Term Rewriting System R:
[x, n, h, t, m, pid, store]
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))
mapf(pid, nil) -> nil
mapf(pid, cons(h, t)) -> app(f(pid, h), mapf(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(mapf(self, nil), store))))
if2(store, m, false) -> process(app(mapf(self, nil), sndsplit(m, store)), m)
if3(store, m, false) -> process(sndsplit(m, app(mapf(self, nil), store)), m)

Innermost Termination of R to be shown.



   R
Dependency Pair Analysis



R contains the following Dependency Pairs:

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)
MAPF(pid, cons(h, t)) -> APP(f(pid, h), mapf(pid, t))
MAPF(pid, cons(h, t)) -> MAPF(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(mapf(self, nil), store))))
IF1(store, m, false) -> EMPTY(fstsplit(m, app(mapf(self, nil), store)))
IF1(store, m, false) -> FSTSPLIT(m, app(mapf(self, nil), store))
IF1(store, m, false) -> APP(mapf(self, nil), store)
IF1(store, m, false) -> MAPF(self, nil)
IF2(store, m, false) -> PROCESS(app(mapf(self, nil), sndsplit(m, store)), m)
IF2(store, m, false) -> APP(mapf(self, nil), sndsplit(m, store))
IF2(store, m, false) -> MAPF(self, nil)
IF2(store, m, false) -> SNDSPLIT(m, store)
IF3(store, m, false) -> PROCESS(sndsplit(m, app(mapf(self, nil), store)), m)
IF3(store, m, false) -> SNDSPLIT(m, app(mapf(self, nil), store))
IF3(store, m, false) -> APP(mapf(self, nil), store)
IF3(store, m, false) -> MAPF(self, nil)

Furthermore, R contains seven SCCs.


   R
DPs
       →DP Problem 1
Polynomial Ordering
       →DP Problem 2
Polo
       →DP Problem 3
Polo
       →DP Problem 4
Polo
       →DP Problem 5
Polo
       →DP Problem 6
Polo
       →DP Problem 7
Rw


Dependency Pair:

FSTSPLIT(s(n), cons(h, t)) -> FSTSPLIT(n, t)


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))
mapf(pid, nil) -> nil
mapf(pid, cons(h, t)) -> app(f(pid, h), mapf(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(mapf(self, nil), store))))
if2(store, m, false) -> process(app(mapf(self, nil), sndsplit(m, store)), m)
if3(store, m, false) -> process(sndsplit(m, app(mapf(self, nil), store)), m)


Strategy:

innermost




The following dependency pair can be strictly oriented:

FSTSPLIT(s(n), cons(h, t)) -> FSTSPLIT(n, t)


There are no usable rules for innermost that need to be oriented.

Used ordering: Polynomial ordering with Polynomial interpretation:
  POL(cons(x1, x2))=  0  
  POL(FSTSPLIT(x1, x2))=  x1  
  POL(s(x1))=  1 + x1  

resulting in one new DP problem.



   R
DPs
       →DP Problem 1
Polo
           →DP Problem 8
Dependency Graph
       →DP Problem 2
Polo
       →DP Problem 3
Polo
       →DP Problem 4
Polo
       →DP Problem 5
Polo
       →DP Problem 6
Polo
       →DP Problem 7
Rw


Dependency Pair:


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))
mapf(pid, nil) -> nil
mapf(pid, cons(h, t)) -> app(f(pid, h), mapf(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(mapf(self, nil), store))))
if2(store, m, false) -> process(app(mapf(self, nil), sndsplit(m, store)), m)
if3(store, m, false) -> process(sndsplit(m, app(mapf(self, nil), store)), m)


Strategy:

innermost




Using the Dependency Graph resulted in no new DP problems.


   R
DPs
       →DP Problem 1
Polo
       →DP Problem 2
Polynomial Ordering
       →DP Problem 3
Polo
       →DP Problem 4
Polo
       →DP Problem 5
Polo
       →DP Problem 6
Polo
       →DP Problem 7
Rw


Dependency Pair:

SNDSPLIT(s(n), cons(h, t)) -> SNDSPLIT(n, t)


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))
mapf(pid, nil) -> nil
mapf(pid, cons(h, t)) -> app(f(pid, h), mapf(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(mapf(self, nil), store))))
if2(store, m, false) -> process(app(mapf(self, nil), sndsplit(m, store)), m)
if3(store, m, false) -> process(sndsplit(m, app(mapf(self, nil), store)), m)


Strategy:

innermost




The following dependency pair can be strictly oriented:

SNDSPLIT(s(n), cons(h, t)) -> SNDSPLIT(n, t)


There are no usable rules for innermost that need to be oriented.

Used ordering: Polynomial ordering with Polynomial interpretation:
  POL(SNDSPLIT(x1, x2))=  x1  
  POL(cons(x1, x2))=  0  
  POL(s(x1))=  1 + x1  

resulting in one new DP problem.



   R
DPs
       →DP Problem 1
Polo
       →DP Problem 2
Polo
           →DP Problem 9
Dependency Graph
       →DP Problem 3
Polo
       →DP Problem 4
Polo
       →DP Problem 5
Polo
       →DP Problem 6
Polo
       →DP Problem 7
Rw


Dependency Pair:


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))
mapf(pid, nil) -> nil
mapf(pid, cons(h, t)) -> app(f(pid, h), mapf(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(mapf(self, nil), store))))
if2(store, m, false) -> process(app(mapf(self, nil), sndsplit(m, store)), m)
if3(store, m, false) -> process(sndsplit(m, app(mapf(self, nil), store)), m)


Strategy:

innermost




Using the Dependency Graph resulted in no new DP problems.


   R
DPs
       →DP Problem 1
Polo
       →DP Problem 2
Polo
       →DP Problem 3
Polynomial Ordering
       →DP Problem 4
Polo
       →DP Problem 5
Polo
       →DP Problem 6
Polo
       →DP Problem 7
Rw


Dependency Pair:

LEQ(s(n), s(m)) -> LEQ(n, m)


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))
mapf(pid, nil) -> nil
mapf(pid, cons(h, t)) -> app(f(pid, h), mapf(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(mapf(self, nil), store))))
if2(store, m, false) -> process(app(mapf(self, nil), sndsplit(m, store)), m)
if3(store, m, false) -> process(sndsplit(m, app(mapf(self, nil), store)), m)


Strategy:

innermost




The following dependency pair can be strictly oriented:

LEQ(s(n), s(m)) -> LEQ(n, m)


There are no usable rules for innermost that need to be oriented.

Used ordering: Polynomial ordering with Polynomial interpretation:
  POL(LEQ(x1, x2))=  x1  
  POL(s(x1))=  1 + x1  

resulting in one new DP problem.



   R
DPs
       →DP Problem 1
Polo
       →DP Problem 2
Polo
       →DP Problem 3
Polo
           →DP Problem 10
Dependency Graph
       →DP Problem 4
Polo
       →DP Problem 5
Polo
       →DP Problem 6
Polo
       →DP Problem 7
Rw


Dependency Pair:


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))
mapf(pid, nil) -> nil
mapf(pid, cons(h, t)) -> app(f(pid, h), mapf(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(mapf(self, nil), store))))
if2(store, m, false) -> process(app(mapf(self, nil), sndsplit(m, store)), m)
if3(store, m, false) -> process(sndsplit(m, app(mapf(self, nil), store)), m)


Strategy:

innermost




Using the Dependency Graph resulted in no new DP problems.


   R
DPs
       →DP Problem 1
Polo
       →DP Problem 2
Polo
       →DP Problem 3
Polo
       →DP Problem 4
Polynomial Ordering
       →DP Problem 5
Polo
       →DP Problem 6
Polo
       →DP Problem 7
Rw


Dependency Pair:

LENGTH(cons(h, t)) -> LENGTH(t)


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))
mapf(pid, nil) -> nil
mapf(pid, cons(h, t)) -> app(f(pid, h), mapf(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(mapf(self, nil), store))))
if2(store, m, false) -> process(app(mapf(self, nil), sndsplit(m, store)), m)
if3(store, m, false) -> process(sndsplit(m, app(mapf(self, nil), store)), m)


Strategy:

innermost




The following dependency pair can be strictly oriented:

LENGTH(cons(h, t)) -> LENGTH(t)


There are no usable rules for innermost that need to be oriented.

Used ordering: Polynomial ordering with Polynomial interpretation:
  POL(cons(x1, x2))=  1 + x2  
  POL(LENGTH(x1))=  x1  

resulting in one new DP problem.



   R
DPs
       →DP Problem 1
Polo
       →DP Problem 2
Polo
       →DP Problem 3
Polo
       →DP Problem 4
Polo
           →DP Problem 11
Dependency Graph
       →DP Problem 5
Polo
       →DP Problem 6
Polo
       →DP Problem 7
Rw


Dependency Pair:


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))
mapf(pid, nil) -> nil
mapf(pid, cons(h, t)) -> app(f(pid, h), mapf(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(mapf(self, nil), store))))
if2(store, m, false) -> process(app(mapf(self, nil), sndsplit(m, store)), m)
if3(store, m, false) -> process(sndsplit(m, app(mapf(self, nil), store)), m)


Strategy:

innermost




Using the Dependency Graph resulted in no new DP problems.


   R
DPs
       →DP Problem 1
Polo
       →DP Problem 2
Polo
       →DP Problem 3
Polo
       →DP Problem 4
Polo
       →DP Problem 5
Polynomial Ordering
       →DP Problem 6
Polo
       →DP Problem 7
Rw


Dependency Pair:

APP(cons(h, t), x) -> APP(t, x)


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))
mapf(pid, nil) -> nil
mapf(pid, cons(h, t)) -> app(f(pid, h), mapf(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(mapf(self, nil), store))))
if2(store, m, false) -> process(app(mapf(self, nil), sndsplit(m, store)), m)
if3(store, m, false) -> process(sndsplit(m, app(mapf(self, nil), store)), m)


Strategy:

innermost




The following dependency pair can be strictly oriented:

APP(cons(h, t), x) -> APP(t, x)


There are no usable rules for innermost that need to be oriented.

Used ordering: Polynomial ordering with Polynomial interpretation:
  POL(cons(x1, x2))=  1 + x2  
  POL(APP(x1, x2))=  x1  

resulting in one new DP problem.



   R
DPs
       →DP Problem 1
Polo
       →DP Problem 2
Polo
       →DP Problem 3
Polo
       →DP Problem 4
Polo
       →DP Problem 5
Polo
           →DP Problem 12
Dependency Graph
       →DP Problem 6
Polo
       →DP Problem 7
Rw


Dependency Pair:


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))
mapf(pid, nil) -> nil
mapf(pid, cons(h, t)) -> app(f(pid, h), mapf(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(mapf(self, nil), store))))
if2(store, m, false) -> process(app(mapf(self, nil), sndsplit(m, store)), m)
if3(store, m, false) -> process(sndsplit(m, app(mapf(self, nil), store)), m)


Strategy:

innermost




Using the Dependency Graph resulted in no new DP problems.


   R
DPs
       →DP Problem 1
Polo
       →DP Problem 2
Polo
       →DP Problem 3
Polo
       →DP Problem 4
Polo
       →DP Problem 5
Polo
       →DP Problem 6
Polynomial Ordering
       →DP Problem 7
Rw


Dependency Pair:

MAPF(pid, cons(h, t)) -> MAPF(pid, t)


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))
mapf(pid, nil) -> nil
mapf(pid, cons(h, t)) -> app(f(pid, h), mapf(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(mapf(self, nil), store))))
if2(store, m, false) -> process(app(mapf(self, nil), sndsplit(m, store)), m)
if3(store, m, false) -> process(sndsplit(m, app(mapf(self, nil), store)), m)


Strategy:

innermost




The following dependency pair can be strictly oriented:

MAPF(pid, cons(h, t)) -> MAPF(pid, t)


There are no usable rules for innermost that need to be oriented.

Used ordering: Polynomial ordering with Polynomial interpretation:
  POL(MAP_F(x1, x2))=  x2  
  POL(cons(x1, x2))=  1 + x2  

resulting in one new DP problem.



   R
DPs
       →DP Problem 1
Polo
       →DP Problem 2
Polo
       →DP Problem 3
Polo
       →DP Problem 4
Polo
       →DP Problem 5
Polo
       →DP Problem 6
Polo
           →DP Problem 13
Dependency Graph
       →DP Problem 7
Rw


Dependency Pair:


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))
mapf(pid, nil) -> nil
mapf(pid, cons(h, t)) -> app(f(pid, h), mapf(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(mapf(self, nil), store))))
if2(store, m, false) -> process(app(mapf(self, nil), sndsplit(m, store)), m)
if3(store, m, false) -> process(sndsplit(m, app(mapf(self, nil), store)), m)


Strategy:

innermost




Using the Dependency Graph resulted in no new DP problems.


   R
DPs
       →DP Problem 1
Polo
       →DP Problem 2
Polo
       →DP Problem 3
Polo
       →DP Problem 4
Polo
       →DP Problem 5
Polo
       →DP Problem 6
Polo
       →DP Problem 7
Rewriting Transformation


Dependency Pairs:

IF3(store, m, false) -> PROCESS(sndsplit(m, app(mapf(self, nil), store)), m)
IF1(store, m, false) -> IF3(store, m, empty(fstsplit(m, app(mapf(self, nil), store))))
IF2(store, m, false) -> PROCESS(app(mapf(self, nil), sndsplit(m, store)), m)
IF1(store, m, true) -> IF2(store, m, empty(fstsplit(m, store)))
PROCESS(store, m) -> IF1(store, m, leq(m, length(store)))


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))
mapf(pid, nil) -> nil
mapf(pid, cons(h, t)) -> app(f(pid, h), mapf(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(mapf(self, nil), store))))
if2(store, m, false) -> process(app(mapf(self, nil), sndsplit(m, store)), m)
if3(store, m, false) -> process(sndsplit(m, app(mapf(self, nil), store)), m)


Strategy:

innermost




On this DP problem, a Rewriting SCC transformation can be performed.
As a result of transforming the rule

IF1(store, m, false) -> IF3(store, m, empty(fstsplit(m, app(mapf(self, nil), store))))
one new Dependency Pair is created:

IF1(store, m, false) -> IF3(store, m, empty(fstsplit(m, app(nil, store))))

The transformation is resulting in one new DP problem:



   R
DPs
       →DP Problem 1
Polo
       →DP Problem 2
Polo
       →DP Problem 3
Polo
       →DP Problem 4
Polo
       →DP Problem 5
Polo
       →DP Problem 6
Polo
       →DP Problem 7
Rw
           →DP Problem 14
Rewriting Transformation


Dependency Pairs:

IF1(store, m, false) -> IF3(store, m, empty(fstsplit(m, app(nil, store))))
IF2(store, m, false) -> PROCESS(app(mapf(self, nil), sndsplit(m, store)), m)
IF1(store, m, true) -> IF2(store, m, empty(fstsplit(m, store)))
PROCESS(store, m) -> IF1(store, m, leq(m, length(store)))
IF3(store, m, false) -> PROCESS(sndsplit(m, app(mapf(self, nil), store)), m)


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))
mapf(pid, nil) -> nil
mapf(pid, cons(h, t)) -> app(f(pid, h), mapf(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(mapf(self, nil), store))))
if2(store, m, false) -> process(app(mapf(self, nil), sndsplit(m, store)), m)
if3(store, m, false) -> process(sndsplit(m, app(mapf(self, nil), store)), m)


Strategy:

innermost




On this DP problem, a Rewriting SCC transformation can be performed.
As a result of transforming the rule

IF2(store, m, false) -> PROCESS(app(mapf(self, nil), sndsplit(m, store)), m)
one new Dependency Pair is created:

IF2(store, m, false) -> PROCESS(app(nil, sndsplit(m, store)), m)

The transformation is resulting in one new DP problem:



   R
DPs
       →DP Problem 1
Polo
       →DP Problem 2
Polo
       →DP Problem 3
Polo
       →DP Problem 4
Polo
       →DP Problem 5
Polo
       →DP Problem 6
Polo
       →DP Problem 7
Rw
           →DP Problem 14
Rw
             ...
               →DP Problem 15
Rewriting Transformation


Dependency Pairs:

IF2(store, m, false) -> PROCESS(app(nil, sndsplit(m, store)), m)
IF1(store, m, true) -> IF2(store, m, empty(fstsplit(m, store)))
PROCESS(store, m) -> IF1(store, m, leq(m, length(store)))
IF3(store, m, false) -> PROCESS(sndsplit(m, app(mapf(self, nil), store)), m)
IF1(store, m, false) -> IF3(store, m, empty(fstsplit(m, app(nil, store))))


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))
mapf(pid, nil) -> nil
mapf(pid, cons(h, t)) -> app(f(pid, h), mapf(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(mapf(self, nil), store))))
if2(store, m, false) -> process(app(mapf(self, nil), sndsplit(m, store)), m)
if3(store, m, false) -> process(sndsplit(m, app(mapf(self, nil), store)), m)


Strategy:

innermost




On this DP problem, a Rewriting SCC transformation can be performed.
As a result of transforming the rule

IF3(store, m, false) -> PROCESS(sndsplit(m, app(mapf(self, nil), store)), m)
one new Dependency Pair is created:

IF3(store, m, false) -> PROCESS(sndsplit(m, app(nil, store)), m)

The transformation is resulting in one new DP problem:



   R
DPs
       →DP Problem 1
Polo
       →DP Problem 2
Polo
       →DP Problem 3
Polo
       →DP Problem 4
Polo
       →DP Problem 5
Polo
       →DP Problem 6
Polo
       →DP Problem 7
Rw
           →DP Problem 14
Rw
             ...
               →DP Problem 16
Rewriting Transformation


Dependency Pairs:

IF3(store, m, false) -> PROCESS(sndsplit(m, app(nil, store)), m)
IF1(store, m, false) -> IF3(store, m, empty(fstsplit(m, app(nil, store))))
IF1(store, m, true) -> IF2(store, m, empty(fstsplit(m, store)))
PROCESS(store, m) -> IF1(store, m, leq(m, length(store)))
IF2(store, m, false) -> PROCESS(app(nil, sndsplit(m, store)), m)


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))
mapf(pid, nil) -> nil
mapf(pid, cons(h, t)) -> app(f(pid, h), mapf(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(mapf(self, nil), store))))
if2(store, m, false) -> process(app(mapf(self, nil), sndsplit(m, store)), m)
if3(store, m, false) -> process(sndsplit(m, app(mapf(self, nil), store)), m)


Strategy:

innermost




On this DP problem, a Rewriting SCC transformation can be performed.
As a result of transforming the rule

IF1(store, m, false) -> IF3(store, m, empty(fstsplit(m, app(nil, store))))
one new Dependency Pair is created:

IF1(store, m, false) -> IF3(store, m, empty(fstsplit(m, store)))

The transformation is resulting in one new DP problem:



   R
DPs
       →DP Problem 1
Polo
       →DP Problem 2
Polo
       →DP Problem 3
Polo
       →DP Problem 4
Polo
       →DP Problem 5
Polo
       →DP Problem 6
Polo
       →DP Problem 7
Rw
           →DP Problem 14
Rw
             ...
               →DP Problem 17
Rewriting Transformation


Dependency Pairs:

IF1(store, m, false) -> IF3(store, m, empty(fstsplit(m, store)))
IF2(store, m, false) -> PROCESS(app(nil, sndsplit(m, store)), m)
IF1(store, m, true) -> IF2(store, m, empty(fstsplit(m, store)))
PROCESS(store, m) -> IF1(store, m, leq(m, length(store)))
IF3(store, m, false) -> PROCESS(sndsplit(m, app(nil, store)), m)


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))
mapf(pid, nil) -> nil
mapf(pid, cons(h, t)) -> app(f(pid, h), mapf(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(mapf(self, nil), store))))
if2(store, m, false) -> process(app(mapf(self, nil), sndsplit(m, store)), m)
if3(store, m, false) -> process(sndsplit(m, app(mapf(self, nil), store)), m)


Strategy:

innermost




On this DP problem, a Rewriting SCC transformation can be performed.
As a result of transforming the rule

IF2(store, m, false) -> PROCESS(app(nil, sndsplit(m, store)), m)
one new Dependency Pair is created:

IF2(store, m, false) -> PROCESS(sndsplit(m, store), m)

The transformation is resulting in one new DP problem:



   R
DPs
       →DP Problem 1
Polo
       →DP Problem 2
Polo
       →DP Problem 3
Polo
       →DP Problem 4
Polo
       →DP Problem 5
Polo
       →DP Problem 6
Polo
       →DP Problem 7
Rw
           →DP Problem 14
Rw
             ...
               →DP Problem 18
Rewriting Transformation


Dependency Pairs:

IF2(store, m, false) -> PROCESS(sndsplit(m, store), m)
IF1(store, m, true) -> IF2(store, m, empty(fstsplit(m, store)))
PROCESS(store, m) -> IF1(store, m, leq(m, length(store)))
IF3(store, m, false) -> PROCESS(sndsplit(m, app(nil, store)), m)
IF1(store, m, false) -> IF3(store, m, empty(fstsplit(m, store)))


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))
mapf(pid, nil) -> nil
mapf(pid, cons(h, t)) -> app(f(pid, h), mapf(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(mapf(self, nil), store))))
if2(store, m, false) -> process(app(mapf(self, nil), sndsplit(m, store)), m)
if3(store, m, false) -> process(sndsplit(m, app(mapf(self, nil), store)), m)


Strategy:

innermost




On this DP problem, a Rewriting SCC transformation can be performed.
As a result of transforming the rule

IF3(store, m, false) -> PROCESS(sndsplit(m, app(nil, store)), m)
one new Dependency Pair is created:

IF3(store, m, false) -> PROCESS(sndsplit(m, store), m)

The transformation is resulting in one new DP problem:



   R
DPs
       →DP Problem 1
Polo
       →DP Problem 2
Polo
       →DP Problem 3
Polo
       →DP Problem 4
Polo
       →DP Problem 5
Polo
       →DP Problem 6
Polo
       →DP Problem 7
Rw
           →DP Problem 14
Rw
             ...
               →DP Problem 19
Narrowing Transformation


Dependency Pairs:

IF3(store, m, false) -> PROCESS(sndsplit(m, store), m)
IF1(store, m, false) -> IF3(store, m, empty(fstsplit(m, store)))
IF1(store, m, true) -> IF2(store, m, empty(fstsplit(m, store)))
PROCESS(store, m) -> IF1(store, m, leq(m, length(store)))
IF2(store, m, false) -> PROCESS(sndsplit(m, store), m)


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))
mapf(pid, nil) -> nil
mapf(pid, cons(h, t)) -> app(f(pid, h), mapf(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(mapf(self, nil), store))))
if2(store, m, false) -> process(app(mapf(self, nil), sndsplit(m, store)), m)
if3(store, m, false) -> process(sndsplit(m, app(mapf(self, nil), store)), m)


Strategy:

innermost




On this DP problem, a Narrowing SCC transformation can be performed.
As a result of transforming the rule

PROCESS(store, m) -> IF1(store, m, leq(m, length(store)))
three new Dependency Pairs are created:

PROCESS(store', 0) -> IF1(store', 0, true)
PROCESS(nil, m) -> IF1(nil, m, leq(m, 0))
PROCESS(cons(h', t'), m) -> IF1(cons(h', t'), m, leq(m, s(length(t'))))

The transformation is resulting in one new DP problem:



   R
DPs
       →DP Problem 1
Polo
       →DP Problem 2
Polo
       →DP Problem 3
Polo
       →DP Problem 4
Polo
       →DP Problem 5
Polo
       →DP Problem 6
Polo
       →DP Problem 7
Rw
           →DP Problem 14
Rw
             ...
               →DP Problem 20
Narrowing Transformation


Dependency Pairs:

PROCESS(cons(h', t'), m) -> IF1(cons(h', t'), m, leq(m, s(length(t'))))
IF1(store, m, false) -> IF3(store, m, empty(fstsplit(m, store)))
PROCESS(nil, m) -> IF1(nil, m, leq(m, 0))
IF2(store, m, false) -> PROCESS(sndsplit(m, store), m)
IF1(store, m, true) -> IF2(store, m, empty(fstsplit(m, store)))
PROCESS(store', 0) -> IF1(store', 0, true)
IF3(store, m, false) -> PROCESS(sndsplit(m, store), m)


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))
mapf(pid, nil) -> nil
mapf(pid, cons(h, t)) -> app(f(pid, h), mapf(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(mapf(self, nil), store))))
if2(store, m, false) -> process(app(mapf(self, nil), sndsplit(m, store)), m)
if3(store, m, false) -> process(sndsplit(m, app(mapf(self, nil), store)), m)


Strategy:

innermost




On this DP problem, a Narrowing SCC transformation can be performed.
As a result of transforming the rule

IF1(store, m, true) -> IF2(store, m, empty(fstsplit(m, store)))
three new Dependency Pairs are created:

IF1(store', 0, true) -> IF2(store', 0, empty(nil))
IF1(nil, s(n'), true) -> IF2(nil, s(n'), empty(nil))
IF1(cons(h', t'), s(n'), true) -> IF2(cons(h', t'), s(n'), empty(cons(h', fstsplit(n', t'))))

The transformation is resulting in one new DP problem:



   R
DPs
       →DP Problem 1
Polo
       →DP Problem 2
Polo
       →DP Problem 3
Polo
       →DP Problem 4
Polo
       →DP Problem 5
Polo
       →DP Problem 6
Polo
       →DP Problem 7
Rw
           →DP Problem 14
Rw
             ...
               →DP Problem 21
Rewriting Transformation


Dependency Pairs:

IF1(cons(h', t'), s(n'), true) -> IF2(cons(h', t'), s(n'), empty(cons(h', fstsplit(n', t'))))
IF1(nil, s(n'), true) -> IF2(nil, s(n'), empty(nil))
PROCESS(nil, m) -> IF1(nil, m, leq(m, 0))
IF2(store, m, false) -> PROCESS(sndsplit(m, store), m)
IF1(store', 0, true) -> IF2(store', 0, empty(nil))
PROCESS(store', 0) -> IF1(store', 0, true)
IF3(store, m, false) -> PROCESS(sndsplit(m, store), m)
IF1(store, m, false) -> IF3(store, m, empty(fstsplit(m, store)))
PROCESS(cons(h', t'), m) -> IF1(cons(h', t'), m, leq(m, s(length(t'))))


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))
mapf(pid, nil) -> nil
mapf(pid, cons(h, t)) -> app(f(pid, h), mapf(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(mapf(self, nil), store))))
if2(store, m, false) -> process(app(mapf(self, nil), sndsplit(m, store)), m)
if3(store, m, false) -> process(sndsplit(m, app(mapf(self, nil), store)), m)


Strategy:

innermost




On this DP problem, a Rewriting SCC transformation can be performed.
As a result of transforming the rule

IF1(store', 0, true) -> IF2(store', 0, empty(nil))
one new Dependency Pair is created:

IF1(store', 0, true) -> IF2(store', 0, true)

The transformation is resulting in one new DP problem:



   R
DPs
       →DP Problem 1
Polo
       →DP Problem 2
Polo
       →DP Problem 3
Polo
       →DP Problem 4
Polo
       →DP Problem 5
Polo
       →DP Problem 6
Polo
       →DP Problem 7
Rw
           →DP Problem 14
Rw
             ...
               →DP Problem 22
Rewriting Transformation


Dependency Pairs:

IF1(nil, s(n'), true) -> IF2(nil, s(n'), empty(nil))
PROCESS(cons(h', t'), m) -> IF1(cons(h', t'), m, leq(m, s(length(t'))))
IF3(store, m, false) -> PROCESS(sndsplit(m, store), m)
IF1(store, m, false) -> IF3(store, m, empty(fstsplit(m, store)))
PROCESS(nil, m) -> IF1(nil, m, leq(m, 0))
IF2(store, m, false) -> PROCESS(sndsplit(m, store), m)
IF1(cons(h', t'), s(n'), true) -> IF2(cons(h', t'), s(n'), empty(cons(h', fstsplit(n', t'))))


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))
mapf(pid, nil) -> nil
mapf(pid, cons(h, t)) -> app(f(pid, h), mapf(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(mapf(self, nil), store))))
if2(store, m, false) -> process(app(mapf(self, nil), sndsplit(m, store)), m)
if3(store, m, false) -> process(sndsplit(m, app(mapf(self, nil), store)), m)


Strategy:

innermost




On this DP problem, a Rewriting SCC transformation can be performed.
As a result of transforming the rule

IF1(nil, s(n'), true) -> IF2(nil, s(n'), empty(nil))
one new Dependency Pair is created:

IF1(nil, s(n'), true) -> IF2(nil, s(n'), true)

The transformation is resulting in one new DP problem:



   R
DPs
       →DP Problem 1
Polo
       →DP Problem 2
Polo
       →DP Problem 3
Polo
       →DP Problem 4
Polo
       →DP Problem 5
Polo
       →DP Problem 6
Polo
       →DP Problem 7
Rw
           →DP Problem 14
Rw
             ...
               →DP Problem 23
Rewriting Transformation


Dependency Pairs:

IF2(store, m, false) -> PROCESS(sndsplit(m, store), m)
IF1(cons(h', t'), s(n'), true) -> IF2(cons(h', t'), s(n'), empty(cons(h', fstsplit(n', t'))))
PROCESS(nil, m) -> IF1(nil, m, leq(m, 0))
IF3(store, m, false) -> PROCESS(sndsplit(m, store), m)
IF1(store, m, false) -> IF3(store, m, empty(fstsplit(m, store)))
PROCESS(cons(h', t'), m) -> IF1(cons(h', t'), m, leq(m, s(length(t'))))


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))
mapf(pid, nil) -> nil
mapf(pid, cons(h, t)) -> app(f(pid, h), mapf(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(mapf(self, nil), store))))
if2(store, m, false) -> process(app(mapf(self, nil), sndsplit(m, store)), m)
if3(store, m, false) -> process(sndsplit(m, app(mapf(self, nil), store)), m)


Strategy:

innermost




On this DP problem, a Rewriting SCC transformation can be performed.
As a result of transforming the rule

IF1(cons(h', t'), s(n'), true) -> IF2(cons(h', t'), s(n'), empty(cons(h', fstsplit(n', t'))))
one new Dependency Pair is created:

IF1(cons(h', t'), s(n'), true) -> IF2(cons(h', t'), s(n'), false)

The transformation is resulting in one new DP problem:



   R
DPs
       →DP Problem 1
Polo
       →DP Problem 2
Polo
       →DP Problem 3
Polo
       →DP Problem 4
Polo
       →DP Problem 5
Polo
       →DP Problem 6
Polo
       →DP Problem 7
Rw
           →DP Problem 14
Rw
             ...
               →DP Problem 24
Narrowing Transformation


Dependency Pairs:

IF1(cons(h', t'), s(n'), true) -> IF2(cons(h', t'), s(n'), false)
PROCESS(cons(h', t'), m) -> IF1(cons(h', t'), m, leq(m, s(length(t'))))
IF3(store, m, false) -> PROCESS(sndsplit(m, store), m)
IF1(store, m, false) -> IF3(store, m, empty(fstsplit(m, store)))
PROCESS(nil, m) -> IF1(nil, m, leq(m, 0))
IF2(store, m, false) -> PROCESS(sndsplit(m, store), m)


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))
mapf(pid, nil) -> nil
mapf(pid, cons(h, t)) -> app(f(pid, h), mapf(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(mapf(self, nil), store))))
if2(store, m, false) -> process(app(mapf(self, nil), sndsplit(m, store)), m)
if3(store, m, false) -> process(sndsplit(m, app(mapf(self, nil), store)), m)


Strategy:

innermost




On this DP problem, a Narrowing SCC transformation can be performed.
As a result of transforming the rule

IF1(store, m, false) -> IF3(store, m, empty(fstsplit(m, store)))
three new Dependency Pairs are created:

IF1(store', 0, false) -> IF3(store', 0, empty(nil))
IF1(nil, s(n'), false) -> IF3(nil, s(n'), empty(nil))
IF1(cons(h', t'), s(n'), false) -> IF3(cons(h', t'), s(n'), empty(cons(h', fstsplit(n', t'))))

The transformation is resulting in one new DP problem:



   R
DPs
       →DP Problem 1
Polo
       →DP Problem 2
Polo
       →DP Problem 3
Polo
       →DP Problem 4
Polo
       →DP Problem 5
Polo
       →DP Problem 6
Polo
       →DP Problem 7
Rw
           →DP Problem 14
Rw
             ...
               →DP Problem 25
Rewriting Transformation


Dependency Pairs:

IF1(nil, s(n'), false) -> IF3(nil, s(n'), empty(nil))
IF1(cons(h', t'), s(n'), false) -> IF3(cons(h', t'), s(n'), empty(cons(h', fstsplit(n', t'))))
PROCESS(cons(h', t'), m) -> IF1(cons(h', t'), m, leq(m, s(length(t'))))
IF3(store, m, false) -> PROCESS(sndsplit(m, store), m)
IF1(store', 0, false) -> IF3(store', 0, empty(nil))
PROCESS(nil, m) -> IF1(nil, m, leq(m, 0))
IF2(store, m, false) -> PROCESS(sndsplit(m, store), m)
IF1(cons(h', t'), s(n'), true) -> IF2(cons(h', t'), s(n'), false)


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))
mapf(pid, nil) -> nil
mapf(pid, cons(h, t)) -> app(f(pid, h), mapf(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(mapf(self, nil), store))))
if2(store, m, false) -> process(app(mapf(self, nil), sndsplit(m, store)), m)
if3(store, m, false) -> process(sndsplit(m, app(mapf(self, nil), store)), m)


Strategy:

innermost




On this DP problem, a Rewriting SCC transformation can be performed.
As a result of transforming the rule

IF1(store', 0, false) -> IF3(store', 0, empty(nil))
one new Dependency Pair is created:

IF1(store', 0, false) -> IF3(store', 0, true)

The transformation is resulting in one new DP problem:



   R
DPs
       →DP Problem 1
Polo
       →DP Problem 2
Polo
       →DP Problem 3
Polo
       →DP Problem 4
Polo
       →DP Problem 5
Polo
       →DP Problem 6
Polo
       →DP Problem 7
Rw
           →DP Problem 14
Rw
             ...
               →DP Problem 26
Rewriting Transformation


Dependency Pairs:

IF1(cons(h', t'), s(n'), false) -> IF3(cons(h', t'), s(n'), empty(cons(h', fstsplit(n', t'))))
IF2(store, m, false) -> PROCESS(sndsplit(m, store), m)
IF1(cons(h', t'), s(n'), true) -> IF2(cons(h', t'), s(n'), false)
PROCESS(cons(h', t'), m) -> IF1(cons(h', t'), m, leq(m, s(length(t'))))
PROCESS(nil, m) -> IF1(nil, m, leq(m, 0))
IF3(store, m, false) -> PROCESS(sndsplit(m, store), m)
IF1(nil, s(n'), false) -> IF3(nil, s(n'), empty(nil))


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))
mapf(pid, nil) -> nil
mapf(pid, cons(h, t)) -> app(f(pid, h), mapf(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(mapf(self, nil), store))))
if2(store, m, false) -> process(app(mapf(self, nil), sndsplit(m, store)), m)
if3(store, m, false) -> process(sndsplit(m, app(mapf(self, nil), store)), m)


Strategy:

innermost




On this DP problem, a Rewriting SCC transformation can be performed.
As a result of transforming the rule

IF1(nil, s(n'), false) -> IF3(nil, s(n'), empty(nil))
one new Dependency Pair is created:

IF1(nil, s(n'), false) -> IF3(nil, s(n'), true)

The transformation is resulting in one new DP problem:



   R
DPs
       →DP Problem 1
Polo
       →DP Problem 2
Polo
       →DP Problem 3
Polo
       →DP Problem 4
Polo
       →DP Problem 5
Polo
       →DP Problem 6
Polo
       →DP Problem 7
Rw
           →DP Problem 14
Rw
             ...
               →DP Problem 27
Rewriting Transformation


Dependency Pairs:

IF2(store, m, false) -> PROCESS(sndsplit(m, store), m)
IF1(cons(h', t'), s(n'), true) -> IF2(cons(h', t'), s(n'), false)
PROCESS(cons(h', t'), m) -> IF1(cons(h', t'), m, leq(m, s(length(t'))))
IF3(store, m, false) -> PROCESS(sndsplit(m, store), m)
IF1(cons(h', t'), s(n'), false) -> IF3(cons(h', t'), s(n'), empty(cons(h', fstsplit(n', t'))))


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))
mapf(pid, nil) -> nil
mapf(pid, cons(h, t)) -> app(f(pid, h), mapf(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(mapf(self, nil), store))))
if2(store, m, false) -> process(app(mapf(self, nil), sndsplit(m, store)), m)
if3(store, m, false) -> process(sndsplit(m, app(mapf(self, nil), store)), m)


Strategy:

innermost




On this DP problem, a Rewriting SCC transformation can be performed.
As a result of transforming the rule

IF1(cons(h', t'), s(n'), false) -> IF3(cons(h', t'), s(n'), empty(cons(h', fstsplit(n', t'))))
one new Dependency Pair is created:

IF1(cons(h', t'), s(n'), false) -> IF3(cons(h', t'), s(n'), false)

The transformation is resulting in one new DP problem:



   R
DPs
       →DP Problem 1
Polo
       →DP Problem 2
Polo
       →DP Problem 3
Polo
       →DP Problem 4
Polo
       →DP Problem 5
Polo
       →DP Problem 6
Polo
       →DP Problem 7
Rw
           →DP Problem 14
Rw
             ...
               →DP Problem 28
Narrowing Transformation


Dependency Pairs:

IF3(store, m, false) -> PROCESS(sndsplit(m, store), m)
IF1(cons(h', t'), s(n'), false) -> IF3(cons(h', t'), s(n'), false)
IF1(cons(h', t'), s(n'), true) -> IF2(cons(h', t'), s(n'), false)
PROCESS(cons(h', t'), m) -> IF1(cons(h', t'), m, leq(m, s(length(t'))))
IF2(store, m, false) -> PROCESS(sndsplit(m, store), m)


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))
mapf(pid, nil) -> nil
mapf(pid, cons(h, t)) -> app(f(pid, h), mapf(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(mapf(self, nil), store))))
if2(store, m, false) -> process(app(mapf(self, nil), sndsplit(m, store)), m)
if3(store, m, false) -> process(sndsplit(m, app(mapf(self, nil), store)), m)


Strategy:

innermost




On this DP problem, a Narrowing SCC transformation can be performed.
As a result of transforming the rule

IF2(store, m, false) -> PROCESS(sndsplit(m, store), m)
three new Dependency Pairs are created:

IF2(store', 0, false) -> PROCESS(store', 0)
IF2(nil, s(n'), false) -> PROCESS(nil, s(n'))
IF2(cons(h', t'), s(n'), false) -> PROCESS(sndsplit(n', t'), s(n'))

The transformation is resulting in one new DP problem:



   R
DPs
       →DP Problem 1
Polo
       →DP Problem 2
Polo
       →DP Problem 3
Polo
       →DP Problem 4
Polo
       →DP Problem 5
Polo
       →DP Problem 6
Polo
       →DP Problem 7
Rw
           →DP Problem 14
Rw
             ...
               →DP Problem 29
Narrowing Transformation


Dependency Pairs:

IF1(cons(h', t'), s(n'), false) -> IF3(cons(h', t'), s(n'), false)
IF2(cons(h', t'), s(n'), false) -> PROCESS(sndsplit(n', t'), s(n'))
IF1(cons(h', t'), s(n'), true) -> IF2(cons(h', t'), s(n'), false)
PROCESS(cons(h', t'), m) -> IF1(cons(h', t'), m, leq(m, s(length(t'))))
IF3(store, m, false) -> PROCESS(sndsplit(m, store), m)


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))
mapf(pid, nil) -> nil
mapf(pid, cons(h, t)) -> app(f(pid, h), mapf(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(mapf(self, nil), store))))
if2(store, m, false) -> process(app(mapf(self, nil), sndsplit(m, store)), m)
if3(store, m, false) -> process(sndsplit(m, app(mapf(self, nil), store)), m)


Strategy:

innermost




On this DP problem, a Narrowing SCC transformation can be performed.
As a result of transforming the rule

IF3(store, m, false) -> PROCESS(sndsplit(m, store), m)
three new Dependency Pairs are created:

IF3(store', 0, false) -> PROCESS(store', 0)
IF3(nil, s(n'), false) -> PROCESS(nil, s(n'))
IF3(cons(h', t'), s(n'), false) -> PROCESS(sndsplit(n', t'), s(n'))

The transformation is resulting in one new DP problem:



   R
DPs
       →DP Problem 1
Polo
       →DP Problem 2
Polo
       →DP Problem 3
Polo
       →DP Problem 4
Polo
       →DP Problem 5
Polo
       →DP Problem 6
Polo
       →DP Problem 7
Rw
           →DP Problem 14
Rw
             ...
               →DP Problem 30
Instantiation Transformation


Dependency Pairs:

IF2(cons(h', t'), s(n'), false) -> PROCESS(sndsplit(n', t'), s(n'))
IF1(cons(h', t'), s(n'), true) -> IF2(cons(h', t'), s(n'), false)
PROCESS(cons(h', t'), m) -> IF1(cons(h', t'), m, leq(m, s(length(t'))))
IF3(cons(h', t'), s(n'), false) -> PROCESS(sndsplit(n', t'), s(n'))
IF1(cons(h', t'), s(n'), false) -> IF3(cons(h', t'), s(n'), false)


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))
mapf(pid, nil) -> nil
mapf(pid, cons(h, t)) -> app(f(pid, h), mapf(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(mapf(self, nil), store))))
if2(store, m, false) -> process(app(mapf(self, nil), sndsplit(m, store)), m)
if3(store, m, false) -> process(sndsplit(m, app(mapf(self, nil), store)), m)


Strategy:

innermost




On this DP problem, an Instantiation SCC transformation can be performed.
As a result of transforming the rule

PROCESS(cons(h', t'), m) -> IF1(cons(h', t'), m, leq(m, s(length(t'))))
one new Dependency Pair is created:

PROCESS(cons(h'', t''), s(n''')) -> IF1(cons(h'', t''), s(n'''), leq(s(n'''), s(length(t''))))

The transformation is resulting in one new DP problem:



   R
DPs
       →DP Problem 1
Polo
       →DP Problem 2
Polo
       →DP Problem 3
Polo
       →DP Problem 4
Polo
       →DP Problem 5
Polo
       →DP Problem 6
Polo
       →DP Problem 7
Rw
           →DP Problem 14
Rw
             ...
               →DP Problem 31
Rewriting Transformation


Dependency Pairs:

IF3(cons(h', t'), s(n'), false) -> PROCESS(sndsplit(n', t'), s(n'))
IF1(cons(h', t'), s(n'), false) -> IF3(cons(h', t'), s(n'), false)
IF1(cons(h', t'), s(n'), true) -> IF2(cons(h', t'), s(n'), false)
PROCESS(cons(h'', t''), s(n''')) -> IF1(cons(h'', t''), s(n'''), leq(s(n'''), s(length(t''))))
IF2(cons(h', t'), s(n'), false) -> PROCESS(sndsplit(n', t'), s(n'))


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))
mapf(pid, nil) -> nil
mapf(pid, cons(h, t)) -> app(f(pid, h), mapf(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(mapf(self, nil), store))))
if2(store, m, false) -> process(app(mapf(self, nil), sndsplit(m, store)), m)
if3(store, m, false) -> process(sndsplit(m, app(mapf(self, nil), store)), m)


Strategy:

innermost




On this DP problem, a Rewriting SCC transformation can be performed.
As a result of transforming the rule

PROCESS(cons(h'', t''), s(n''')) -> IF1(cons(h'', t''), s(n'''), leq(s(n'''), s(length(t''))))
one new Dependency Pair is created:

PROCESS(cons(h'', t''), s(n''')) -> IF1(cons(h'', t''), s(n'''), leq(n''', length(t'')))

The transformation is resulting in one new DP problem:



   R
DPs
       →DP Problem 1
Polo
       →DP Problem 2
Polo
       →DP Problem 3
Polo
       →DP Problem 4
Polo
       →DP Problem 5
Polo
       →DP Problem 6
Polo
       →DP Problem 7
Rw
           →DP Problem 14
Rw
             ...
               →DP Problem 32
Polynomial Ordering


Dependency Pairs:

IF1(cons(h', t'), s(n'), false) -> IF3(cons(h', t'), s(n'), false)
IF2(cons(h', t'), s(n'), false) -> PROCESS(sndsplit(n', t'), s(n'))
IF1(cons(h', t'), s(n'), true) -> IF2(cons(h', t'), s(n'), false)
PROCESS(cons(h'', t''), s(n''')) -> IF1(cons(h'', t''), s(n'''), leq(n''', length(t'')))
IF3(cons(h', t'), s(n'), false) -> PROCESS(sndsplit(n', t'), s(n'))


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))
mapf(pid, nil) -> nil
mapf(pid, cons(h, t)) -> app(f(pid, h), mapf(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(mapf(self, nil), store))))
if2(store, m, false) -> process(app(mapf(self, nil), sndsplit(m, store)), m)
if3(store, m, false) -> process(sndsplit(m, app(mapf(self, nil), store)), m)


Strategy:

innermost




The following dependency pairs can be strictly oriented:

IF2(cons(h', t'), s(n'), false) -> PROCESS(sndsplit(n', t'), s(n'))
IF3(cons(h', t'), s(n'), false) -> PROCESS(sndsplit(n', t'), s(n'))


Additionally, the following usable rules for innermost can be oriented:

sndsplit(0, x) -> x
sndsplit(s(n), nil) -> nil
sndsplit(s(n), cons(h, t)) -> sndsplit(n, t)
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))


Used ordering: Polynomial ordering with Polynomial interpretation:
  POL(IF3(x1, x2, x3))=  x1  
  POL(PROCESS(x1, x2))=  x1  
  POL(sndsplit(x1, x2))=  x2  
  POL(false)=  0  
  POL(true)=  0  
  POL(IF2(x1, x2, x3))=  x1  
  POL(0)=  0  
  POL(IF1(x1, x2, x3))=  x1  
  POL(cons(x1, x2))=  1 + x2  
  POL(leq(x1, x2))=  0  
  POL(nil)=  0  
  POL(s(x1))=  0  
  POL(length(x1))=  0  

resulting in one new DP problem.



   R
DPs
       →DP Problem 1
Polo
       →DP Problem 2
Polo
       →DP Problem 3
Polo
       →DP Problem 4
Polo
       →DP Problem 5
Polo
       →DP Problem 6
Polo
       →DP Problem 7
Rw
           →DP Problem 14
Rw
             ...
               →DP Problem 33
Dependency Graph


Dependency Pairs:

IF1(cons(h', t'), s(n'), false) -> IF3(cons(h', t'), s(n'), false)
IF1(cons(h', t'), s(n'), true) -> IF2(cons(h', t'), s(n'), false)
PROCESS(cons(h'', t''), s(n''')) -> IF1(cons(h'', t''), s(n'''), leq(n''', length(t'')))


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))
mapf(pid, nil) -> nil
mapf(pid, cons(h, t)) -> app(f(pid, h), mapf(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(mapf(self, nil), store))))
if2(store, m, false) -> process(app(mapf(self, nil), sndsplit(m, store)), m)
if3(store, m, false) -> process(sndsplit(m, app(mapf(self, nil), store)), m)


Strategy:

innermost




Using the Dependency Graph resulted in no new DP problems.

Innermost Termination of R successfully shown.
Duration:
0:01 minutes