Termination w.r.t. Q proof of /tmp/tmpQpMYla/AAECC.trs

Termination w.r.t. Q of the given QTRS could not be shown:

0 QTRS

↳1 AAECC Innermost (⇔)

↳2 QTRS

↳3 DependencyPairsProof (⇔)

↳4 QDP

↳5 DependencyGraphProof (⇔)

↳6 AND

  ↳7 QDP

    ↳8 QDPOrderProof (⇔)

    ↳9 QDP

    ↳10 PisEmptyProof (⇔)

    ↳11 TRUE

  ↳12 QDP

    ↳13 QDPOrderProof (⇔)

    ↳14 QDP

    ↳15 PisEmptyProof (⇔)

    ↳16 TRUE

  ↳17 QDP

    ↳18 QDPOrderProof (⇔)

    ↳19 QDP

    ↳20 PisEmptyProof (⇔)

    ↳21 TRUE

  ↳22 QDP

    ↳23 QDPOrderProof (⇔)

    ↳24 QDP

    ↳25 PisEmptyProof (⇔)

    ↳26 TRUE

  ↳27 QDP

    ↳28 QDPOrderProof (⇔)

    ↳29 QDP

    ↳30 PisEmptyProof (⇔)

    ↳31 TRUE

  ↳32 QDP

    ↳33 QDPOrderProof (⇔)

    ↳34 QDP

    ↳35 PisEmptyProof (⇔)

    ↳36 TRUE

  ↳37 QDP

(0) Obligation:

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

fstsplit(0, x) → nil
fstsplit(s(n), nil) → nil
fstsplit(s(n), cons(h, t)) → cons(h, fstsplit(n, t))
sndsplit(0, x) → x
sndsplit(s(n), nil) → nil
sndsplit(s(n), cons(h, t)) → sndsplit(n, t)
empty(nil) → true
empty(cons(h, t)) → false
leq(0, m) → true
leq(s(n), 0) → false
leq(s(n), s(m)) → leq(n, m)
length(nil) → 0
length(cons(h, t)) → s(length(t))
app(nil, x) → x
app(cons(h, t), x) → cons(h, app(t, x))
map_f(pid, nil) → nil
map_f(pid, cons(h, t)) → app(f(pid, h), map_f(pid, t))
process(store, m) → if1(store, m, leq(m, length(store)))
if1(store, m, true) → if2(store, m, empty(fstsplit(m, store)))
if1(store, m, false) → if3(store, m, empty(fstsplit(m, app(map_f(self, nil), store))))
if2(store, m, false) → process(app(map_f(self, nil), sndsplit(m, store)), m)
if3(store, m, false) → process(sndsplit(m, app(map_f(self, nil), store)), m)

Q is empty.

(1) AAECC Innermost (EQUIVALENT transformation)

We have applied [NOC,AAECCNOC] to switch to innermost. The TRS R 1 is

sndsplit(0, x) → x
sndsplit(s(n), nil) → nil
sndsplit(s(n), cons(h, t)) → sndsplit(n, t)
length(nil) → 0
length(cons(h, t)) → s(length(t))
app(nil, x) → x
app(cons(h, t), x) → cons(h, app(t, x))
map_f(pid, nil) → nil
map_f(pid, cons(h, t)) → app(f(pid, h), map_f(pid, t))
fstsplit(0, x) → nil
fstsplit(s(n), nil) → nil
fstsplit(s(n), cons(h, t)) → cons(h, fstsplit(n, t))
empty(nil) → true
empty(cons(h, t)) → false
leq(0, m) → true
leq(s(n), 0) → false
leq(s(n), s(m)) → leq(n, m)

The TRS R 2 is

process(store, m) → if1(store, m, leq(m, length(store)))
if1(store, m, true) → if2(store, m, empty(fstsplit(m, store)))
if1(store, m, false) → if3(store, m, empty(fstsplit(m, app(map_f(self, nil), store))))
if2(store, m, false) → process(app(map_f(self, nil), sndsplit(m, store)), m)
if3(store, m, false) → process(sndsplit(m, app(map_f(self, nil), store)), m)

The signature Sigma is {process, if2, if3, if1}

(2) Obligation:

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

fstsplit(0, x) → nil
fstsplit(s(n), nil) → nil
fstsplit(s(n), cons(h, t)) → cons(h, fstsplit(n, t))
sndsplit(0, x) → x
sndsplit(s(n), nil) → nil
sndsplit(s(n), cons(h, t)) → sndsplit(n, t)
empty(nil) → true
empty(cons(h, t)) → false
leq(0, m) → true
leq(s(n), 0) → false
leq(s(n), s(m)) → leq(n, m)
length(nil) → 0
length(cons(h, t)) → s(length(t))
app(nil, x) → x
app(cons(h, t), x) → cons(h, app(t, x))
map_f(pid, nil) → nil
map_f(pid, cons(h, t)) → app(f(pid, h), map_f(pid, t))
process(store, m) → if1(store, m, leq(m, length(store)))
if1(store, m, true) → if2(store, m, empty(fstsplit(m, store)))
if1(store, m, false) → if3(store, m, empty(fstsplit(m, app(map_f(self, nil), store))))
if2(store, m, false) → process(app(map_f(self, nil), sndsplit(m, store)), m)
if3(store, m, false) → process(sndsplit(m, app(map_f(self, nil), store)), m)

The set Q consists of the following terms:

fstsplit(0, x0)
fstsplit(s(x0), nil)
fstsplit(s(x0), cons(x1, x2))
sndsplit(0, x0)
sndsplit(s(x0), nil)
sndsplit(s(x0), cons(x1, x2))
empty(nil)
empty(cons(x0, x1))
leq(0, x0)
leq(s(x0), 0)
leq(s(x0), s(x1))
length(nil)
length(cons(x0, x1))
app(nil, x0)
app(cons(x0, x1), x2)
map_f(x0, nil)
map_f(x0, cons(x1, x2))
process(x0, x1)
if1(x0, x1, true)
if1(x0, x1, false)
if2(x0, x1, false)
if3(x0, x1, false)

(3) DependencyPairsProof (EQUIVALENT transformation)

Using Dependency Pairs [AG00,LPAR04] we result in the following initial DP problem.

(4) Obligation:

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

FSTSPLIT(s(n), cons(h, t)) → FSTSPLIT(n, t)
SNDSPLIT(s(n), cons(h, t)) → SNDSPLIT(n, t)
LEQ(s(n), s(m)) → LEQ(n, m)
LENGTH(cons(h, t)) → LENGTH(t)
APP(cons(h, t), x) → APP(t, x)
MAP_F(pid, cons(h, t)) → APP(f(pid, h), map_f(pid, t))
MAP_F(pid, cons(h, t)) → MAP_F(pid, t)
PROCESS(store, m) → IF1(store, m, leq(m, length(store)))
PROCESS(store, m) → LEQ(m, length(store))
PROCESS(store, m) → LENGTH(store)
IF1(store, m, true) → IF2(store, m, empty(fstsplit(m, store)))
IF1(store, m, true) → EMPTY(fstsplit(m, store))
IF1(store, m, true) → FSTSPLIT(m, store)
IF1(store, m, false) → IF3(store, m, empty(fstsplit(m, app(map_f(self, nil), store))))
IF1(store, m, false) → EMPTY(fstsplit(m, app(map_f(self, nil), store)))
IF1(store, m, false) → FSTSPLIT(m, app(map_f(self, nil), store))
IF1(store, m, false) → APP(map_f(self, nil), store)
IF1(store, m, false) → MAP_F(self, nil)
IF2(store, m, false) → PROCESS(app(map_f(self, nil), sndsplit(m, store)), m)
IF2(store, m, false) → APP(map_f(self, nil), sndsplit(m, store))
IF2(store, m, false) → MAP_F(self, nil)
IF2(store, m, false) → SNDSPLIT(m, store)
IF3(store, m, false) → PROCESS(sndsplit(m, app(map_f(self, nil), store)), m)
IF3(store, m, false) → SNDSPLIT(m, app(map_f(self, nil), store))
IF3(store, m, false) → APP(map_f(self, nil), store)
IF3(store, m, false) → MAP_F(self, nil)

The TRS R consists of the following rules:

fstsplit(0, x) → nil
fstsplit(s(n), nil) → nil
fstsplit(s(n), cons(h, t)) → cons(h, fstsplit(n, t))
sndsplit(0, x) → x
sndsplit(s(n), nil) → nil
sndsplit(s(n), cons(h, t)) → sndsplit(n, t)
empty(nil) → true
empty(cons(h, t)) → false
leq(0, m) → true
leq(s(n), 0) → false
leq(s(n), s(m)) → leq(n, m)
length(nil) → 0
length(cons(h, t)) → s(length(t))
app(nil, x) → x
app(cons(h, t), x) → cons(h, app(t, x))
map_f(pid, nil) → nil
map_f(pid, cons(h, t)) → app(f(pid, h), map_f(pid, t))
process(store, m) → if1(store, m, leq(m, length(store)))
if1(store, m, true) → if2(store, m, empty(fstsplit(m, store)))
if1(store, m, false) → if3(store, m, empty(fstsplit(m, app(map_f(self, nil), store))))
if2(store, m, false) → process(app(map_f(self, nil), sndsplit(m, store)), m)
if3(store, m, false) → process(sndsplit(m, app(map_f(self, nil), store)), m)

The set Q consists of the following terms:

fstsplit(0, x0)
fstsplit(s(x0), nil)
fstsplit(s(x0), cons(x1, x2))
sndsplit(0, x0)
sndsplit(s(x0), nil)
sndsplit(s(x0), cons(x1, x2))
empty(nil)
empty(cons(x0, x1))
leq(0, x0)
leq(s(x0), 0)
leq(s(x0), s(x1))
length(nil)
length(cons(x0, x1))
app(nil, x0)
app(cons(x0, x1), x2)
map_f(x0, nil)
map_f(x0, cons(x1, x2))
process(x0, x1)
if1(x0, x1, true)
if1(x0, x1, false)
if2(x0, x1, false)
if3(x0, x1, false)

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

(5) DependencyGraphProof (EQUIVALENT transformation)

The approximation of the Dependency Graph [LPAR04,FROCOS05,EDGSTAR] contains 7 SCCs with 15 less nodes.

(6) Complex Obligation (AND)

(7) Obligation:

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

APP(cons(h, t), x) → APP(t, x)

The TRS R consists of the following rules:

fstsplit(0, x) → nil
fstsplit(s(n), nil) → nil
fstsplit(s(n), cons(h, t)) → cons(h, fstsplit(n, t))
sndsplit(0, x) → x
sndsplit(s(n), nil) → nil
sndsplit(s(n), cons(h, t)) → sndsplit(n, t)
empty(nil) → true
empty(cons(h, t)) → false
leq(0, m) → true
leq(s(n), 0) → false
leq(s(n), s(m)) → leq(n, m)
length(nil) → 0
length(cons(h, t)) → s(length(t))
app(nil, x) → x
app(cons(h, t), x) → cons(h, app(t, x))
map_f(pid, nil) → nil
map_f(pid, cons(h, t)) → app(f(pid, h), map_f(pid, t))
process(store, m) → if1(store, m, leq(m, length(store)))
if1(store, m, true) → if2(store, m, empty(fstsplit(m, store)))
if1(store, m, false) → if3(store, m, empty(fstsplit(m, app(map_f(self, nil), store))))
if2(store, m, false) → process(app(map_f(self, nil), sndsplit(m, store)), m)
if3(store, m, false) → process(sndsplit(m, app(map_f(self, nil), store)), m)

The set Q consists of the following terms:

fstsplit(0, x0)
fstsplit(s(x0), nil)
fstsplit(s(x0), cons(x1, x2))
sndsplit(0, x0)
sndsplit(s(x0), nil)
sndsplit(s(x0), cons(x1, x2))
empty(nil)
empty(cons(x0, x1))
leq(0, x0)
leq(s(x0), 0)
leq(s(x0), s(x1))
length(nil)
length(cons(x0, x1))
app(nil, x0)
app(cons(x0, x1), x2)
map_f(x0, nil)
map_f(x0, cons(x1, x2))
process(x0, x1)
if1(x0, x1, true)
if1(x0, x1, false)
if2(x0, x1, false)
if3(x0, x1, false)

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

(8) QDPOrderProof (EQUIVALENT transformation)

We use the reduction pair processor [LPAR04].

The following pairs can be oriented strictly and are deleted.

APP(cons(h, t), x) → APP(t, x)

The remaining pairs can at least be oriented weakly.
Used ordering: Combined order from the following AFS and order.
APP(x1, x2) = x1
cons(x1, x2) = cons(x1, x2)
fstsplit(x1, x2) = x2
0 = 0
nil = nil
s(x1) = x1
sndsplit(x1, x2) = sndsplit(x1, x2)
empty(x1) = empty(x1)
true = true
false = false
leq(x1, x2) = leq
length(x1) = length
app(x1, x2) = app(x1, x2)
map_f(x1, x2) = map_f(x1, x2)
f(x1, x2) = f(x1, x2)
process(x1, x2) = process(x2)
if1(x1, x2, x3) = if1(x2)
if2(x1, x2, x3) = if2(x2)
if3(x1, x2, x3) = if3(x2)
self = self

Lexicographic path order with status [LPO].
Quasi-Precedence:

sndsplit₂ > [nil, process₁, if1₁, if2₁, if3₁]
empty₁ > [false, leq, app₂] > cons₂ > [0, length] > [nil, process₁, if1₁, if2₁, if3₁]
empty₁ > [false, leq, app₂] > true > [nil, process₁, if1₁, if2₁, if3₁]
map_f₂ > [false, leq, app₂] > cons₂ > [0, length] > [nil, process₁, if1₁, if2₁, if3₁]
map_f₂ > [false, leq, app₂] > true > [nil, process₁, if1₁, if2₁, if3₁]
map_f₂ > f₂ > [nil, process₁, if1₁, if2₁, if3₁]
self > [nil, process₁, if1₁, if2₁, if3₁]

Status:

empty₁: [1]
f₂: [2,1]
true: []
if3₁: [1]
if2₁: [1]
if1₁: [1]
length: []
0: []
map_f₂: [1,2]
sndsplit₂: [1,2]
cons₂: [2,1]
self: []
false: []
app₂: [2,1]
leq: []
nil: []
process₁: [1]

The following usable rules [FROCOS05] were oriented:

fstsplit(0, x) → nil
fstsplit(s(n), nil) → nil
fstsplit(s(n), cons(h, t)) → cons(h, fstsplit(n, t))
sndsplit(0, x) → x
sndsplit(s(n), nil) → nil
sndsplit(s(n), cons(h, t)) → sndsplit(n, t)
empty(nil) → true
empty(cons(h, t)) → false
leq(0, m) → true
leq(s(n), 0) → false
leq(s(n), s(m)) → leq(n, m)
length(nil) → 0
length(cons(h, t)) → s(length(t))
app(nil, x) → x
app(cons(h, t), x) → cons(h, app(t, x))
map_f(pid, nil) → nil
map_f(pid, cons(h, t)) → app(f(pid, h), map_f(pid, t))
process(store, m) → if1(store, m, leq(m, length(store)))
if1(store, m, true) → if2(store, m, empty(fstsplit(m, store)))
if1(store, m, false) → if3(store, m, empty(fstsplit(m, app(map_f(self, nil), store))))
if2(store, m, false) → process(app(map_f(self, nil), sndsplit(m, store)), m)
if3(store, m, false) → process(sndsplit(m, app(map_f(self, nil), store)), m)

(9) Obligation:

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

fstsplit(0, x) → nil
fstsplit(s(n), nil) → nil
fstsplit(s(n), cons(h, t)) → cons(h, fstsplit(n, t))
sndsplit(0, x) → x
sndsplit(s(n), nil) → nil
sndsplit(s(n), cons(h, t)) → sndsplit(n, t)
empty(nil) → true
empty(cons(h, t)) → false
leq(0, m) → true
leq(s(n), 0) → false
leq(s(n), s(m)) → leq(n, m)
length(nil) → 0
length(cons(h, t)) → s(length(t))
app(nil, x) → x
app(cons(h, t), x) → cons(h, app(t, x))
map_f(pid, nil) → nil
map_f(pid, cons(h, t)) → app(f(pid, h), map_f(pid, t))
process(store, m) → if1(store, m, leq(m, length(store)))
if1(store, m, true) → if2(store, m, empty(fstsplit(m, store)))
if1(store, m, false) → if3(store, m, empty(fstsplit(m, app(map_f(self, nil), store))))
if2(store, m, false) → process(app(map_f(self, nil), sndsplit(m, store)), m)
if3(store, m, false) → process(sndsplit(m, app(map_f(self, nil), store)), m)

The set Q consists of the following terms:

fstsplit(0, x0)
fstsplit(s(x0), nil)
fstsplit(s(x0), cons(x1, x2))
sndsplit(0, x0)
sndsplit(s(x0), nil)
sndsplit(s(x0), cons(x1, x2))
empty(nil)
empty(cons(x0, x1))
leq(0, x0)
leq(s(x0), 0)
leq(s(x0), s(x1))
length(nil)
length(cons(x0, x1))
app(nil, x0)
app(cons(x0, x1), x2)
map_f(x0, nil)
map_f(x0, cons(x1, x2))
process(x0, x1)
if1(x0, x1, true)
if1(x0, x1, false)
if2(x0, x1, false)
if3(x0, x1, false)

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

(10) PisEmptyProof (EQUIVALENT transformation)

The TRS P is empty. Hence, there is no (P,Q,R) chain.

(11) TRUE

(12) Obligation:

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

MAP_F(pid, cons(h, t)) → MAP_F(pid, t)

The TRS R consists of the following rules:

fstsplit(0, x) → nil
fstsplit(s(n), nil) → nil
fstsplit(s(n), cons(h, t)) → cons(h, fstsplit(n, t))
sndsplit(0, x) → x
sndsplit(s(n), nil) → nil
sndsplit(s(n), cons(h, t)) → sndsplit(n, t)
empty(nil) → true
empty(cons(h, t)) → false
leq(0, m) → true
leq(s(n), 0) → false
leq(s(n), s(m)) → leq(n, m)
length(nil) → 0
length(cons(h, t)) → s(length(t))
app(nil, x) → x
app(cons(h, t), x) → cons(h, app(t, x))
map_f(pid, nil) → nil
map_f(pid, cons(h, t)) → app(f(pid, h), map_f(pid, t))
process(store, m) → if1(store, m, leq(m, length(store)))
if1(store, m, true) → if2(store, m, empty(fstsplit(m, store)))
if1(store, m, false) → if3(store, m, empty(fstsplit(m, app(map_f(self, nil), store))))
if2(store, m, false) → process(app(map_f(self, nil), sndsplit(m, store)), m)
if3(store, m, false) → process(sndsplit(m, app(map_f(self, nil), store)), m)

The set Q consists of the following terms:

fstsplit(0, x0)
fstsplit(s(x0), nil)
fstsplit(s(x0), cons(x1, x2))
sndsplit(0, x0)
sndsplit(s(x0), nil)
sndsplit(s(x0), cons(x1, x2))
empty(nil)
empty(cons(x0, x1))
leq(0, x0)
leq(s(x0), 0)
leq(s(x0), s(x1))
length(nil)
length(cons(x0, x1))
app(nil, x0)
app(cons(x0, x1), x2)
map_f(x0, nil)
map_f(x0, cons(x1, x2))
process(x0, x1)
if1(x0, x1, true)
if1(x0, x1, false)
if2(x0, x1, false)
if3(x0, x1, false)

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

(13) QDPOrderProof (EQUIVALENT transformation)

We use the reduction pair processor [LPAR04].

The following pairs can be oriented strictly and are deleted.

MAP_F(pid, cons(h, t)) → MAP_F(pid, t)

The remaining pairs can at least be oriented weakly.
Used ordering: Combined order from the following AFS and order.
MAP_F(x1, x2) = x2
cons(x1, x2) = cons(x1, x2)
fstsplit(x1, x2) = fstsplit(x2)
0 = 0
nil = nil
s(x1) = s
sndsplit(x1, x2) = sndsplit(x2)
empty(x1) = empty
true = true
false = false
leq(x1, x2) = leq
length(x1) = x1
app(x1, x2) = app(x1, x2)
map_f(x1, x2) = map_f(x1, x2)
f(x1, x2) = f(x1, x2)
process(x1, x2) = process
if1(x1, x2, x3) = x3
if2(x1, x2, x3) = if2
if3(x1, x2, x3) = x3
self = self

Lexicographic path order with status [LPO].
Quasi-Precedence:

fstsplit₁ > cons₂ > [0, s, empty, true, false, leq, process, if2]
fstsplit₁ > nil > [0, s, empty, true, false, leq, process, if2]
sndsplit₁ > nil > [0, s, empty, true, false, leq, process, if2]
map_f₂ > nil > [0, s, empty, true, false, leq, process, if2]
map_f₂ > app₂ > cons₂ > [0, s, empty, true, false, leq, process, if2]
map_f₂ > f₂ > [0, s, empty, true, false, leq, process, if2]
self > [0, s, empty, true, false, leq, process, if2]

Status:

f₂: [2,1]
true: []
empty: []
s: []
fstsplit₁: [1]
0: []
map_f₂: [1,2]
process: []
cons₂: [2,1]
self: []
sndsplit₁: [1]
false: []
app₂: [2,1]
if2: []
leq: []
nil: []

The following usable rules [FROCOS05] were oriented:

fstsplit(0, x) → nil
fstsplit(s(n), nil) → nil
fstsplit(s(n), cons(h, t)) → cons(h, fstsplit(n, t))
sndsplit(0, x) → x
sndsplit(s(n), nil) → nil
sndsplit(s(n), cons(h, t)) → sndsplit(n, t)
empty(nil) → true
empty(cons(h, t)) → false
leq(0, m) → true
leq(s(n), 0) → false
leq(s(n), s(m)) → leq(n, m)
length(nil) → 0
length(cons(h, t)) → s(length(t))
app(nil, x) → x
app(cons(h, t), x) → cons(h, app(t, x))
map_f(pid, nil) → nil
map_f(pid, cons(h, t)) → app(f(pid, h), map_f(pid, t))
process(store, m) → if1(store, m, leq(m, length(store)))
if1(store, m, true) → if2(store, m, empty(fstsplit(m, store)))
if1(store, m, false) → if3(store, m, empty(fstsplit(m, app(map_f(self, nil), store))))
if2(store, m, false) → process(app(map_f(self, nil), sndsplit(m, store)), m)
if3(store, m, false) → process(sndsplit(m, app(map_f(self, nil), store)), m)

(14) Obligation:

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

fstsplit(0, x) → nil
fstsplit(s(n), nil) → nil
fstsplit(s(n), cons(h, t)) → cons(h, fstsplit(n, t))
sndsplit(0, x) → x
sndsplit(s(n), nil) → nil
sndsplit(s(n), cons(h, t)) → sndsplit(n, t)
empty(nil) → true
empty(cons(h, t)) → false
leq(0, m) → true
leq(s(n), 0) → false
leq(s(n), s(m)) → leq(n, m)
length(nil) → 0
length(cons(h, t)) → s(length(t))
app(nil, x) → x
app(cons(h, t), x) → cons(h, app(t, x))
map_f(pid, nil) → nil
map_f(pid, cons(h, t)) → app(f(pid, h), map_f(pid, t))
process(store, m) → if1(store, m, leq(m, length(store)))
if1(store, m, true) → if2(store, m, empty(fstsplit(m, store)))
if1(store, m, false) → if3(store, m, empty(fstsplit(m, app(map_f(self, nil), store))))
if2(store, m, false) → process(app(map_f(self, nil), sndsplit(m, store)), m)
if3(store, m, false) → process(sndsplit(m, app(map_f(self, nil), store)), m)

The set Q consists of the following terms:

fstsplit(0, x0)
fstsplit(s(x0), nil)
fstsplit(s(x0), cons(x1, x2))
sndsplit(0, x0)
sndsplit(s(x0), nil)
sndsplit(s(x0), cons(x1, x2))
empty(nil)
empty(cons(x0, x1))
leq(0, x0)
leq(s(x0), 0)
leq(s(x0), s(x1))
length(nil)
length(cons(x0, x1))
app(nil, x0)
app(cons(x0, x1), x2)
map_f(x0, nil)
map_f(x0, cons(x1, x2))
process(x0, x1)
if1(x0, x1, true)
if1(x0, x1, false)
if2(x0, x1, false)
if3(x0, x1, false)

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

(15) PisEmptyProof (EQUIVALENT transformation)

The TRS P is empty. Hence, there is no (P,Q,R) chain.

(16) TRUE

(17) Obligation:

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

LENGTH(cons(h, t)) → LENGTH(t)

The TRS R consists of the following rules:

fstsplit(0, x) → nil
fstsplit(s(n), nil) → nil
fstsplit(s(n), cons(h, t)) → cons(h, fstsplit(n, t))
sndsplit(0, x) → x
sndsplit(s(n), nil) → nil
sndsplit(s(n), cons(h, t)) → sndsplit(n, t)
empty(nil) → true
empty(cons(h, t)) → false
leq(0, m) → true
leq(s(n), 0) → false
leq(s(n), s(m)) → leq(n, m)
length(nil) → 0
length(cons(h, t)) → s(length(t))
app(nil, x) → x
app(cons(h, t), x) → cons(h, app(t, x))
map_f(pid, nil) → nil
map_f(pid, cons(h, t)) → app(f(pid, h), map_f(pid, t))
process(store, m) → if1(store, m, leq(m, length(store)))
if1(store, m, true) → if2(store, m, empty(fstsplit(m, store)))
if1(store, m, false) → if3(store, m, empty(fstsplit(m, app(map_f(self, nil), store))))
if2(store, m, false) → process(app(map_f(self, nil), sndsplit(m, store)), m)
if3(store, m, false) → process(sndsplit(m, app(map_f(self, nil), store)), m)

The set Q consists of the following terms:

fstsplit(0, x0)
fstsplit(s(x0), nil)
fstsplit(s(x0), cons(x1, x2))
sndsplit(0, x0)
sndsplit(s(x0), nil)
sndsplit(s(x0), cons(x1, x2))
empty(nil)
empty(cons(x0, x1))
leq(0, x0)
leq(s(x0), 0)
leq(s(x0), s(x1))
length(nil)
length(cons(x0, x1))
app(nil, x0)
app(cons(x0, x1), x2)
map_f(x0, nil)
map_f(x0, cons(x1, x2))
process(x0, x1)
if1(x0, x1, true)
if1(x0, x1, false)
if2(x0, x1, false)
if3(x0, x1, false)

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

(18) QDPOrderProof (EQUIVALENT transformation)

We use the reduction pair processor [LPAR04].

The following pairs can be oriented strictly and are deleted.

LENGTH(cons(h, t)) → LENGTH(t)

The remaining pairs can at least be oriented weakly.
Used ordering: Combined order from the following AFS and order.
LENGTH(x1) = LENGTH(x1)
cons(x1, x2) = cons(x1, x2)
fstsplit(x1, x2) = fstsplit(x2)
0 = 0
nil = nil
s(x1) = x1
sndsplit(x1, x2) = sndsplit(x1, x2)
empty(x1) = empty(x1)
true = true
false = false
leq(x1, x2) = leq
length(x1) = length(x1)
app(x1, x2) = app(x1, x2)
map_f(x1, x2) = map_f(x1, x2)
f(x1, x2) = f
process(x1, x2) = x2
if1(x1, x2, x3) = x2
if2(x1, x2, x3) = x2
if3(x1, x2, x3) = x2
self = self

Lexicographic path order with status [LPO].
Quasi-Precedence:

[empty₁, false, leq, map_f₂] > fstsplit₁ > cons₂
[empty₁, false, leq, map_f₂] > fstsplit₁ > nil
[empty₁, false, leq, map_f₂] > sndsplit₂ > nil
[empty₁, false, leq, map_f₂] > true
[empty₁, false, leq, map_f₂] > app₂ > cons₂
[empty₁, false, leq, map_f₂] > f
[empty₁, false, leq, map_f₂] > self
length₁ > 0 > nil
length₁ > 0 > true

Status:

empty₁: [1]
LENGTH₁: [1]
f: []
true: []
fstsplit₁: [1]
0: []
map_f₂: [1,2]
sndsplit₂: [1,2]
cons₂: [2,1]
self: []
false: []
app₂: [1,2]
length₁: [1]
leq: []
nil: []

The following usable rules [FROCOS05] were oriented:

fstsplit(0, x) → nil
fstsplit(s(n), nil) → nil
fstsplit(s(n), cons(h, t)) → cons(h, fstsplit(n, t))
sndsplit(0, x) → x
sndsplit(s(n), nil) → nil
sndsplit(s(n), cons(h, t)) → sndsplit(n, t)
empty(nil) → true
empty(cons(h, t)) → false
leq(0, m) → true
leq(s(n), 0) → false
leq(s(n), s(m)) → leq(n, m)
length(nil) → 0
length(cons(h, t)) → s(length(t))
app(nil, x) → x
app(cons(h, t), x) → cons(h, app(t, x))
map_f(pid, nil) → nil
map_f(pid, cons(h, t)) → app(f(pid, h), map_f(pid, t))
process(store, m) → if1(store, m, leq(m, length(store)))
if1(store, m, true) → if2(store, m, empty(fstsplit(m, store)))
if1(store, m, false) → if3(store, m, empty(fstsplit(m, app(map_f(self, nil), store))))
if2(store, m, false) → process(app(map_f(self, nil), sndsplit(m, store)), m)
if3(store, m, false) → process(sndsplit(m, app(map_f(self, nil), store)), m)

(19) Obligation:

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

fstsplit(0, x) → nil
fstsplit(s(n), nil) → nil
fstsplit(s(n), cons(h, t)) → cons(h, fstsplit(n, t))
sndsplit(0, x) → x
sndsplit(s(n), nil) → nil
sndsplit(s(n), cons(h, t)) → sndsplit(n, t)
empty(nil) → true
empty(cons(h, t)) → false
leq(0, m) → true
leq(s(n), 0) → false
leq(s(n), s(m)) → leq(n, m)
length(nil) → 0
length(cons(h, t)) → s(length(t))
app(nil, x) → x
app(cons(h, t), x) → cons(h, app(t, x))
map_f(pid, nil) → nil
map_f(pid, cons(h, t)) → app(f(pid, h), map_f(pid, t))
process(store, m) → if1(store, m, leq(m, length(store)))
if1(store, m, true) → if2(store, m, empty(fstsplit(m, store)))
if1(store, m, false) → if3(store, m, empty(fstsplit(m, app(map_f(self, nil), store))))
if2(store, m, false) → process(app(map_f(self, nil), sndsplit(m, store)), m)
if3(store, m, false) → process(sndsplit(m, app(map_f(self, nil), store)), m)

The set Q consists of the following terms:

fstsplit(0, x0)
fstsplit(s(x0), nil)
fstsplit(s(x0), cons(x1, x2))
sndsplit(0, x0)
sndsplit(s(x0), nil)
sndsplit(s(x0), cons(x1, x2))
empty(nil)
empty(cons(x0, x1))
leq(0, x0)
leq(s(x0), 0)
leq(s(x0), s(x1))
length(nil)
length(cons(x0, x1))
app(nil, x0)
app(cons(x0, x1), x2)
map_f(x0, nil)
map_f(x0, cons(x1, x2))
process(x0, x1)
if1(x0, x1, true)
if1(x0, x1, false)
if2(x0, x1, false)
if3(x0, x1, false)

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

(20) PisEmptyProof (EQUIVALENT transformation)

The TRS P is empty. Hence, there is no (P,Q,R) chain.

(21) TRUE

(22) Obligation:

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

LEQ(s(n), s(m)) → LEQ(n, m)

The TRS R consists of the following rules:

fstsplit(0, x) → nil
fstsplit(s(n), nil) → nil
fstsplit(s(n), cons(h, t)) → cons(h, fstsplit(n, t))
sndsplit(0, x) → x
sndsplit(s(n), nil) → nil
sndsplit(s(n), cons(h, t)) → sndsplit(n, t)
empty(nil) → true
empty(cons(h, t)) → false
leq(0, m) → true
leq(s(n), 0) → false
leq(s(n), s(m)) → leq(n, m)
length(nil) → 0
length(cons(h, t)) → s(length(t))
app(nil, x) → x
app(cons(h, t), x) → cons(h, app(t, x))
map_f(pid, nil) → nil
map_f(pid, cons(h, t)) → app(f(pid, h), map_f(pid, t))
process(store, m) → if1(store, m, leq(m, length(store)))
if1(store, m, true) → if2(store, m, empty(fstsplit(m, store)))
if1(store, m, false) → if3(store, m, empty(fstsplit(m, app(map_f(self, nil), store))))
if2(store, m, false) → process(app(map_f(self, nil), sndsplit(m, store)), m)
if3(store, m, false) → process(sndsplit(m, app(map_f(self, nil), store)), m)

The set Q consists of the following terms:

fstsplit(0, x0)
fstsplit(s(x0), nil)
fstsplit(s(x0), cons(x1, x2))
sndsplit(0, x0)
sndsplit(s(x0), nil)
sndsplit(s(x0), cons(x1, x2))
empty(nil)
empty(cons(x0, x1))
leq(0, x0)
leq(s(x0), 0)
leq(s(x0), s(x1))
length(nil)
length(cons(x0, x1))
app(nil, x0)
app(cons(x0, x1), x2)
map_f(x0, nil)
map_f(x0, cons(x1, x2))
process(x0, x1)
if1(x0, x1, true)
if1(x0, x1, false)
if2(x0, x1, false)
if3(x0, x1, false)

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

(23) QDPOrderProof (EQUIVALENT transformation)

We use the reduction pair processor [LPAR04].

The following pairs can be oriented strictly and are deleted.

LEQ(s(n), s(m)) → LEQ(n, m)

The remaining pairs can at least be oriented weakly.
Used ordering: Combined order from the following AFS and order.
LEQ(x1, x2) = x1
s(x1) = s(x1)
fstsplit(x1, x2) = fstsplit(x1, x2)
0 = 0
nil = nil
cons(x1, x2) = cons(x1, x2)
sndsplit(x1, x2) = sndsplit(x1, x2)
empty(x1) = empty(x1)
true = true
false = false
leq(x1, x2) = leq
length(x1) = length(x1)
app(x1, x2) = app(x1, x2)
map_f(x1, x2) = map_f(x2)
f(x1, x2) = x2
process(x1, x2) = process(x2)
if1(x1, x2, x3) = if1(x2)
if2(x1, x2, x3) = if2(x2)
if3(x1, x2, x3) = if3(x2)
self = self

Lexicographic path order with status [LPO].
Quasi-Precedence:

fstsplit₂ > cons₂ > s₁ > nil > 0
empty₁ > true > [sndsplit₂, process₁, if1₁, if2₁, if3₁, self] > nil > 0
empty₁ > false > [sndsplit₂, process₁, if1₁, if2₁, if3₁, self] > nil > 0
empty₁ > false > [app₂, map_f₁] > cons₂ > s₁ > nil > 0
leq > true > [sndsplit₂, process₁, if1₁, if2₁, if3₁, self] > nil > 0
leq > false > [sndsplit₂, process₁, if1₁, if2₁, if3₁, self] > nil > 0
leq > false > [app₂, map_f₁] > cons₂ > s₁ > nil > 0
length₁ > s₁ > nil > 0

Status:

empty₁: [1]
true: []
if3₁: [1]
if2₁: [1]
if1₁: [1]
0: []
sndsplit₂: [2,1]
cons₂: [2,1]
self: []
false: []
s₁: [1]
fstsplit₂: [1,2]
app₂: [1,2]
length₁: [1]
leq: []
map_f₁: [1]
nil: []
process₁: [1]

The following usable rules [FROCOS05] were oriented:

fstsplit(0, x) → nil
fstsplit(s(n), nil) → nil
fstsplit(s(n), cons(h, t)) → cons(h, fstsplit(n, t))
sndsplit(0, x) → x
sndsplit(s(n), nil) → nil
sndsplit(s(n), cons(h, t)) → sndsplit(n, t)
empty(nil) → true
empty(cons(h, t)) → false
leq(0, m) → true
leq(s(n), 0) → false
leq(s(n), s(m)) → leq(n, m)
length(nil) → 0
length(cons(h, t)) → s(length(t))
app(nil, x) → x
app(cons(h, t), x) → cons(h, app(t, x))
map_f(pid, nil) → nil
map_f(pid, cons(h, t)) → app(f(pid, h), map_f(pid, t))
process(store, m) → if1(store, m, leq(m, length(store)))
if1(store, m, true) → if2(store, m, empty(fstsplit(m, store)))
if1(store, m, false) → if3(store, m, empty(fstsplit(m, app(map_f(self, nil), store))))
if2(store, m, false) → process(app(map_f(self, nil), sndsplit(m, store)), m)
if3(store, m, false) → process(sndsplit(m, app(map_f(self, nil), store)), m)

(24) Obligation:

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

fstsplit(0, x) → nil
fstsplit(s(n), nil) → nil
fstsplit(s(n), cons(h, t)) → cons(h, fstsplit(n, t))
sndsplit(0, x) → x
sndsplit(s(n), nil) → nil
sndsplit(s(n), cons(h, t)) → sndsplit(n, t)
empty(nil) → true
empty(cons(h, t)) → false
leq(0, m) → true
leq(s(n), 0) → false
leq(s(n), s(m)) → leq(n, m)
length(nil) → 0
length(cons(h, t)) → s(length(t))
app(nil, x) → x
app(cons(h, t), x) → cons(h, app(t, x))
map_f(pid, nil) → nil
map_f(pid, cons(h, t)) → app(f(pid, h), map_f(pid, t))
process(store, m) → if1(store, m, leq(m, length(store)))
if1(store, m, true) → if2(store, m, empty(fstsplit(m, store)))
if1(store, m, false) → if3(store, m, empty(fstsplit(m, app(map_f(self, nil), store))))
if2(store, m, false) → process(app(map_f(self, nil), sndsplit(m, store)), m)
if3(store, m, false) → process(sndsplit(m, app(map_f(self, nil), store)), m)

The set Q consists of the following terms:

fstsplit(0, x0)
fstsplit(s(x0), nil)
fstsplit(s(x0), cons(x1, x2))
sndsplit(0, x0)
sndsplit(s(x0), nil)
sndsplit(s(x0), cons(x1, x2))
empty(nil)
empty(cons(x0, x1))
leq(0, x0)
leq(s(x0), 0)
leq(s(x0), s(x1))
length(nil)
length(cons(x0, x1))
app(nil, x0)
app(cons(x0, x1), x2)
map_f(x0, nil)
map_f(x0, cons(x1, x2))
process(x0, x1)
if1(x0, x1, true)
if1(x0, x1, false)
if2(x0, x1, false)
if3(x0, x1, false)

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

(25) PisEmptyProof (EQUIVALENT transformation)

The TRS P is empty. Hence, there is no (P,Q,R) chain.

(26) TRUE

(27) Obligation:

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

SNDSPLIT(s(n), cons(h, t)) → SNDSPLIT(n, t)

The TRS R consists of the following rules:

fstsplit(0, x) → nil
fstsplit(s(n), nil) → nil
fstsplit(s(n), cons(h, t)) → cons(h, fstsplit(n, t))
sndsplit(0, x) → x
sndsplit(s(n), nil) → nil
sndsplit(s(n), cons(h, t)) → sndsplit(n, t)
empty(nil) → true
empty(cons(h, t)) → false
leq(0, m) → true
leq(s(n), 0) → false
leq(s(n), s(m)) → leq(n, m)
length(nil) → 0
length(cons(h, t)) → s(length(t))
app(nil, x) → x
app(cons(h, t), x) → cons(h, app(t, x))
map_f(pid, nil) → nil
map_f(pid, cons(h, t)) → app(f(pid, h), map_f(pid, t))
process(store, m) → if1(store, m, leq(m, length(store)))
if1(store, m, true) → if2(store, m, empty(fstsplit(m, store)))
if1(store, m, false) → if3(store, m, empty(fstsplit(m, app(map_f(self, nil), store))))
if2(store, m, false) → process(app(map_f(self, nil), sndsplit(m, store)), m)
if3(store, m, false) → process(sndsplit(m, app(map_f(self, nil), store)), m)

The set Q consists of the following terms:

fstsplit(0, x0)
fstsplit(s(x0), nil)
fstsplit(s(x0), cons(x1, x2))
sndsplit(0, x0)
sndsplit(s(x0), nil)
sndsplit(s(x0), cons(x1, x2))
empty(nil)
empty(cons(x0, x1))
leq(0, x0)
leq(s(x0), 0)
leq(s(x0), s(x1))
length(nil)
length(cons(x0, x1))
app(nil, x0)
app(cons(x0, x1), x2)
map_f(x0, nil)
map_f(x0, cons(x1, x2))
process(x0, x1)
if1(x0, x1, true)
if1(x0, x1, false)
if2(x0, x1, false)
if3(x0, x1, false)

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

(28) QDPOrderProof (EQUIVALENT transformation)

We use the reduction pair processor [LPAR04].

The following pairs can be oriented strictly and are deleted.

SNDSPLIT(s(n), cons(h, t)) → SNDSPLIT(n, t)

The remaining pairs can at least be oriented weakly.
Used ordering: Combined order from the following AFS and order.
SNDSPLIT(x1, x2) = SNDSPLIT(x2)
s(x1) = x1
cons(x1, x2) = cons(x1, x2)
fstsplit(x1, x2) = fstsplit(x1, x2)
0 = 0
nil = nil
sndsplit(x1, x2) = sndsplit(x1, x2)
empty(x1) = empty(x1)
true = true
false = false
leq(x1, x2) = leq
length(x1) = length(x1)
app(x1, x2) = app(x1, x2)
map_f(x1, x2) = map_f(x1, x2)
f(x1, x2) = x2
process(x1, x2) = process
if1(x1, x2, x3) = x3
if2(x1, x2, x3) = if2
if3(x1, x2, x3) = if3
self = self

Lexicographic path order with status [LPO].
Quasi-Precedence:

[fstsplit₂, empty₁, true, false, leq, process, if2, if3, self] > nil > 0
[fstsplit₂, empty₁, true, false, leq, process, if2, if3, self] > sndsplit₂
[fstsplit₂, empty₁, true, false, leq, process, if2, if3, self] > length₁
[fstsplit₂, empty₁, true, false, leq, process, if2, if3, self] > map_f₂ > app₂ > cons₂

Status:

empty₁: [1]
if3: []
true: []
0: []
map_f₂: [1,2]
sndsplit₂: [1,2]
process: []
cons₂: [2,1]
self: []
false: []
fstsplit₂: [2,1]
app₂: [1,2]
SNDSPLIT₁: [1]
if2: []
length₁: [1]
leq: []
nil: []

The following usable rules [FROCOS05] were oriented:

fstsplit(0, x) → nil
fstsplit(s(n), nil) → nil
fstsplit(s(n), cons(h, t)) → cons(h, fstsplit(n, t))
sndsplit(0, x) → x
sndsplit(s(n), nil) → nil
sndsplit(s(n), cons(h, t)) → sndsplit(n, t)
empty(nil) → true
empty(cons(h, t)) → false
leq(0, m) → true
leq(s(n), 0) → false
leq(s(n), s(m)) → leq(n, m)
length(nil) → 0
length(cons(h, t)) → s(length(t))
app(nil, x) → x
app(cons(h, t), x) → cons(h, app(t, x))
map_f(pid, nil) → nil
map_f(pid, cons(h, t)) → app(f(pid, h), map_f(pid, t))
process(store, m) → if1(store, m, leq(m, length(store)))
if1(store, m, true) → if2(store, m, empty(fstsplit(m, store)))
if1(store, m, false) → if3(store, m, empty(fstsplit(m, app(map_f(self, nil), store))))
if2(store, m, false) → process(app(map_f(self, nil), sndsplit(m, store)), m)
if3(store, m, false) → process(sndsplit(m, app(map_f(self, nil), store)), m)

(29) Obligation:

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

fstsplit(0, x) → nil
fstsplit(s(n), nil) → nil
fstsplit(s(n), cons(h, t)) → cons(h, fstsplit(n, t))
sndsplit(0, x) → x
sndsplit(s(n), nil) → nil
sndsplit(s(n), cons(h, t)) → sndsplit(n, t)
empty(nil) → true
empty(cons(h, t)) → false
leq(0, m) → true
leq(s(n), 0) → false
leq(s(n), s(m)) → leq(n, m)
length(nil) → 0
length(cons(h, t)) → s(length(t))
app(nil, x) → x
app(cons(h, t), x) → cons(h, app(t, x))
map_f(pid, nil) → nil
map_f(pid, cons(h, t)) → app(f(pid, h), map_f(pid, t))
process(store, m) → if1(store, m, leq(m, length(store)))
if1(store, m, true) → if2(store, m, empty(fstsplit(m, store)))
if1(store, m, false) → if3(store, m, empty(fstsplit(m, app(map_f(self, nil), store))))
if2(store, m, false) → process(app(map_f(self, nil), sndsplit(m, store)), m)
if3(store, m, false) → process(sndsplit(m, app(map_f(self, nil), store)), m)

The set Q consists of the following terms:

fstsplit(0, x0)
fstsplit(s(x0), nil)
fstsplit(s(x0), cons(x1, x2))
sndsplit(0, x0)
sndsplit(s(x0), nil)
sndsplit(s(x0), cons(x1, x2))
empty(nil)
empty(cons(x0, x1))
leq(0, x0)
leq(s(x0), 0)
leq(s(x0), s(x1))
length(nil)
length(cons(x0, x1))
app(nil, x0)
app(cons(x0, x1), x2)
map_f(x0, nil)
map_f(x0, cons(x1, x2))
process(x0, x1)
if1(x0, x1, true)
if1(x0, x1, false)
if2(x0, x1, false)
if3(x0, x1, false)

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

(30) PisEmptyProof (EQUIVALENT transformation)

The TRS P is empty. Hence, there is no (P,Q,R) chain.

(31) TRUE

(32) Obligation:

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

FSTSPLIT(s(n), cons(h, t)) → FSTSPLIT(n, t)

The TRS R consists of the following rules:

fstsplit(0, x) → nil
fstsplit(s(n), nil) → nil
fstsplit(s(n), cons(h, t)) → cons(h, fstsplit(n, t))
sndsplit(0, x) → x
sndsplit(s(n), nil) → nil
sndsplit(s(n), cons(h, t)) → sndsplit(n, t)
empty(nil) → true
empty(cons(h, t)) → false
leq(0, m) → true
leq(s(n), 0) → false
leq(s(n), s(m)) → leq(n, m)
length(nil) → 0
length(cons(h, t)) → s(length(t))
app(nil, x) → x
app(cons(h, t), x) → cons(h, app(t, x))
map_f(pid, nil) → nil
map_f(pid, cons(h, t)) → app(f(pid, h), map_f(pid, t))
process(store, m) → if1(store, m, leq(m, length(store)))
if1(store, m, true) → if2(store, m, empty(fstsplit(m, store)))
if1(store, m, false) → if3(store, m, empty(fstsplit(m, app(map_f(self, nil), store))))
if2(store, m, false) → process(app(map_f(self, nil), sndsplit(m, store)), m)
if3(store, m, false) → process(sndsplit(m, app(map_f(self, nil), store)), m)

The set Q consists of the following terms:

fstsplit(0, x0)
fstsplit(s(x0), nil)
fstsplit(s(x0), cons(x1, x2))
sndsplit(0, x0)
sndsplit(s(x0), nil)
sndsplit(s(x0), cons(x1, x2))
empty(nil)
empty(cons(x0, x1))
leq(0, x0)
leq(s(x0), 0)
leq(s(x0), s(x1))
length(nil)
length(cons(x0, x1))
app(nil, x0)
app(cons(x0, x1), x2)
map_f(x0, nil)
map_f(x0, cons(x1, x2))
process(x0, x1)
if1(x0, x1, true)
if1(x0, x1, false)
if2(x0, x1, false)
if3(x0, x1, false)

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

(33) QDPOrderProof (EQUIVALENT transformation)

We use the reduction pair processor [LPAR04].

The following pairs can be oriented strictly and are deleted.

FSTSPLIT(s(n), cons(h, t)) → FSTSPLIT(n, t)

The remaining pairs can at least be oriented weakly.
Used ordering: Combined order from the following AFS and order.
FSTSPLIT(x1, x2) = x2
s(x1) = x1
cons(x1, x2) = cons(x1, x2)
fstsplit(x1, x2) = x2
0 = 0
nil = nil
sndsplit(x1, x2) = sndsplit(x1, x2)
empty(x1) = empty
true = true
false = false
leq(x1, x2) = leq
length(x1) = length
app(x1, x2) = app(x1, x2)
map_f(x1, x2) = map_f(x1, x2)
f(x1, x2) = x1
process(x1, x2) = process
if1(x1, x2, x3) = x3
if2(x1, x2, x3) = x3
if3(x1, x2, x3) = if3
self = self

Lexicographic path order with status [LPO].
Quasi-Precedence:

length > [0, nil, sndsplit₂, empty, true, false, leq, process, if3]
map_f₂ > app₂ > cons₂ > [0, nil, sndsplit₂, empty, true, false, leq, process, if3]
self > [0, nil, sndsplit₂, empty, true, false, leq, process, if3]

Status:

if3: []
true: []
empty: []
length: []
0: []
map_f₂: [2,1]
process: []
sndsplit₂: [1,2]
cons₂: [2,1]
self: []
false: []
app₂: [2,1]
leq: []
nil: []

The following usable rules [FROCOS05] were oriented:

fstsplit(0, x) → nil
fstsplit(s(n), nil) → nil
fstsplit(s(n), cons(h, t)) → cons(h, fstsplit(n, t))
sndsplit(0, x) → x
sndsplit(s(n), nil) → nil
sndsplit(s(n), cons(h, t)) → sndsplit(n, t)
empty(nil) → true
empty(cons(h, t)) → false
leq(0, m) → true
leq(s(n), 0) → false
leq(s(n), s(m)) → leq(n, m)
length(nil) → 0
length(cons(h, t)) → s(length(t))
app(nil, x) → x
app(cons(h, t), x) → cons(h, app(t, x))
map_f(pid, nil) → nil
map_f(pid, cons(h, t)) → app(f(pid, h), map_f(pid, t))
process(store, m) → if1(store, m, leq(m, length(store)))
if1(store, m, true) → if2(store, m, empty(fstsplit(m, store)))
if1(store, m, false) → if3(store, m, empty(fstsplit(m, app(map_f(self, nil), store))))
if2(store, m, false) → process(app(map_f(self, nil), sndsplit(m, store)), m)
if3(store, m, false) → process(sndsplit(m, app(map_f(self, nil), store)), m)

(34) Obligation:

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

fstsplit(0, x) → nil
fstsplit(s(n), nil) → nil
fstsplit(s(n), cons(h, t)) → cons(h, fstsplit(n, t))
sndsplit(0, x) → x
sndsplit(s(n), nil) → nil
sndsplit(s(n), cons(h, t)) → sndsplit(n, t)
empty(nil) → true
empty(cons(h, t)) → false
leq(0, m) → true
leq(s(n), 0) → false
leq(s(n), s(m)) → leq(n, m)
length(nil) → 0
length(cons(h, t)) → s(length(t))
app(nil, x) → x
app(cons(h, t), x) → cons(h, app(t, x))
map_f(pid, nil) → nil
map_f(pid, cons(h, t)) → app(f(pid, h), map_f(pid, t))
process(store, m) → if1(store, m, leq(m, length(store)))
if1(store, m, true) → if2(store, m, empty(fstsplit(m, store)))
if1(store, m, false) → if3(store, m, empty(fstsplit(m, app(map_f(self, nil), store))))
if2(store, m, false) → process(app(map_f(self, nil), sndsplit(m, store)), m)
if3(store, m, false) → process(sndsplit(m, app(map_f(self, nil), store)), m)

The set Q consists of the following terms:

fstsplit(0, x0)
fstsplit(s(x0), nil)
fstsplit(s(x0), cons(x1, x2))
sndsplit(0, x0)
sndsplit(s(x0), nil)
sndsplit(s(x0), cons(x1, x2))
empty(nil)
empty(cons(x0, x1))
leq(0, x0)
leq(s(x0), 0)
leq(s(x0), s(x1))
length(nil)
length(cons(x0, x1))
app(nil, x0)
app(cons(x0, x1), x2)
map_f(x0, nil)
map_f(x0, cons(x1, x2))
process(x0, x1)
if1(x0, x1, true)
if1(x0, x1, false)
if2(x0, x1, false)
if3(x0, x1, false)

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

(35) PisEmptyProof (EQUIVALENT transformation)

The TRS P is empty. Hence, there is no (P,Q,R) chain.

(36) TRUE

(37) Obligation:

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

IF1(store, m, true) → IF2(store, m, empty(fstsplit(m, store)))
IF2(store, m, false) → PROCESS(app(map_f(self, nil), sndsplit(m, store)), m)
PROCESS(store, m) → IF1(store, m, leq(m, length(store)))
IF1(store, m, false) → IF3(store, m, empty(fstsplit(m, app(map_f(self, nil), store))))
IF3(store, m, false) → PROCESS(sndsplit(m, app(map_f(self, nil), store)), m)

The TRS R consists of the following rules:

fstsplit(0, x) → nil
fstsplit(s(n), nil) → nil
fstsplit(s(n), cons(h, t)) → cons(h, fstsplit(n, t))
sndsplit(0, x) → x
sndsplit(s(n), nil) → nil
sndsplit(s(n), cons(h, t)) → sndsplit(n, t)
empty(nil) → true
empty(cons(h, t)) → false
leq(0, m) → true
leq(s(n), 0) → false
leq(s(n), s(m)) → leq(n, m)
length(nil) → 0
length(cons(h, t)) → s(length(t))
app(nil, x) → x
app(cons(h, t), x) → cons(h, app(t, x))
map_f(pid, nil) → nil
map_f(pid, cons(h, t)) → app(f(pid, h), map_f(pid, t))
process(store, m) → if1(store, m, leq(m, length(store)))
if1(store, m, true) → if2(store, m, empty(fstsplit(m, store)))
if1(store, m, false) → if3(store, m, empty(fstsplit(m, app(map_f(self, nil), store))))
if2(store, m, false) → process(app(map_f(self, nil), sndsplit(m, store)), m)
if3(store, m, false) → process(sndsplit(m, app(map_f(self, nil), store)), m)

The set Q consists of the following terms:

fstsplit(0, x0)
fstsplit(s(x0), nil)
fstsplit(s(x0), cons(x1, x2))
sndsplit(0, x0)
sndsplit(s(x0), nil)
sndsplit(s(x0), cons(x1, x2))
empty(nil)
empty(cons(x0, x1))
leq(0, x0)
leq(s(x0), 0)
leq(s(x0), s(x1))
length(nil)
length(cons(x0, x1))
app(nil, x0)
app(cons(x0, x1), x2)
map_f(x0, nil)
map_f(x0, cons(x1, x2))
process(x0, x1)
if1(x0, x1, true)
if1(x0, x1, false)
if2(x0, x1, false)
if3(x0, x1, false)

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