Termination w.r.t. Q proof of /tmp/tmpgskjsw/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) = APP(x1)
cons(x1, x2) = cons(x1, x2)

Lexicographic path order with status [LPO].
Precedence:

cons₂ > APP₁

Status:

APP₁: [1]
cons₂: [1,2]

The following usable rules [FROCOS05] were oriented: none

(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: Lexicographic path order with status [LPO].
Precedence:

cons₂ > MAP_F₂

Status:

cons₂: [1,2]
MAP_F₂: [1,2]

The following usable rules [FROCOS05] were oriented: none

(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) = x1
cons(x1, x2) = cons(x1, x2)

Lexicographic path order with status [LPO].
Precedence:

trivial

Status:

cons₂: [1,2]

The following usable rules [FROCOS05] were oriented: none

(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) = LEQ(x1)
s(x1) = s(x1)

Lexicographic path order with status [LPO].
Precedence:

s₁ > LEQ₁

Status:

LEQ₁: [1]
s₁: [1]

The following usable rules [FROCOS05] were oriented: none

(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(x1)
s(x1) = s(x1)
cons(x1, x2) = cons(x1, x2)

Lexicographic path order with status [LPO].
Precedence:

s₁ > SNDSPLIT₁
cons₂ > SNDSPLIT₁

Status:

cons₂: [1,2]
s₁: [1]
SNDSPLIT₁: [1]

The following usable rules [FROCOS05] were oriented: none

(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) = FSTSPLIT(x1)
s(x1) = s(x1)
cons(x1, x2) = cons(x1, x2)

Lexicographic path order with status [LPO].
Precedence:

s₁ > FSTSPLIT₁
cons₂ > FSTSPLIT₁

Status:

cons₂: [1,2]
FSTSPLIT₁: [1]
s₁: [1]

The following usable rules [FROCOS05] were oriented: none

(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.