRS

  ↳ Non-Overlap Check

    ↳ QTRS

      ↳ DependencyPairsProof

        ↳ QDP

          ↳ DependencyGraphProof

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

IF3₃(store, m, false) -> SNDSPLIT₂(m, app₂(map_f₂(self, nil), store))
PROCESS₂(store, m) -> LENGTH₁(store)
SNDSPLIT₂(s₁(n), cons₂(h, t)) -> SNDSPLIT₂(n, t)
IF2₃(store, m, false) -> PROCESS₂(app₂(map_f₂(self, nil), sndsplit₂(m, store)), m)
IF3₃(store, m, false) -> PROCESS₂(sndsplit₂(m, app₂(map_f₂(self, nil), store)), m)
IF2₃(store, m, false) -> APP₂(map_f₂(self, nil), sndsplit₂(m, store))
IF1₃(store, m, true) -> FSTSPLIT₂(m, store)
IF2₃(store, m, false) -> MAP_F₂(self, nil)
MAP_F₂(pid, cons₂(h, t)) -> APP₂(f₂(pid, h), map_f₂(pid, t))
IF1₃(store, m, true) -> EMPTY₁(fstsplit₂(m, store))
PROCESS₂(store, m) -> IF1₃(store, m, leq₂(m, length₁(store)))
IF3₃(store, m, false) -> APP₂(map_f₂(self, nil), store)
FSTSPLIT₂(s₁(n), cons₂(h, t)) -> FSTSPLIT₂(n, t)
IF1₃(store, m, true) -> IF2₃(store, m, empty₁(fstsplit₂(m, store)))
IF1₃(store, m, false) -> MAP_F₂(self, nil)
IF1₃(store, m, false) -> FSTSPLIT₂(m, app₂(map_f₂(self, nil), store))
IF1₃(store, m, false) -> EMPTY₁(fstsplit₂(m, app₂(map_f₂(self, nil), store)))
APP₂(cons₂(h, t), x) -> APP₂(t, x)
IF3₃(store, m, false) -> MAP_F₂(self, nil)
LENGTH₁(cons₂(h, t)) -> LENGTH₁(t)
IF2₃(store, m, false) -> SNDSPLIT₂(m, store)
PROCESS₂(store, m) -> LEQ₂(m, length₁(store))
IF1₃(store, m, false) -> IF3₃(store, m, empty₁(fstsplit₂(m, app₂(map_f₂(self, nil), store))))
LEQ₂(s₁(n), s₁(m)) -> LEQ₂(n, m)
MAP_F₂(pid, cons₂(h, t)) -> MAP_F₂(pid, t)
IF1₃(store, m, false) -> APP₂(map_f₂(self, nil), store)

The TRS R consists of the following rules:

fstsplit₂(0, x) -> nil
fstsplit₂(s₁(n), nil) -> nil
fstsplit₂(s₁(n), cons₂(h, t)) -> cons₂(h, fstsplit₂(n, t))
sndsplit₂(0, x) -> x
sndsplit₂(s₁(n), nil) -> nil
sndsplit₂(s₁(n), cons₂(h, t)) -> sndsplit₂(n, t)
empty₁(nil) -> true
empty₁(cons₂(h, t)) -> false
leq₂(0, m) -> true
leq₂(s₁(n), 0) -> false
leq₂(s₁(n), s₁(m)) -> leq₂(n, m)
length₁(nil) -> 0
length₁(cons₂(h, t)) -> s₁(length₁(t))
app₂(nil, x) -> x
app₂(cons₂(h, t), x) -> cons₂(h, app₂(t, x))
map_f₂(pid, nil) -> nil
map_f₂(pid, cons₂(h, t)) -> app₂(f₂(pid, h), map_f₂(pid, t))
process₂(store, m) -> if1₃(store, m, leq₂(m, length₁(store)))
if1₃(store, m, true) -> if2₃(store, m, empty₁(fstsplit₂(m, store)))
if1₃(store, m, false) -> if3₃(store, m, empty₁(fstsplit₂(m, app₂(map_f₂(self, nil), store))))
if2₃(store, m, false) -> process₂(app₂(map_f₂(self, nil), sndsplit₂(m, store)), m)
if3₃(store, m, false) -> process₂(sndsplit₂(m, app₂(map_f₂(self, nil), store)), m)

The set Q consists of the following terms:

fstsplit₂(0, x0)
fstsplit₂(s₁(x0), nil)
fstsplit₂(s₁(x0), cons₂(x1, x2))
sndsplit₂(0, x0)
sndsplit₂(s₁(x0), nil)
sndsplit₂(s₁(x0), cons₂(x1, x2))
empty₁(nil)
empty₁(cons₂(x0, x1))
leq₂(0, x0)
leq₂(s₁(x0), 0)
leq₂(s₁(x0), s₁(x1))
length₁(nil)
length₁(cons₂(x0, x1))
app₂(nil, x0)
app₂(cons₂(x0, x1), x2)
map_f₂(x0, nil)
map_f₂(x0, cons₂(x1, x2))
process₂(x0, x1)
if1₃(x0, x1, true)
if1₃(x0, x1, false)
if2₃(x0, x1, false)
if3₃(x0, x1, false)

We have to consider all minimal (P,Q,R)-chains.
The approximation of the Dependency Graph [13,14,18] contains 7 SCCs with 15 less nodes.

↳ QTRS

  ↳ Non-Overlap Check

    ↳ QTRS

      ↳ DependencyPairsProof

        ↳ QDP

          ↳ DependencyGraphProof

            ↳ AND

              ↳ QDP

                ↳ QDPOrderProof

              ↳ QDP

              ↳ QDP

              ↳ QDP

              ↳ QDP

              ↳ QDP

              ↳ QDP

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

APP₂(cons₂(h, t), x) -> APP₂(t, x)

The TRS R consists of the following rules:

fstsplit₂(0, x) -> nil
fstsplit₂(s₁(n), nil) -> nil
fstsplit₂(s₁(n), cons₂(h, t)) -> cons₂(h, fstsplit₂(n, t))
sndsplit₂(0, x) -> x
sndsplit₂(s₁(n), nil) -> nil
sndsplit₂(s₁(n), cons₂(h, t)) -> sndsplit₂(n, t)
empty₁(nil) -> true
empty₁(cons₂(h, t)) -> false
leq₂(0, m) -> true
leq₂(s₁(n), 0) -> false
leq₂(s₁(n), s₁(m)) -> leq₂(n, m)
length₁(nil) -> 0
length₁(cons₂(h, t)) -> s₁(length₁(t))
app₂(nil, x) -> x
app₂(cons₂(h, t), x) -> cons₂(h, app₂(t, x))
map_f₂(pid, nil) -> nil
map_f₂(pid, cons₂(h, t)) -> app₂(f₂(pid, h), map_f₂(pid, t))
process₂(store, m) -> if1₃(store, m, leq₂(m, length₁(store)))
if1₃(store, m, true) -> if2₃(store, m, empty₁(fstsplit₂(m, store)))
if1₃(store, m, false) -> if3₃(store, m, empty₁(fstsplit₂(m, app₂(map_f₂(self, nil), store))))
if2₃(store, m, false) -> process₂(app₂(map_f₂(self, nil), sndsplit₂(m, store)), m)
if3₃(store, m, false) -> process₂(sndsplit₂(m, app₂(map_f₂(self, nil), store)), m)

The set Q consists of the following terms:

fstsplit₂(0, x0)
fstsplit₂(s₁(x0), nil)
fstsplit₂(s₁(x0), cons₂(x1, x2))
sndsplit₂(0, x0)
sndsplit₂(s₁(x0), nil)
sndsplit₂(s₁(x0), cons₂(x1, x2))
empty₁(nil)
empty₁(cons₂(x0, x1))
leq₂(0, x0)
leq₂(s₁(x0), 0)
leq₂(s₁(x0), s₁(x1))
length₁(nil)
length₁(cons₂(x0, x1))
app₂(nil, x0)
app₂(cons₂(x0, x1), x2)
map_f₂(x0, nil)
map_f₂(x0, cons₂(x1, x2))
process₂(x0, x1)
if1₃(x0, x1, true)
if1₃(x0, x1, false)
if2₃(x0, x1, false)
if3₃(x0, x1, false)

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

The following pairs can be strictly oriented and are deleted.

APP₂(cons₂(h, t), x) -> APP₂(t, x)

The remaining pairs can at least by weakly be oriented.
none
Used ordering: Combined order from the following AFS and order.
APP₂(x1, x2) = APP₁(x1)
cons₂(x1, x2) = cons₁(x2)

Lexicographic Path Order [19].
Precedence:

[APP₁, cons_1]

The following usable rules [14] were oriented: none

↳ QTRS

  ↳ Non-Overlap Check

    ↳ QTRS

      ↳ DependencyPairsProof

        ↳ QDP

          ↳ DependencyGraphProof

            ↳ AND

              ↳ QDP

                ↳ QDPOrderProof

                  ↳ QDP

                    ↳ PisEmptyProof

              ↳ QDP

              ↳ QDP

              ↳ QDP

              ↳ QDP

              ↳ QDP

              ↳ QDP

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

fstsplit₂(0, x) -> nil
fstsplit₂(s₁(n), nil) -> nil
fstsplit₂(s₁(n), cons₂(h, t)) -> cons₂(h, fstsplit₂(n, t))
sndsplit₂(0, x) -> x
sndsplit₂(s₁(n), nil) -> nil
sndsplit₂(s₁(n), cons₂(h, t)) -> sndsplit₂(n, t)
empty₁(nil) -> true
empty₁(cons₂(h, t)) -> false
leq₂(0, m) -> true
leq₂(s₁(n), 0) -> false
leq₂(s₁(n), s₁(m)) -> leq₂(n, m)
length₁(nil) -> 0
length₁(cons₂(h, t)) -> s₁(length₁(t))
app₂(nil, x) -> x
app₂(cons₂(h, t), x) -> cons₂(h, app₂(t, x))
map_f₂(pid, nil) -> nil
map_f₂(pid, cons₂(h, t)) -> app₂(f₂(pid, h), map_f₂(pid, t))
process₂(store, m) -> if1₃(store, m, leq₂(m, length₁(store)))
if1₃(store, m, true) -> if2₃(store, m, empty₁(fstsplit₂(m, store)))
if1₃(store, m, false) -> if3₃(store, m, empty₁(fstsplit₂(m, app₂(map_f₂(self, nil), store))))
if2₃(store, m, false) -> process₂(app₂(map_f₂(self, nil), sndsplit₂(m, store)), m)
if3₃(store, m, false) -> process₂(sndsplit₂(m, app₂(map_f₂(self, nil), store)), m)

The set Q consists of the following terms:

fstsplit₂(0, x0)
fstsplit₂(s₁(x0), nil)
fstsplit₂(s₁(x0), cons₂(x1, x2))
sndsplit₂(0, x0)
sndsplit₂(s₁(x0), nil)
sndsplit₂(s₁(x0), cons₂(x1, x2))
empty₁(nil)
empty₁(cons₂(x0, x1))
leq₂(0, x0)
leq₂(s₁(x0), 0)
leq₂(s₁(x0), s₁(x1))
length₁(nil)
length₁(cons₂(x0, x1))
app₂(nil, x0)
app₂(cons₂(x0, x1), x2)
map_f₂(x0, nil)
map_f₂(x0, cons₂(x1, x2))
process₂(x0, x1)
if1₃(x0, x1, true)
if1₃(x0, x1, false)
if2₃(x0, x1, false)
if3₃(x0, x1, false)

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

↳ QTRS

  ↳ Non-Overlap Check

    ↳ QTRS

      ↳ DependencyPairsProof

        ↳ QDP

          ↳ DependencyGraphProof

            ↳ AND

              ↳ QDP

              ↳ QDP

                ↳ QDPOrderProof

              ↳ QDP

              ↳ QDP

              ↳ QDP

              ↳ QDP

              ↳ QDP

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

MAP_F₂(pid, cons₂(h, t)) -> MAP_F₂(pid, t)

The TRS R consists of the following rules:

fstsplit₂(0, x) -> nil
fstsplit₂(s₁(n), nil) -> nil
fstsplit₂(s₁(n), cons₂(h, t)) -> cons₂(h, fstsplit₂(n, t))
sndsplit₂(0, x) -> x
sndsplit₂(s₁(n), nil) -> nil
sndsplit₂(s₁(n), cons₂(h, t)) -> sndsplit₂(n, t)
empty₁(nil) -> true
empty₁(cons₂(h, t)) -> false
leq₂(0, m) -> true
leq₂(s₁(n), 0) -> false
leq₂(s₁(n), s₁(m)) -> leq₂(n, m)
length₁(nil) -> 0
length₁(cons₂(h, t)) -> s₁(length₁(t))
app₂(nil, x) -> x
app₂(cons₂(h, t), x) -> cons₂(h, app₂(t, x))
map_f₂(pid, nil) -> nil
map_f₂(pid, cons₂(h, t)) -> app₂(f₂(pid, h), map_f₂(pid, t))
process₂(store, m) -> if1₃(store, m, leq₂(m, length₁(store)))
if1₃(store, m, true) -> if2₃(store, m, empty₁(fstsplit₂(m, store)))
if1₃(store, m, false) -> if3₃(store, m, empty₁(fstsplit₂(m, app₂(map_f₂(self, nil), store))))
if2₃(store, m, false) -> process₂(app₂(map_f₂(self, nil), sndsplit₂(m, store)), m)
if3₃(store, m, false) -> process₂(sndsplit₂(m, app₂(map_f₂(self, nil), store)), m)

The set Q consists of the following terms:

fstsplit₂(0, x0)
fstsplit₂(s₁(x0), nil)
fstsplit₂(s₁(x0), cons₂(x1, x2))
sndsplit₂(0, x0)
sndsplit₂(s₁(x0), nil)
sndsplit₂(s₁(x0), cons₂(x1, x2))
empty₁(nil)
empty₁(cons₂(x0, x1))
leq₂(0, x0)
leq₂(s₁(x0), 0)
leq₂(s₁(x0), s₁(x1))
length₁(nil)
length₁(cons₂(x0, x1))
app₂(nil, x0)
app₂(cons₂(x0, x1), x2)
map_f₂(x0, nil)
map_f₂(x0, cons₂(x1, x2))
process₂(x0, x1)
if1₃(x0, x1, true)
if1₃(x0, x1, false)
if2₃(x0, x1, false)
if3₃(x0, x1, false)

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

The following pairs can be strictly oriented and are deleted.

MAP_F₂(pid, cons₂(h, t)) -> MAP_F₂(pid, t)

The remaining pairs can at least by weakly be oriented.
none
Used ordering: Combined order from the following AFS and order.
MAP_F₂(x1, x2) = MAP_F₁(x2)
cons₂(x1, x2) = cons₁(x2)

Lexicographic Path Order [19].
Precedence:

[MAP_F₁, cons_1]

The following usable rules [14] were oriented: none

↳ QTRS

  ↳ Non-Overlap Check

    ↳ QTRS

      ↳ DependencyPairsProof

        ↳ QDP

          ↳ DependencyGraphProof

            ↳ AND

              ↳ QDP

              ↳ QDP

                ↳ QDPOrderProof

                  ↳ QDP

                    ↳ PisEmptyProof

              ↳ QDP

              ↳ QDP

              ↳ QDP

              ↳ QDP

              ↳ QDP

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

fstsplit₂(0, x) -> nil
fstsplit₂(s₁(n), nil) -> nil
fstsplit₂(s₁(n), cons₂(h, t)) -> cons₂(h, fstsplit₂(n, t))
sndsplit₂(0, x) -> x
sndsplit₂(s₁(n), nil) -> nil
sndsplit₂(s₁(n), cons₂(h, t)) -> sndsplit₂(n, t)
empty₁(nil) -> true
empty₁(cons₂(h, t)) -> false
leq₂(0, m) -> true
leq₂(s₁(n), 0) -> false
leq₂(s₁(n), s₁(m)) -> leq₂(n, m)
length₁(nil) -> 0
length₁(cons₂(h, t)) -> s₁(length₁(t))
app₂(nil, x) -> x
app₂(cons₂(h, t), x) -> cons₂(h, app₂(t, x))
map_f₂(pid, nil) -> nil
map_f₂(pid, cons₂(h, t)) -> app₂(f₂(pid, h), map_f₂(pid, t))
process₂(store, m) -> if1₃(store, m, leq₂(m, length₁(store)))
if1₃(store, m, true) -> if2₃(store, m, empty₁(fstsplit₂(m, store)))
if1₃(store, m, false) -> if3₃(store, m, empty₁(fstsplit₂(m, app₂(map_f₂(self, nil), store))))
if2₃(store, m, false) -> process₂(app₂(map_f₂(self, nil), sndsplit₂(m, store)), m)
if3₃(store, m, false) -> process₂(sndsplit₂(m, app₂(map_f₂(self, nil), store)), m)

The set Q consists of the following terms:

fstsplit₂(0, x0)
fstsplit₂(s₁(x0), nil)
fstsplit₂(s₁(x0), cons₂(x1, x2))
sndsplit₂(0, x0)
sndsplit₂(s₁(x0), nil)
sndsplit₂(s₁(x0), cons₂(x1, x2))
empty₁(nil)
empty₁(cons₂(x0, x1))
leq₂(0, x0)
leq₂(s₁(x0), 0)
leq₂(s₁(x0), s₁(x1))
length₁(nil)
length₁(cons₂(x0, x1))
app₂(nil, x0)
app₂(cons₂(x0, x1), x2)
map_f₂(x0, nil)
map_f₂(x0, cons₂(x1, x2))
process₂(x0, x1)
if1₃(x0, x1, true)
if1₃(x0, x1, false)
if2₃(x0, x1, false)
if3₃(x0, x1, false)

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

↳ QTRS

  ↳ Non-Overlap Check

    ↳ QTRS

      ↳ DependencyPairsProof

        ↳ QDP

          ↳ DependencyGraphProof

            ↳ AND

              ↳ QDP

              ↳ QDP

              ↳ QDP

                ↳ QDPOrderProof

              ↳ QDP

              ↳ QDP

              ↳ QDP

              ↳ QDP

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

LENGTH₁(cons₂(h, t)) -> LENGTH₁(t)

The TRS R consists of the following rules:

fstsplit₂(0, x) -> nil
fstsplit₂(s₁(n), nil) -> nil
fstsplit₂(s₁(n), cons₂(h, t)) -> cons₂(h, fstsplit₂(n, t))
sndsplit₂(0, x) -> x
sndsplit₂(s₁(n), nil) -> nil
sndsplit₂(s₁(n), cons₂(h, t)) -> sndsplit₂(n, t)
empty₁(nil) -> true
empty₁(cons₂(h, t)) -> false
leq₂(0, m) -> true
leq₂(s₁(n), 0) -> false
leq₂(s₁(n), s₁(m)) -> leq₂(n, m)
length₁(nil) -> 0
length₁(cons₂(h, t)) -> s₁(length₁(t))
app₂(nil, x) -> x
app₂(cons₂(h, t), x) -> cons₂(h, app₂(t, x))
map_f₂(pid, nil) -> nil
map_f₂(pid, cons₂(h, t)) -> app₂(f₂(pid, h), map_f₂(pid, t))
process₂(store, m) -> if1₃(store, m, leq₂(m, length₁(store)))
if1₃(store, m, true) -> if2₃(store, m, empty₁(fstsplit₂(m, store)))
if1₃(store, m, false) -> if3₃(store, m, empty₁(fstsplit₂(m, app₂(map_f₂(self, nil), store))))
if2₃(store, m, false) -> process₂(app₂(map_f₂(self, nil), sndsplit₂(m, store)), m)
if3₃(store, m, false) -> process₂(sndsplit₂(m, app₂(map_f₂(self, nil), store)), m)

The set Q consists of the following terms:

fstsplit₂(0, x0)
fstsplit₂(s₁(x0), nil)
fstsplit₂(s₁(x0), cons₂(x1, x2))
sndsplit₂(0, x0)
sndsplit₂(s₁(x0), nil)
sndsplit₂(s₁(x0), cons₂(x1, x2))
empty₁(nil)
empty₁(cons₂(x0, x1))
leq₂(0, x0)
leq₂(s₁(x0), 0)
leq₂(s₁(x0), s₁(x1))
length₁(nil)
length₁(cons₂(x0, x1))
app₂(nil, x0)
app₂(cons₂(x0, x1), x2)
map_f₂(x0, nil)
map_f₂(x0, cons₂(x1, x2))
process₂(x0, x1)
if1₃(x0, x1, true)
if1₃(x0, x1, false)
if2₃(x0, x1, false)
if3₃(x0, x1, false)

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

The following pairs can be strictly oriented and are deleted.

LENGTH₁(cons₂(h, t)) -> LENGTH₁(t)

The remaining pairs can at least by weakly be oriented.
none
Used ordering: Combined order from the following AFS and order.
LENGTH₁(x1) = LENGTH₁(x1)
cons₂(x1, x2) = cons₂(x1, x2)

Lexicographic Path Order [19].
Precedence:

cons₂ > LENGTH₁

The following usable rules [14] were oriented: none

↳ QTRS

  ↳ Non-Overlap Check

    ↳ QTRS

      ↳ DependencyPairsProof

        ↳ QDP

          ↳ DependencyGraphProof

            ↳ AND

              ↳ QDP

              ↳ QDP

              ↳ QDP

                ↳ QDPOrderProof

                  ↳ QDP

                    ↳ PisEmptyProof

              ↳ QDP

              ↳ QDP

              ↳ QDP

              ↳ QDP

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

fstsplit₂(0, x) -> nil
fstsplit₂(s₁(n), nil) -> nil
fstsplit₂(s₁(n), cons₂(h, t)) -> cons₂(h, fstsplit₂(n, t))
sndsplit₂(0, x) -> x
sndsplit₂(s₁(n), nil) -> nil
sndsplit₂(s₁(n), cons₂(h, t)) -> sndsplit₂(n, t)
empty₁(nil) -> true
empty₁(cons₂(h, t)) -> false
leq₂(0, m) -> true
leq₂(s₁(n), 0) -> false
leq₂(s₁(n), s₁(m)) -> leq₂(n, m)
length₁(nil) -> 0
length₁(cons₂(h, t)) -> s₁(length₁(t))
app₂(nil, x) -> x
app₂(cons₂(h, t), x) -> cons₂(h, app₂(t, x))
map_f₂(pid, nil) -> nil
map_f₂(pid, cons₂(h, t)) -> app₂(f₂(pid, h), map_f₂(pid, t))
process₂(store, m) -> if1₃(store, m, leq₂(m, length₁(store)))
if1₃(store, m, true) -> if2₃(store, m, empty₁(fstsplit₂(m, store)))
if1₃(store, m, false) -> if3₃(store, m, empty₁(fstsplit₂(m, app₂(map_f₂(self, nil), store))))
if2₃(store, m, false) -> process₂(app₂(map_f₂(self, nil), sndsplit₂(m, store)), m)
if3₃(store, m, false) -> process₂(sndsplit₂(m, app₂(map_f₂(self, nil), store)), m)

The set Q consists of the following terms:

fstsplit₂(0, x0)
fstsplit₂(s₁(x0), nil)
fstsplit₂(s₁(x0), cons₂(x1, x2))
sndsplit₂(0, x0)
sndsplit₂(s₁(x0), nil)
sndsplit₂(s₁(x0), cons₂(x1, x2))
empty₁(nil)
empty₁(cons₂(x0, x1))
leq₂(0, x0)
leq₂(s₁(x0), 0)
leq₂(s₁(x0), s₁(x1))
length₁(nil)
length₁(cons₂(x0, x1))
app₂(nil, x0)
app₂(cons₂(x0, x1), x2)
map_f₂(x0, nil)
map_f₂(x0, cons₂(x1, x2))
process₂(x0, x1)
if1₃(x0, x1, true)
if1₃(x0, x1, false)
if2₃(x0, x1, false)
if3₃(x0, x1, false)

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

↳ QTRS

  ↳ Non-Overlap Check

    ↳ QTRS

      ↳ DependencyPairsProof

        ↳ QDP

          ↳ DependencyGraphProof

            ↳ AND

              ↳ QDP

              ↳ QDP

              ↳ QDP

              ↳ QDP

                ↳ QDPOrderProof

              ↳ QDP

              ↳ QDP

              ↳ QDP

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

LEQ₂(s₁(n), s₁(m)) -> LEQ₂(n, m)

The TRS R consists of the following rules:

fstsplit₂(0, x) -> nil
fstsplit₂(s₁(n), nil) -> nil
fstsplit₂(s₁(n), cons₂(h, t)) -> cons₂(h, fstsplit₂(n, t))
sndsplit₂(0, x) -> x
sndsplit₂(s₁(n), nil) -> nil
sndsplit₂(s₁(n), cons₂(h, t)) -> sndsplit₂(n, t)
empty₁(nil) -> true
empty₁(cons₂(h, t)) -> false
leq₂(0, m) -> true
leq₂(s₁(n), 0) -> false
leq₂(s₁(n), s₁(m)) -> leq₂(n, m)
length₁(nil) -> 0
length₁(cons₂(h, t)) -> s₁(length₁(t))
app₂(nil, x) -> x
app₂(cons₂(h, t), x) -> cons₂(h, app₂(t, x))
map_f₂(pid, nil) -> nil
map_f₂(pid, cons₂(h, t)) -> app₂(f₂(pid, h), map_f₂(pid, t))
process₂(store, m) -> if1₃(store, m, leq₂(m, length₁(store)))
if1₃(store, m, true) -> if2₃(store, m, empty₁(fstsplit₂(m, store)))
if1₃(store, m, false) -> if3₃(store, m, empty₁(fstsplit₂(m, app₂(map_f₂(self, nil), store))))
if2₃(store, m, false) -> process₂(app₂(map_f₂(self, nil), sndsplit₂(m, store)), m)
if3₃(store, m, false) -> process₂(sndsplit₂(m, app₂(map_f₂(self, nil), store)), m)

The set Q consists of the following terms:

fstsplit₂(0, x0)
fstsplit₂(s₁(x0), nil)
fstsplit₂(s₁(x0), cons₂(x1, x2))
sndsplit₂(0, x0)
sndsplit₂(s₁(x0), nil)
sndsplit₂(s₁(x0), cons₂(x1, x2))
empty₁(nil)
empty₁(cons₂(x0, x1))
leq₂(0, x0)
leq₂(s₁(x0), 0)
leq₂(s₁(x0), s₁(x1))
length₁(nil)
length₁(cons₂(x0, x1))
app₂(nil, x0)
app₂(cons₂(x0, x1), x2)
map_f₂(x0, nil)
map_f₂(x0, cons₂(x1, x2))
process₂(x0, x1)
if1₃(x0, x1, true)
if1₃(x0, x1, false)
if2₃(x0, x1, false)
if3₃(x0, x1, false)

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

The following pairs can be strictly oriented and are deleted.

LEQ₂(s₁(n), s₁(m)) -> LEQ₂(n, m)

The remaining pairs can at least by weakly be oriented.
none
Used ordering: Combined order from the following AFS and order.
LEQ₂(x1, x2) = LEQ₁(x1)
s₁(x1) = s₁(x1)

Lexicographic Path Order [19].
Precedence:

trivial

The following usable rules [14] were oriented: none

↳ QTRS

  ↳ Non-Overlap Check

    ↳ QTRS

      ↳ DependencyPairsProof

        ↳ QDP

          ↳ DependencyGraphProof

            ↳ AND

              ↳ QDP

              ↳ QDP

              ↳ QDP

              ↳ QDP

                ↳ QDPOrderProof

                  ↳ QDP

                    ↳ PisEmptyProof

              ↳ QDP

              ↳ QDP

              ↳ QDP

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

fstsplit₂(0, x) -> nil
fstsplit₂(s₁(n), nil) -> nil
fstsplit₂(s₁(n), cons₂(h, t)) -> cons₂(h, fstsplit₂(n, t))
sndsplit₂(0, x) -> x
sndsplit₂(s₁(n), nil) -> nil
sndsplit₂(s₁(n), cons₂(h, t)) -> sndsplit₂(n, t)
empty₁(nil) -> true
empty₁(cons₂(h, t)) -> false
leq₂(0, m) -> true
leq₂(s₁(n), 0) -> false
leq₂(s₁(n), s₁(m)) -> leq₂(n, m)
length₁(nil) -> 0
length₁(cons₂(h, t)) -> s₁(length₁(t))
app₂(nil, x) -> x
app₂(cons₂(h, t), x) -> cons₂(h, app₂(t, x))
map_f₂(pid, nil) -> nil
map_f₂(pid, cons₂(h, t)) -> app₂(f₂(pid, h), map_f₂(pid, t))
process₂(store, m) -> if1₃(store, m, leq₂(m, length₁(store)))
if1₃(store, m, true) -> if2₃(store, m, empty₁(fstsplit₂(m, store)))
if1₃(store, m, false) -> if3₃(store, m, empty₁(fstsplit₂(m, app₂(map_f₂(self, nil), store))))
if2₃(store, m, false) -> process₂(app₂(map_f₂(self, nil), sndsplit₂(m, store)), m)
if3₃(store, m, false) -> process₂(sndsplit₂(m, app₂(map_f₂(self, nil), store)), m)

The set Q consists of the following terms:

fstsplit₂(0, x0)
fstsplit₂(s₁(x0), nil)
fstsplit₂(s₁(x0), cons₂(x1, x2))
sndsplit₂(0, x0)
sndsplit₂(s₁(x0), nil)
sndsplit₂(s₁(x0), cons₂(x1, x2))
empty₁(nil)
empty₁(cons₂(x0, x1))
leq₂(0, x0)
leq₂(s₁(x0), 0)
leq₂(s₁(x0), s₁(x1))
length₁(nil)
length₁(cons₂(x0, x1))
app₂(nil, x0)
app₂(cons₂(x0, x1), x2)
map_f₂(x0, nil)
map_f₂(x0, cons₂(x1, x2))
process₂(x0, x1)
if1₃(x0, x1, true)
if1₃(x0, x1, false)
if2₃(x0, x1, false)
if3₃(x0, x1, false)

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

↳ QTRS

  ↳ Non-Overlap Check

    ↳ QTRS

      ↳ DependencyPairsProof

        ↳ QDP

          ↳ DependencyGraphProof

            ↳ AND

              ↳ QDP

              ↳ QDP

              ↳ QDP

              ↳ QDP

              ↳ QDP

                ↳ QDPOrderProof

              ↳ QDP

              ↳ QDP

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

SNDSPLIT₂(s₁(n), cons₂(h, t)) -> SNDSPLIT₂(n, t)

The TRS R consists of the following rules:

fstsplit₂(0, x) -> nil
fstsplit₂(s₁(n), nil) -> nil
fstsplit₂(s₁(n), cons₂(h, t)) -> cons₂(h, fstsplit₂(n, t))
sndsplit₂(0, x) -> x
sndsplit₂(s₁(n), nil) -> nil
sndsplit₂(s₁(n), cons₂(h, t)) -> sndsplit₂(n, t)
empty₁(nil) -> true
empty₁(cons₂(h, t)) -> false
leq₂(0, m) -> true
leq₂(s₁(n), 0) -> false
leq₂(s₁(n), s₁(m)) -> leq₂(n, m)
length₁(nil) -> 0
length₁(cons₂(h, t)) -> s₁(length₁(t))
app₂(nil, x) -> x
app₂(cons₂(h, t), x) -> cons₂(h, app₂(t, x))
map_f₂(pid, nil) -> nil
map_f₂(pid, cons₂(h, t)) -> app₂(f₂(pid, h), map_f₂(pid, t))
process₂(store, m) -> if1₃(store, m, leq₂(m, length₁(store)))
if1₃(store, m, true) -> if2₃(store, m, empty₁(fstsplit₂(m, store)))
if1₃(store, m, false) -> if3₃(store, m, empty₁(fstsplit₂(m, app₂(map_f₂(self, nil), store))))
if2₃(store, m, false) -> process₂(app₂(map_f₂(self, nil), sndsplit₂(m, store)), m)
if3₃(store, m, false) -> process₂(sndsplit₂(m, app₂(map_f₂(self, nil), store)), m)

The set Q consists of the following terms:

fstsplit₂(0, x0)
fstsplit₂(s₁(x0), nil)
fstsplit₂(s₁(x0), cons₂(x1, x2))
sndsplit₂(0, x0)
sndsplit₂(s₁(x0), nil)
sndsplit₂(s₁(x0), cons₂(x1, x2))
empty₁(nil)
empty₁(cons₂(x0, x1))
leq₂(0, x0)
leq₂(s₁(x0), 0)
leq₂(s₁(x0), s₁(x1))
length₁(nil)
length₁(cons₂(x0, x1))
app₂(nil, x0)
app₂(cons₂(x0, x1), x2)
map_f₂(x0, nil)
map_f₂(x0, cons₂(x1, x2))
process₂(x0, x1)
if1₃(x0, x1, true)
if1₃(x0, x1, false)
if2₃(x0, x1, false)
if3₃(x0, x1, false)

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

The following pairs can be strictly oriented and are deleted.

SNDSPLIT₂(s₁(n), cons₂(h, t)) -> SNDSPLIT₂(n, t)

The remaining pairs can at least by weakly be oriented.
none
Used ordering: Combined order from the following AFS and order.
SNDSPLIT₂(x1, x2) = x1
s₁(x1) = s₁(x1)
cons₂(x1, x2) = x1

Lexicographic Path Order [19].
Precedence:

trivial

The following usable rules [14] were oriented: none

↳ QTRS

  ↳ Non-Overlap Check

    ↳ QTRS

      ↳ DependencyPairsProof

        ↳ QDP

          ↳ DependencyGraphProof

            ↳ AND

              ↳ QDP

              ↳ QDP

              ↳ QDP

              ↳ QDP

              ↳ QDP

                ↳ QDPOrderProof

                  ↳ QDP

                    ↳ PisEmptyProof

              ↳ QDP

              ↳ QDP

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

fstsplit₂(0, x) -> nil
fstsplit₂(s₁(n), nil) -> nil
fstsplit₂(s₁(n), cons₂(h, t)) -> cons₂(h, fstsplit₂(n, t))
sndsplit₂(0, x) -> x
sndsplit₂(s₁(n), nil) -> nil
sndsplit₂(s₁(n), cons₂(h, t)) -> sndsplit₂(n, t)
empty₁(nil) -> true
empty₁(cons₂(h, t)) -> false
leq₂(0, m) -> true
leq₂(s₁(n), 0) -> false
leq₂(s₁(n), s₁(m)) -> leq₂(n, m)
length₁(nil) -> 0
length₁(cons₂(h, t)) -> s₁(length₁(t))
app₂(nil, x) -> x
app₂(cons₂(h, t), x) -> cons₂(h, app₂(t, x))
map_f₂(pid, nil) -> nil
map_f₂(pid, cons₂(h, t)) -> app₂(f₂(pid, h), map_f₂(pid, t))
process₂(store, m) -> if1₃(store, m, leq₂(m, length₁(store)))
if1₃(store, m, true) -> if2₃(store, m, empty₁(fstsplit₂(m, store)))
if1₃(store, m, false) -> if3₃(store, m, empty₁(fstsplit₂(m, app₂(map_f₂(self, nil), store))))
if2₃(store, m, false) -> process₂(app₂(map_f₂(self, nil), sndsplit₂(m, store)), m)
if3₃(store, m, false) -> process₂(sndsplit₂(m, app₂(map_f₂(self, nil), store)), m)

The set Q consists of the following terms:

fstsplit₂(0, x0)
fstsplit₂(s₁(x0), nil)
fstsplit₂(s₁(x0), cons₂(x1, x2))
sndsplit₂(0, x0)
sndsplit₂(s₁(x0), nil)
sndsplit₂(s₁(x0), cons₂(x1, x2))
empty₁(nil)
empty₁(cons₂(x0, x1))
leq₂(0, x0)
leq₂(s₁(x0), 0)
leq₂(s₁(x0), s₁(x1))
length₁(nil)
length₁(cons₂(x0, x1))
app₂(nil, x0)
app₂(cons₂(x0, x1), x2)
map_f₂(x0, nil)
map_f₂(x0, cons₂(x1, x2))
process₂(x0, x1)
if1₃(x0, x1, true)
if1₃(x0, x1, false)
if2₃(x0, x1, false)
if3₃(x0, x1, false)

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

↳ QTRS

  ↳ Non-Overlap Check

    ↳ QTRS

      ↳ DependencyPairsProof

        ↳ QDP

          ↳ DependencyGraphProof

            ↳ AND

              ↳ QDP

              ↳ QDP

              ↳ QDP

              ↳ QDP

              ↳ QDP

              ↳ QDP

                ↳ QDPOrderProof

              ↳ QDP

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

FSTSPLIT₂(s₁(n), cons₂(h, t)) -> FSTSPLIT₂(n, t)

The TRS R consists of the following rules:

fstsplit₂(0, x) -> nil
fstsplit₂(s₁(n), nil) -> nil
fstsplit₂(s₁(n), cons₂(h, t)) -> cons₂(h, fstsplit₂(n, t))
sndsplit₂(0, x) -> x
sndsplit₂(s₁(n), nil) -> nil
sndsplit₂(s₁(n), cons₂(h, t)) -> sndsplit₂(n, t)
empty₁(nil) -> true
empty₁(cons₂(h, t)) -> false
leq₂(0, m) -> true
leq₂(s₁(n), 0) -> false
leq₂(s₁(n), s₁(m)) -> leq₂(n, m)
length₁(nil) -> 0
length₁(cons₂(h, t)) -> s₁(length₁(t))
app₂(nil, x) -> x
app₂(cons₂(h, t), x) -> cons₂(h, app₂(t, x))
map_f₂(pid, nil) -> nil
map_f₂(pid, cons₂(h, t)) -> app₂(f₂(pid, h), map_f₂(pid, t))
process₂(store, m) -> if1₃(store, m, leq₂(m, length₁(store)))
if1₃(store, m, true) -> if2₃(store, m, empty₁(fstsplit₂(m, store)))
if1₃(store, m, false) -> if3₃(store, m, empty₁(fstsplit₂(m, app₂(map_f₂(self, nil), store))))
if2₃(store, m, false) -> process₂(app₂(map_f₂(self, nil), sndsplit₂(m, store)), m)
if3₃(store, m, false) -> process₂(sndsplit₂(m, app₂(map_f₂(self, nil), store)), m)

The set Q consists of the following terms:

fstsplit₂(0, x0)
fstsplit₂(s₁(x0), nil)
fstsplit₂(s₁(x0), cons₂(x1, x2))
sndsplit₂(0, x0)
sndsplit₂(s₁(x0), nil)
sndsplit₂(s₁(x0), cons₂(x1, x2))
empty₁(nil)
empty₁(cons₂(x0, x1))
leq₂(0, x0)
leq₂(s₁(x0), 0)
leq₂(s₁(x0), s₁(x1))
length₁(nil)
length₁(cons₂(x0, x1))
app₂(nil, x0)
app₂(cons₂(x0, x1), x2)
map_f₂(x0, nil)
map_f₂(x0, cons₂(x1, x2))
process₂(x0, x1)
if1₃(x0, x1, true)
if1₃(x0, x1, false)
if2₃(x0, x1, false)
if3₃(x0, x1, false)

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

The following pairs can be strictly oriented and are deleted.

FSTSPLIT₂(s₁(n), cons₂(h, t)) -> FSTSPLIT₂(n, t)

The remaining pairs can at least by weakly be oriented.
none
Used ordering: Combined order from the following AFS and order.
FSTSPLIT₂(x1, x2) = x1
s₁(x1) = s₁(x1)
cons₂(x1, x2) = x1

Lexicographic Path Order [19].
Precedence:

trivial

The following usable rules [14] were oriented: none

↳ QTRS

  ↳ Non-Overlap Check

    ↳ QTRS

      ↳ DependencyPairsProof

        ↳ QDP

          ↳ DependencyGraphProof

            ↳ AND

              ↳ QDP

              ↳ QDP

              ↳ QDP

              ↳ QDP

              ↳ QDP

              ↳ QDP

                ↳ QDPOrderProof

                  ↳ QDP

                    ↳ PisEmptyProof

              ↳ QDP

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

fstsplit₂(0, x) -> nil
fstsplit₂(s₁(n), nil) -> nil
fstsplit₂(s₁(n), cons₂(h, t)) -> cons₂(h, fstsplit₂(n, t))
sndsplit₂(0, x) -> x
sndsplit₂(s₁(n), nil) -> nil
sndsplit₂(s₁(n), cons₂(h, t)) -> sndsplit₂(n, t)
empty₁(nil) -> true
empty₁(cons₂(h, t)) -> false
leq₂(0, m) -> true
leq₂(s₁(n), 0) -> false
leq₂(s₁(n), s₁(m)) -> leq₂(n, m)
length₁(nil) -> 0
length₁(cons₂(h, t)) -> s₁(length₁(t))
app₂(nil, x) -> x
app₂(cons₂(h, t), x) -> cons₂(h, app₂(t, x))
map_f₂(pid, nil) -> nil
map_f₂(pid, cons₂(h, t)) -> app₂(f₂(pid, h), map_f₂(pid, t))
process₂(store, m) -> if1₃(store, m, leq₂(m, length₁(store)))
if1₃(store, m, true) -> if2₃(store, m, empty₁(fstsplit₂(m, store)))
if1₃(store, m, false) -> if3₃(store, m, empty₁(fstsplit₂(m, app₂(map_f₂(self, nil), store))))
if2₃(store, m, false) -> process₂(app₂(map_f₂(self, nil), sndsplit₂(m, store)), m)
if3₃(store, m, false) -> process₂(sndsplit₂(m, app₂(map_f₂(self, nil), store)), m)

The set Q consists of the following terms:

fstsplit₂(0, x0)
fstsplit₂(s₁(x0), nil)
fstsplit₂(s₁(x0), cons₂(x1, x2))
sndsplit₂(0, x0)
sndsplit₂(s₁(x0), nil)
sndsplit₂(s₁(x0), cons₂(x1, x2))
empty₁(nil)
empty₁(cons₂(x0, x1))
leq₂(0, x0)
leq₂(s₁(x0), 0)
leq₂(s₁(x0), s₁(x1))
length₁(nil)
length₁(cons₂(x0, x1))
app₂(nil, x0)
app₂(cons₂(x0, x1), x2)
map_f₂(x0, nil)
map_f₂(x0, cons₂(x1, x2))
process₂(x0, x1)
if1₃(x0, x1, true)
if1₃(x0, x1, false)
if2₃(x0, x1, false)
if3₃(x0, x1, false)

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

↳ QTRS

  ↳ Non-Overlap Check

    ↳ QTRS

      ↳ DependencyPairsProof

        ↳ QDP

          ↳ DependencyGraphProof

            ↳ AND

              ↳ QDP

              ↳ QDP

              ↳ QDP

              ↳ QDP

              ↳ QDP

              ↳ QDP

              ↳ QDP

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

PROCESS₂(store, m) -> IF1₃(store, m, leq₂(m, length₁(store)))
IF1₃(store, m, false) -> IF3₃(store, m, empty₁(fstsplit₂(m, app₂(map_f₂(self, nil), store))))
IF3₃(store, m, false) -> PROCESS₂(sndsplit₂(m, app₂(map_f₂(self, nil), store)), m)
IF2₃(store, m, false) -> PROCESS₂(app₂(map_f₂(self, nil), sndsplit₂(m, store)), m)
IF1₃(store, m, true) -> IF2₃(store, m, empty₁(fstsplit₂(m, store)))

The TRS R consists of the following rules:

fstsplit₂(0, x) -> nil
fstsplit₂(s₁(n), nil) -> nil
fstsplit₂(s₁(n), cons₂(h, t)) -> cons₂(h, fstsplit₂(n, t))
sndsplit₂(0, x) -> x
sndsplit₂(s₁(n), nil) -> nil
sndsplit₂(s₁(n), cons₂(h, t)) -> sndsplit₂(n, t)
empty₁(nil) -> true
empty₁(cons₂(h, t)) -> false
leq₂(0, m) -> true
leq₂(s₁(n), 0) -> false
leq₂(s₁(n), s₁(m)) -> leq₂(n, m)
length₁(nil) -> 0
length₁(cons₂(h, t)) -> s₁(length₁(t))
app₂(nil, x) -> x
app₂(cons₂(h, t), x) -> cons₂(h, app₂(t, x))
map_f₂(pid, nil) -> nil
map_f₂(pid, cons₂(h, t)) -> app₂(f₂(pid, h), map_f₂(pid, t))
process₂(store, m) -> if1₃(store, m, leq₂(m, length₁(store)))
if1₃(store, m, true) -> if2₃(store, m, empty₁(fstsplit₂(m, store)))
if1₃(store, m, false) -> if3₃(store, m, empty₁(fstsplit₂(m, app₂(map_f₂(self, nil), store))))
if2₃(store, m, false) -> process₂(app₂(map_f₂(self, nil), sndsplit₂(m, store)), m)
if3₃(store, m, false) -> process₂(sndsplit₂(m, app₂(map_f₂(self, nil), store)), m)

The set Q consists of the following terms:

fstsplit₂(0, x0)
fstsplit₂(s₁(x0), nil)
fstsplit₂(s₁(x0), cons₂(x1, x2))
sndsplit₂(0, x0)
sndsplit₂(s₁(x0), nil)
sndsplit₂(s₁(x0), cons₂(x1, x2))
empty₁(nil)
empty₁(cons₂(x0, x1))
leq₂(0, x0)
leq₂(s₁(x0), 0)
leq₂(s₁(x0), s₁(x1))
length₁(nil)
length₁(cons₂(x0, x1))
app₂(nil, x0)
app₂(cons₂(x0, x1), x2)
map_f₂(x0, nil)
map_f₂(x0, cons₂(x1, x2))
process₂(x0, x1)
if1₃(x0, x1, true)
if1₃(x0, x1, false)
if2₃(x0, x1, false)
if3₃(x0, x1, false)

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