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))
head(cons(h, t)) → h
tail(cons(h, t)) → t
ring(st_1, in_2, st_2, in_3, st_3, m) → if_1(st_1, in_2, st_2, in_3, st_3, m, empty(fstsplit(m, st_1)))
if_1(st_1, in_2, st_2, in_3, st_3, m, false) → ring(sndsplit(m, st_1), cons(fstsplit(m, st_1), in_2), st_2, in_3, st_3, m)
ring(st_1, in_2, st_2, in_3, st_3, m) → if_2(st_1, in_2, st_2, in_3, st_3, m, leq(m, length(st_2)))
if_2(st_1, in_2, st_2, in_3, st_3, m, true) → if_3(st_1, in_2, st_2, in_3, st_3, m, empty(fstsplit(m, st_2)))
if_3(st_1, in_2, st_2, in_3, st_3, m, false) → ring(st_1, in_2, sndsplit(m, st_2), cons(fstsplit(m, st_2), in_3), st_3, m)
if_2(st_1, in_2, st_2, in_3, st_3, m, false) → if_4(st_1, in_2, st_2, in_3, st_3, m, empty(fstsplit(m, app(map_f(two, head(in_2)), st_2))))
if_4(st_1, in_2, st_2, in_3, st_3, m, false) → ring(st_1, tail(in_2), sndsplit(m, app(map_f(two, head(in_2)), st_2)), cons(fstsplit(m, app(map_f(two, head(in_2)), st_2)), in_3), st_3, m)
ring(st_1, in_2, st_2, in_3, st_3, m) → if_5(st_1, in_2, st_2, in_3, st_3, m, empty(map_f(two, head(in_2))))
if_5(st_1, in_2, st_2, in_3, st_3, m, true) → ring(st_1, tail(in_2), st_2, in_3, st_3, m)
ring(st_1, in_2, st_2, in_3, st_3, m) → if_6(st_1, in_2, st_2, in_3, st_3, m, leq(m, length(st_3)))
if_6(st_1, in_2, st_2, in_3, st_3, m, true) → if_7(st_1, in_2, st_2, in_3, st_3, m, empty(fstsplit(m, st_3)))
if_7(st_1, in_2, st_2, in_3, st_3, m, false) → ring(st_1, in_2, st_2, in_3, sndsplit(m, st_3), m)
if_6(st_1, in_2, st_2, in_3, st_3, m, false) → if_8(st_1, in_2, st_2, in_3, st_3, m, empty(fstsplit(m, app(map_f(three, head(in_3)), st_3))))
if_8(st_1, in_2, st_2, in_3, st_3, m, false) → ring(st_1, in_2, st_2, tail(in_3), sndsplit(m, app(map_f(three, head(in_3)), st_3)), m)
ring(st_1, in_2, st_2, in_3, st_3, m) → if_9(st_1, in_2, st_2, in_3, st_3, m, empty(map_f(three, head(in_3))))
if_9(st_1, in_2, st_2, in_3, st_3, m, true) → ring(st_1, in_2, st_2, tail(in_3), st_3, 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))
head(cons(h, t)) → h
tail(cons(h, t)) → 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

ring(st_1, in_2, st_2, in_3, st_3, m) → if_1(st_1, in_2, st_2, in_3, st_3, m, empty(fstsplit(m, st_1)))
if_1(st_1, in_2, st_2, in_3, st_3, m, false) → ring(sndsplit(m, st_1), cons(fstsplit(m, st_1), in_2), st_2, in_3, st_3, m)
ring(st_1, in_2, st_2, in_3, st_3, m) → if_2(st_1, in_2, st_2, in_3, st_3, m, leq(m, length(st_2)))
if_2(st_1, in_2, st_2, in_3, st_3, m, true) → if_3(st_1, in_2, st_2, in_3, st_3, m, empty(fstsplit(m, st_2)))
if_3(st_1, in_2, st_2, in_3, st_3, m, false) → ring(st_1, in_2, sndsplit(m, st_2), cons(fstsplit(m, st_2), in_3), st_3, m)
if_2(st_1, in_2, st_2, in_3, st_3, m, false) → if_4(st_1, in_2, st_2, in_3, st_3, m, empty(fstsplit(m, app(map_f(two, head(in_2)), st_2))))
if_4(st_1, in_2, st_2, in_3, st_3, m, false) → ring(st_1, tail(in_2), sndsplit(m, app(map_f(two, head(in_2)), st_2)), cons(fstsplit(m, app(map_f(two, head(in_2)), st_2)), in_3), st_3, m)
ring(st_1, in_2, st_2, in_3, st_3, m) → if_5(st_1, in_2, st_2, in_3, st_3, m, empty(map_f(two, head(in_2))))
if_5(st_1, in_2, st_2, in_3, st_3, m, true) → ring(st_1, tail(in_2), st_2, in_3, st_3, m)
ring(st_1, in_2, st_2, in_3, st_3, m) → if_6(st_1, in_2, st_2, in_3, st_3, m, leq(m, length(st_3)))
if_6(st_1, in_2, st_2, in_3, st_3, m, true) → if_7(st_1, in_2, st_2, in_3, st_3, m, empty(fstsplit(m, st_3)))
if_7(st_1, in_2, st_2, in_3, st_3, m, false) → ring(st_1, in_2, st_2, in_3, sndsplit(m, st_3), m)
if_6(st_1, in_2, st_2, in_3, st_3, m, false) → if_8(st_1, in_2, st_2, in_3, st_3, m, empty(fstsplit(m, app(map_f(three, head(in_3)), st_3))))
if_8(st_1, in_2, st_2, in_3, st_3, m, false) → ring(st_1, in_2, st_2, tail(in_3), sndsplit(m, app(map_f(three, head(in_3)), st_3)), m)
ring(st_1, in_2, st_2, in_3, st_3, m) → if_9(st_1, in_2, st_2, in_3, st_3, m, empty(map_f(three, head(in_3))))
if_9(st_1, in_2, st_2, in_3, st_3, m, true) → ring(st_1, in_2, st_2, tail(in_3), st_3, m)

The signature Sigma is {ring, if_1, if_2, if_3, if_4, if_5, if_6, if_7, if_8, if_9}

(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))
head(cons(h, t)) → h
tail(cons(h, t)) → t
ring(st_1, in_2, st_2, in_3, st_3, m) → if_1(st_1, in_2, st_2, in_3, st_3, m, empty(fstsplit(m, st_1)))
if_1(st_1, in_2, st_2, in_3, st_3, m, false) → ring(sndsplit(m, st_1), cons(fstsplit(m, st_1), in_2), st_2, in_3, st_3, m)
ring(st_1, in_2, st_2, in_3, st_3, m) → if_2(st_1, in_2, st_2, in_3, st_3, m, leq(m, length(st_2)))
if_2(st_1, in_2, st_2, in_3, st_3, m, true) → if_3(st_1, in_2, st_2, in_3, st_3, m, empty(fstsplit(m, st_2)))
if_3(st_1, in_2, st_2, in_3, st_3, m, false) → ring(st_1, in_2, sndsplit(m, st_2), cons(fstsplit(m, st_2), in_3), st_3, m)
if_2(st_1, in_2, st_2, in_3, st_3, m, false) → if_4(st_1, in_2, st_2, in_3, st_3, m, empty(fstsplit(m, app(map_f(two, head(in_2)), st_2))))
if_4(st_1, in_2, st_2, in_3, st_3, m, false) → ring(st_1, tail(in_2), sndsplit(m, app(map_f(two, head(in_2)), st_2)), cons(fstsplit(m, app(map_f(two, head(in_2)), st_2)), in_3), st_3, m)
ring(st_1, in_2, st_2, in_3, st_3, m) → if_5(st_1, in_2, st_2, in_3, st_3, m, empty(map_f(two, head(in_2))))
if_5(st_1, in_2, st_2, in_3, st_3, m, true) → ring(st_1, tail(in_2), st_2, in_3, st_3, m)
ring(st_1, in_2, st_2, in_3, st_3, m) → if_6(st_1, in_2, st_2, in_3, st_3, m, leq(m, length(st_3)))
if_6(st_1, in_2, st_2, in_3, st_3, m, true) → if_7(st_1, in_2, st_2, in_3, st_3, m, empty(fstsplit(m, st_3)))
if_7(st_1, in_2, st_2, in_3, st_3, m, false) → ring(st_1, in_2, st_2, in_3, sndsplit(m, st_3), m)
if_6(st_1, in_2, st_2, in_3, st_3, m, false) → if_8(st_1, in_2, st_2, in_3, st_3, m, empty(fstsplit(m, app(map_f(three, head(in_3)), st_3))))
if_8(st_1, in_2, st_2, in_3, st_3, m, false) → ring(st_1, in_2, st_2, tail(in_3), sndsplit(m, app(map_f(three, head(in_3)), st_3)), m)
ring(st_1, in_2, st_2, in_3, st_3, m) → if_9(st_1, in_2, st_2, in_3, st_3, m, empty(map_f(three, head(in_3))))
if_9(st_1, in_2, st_2, in_3, st_3, m, true) → ring(st_1, in_2, st_2, tail(in_3), st_3, 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))
head(cons(x0, x1))
tail(cons(x0, x1))
ring(x0, x1, x2, x3, x4, x5)
if_1(x0, x1, x2, x3, x4, x5, false)
if_2(x0, x1, x2, x3, x4, x5, true)
if_3(x0, x1, x2, x3, x4, x5, false)
if_2(x0, x1, x2, x3, x4, x5, false)
if_4(x0, x1, x2, x3, x4, x5, false)
if_5(x0, x1, x2, x3, x4, x5, true)
if_6(x0, x1, x2, x3, x4, x5, true)
if_7(x0, x1, x2, x3, x4, x5, false)
if_6(x0, x1, x2, x3, x4, x5, false)
if_8(x0, x1, x2, x3, x4, x5, false)
if_9(x0, x1, x2, x3, x4, x5, true)

(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)
RING(st_1, in_2, st_2, in_3, st_3, m) → IF_1(st_1, in_2, st_2, in_3, st_3, m, empty(fstsplit(m, st_1)))
RING(st_1, in_2, st_2, in_3, st_3, m) → EMPTY(fstsplit(m, st_1))
RING(st_1, in_2, st_2, in_3, st_3, m) → FSTSPLIT(m, st_1)
IF_1(st_1, in_2, st_2, in_3, st_3, m, false) → RING(sndsplit(m, st_1), cons(fstsplit(m, st_1), in_2), st_2, in_3, st_3, m)
IF_1(st_1, in_2, st_2, in_3, st_3, m, false) → SNDSPLIT(m, st_1)
IF_1(st_1, in_2, st_2, in_3, st_3, m, false) → FSTSPLIT(m, st_1)
RING(st_1, in_2, st_2, in_3, st_3, m) → IF_2(st_1, in_2, st_2, in_3, st_3, m, leq(m, length(st_2)))
RING(st_1, in_2, st_2, in_3, st_3, m) → LEQ(m, length(st_2))
RING(st_1, in_2, st_2, in_3, st_3, m) → LENGTH(st_2)
IF_2(st_1, in_2, st_2, in_3, st_3, m, true) → IF_3(st_1, in_2, st_2, in_3, st_3, m, empty(fstsplit(m, st_2)))
IF_2(st_1, in_2, st_2, in_3, st_3, m, true) → EMPTY(fstsplit(m, st_2))
IF_2(st_1, in_2, st_2, in_3, st_3, m, true) → FSTSPLIT(m, st_2)
IF_3(st_1, in_2, st_2, in_3, st_3, m, false) → RING(st_1, in_2, sndsplit(m, st_2), cons(fstsplit(m, st_2), in_3), st_3, m)
IF_3(st_1, in_2, st_2, in_3, st_3, m, false) → SNDSPLIT(m, st_2)
IF_3(st_1, in_2, st_2, in_3, st_3, m, false) → FSTSPLIT(m, st_2)
IF_2(st_1, in_2, st_2, in_3, st_3, m, false) → IF_4(st_1, in_2, st_2, in_3, st_3, m, empty(fstsplit(m, app(map_f(two, head(in_2)), st_2))))
IF_2(st_1, in_2, st_2, in_3, st_3, m, false) → EMPTY(fstsplit(m, app(map_f(two, head(in_2)), st_2)))
IF_2(st_1, in_2, st_2, in_3, st_3, m, false) → FSTSPLIT(m, app(map_f(two, head(in_2)), st_2))
IF_2(st_1, in_2, st_2, in_3, st_3, m, false) → APP(map_f(two, head(in_2)), st_2)
IF_2(st_1, in_2, st_2, in_3, st_3, m, false) → MAP_F(two, head(in_2))
IF_2(st_1, in_2, st_2, in_3, st_3, m, false) → HEAD(in_2)
IF_4(st_1, in_2, st_2, in_3, st_3, m, false) → RING(st_1, tail(in_2), sndsplit(m, app(map_f(two, head(in_2)), st_2)), cons(fstsplit(m, app(map_f(two, head(in_2)), st_2)), in_3), st_3, m)
IF_4(st_1, in_2, st_2, in_3, st_3, m, false) → TAIL(in_2)
IF_4(st_1, in_2, st_2, in_3, st_3, m, false) → SNDSPLIT(m, app(map_f(two, head(in_2)), st_2))
IF_4(st_1, in_2, st_2, in_3, st_3, m, false) → APP(map_f(two, head(in_2)), st_2)
IF_4(st_1, in_2, st_2, in_3, st_3, m, false) → MAP_F(two, head(in_2))
IF_4(st_1, in_2, st_2, in_3, st_3, m, false) → HEAD(in_2)
IF_4(st_1, in_2, st_2, in_3, st_3, m, false) → FSTSPLIT(m, app(map_f(two, head(in_2)), st_2))
RING(st_1, in_2, st_2, in_3, st_3, m) → IF_5(st_1, in_2, st_2, in_3, st_3, m, empty(map_f(two, head(in_2))))
RING(st_1, in_2, st_2, in_3, st_3, m) → EMPTY(map_f(two, head(in_2)))
RING(st_1, in_2, st_2, in_3, st_3, m) → MAP_F(two, head(in_2))
RING(st_1, in_2, st_2, in_3, st_3, m) → HEAD(in_2)
IF_5(st_1, in_2, st_2, in_3, st_3, m, true) → RING(st_1, tail(in_2), st_2, in_3, st_3, m)
IF_5(st_1, in_2, st_2, in_3, st_3, m, true) → TAIL(in_2)
RING(st_1, in_2, st_2, in_3, st_3, m) → IF_6(st_1, in_2, st_2, in_3, st_3, m, leq(m, length(st_3)))
RING(st_1, in_2, st_2, in_3, st_3, m) → LEQ(m, length(st_3))
RING(st_1, in_2, st_2, in_3, st_3, m) → LENGTH(st_3)
IF_6(st_1, in_2, st_2, in_3, st_3, m, true) → IF_7(st_1, in_2, st_2, in_3, st_3, m, empty(fstsplit(m, st_3)))
IF_6(st_1, in_2, st_2, in_3, st_3, m, true) → EMPTY(fstsplit(m, st_3))
IF_6(st_1, in_2, st_2, in_3, st_3, m, true) → FSTSPLIT(m, st_3)
IF_7(st_1, in_2, st_2, in_3, st_3, m, false) → RING(st_1, in_2, st_2, in_3, sndsplit(m, st_3), m)
IF_7(st_1, in_2, st_2, in_3, st_3, m, false) → SNDSPLIT(m, st_3)
IF_6(st_1, in_2, st_2, in_3, st_3, m, false) → IF_8(st_1, in_2, st_2, in_3, st_3, m, empty(fstsplit(m, app(map_f(three, head(in_3)), st_3))))
IF_6(st_1, in_2, st_2, in_3, st_3, m, false) → EMPTY(fstsplit(m, app(map_f(three, head(in_3)), st_3)))
IF_6(st_1, in_2, st_2, in_3, st_3, m, false) → FSTSPLIT(m, app(map_f(three, head(in_3)), st_3))
IF_6(st_1, in_2, st_2, in_3, st_3, m, false) → APP(map_f(three, head(in_3)), st_3)
IF_6(st_1, in_2, st_2, in_3, st_3, m, false) → MAP_F(three, head(in_3))
IF_6(st_1, in_2, st_2, in_3, st_3, m, false) → HEAD(in_3)
IF_8(st_1, in_2, st_2, in_3, st_3, m, false) → RING(st_1, in_2, st_2, tail(in_3), sndsplit(m, app(map_f(three, head(in_3)), st_3)), m)
IF_8(st_1, in_2, st_2, in_3, st_3, m, false) → TAIL(in_3)
IF_8(st_1, in_2, st_2, in_3, st_3, m, false) → SNDSPLIT(m, app(map_f(three, head(in_3)), st_3))
IF_8(st_1, in_2, st_2, in_3, st_3, m, false) → APP(map_f(three, head(in_3)), st_3)
IF_8(st_1, in_2, st_2, in_3, st_3, m, false) → MAP_F(three, head(in_3))
IF_8(st_1, in_2, st_2, in_3, st_3, m, false) → HEAD(in_3)
RING(st_1, in_2, st_2, in_3, st_3, m) → IF_9(st_1, in_2, st_2, in_3, st_3, m, empty(map_f(three, head(in_3))))
RING(st_1, in_2, st_2, in_3, st_3, m) → EMPTY(map_f(three, head(in_3)))
RING(st_1, in_2, st_2, in_3, st_3, m) → MAP_F(three, head(in_3))
RING(st_1, in_2, st_2, in_3, st_3, m) → HEAD(in_3)
IF_9(st_1, in_2, st_2, in_3, st_3, m, true) → RING(st_1, in_2, st_2, tail(in_3), st_3, m)
IF_9(st_1, in_2, st_2, in_3, st_3, m, true) → TAIL(in_3)

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))
head(cons(h, t)) → h
tail(cons(h, t)) → t
ring(st_1, in_2, st_2, in_3, st_3, m) → if_1(st_1, in_2, st_2, in_3, st_3, m, empty(fstsplit(m, st_1)))
if_1(st_1, in_2, st_2, in_3, st_3, m, false) → ring(sndsplit(m, st_1), cons(fstsplit(m, st_1), in_2), st_2, in_3, st_3, m)
ring(st_1, in_2, st_2, in_3, st_3, m) → if_2(st_1, in_2, st_2, in_3, st_3, m, leq(m, length(st_2)))
if_2(st_1, in_2, st_2, in_3, st_3, m, true) → if_3(st_1, in_2, st_2, in_3, st_3, m, empty(fstsplit(m, st_2)))
if_3(st_1, in_2, st_2, in_3, st_3, m, false) → ring(st_1, in_2, sndsplit(m, st_2), cons(fstsplit(m, st_2), in_3), st_3, m)
if_2(st_1, in_2, st_2, in_3, st_3, m, false) → if_4(st_1, in_2, st_2, in_3, st_3, m, empty(fstsplit(m, app(map_f(two, head(in_2)), st_2))))
if_4(st_1, in_2, st_2, in_3, st_3, m, false) → ring(st_1, tail(in_2), sndsplit(m, app(map_f(two, head(in_2)), st_2)), cons(fstsplit(m, app(map_f(two, head(in_2)), st_2)), in_3), st_3, m)
ring(st_1, in_2, st_2, in_3, st_3, m) → if_5(st_1, in_2, st_2, in_3, st_3, m, empty(map_f(two, head(in_2))))
if_5(st_1, in_2, st_2, in_3, st_3, m, true) → ring(st_1, tail(in_2), st_2, in_3, st_3, m)
ring(st_1, in_2, st_2, in_3, st_3, m) → if_6(st_1, in_2, st_2, in_3, st_3, m, leq(m, length(st_3)))
if_6(st_1, in_2, st_2, in_3, st_3, m, true) → if_7(st_1, in_2, st_2, in_3, st_3, m, empty(fstsplit(m, st_3)))
if_7(st_1, in_2, st_2, in_3, st_3, m, false) → ring(st_1, in_2, st_2, in_3, sndsplit(m, st_3), m)
if_6(st_1, in_2, st_2, in_3, st_3, m, false) → if_8(st_1, in_2, st_2, in_3, st_3, m, empty(fstsplit(m, app(map_f(three, head(in_3)), st_3))))
if_8(st_1, in_2, st_2, in_3, st_3, m, false) → ring(st_1, in_2, st_2, tail(in_3), sndsplit(m, app(map_f(three, head(in_3)), st_3)), m)
ring(st_1, in_2, st_2, in_3, st_3, m) → if_9(st_1, in_2, st_2, in_3, st_3, m, empty(map_f(three, head(in_3))))
if_9(st_1, in_2, st_2, in_3, st_3, m, true) → ring(st_1, in_2, st_2, tail(in_3), st_3, 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))
head(cons(x0, x1))
tail(cons(x0, x1))
ring(x0, x1, x2, x3, x4, x5)
if_1(x0, x1, x2, x3, x4, x5, false)
if_2(x0, x1, x2, x3, x4, x5, true)
if_3(x0, x1, x2, x3, x4, x5, false)
if_2(x0, x1, x2, x3, x4, x5, false)
if_4(x0, x1, x2, x3, x4, x5, false)
if_5(x0, x1, x2, x3, x4, x5, true)
if_6(x0, x1, x2, x3, x4, x5, true)
if_7(x0, x1, x2, x3, x4, x5, false)
if_6(x0, x1, x2, x3, x4, x5, false)
if_8(x0, x1, x2, x3, x4, x5, false)
if_9(x0, x1, x2, x3, x4, x5, true)

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 45 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))
head(cons(h, t)) → h
tail(cons(h, t)) → t
ring(st_1, in_2, st_2, in_3, st_3, m) → if_1(st_1, in_2, st_2, in_3, st_3, m, empty(fstsplit(m, st_1)))
if_1(st_1, in_2, st_2, in_3, st_3, m, false) → ring(sndsplit(m, st_1), cons(fstsplit(m, st_1), in_2), st_2, in_3, st_3, m)
ring(st_1, in_2, st_2, in_3, st_3, m) → if_2(st_1, in_2, st_2, in_3, st_3, m, leq(m, length(st_2)))
if_2(st_1, in_2, st_2, in_3, st_3, m, true) → if_3(st_1, in_2, st_2, in_3, st_3, m, empty(fstsplit(m, st_2)))
if_3(st_1, in_2, st_2, in_3, st_3, m, false) → ring(st_1, in_2, sndsplit(m, st_2), cons(fstsplit(m, st_2), in_3), st_3, m)
if_2(st_1, in_2, st_2, in_3, st_3, m, false) → if_4(st_1, in_2, st_2, in_3, st_3, m, empty(fstsplit(m, app(map_f(two, head(in_2)), st_2))))
if_4(st_1, in_2, st_2, in_3, st_3, m, false) → ring(st_1, tail(in_2), sndsplit(m, app(map_f(two, head(in_2)), st_2)), cons(fstsplit(m, app(map_f(two, head(in_2)), st_2)), in_3), st_3, m)
ring(st_1, in_2, st_2, in_3, st_3, m) → if_5(st_1, in_2, st_2, in_3, st_3, m, empty(map_f(two, head(in_2))))
if_5(st_1, in_2, st_2, in_3, st_3, m, true) → ring(st_1, tail(in_2), st_2, in_3, st_3, m)
ring(st_1, in_2, st_2, in_3, st_3, m) → if_6(st_1, in_2, st_2, in_3, st_3, m, leq(m, length(st_3)))
if_6(st_1, in_2, st_2, in_3, st_3, m, true) → if_7(st_1, in_2, st_2, in_3, st_3, m, empty(fstsplit(m, st_3)))
if_7(st_1, in_2, st_2, in_3, st_3, m, false) → ring(st_1, in_2, st_2, in_3, sndsplit(m, st_3), m)
if_6(st_1, in_2, st_2, in_3, st_3, m, false) → if_8(st_1, in_2, st_2, in_3, st_3, m, empty(fstsplit(m, app(map_f(three, head(in_3)), st_3))))
if_8(st_1, in_2, st_2, in_3, st_3, m, false) → ring(st_1, in_2, st_2, tail(in_3), sndsplit(m, app(map_f(three, head(in_3)), st_3)), m)
ring(st_1, in_2, st_2, in_3, st_3, m) → if_9(st_1, in_2, st_2, in_3, st_3, m, empty(map_f(three, head(in_3))))
if_9(st_1, in_2, st_2, in_3, st_3, m, true) → ring(st_1, in_2, st_2, tail(in_3), st_3, 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))
head(cons(x0, x1))
tail(cons(x0, x1))
ring(x0, x1, x2, x3, x4, x5)
if_1(x0, x1, x2, x3, x4, x5, false)
if_2(x0, x1, x2, x3, x4, x5, true)
if_3(x0, x1, x2, x3, x4, x5, false)
if_2(x0, x1, x2, x3, x4, x5, false)
if_4(x0, x1, x2, x3, x4, x5, false)
if_5(x0, x1, x2, x3, x4, x5, true)
if_6(x0, x1, x2, x3, x4, x5, true)
if_7(x0, x1, x2, x3, x4, x5, false)
if_6(x0, x1, x2, x3, x4, x5, false)
if_8(x0, x1, x2, x3, x4, x5, false)
if_9(x0, x1, x2, x3, x4, x5, true)

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 [LPO].
Precedence:
cons2 > APP1

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))
head(cons(h, t)) → h
tail(cons(h, t)) → t
ring(st_1, in_2, st_2, in_3, st_3, m) → if_1(st_1, in_2, st_2, in_3, st_3, m, empty(fstsplit(m, st_1)))
if_1(st_1, in_2, st_2, in_3, st_3, m, false) → ring(sndsplit(m, st_1), cons(fstsplit(m, st_1), in_2), st_2, in_3, st_3, m)
ring(st_1, in_2, st_2, in_3, st_3, m) → if_2(st_1, in_2, st_2, in_3, st_3, m, leq(m, length(st_2)))
if_2(st_1, in_2, st_2, in_3, st_3, m, true) → if_3(st_1, in_2, st_2, in_3, st_3, m, empty(fstsplit(m, st_2)))
if_3(st_1, in_2, st_2, in_3, st_3, m, false) → ring(st_1, in_2, sndsplit(m, st_2), cons(fstsplit(m, st_2), in_3), st_3, m)
if_2(st_1, in_2, st_2, in_3, st_3, m, false) → if_4(st_1, in_2, st_2, in_3, st_3, m, empty(fstsplit(m, app(map_f(two, head(in_2)), st_2))))
if_4(st_1, in_2, st_2, in_3, st_3, m, false) → ring(st_1, tail(in_2), sndsplit(m, app(map_f(two, head(in_2)), st_2)), cons(fstsplit(m, app(map_f(two, head(in_2)), st_2)), in_3), st_3, m)
ring(st_1, in_2, st_2, in_3, st_3, m) → if_5(st_1, in_2, st_2, in_3, st_3, m, empty(map_f(two, head(in_2))))
if_5(st_1, in_2, st_2, in_3, st_3, m, true) → ring(st_1, tail(in_2), st_2, in_3, st_3, m)
ring(st_1, in_2, st_2, in_3, st_3, m) → if_6(st_1, in_2, st_2, in_3, st_3, m, leq(m, length(st_3)))
if_6(st_1, in_2, st_2, in_3, st_3, m, true) → if_7(st_1, in_2, st_2, in_3, st_3, m, empty(fstsplit(m, st_3)))
if_7(st_1, in_2, st_2, in_3, st_3, m, false) → ring(st_1, in_2, st_2, in_3, sndsplit(m, st_3), m)
if_6(st_1, in_2, st_2, in_3, st_3, m, false) → if_8(st_1, in_2, st_2, in_3, st_3, m, empty(fstsplit(m, app(map_f(three, head(in_3)), st_3))))
if_8(st_1, in_2, st_2, in_3, st_3, m, false) → ring(st_1, in_2, st_2, tail(in_3), sndsplit(m, app(map_f(three, head(in_3)), st_3)), m)
ring(st_1, in_2, st_2, in_3, st_3, m) → if_9(st_1, in_2, st_2, in_3, st_3, m, empty(map_f(three, head(in_3))))
if_9(st_1, in_2, st_2, in_3, st_3, m, true) → ring(st_1, in_2, st_2, tail(in_3), st_3, 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))
head(cons(x0, x1))
tail(cons(x0, x1))
ring(x0, x1, x2, x3, x4, x5)
if_1(x0, x1, x2, x3, x4, x5, false)
if_2(x0, x1, x2, x3, x4, x5, true)
if_3(x0, x1, x2, x3, x4, x5, false)
if_2(x0, x1, x2, x3, x4, x5, false)
if_4(x0, x1, x2, x3, x4, x5, false)
if_5(x0, x1, x2, x3, x4, x5, true)
if_6(x0, x1, x2, x3, x4, x5, true)
if_7(x0, x1, x2, x3, x4, x5, false)
if_6(x0, x1, x2, x3, x4, x5, false)
if_8(x0, x1, x2, x3, x4, x5, false)
if_9(x0, x1, x2, x3, x4, x5, true)

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))
head(cons(h, t)) → h
tail(cons(h, t)) → t
ring(st_1, in_2, st_2, in_3, st_3, m) → if_1(st_1, in_2, st_2, in_3, st_3, m, empty(fstsplit(m, st_1)))
if_1(st_1, in_2, st_2, in_3, st_3, m, false) → ring(sndsplit(m, st_1), cons(fstsplit(m, st_1), in_2), st_2, in_3, st_3, m)
ring(st_1, in_2, st_2, in_3, st_3, m) → if_2(st_1, in_2, st_2, in_3, st_3, m, leq(m, length(st_2)))
if_2(st_1, in_2, st_2, in_3, st_3, m, true) → if_3(st_1, in_2, st_2, in_3, st_3, m, empty(fstsplit(m, st_2)))
if_3(st_1, in_2, st_2, in_3, st_3, m, false) → ring(st_1, in_2, sndsplit(m, st_2), cons(fstsplit(m, st_2), in_3), st_3, m)
if_2(st_1, in_2, st_2, in_3, st_3, m, false) → if_4(st_1, in_2, st_2, in_3, st_3, m, empty(fstsplit(m, app(map_f(two, head(in_2)), st_2))))
if_4(st_1, in_2, st_2, in_3, st_3, m, false) → ring(st_1, tail(in_2), sndsplit(m, app(map_f(two, head(in_2)), st_2)), cons(fstsplit(m, app(map_f(two, head(in_2)), st_2)), in_3), st_3, m)
ring(st_1, in_2, st_2, in_3, st_3, m) → if_5(st_1, in_2, st_2, in_3, st_3, m, empty(map_f(two, head(in_2))))
if_5(st_1, in_2, st_2, in_3, st_3, m, true) → ring(st_1, tail(in_2), st_2, in_3, st_3, m)
ring(st_1, in_2, st_2, in_3, st_3, m) → if_6(st_1, in_2, st_2, in_3, st_3, m, leq(m, length(st_3)))
if_6(st_1, in_2, st_2, in_3, st_3, m, true) → if_7(st_1, in_2, st_2, in_3, st_3, m, empty(fstsplit(m, st_3)))
if_7(st_1, in_2, st_2, in_3, st_3, m, false) → ring(st_1, in_2, st_2, in_3, sndsplit(m, st_3), m)
if_6(st_1, in_2, st_2, in_3, st_3, m, false) → if_8(st_1, in_2, st_2, in_3, st_3, m, empty(fstsplit(m, app(map_f(three, head(in_3)), st_3))))
if_8(st_1, in_2, st_2, in_3, st_3, m, false) → ring(st_1, in_2, st_2, tail(in_3), sndsplit(m, app(map_f(three, head(in_3)), st_3)), m)
ring(st_1, in_2, st_2, in_3, st_3, m) → if_9(st_1, in_2, st_2, in_3, st_3, m, empty(map_f(three, head(in_3))))
if_9(st_1, in_2, st_2, in_3, st_3, m, true) → ring(st_1, in_2, st_2, tail(in_3), st_3, 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))
head(cons(x0, x1))
tail(cons(x0, x1))
ring(x0, x1, x2, x3, x4, x5)
if_1(x0, x1, x2, x3, x4, x5, false)
if_2(x0, x1, x2, x3, x4, x5, true)
if_3(x0, x1, x2, x3, x4, x5, false)
if_2(x0, x1, x2, x3, x4, x5, false)
if_4(x0, x1, x2, x3, x4, x5, false)
if_5(x0, x1, x2, x3, x4, x5, true)
if_6(x0, x1, x2, x3, x4, x5, true)
if_7(x0, x1, x2, x3, x4, x5, false)
if_6(x0, x1, x2, x3, x4, x5, false)
if_8(x0, x1, x2, x3, x4, x5, false)
if_9(x0, x1, x2, x3, x4, x5, true)

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

Lexicographic Path Order [LPO].
Precedence:
cons2 > MAPF1

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))
head(cons(h, t)) → h
tail(cons(h, t)) → t
ring(st_1, in_2, st_2, in_3, st_3, m) → if_1(st_1, in_2, st_2, in_3, st_3, m, empty(fstsplit(m, st_1)))
if_1(st_1, in_2, st_2, in_3, st_3, m, false) → ring(sndsplit(m, st_1), cons(fstsplit(m, st_1), in_2), st_2, in_3, st_3, m)
ring(st_1, in_2, st_2, in_3, st_3, m) → if_2(st_1, in_2, st_2, in_3, st_3, m, leq(m, length(st_2)))
if_2(st_1, in_2, st_2, in_3, st_3, m, true) → if_3(st_1, in_2, st_2, in_3, st_3, m, empty(fstsplit(m, st_2)))
if_3(st_1, in_2, st_2, in_3, st_3, m, false) → ring(st_1, in_2, sndsplit(m, st_2), cons(fstsplit(m, st_2), in_3), st_3, m)
if_2(st_1, in_2, st_2, in_3, st_3, m, false) → if_4(st_1, in_2, st_2, in_3, st_3, m, empty(fstsplit(m, app(map_f(two, head(in_2)), st_2))))
if_4(st_1, in_2, st_2, in_3, st_3, m, false) → ring(st_1, tail(in_2), sndsplit(m, app(map_f(two, head(in_2)), st_2)), cons(fstsplit(m, app(map_f(two, head(in_2)), st_2)), in_3), st_3, m)
ring(st_1, in_2, st_2, in_3, st_3, m) → if_5(st_1, in_2, st_2, in_3, st_3, m, empty(map_f(two, head(in_2))))
if_5(st_1, in_2, st_2, in_3, st_3, m, true) → ring(st_1, tail(in_2), st_2, in_3, st_3, m)
ring(st_1, in_2, st_2, in_3, st_3, m) → if_6(st_1, in_2, st_2, in_3, st_3, m, leq(m, length(st_3)))
if_6(st_1, in_2, st_2, in_3, st_3, m, true) → if_7(st_1, in_2, st_2, in_3, st_3, m, empty(fstsplit(m, st_3)))
if_7(st_1, in_2, st_2, in_3, st_3, m, false) → ring(st_1, in_2, st_2, in_3, sndsplit(m, st_3), m)
if_6(st_1, in_2, st_2, in_3, st_3, m, false) → if_8(st_1, in_2, st_2, in_3, st_3, m, empty(fstsplit(m, app(map_f(three, head(in_3)), st_3))))
if_8(st_1, in_2, st_2, in_3, st_3, m, false) → ring(st_1, in_2, st_2, tail(in_3), sndsplit(m, app(map_f(three, head(in_3)), st_3)), m)
ring(st_1, in_2, st_2, in_3, st_3, m) → if_9(st_1, in_2, st_2, in_3, st_3, m, empty(map_f(three, head(in_3))))
if_9(st_1, in_2, st_2, in_3, st_3, m, true) → ring(st_1, in_2, st_2, tail(in_3), st_3, 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))
head(cons(x0, x1))
tail(cons(x0, x1))
ring(x0, x1, x2, x3, x4, x5)
if_1(x0, x1, x2, x3, x4, x5, false)
if_2(x0, x1, x2, x3, x4, x5, true)
if_3(x0, x1, x2, x3, x4, x5, false)
if_2(x0, x1, x2, x3, x4, x5, false)
if_4(x0, x1, x2, x3, x4, x5, false)
if_5(x0, x1, x2, x3, x4, x5, true)
if_6(x0, x1, x2, x3, x4, x5, true)
if_7(x0, x1, x2, x3, x4, x5, false)
if_6(x0, x1, x2, x3, x4, x5, false)
if_8(x0, x1, x2, x3, x4, x5, false)
if_9(x0, x1, x2, x3, x4, x5, true)

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))
head(cons(h, t)) → h
tail(cons(h, t)) → t
ring(st_1, in_2, st_2, in_3, st_3, m) → if_1(st_1, in_2, st_2, in_3, st_3, m, empty(fstsplit(m, st_1)))
if_1(st_1, in_2, st_2, in_3, st_3, m, false) → ring(sndsplit(m, st_1), cons(fstsplit(m, st_1), in_2), st_2, in_3, st_3, m)
ring(st_1, in_2, st_2, in_3, st_3, m) → if_2(st_1, in_2, st_2, in_3, st_3, m, leq(m, length(st_2)))
if_2(st_1, in_2, st_2, in_3, st_3, m, true) → if_3(st_1, in_2, st_2, in_3, st_3, m, empty(fstsplit(m, st_2)))
if_3(st_1, in_2, st_2, in_3, st_3, m, false) → ring(st_1, in_2, sndsplit(m, st_2), cons(fstsplit(m, st_2), in_3), st_3, m)
if_2(st_1, in_2, st_2, in_3, st_3, m, false) → if_4(st_1, in_2, st_2, in_3, st_3, m, empty(fstsplit(m, app(map_f(two, head(in_2)), st_2))))
if_4(st_1, in_2, st_2, in_3, st_3, m, false) → ring(st_1, tail(in_2), sndsplit(m, app(map_f(two, head(in_2)), st_2)), cons(fstsplit(m, app(map_f(two, head(in_2)), st_2)), in_3), st_3, m)
ring(st_1, in_2, st_2, in_3, st_3, m) → if_5(st_1, in_2, st_2, in_3, st_3, m, empty(map_f(two, head(in_2))))
if_5(st_1, in_2, st_2, in_3, st_3, m, true) → ring(st_1, tail(in_2), st_2, in_3, st_3, m)
ring(st_1, in_2, st_2, in_3, st_3, m) → if_6(st_1, in_2, st_2, in_3, st_3, m, leq(m, length(st_3)))
if_6(st_1, in_2, st_2, in_3, st_3, m, true) → if_7(st_1, in_2, st_2, in_3, st_3, m, empty(fstsplit(m, st_3)))
if_7(st_1, in_2, st_2, in_3, st_3, m, false) → ring(st_1, in_2, st_2, in_3, sndsplit(m, st_3), m)
if_6(st_1, in_2, st_2, in_3, st_3, m, false) → if_8(st_1, in_2, st_2, in_3, st_3, m, empty(fstsplit(m, app(map_f(three, head(in_3)), st_3))))
if_8(st_1, in_2, st_2, in_3, st_3, m, false) → ring(st_1, in_2, st_2, tail(in_3), sndsplit(m, app(map_f(three, head(in_3)), st_3)), m)
ring(st_1, in_2, st_2, in_3, st_3, m) → if_9(st_1, in_2, st_2, in_3, st_3, m, empty(map_f(three, head(in_3))))
if_9(st_1, in_2, st_2, in_3, st_3, m, true) → ring(st_1, in_2, st_2, tail(in_3), st_3, 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))
head(cons(x0, x1))
tail(cons(x0, x1))
ring(x0, x1, x2, x3, x4, x5)
if_1(x0, x1, x2, x3, x4, x5, false)
if_2(x0, x1, x2, x3, x4, x5, true)
if_3(x0, x1, x2, x3, x4, x5, false)
if_2(x0, x1, x2, x3, x4, x5, false)
if_4(x0, x1, x2, x3, x4, x5, false)
if_5(x0, x1, x2, x3, x4, x5, true)
if_6(x0, x1, x2, x3, x4, x5, true)
if_7(x0, x1, x2, x3, x4, x5, false)
if_6(x0, x1, x2, x3, x4, x5, false)
if_8(x0, x1, x2, x3, x4, x5, false)
if_9(x0, x1, x2, x3, x4, x5, true)

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: Lexicographic Path Order [LPO].
Precedence:
cons2 > LENGTH1

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))
head(cons(h, t)) → h
tail(cons(h, t)) → t
ring(st_1, in_2, st_2, in_3, st_3, m) → if_1(st_1, in_2, st_2, in_3, st_3, m, empty(fstsplit(m, st_1)))
if_1(st_1, in_2, st_2, in_3, st_3, m, false) → ring(sndsplit(m, st_1), cons(fstsplit(m, st_1), in_2), st_2, in_3, st_3, m)
ring(st_1, in_2, st_2, in_3, st_3, m) → if_2(st_1, in_2, st_2, in_3, st_3, m, leq(m, length(st_2)))
if_2(st_1, in_2, st_2, in_3, st_3, m, true) → if_3(st_1, in_2, st_2, in_3, st_3, m, empty(fstsplit(m, st_2)))
if_3(st_1, in_2, st_2, in_3, st_3, m, false) → ring(st_1, in_2, sndsplit(m, st_2), cons(fstsplit(m, st_2), in_3), st_3, m)
if_2(st_1, in_2, st_2, in_3, st_3, m, false) → if_4(st_1, in_2, st_2, in_3, st_3, m, empty(fstsplit(m, app(map_f(two, head(in_2)), st_2))))
if_4(st_1, in_2, st_2, in_3, st_3, m, false) → ring(st_1, tail(in_2), sndsplit(m, app(map_f(two, head(in_2)), st_2)), cons(fstsplit(m, app(map_f(two, head(in_2)), st_2)), in_3), st_3, m)
ring(st_1, in_2, st_2, in_3, st_3, m) → if_5(st_1, in_2, st_2, in_3, st_3, m, empty(map_f(two, head(in_2))))
if_5(st_1, in_2, st_2, in_3, st_3, m, true) → ring(st_1, tail(in_2), st_2, in_3, st_3, m)
ring(st_1, in_2, st_2, in_3, st_3, m) → if_6(st_1, in_2, st_2, in_3, st_3, m, leq(m, length(st_3)))
if_6(st_1, in_2, st_2, in_3, st_3, m, true) → if_7(st_1, in_2, st_2, in_3, st_3, m, empty(fstsplit(m, st_3)))
if_7(st_1, in_2, st_2, in_3, st_3, m, false) → ring(st_1, in_2, st_2, in_3, sndsplit(m, st_3), m)
if_6(st_1, in_2, st_2, in_3, st_3, m, false) → if_8(st_1, in_2, st_2, in_3, st_3, m, empty(fstsplit(m, app(map_f(three, head(in_3)), st_3))))
if_8(st_1, in_2, st_2, in_3, st_3, m, false) → ring(st_1, in_2, st_2, tail(in_3), sndsplit(m, app(map_f(three, head(in_3)), st_3)), m)
ring(st_1, in_2, st_2, in_3, st_3, m) → if_9(st_1, in_2, st_2, in_3, st_3, m, empty(map_f(three, head(in_3))))
if_9(st_1, in_2, st_2, in_3, st_3, m, true) → ring(st_1, in_2, st_2, tail(in_3), st_3, 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))
head(cons(x0, x1))
tail(cons(x0, x1))
ring(x0, x1, x2, x3, x4, x5)
if_1(x0, x1, x2, x3, x4, x5, false)
if_2(x0, x1, x2, x3, x4, x5, true)
if_3(x0, x1, x2, x3, x4, x5, false)
if_2(x0, x1, x2, x3, x4, x5, false)
if_4(x0, x1, x2, x3, x4, x5, false)
if_5(x0, x1, x2, x3, x4, x5, true)
if_6(x0, x1, x2, x3, x4, x5, true)
if_7(x0, x1, x2, x3, x4, x5, false)
if_6(x0, x1, x2, x3, x4, x5, false)
if_8(x0, x1, x2, x3, x4, x5, false)
if_9(x0, x1, x2, x3, x4, x5, true)

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))
head(cons(h, t)) → h
tail(cons(h, t)) → t
ring(st_1, in_2, st_2, in_3, st_3, m) → if_1(st_1, in_2, st_2, in_3, st_3, m, empty(fstsplit(m, st_1)))
if_1(st_1, in_2, st_2, in_3, st_3, m, false) → ring(sndsplit(m, st_1), cons(fstsplit(m, st_1), in_2), st_2, in_3, st_3, m)
ring(st_1, in_2, st_2, in_3, st_3, m) → if_2(st_1, in_2, st_2, in_3, st_3, m, leq(m, length(st_2)))
if_2(st_1, in_2, st_2, in_3, st_3, m, true) → if_3(st_1, in_2, st_2, in_3, st_3, m, empty(fstsplit(m, st_2)))
if_3(st_1, in_2, st_2, in_3, st_3, m, false) → ring(st_1, in_2, sndsplit(m, st_2), cons(fstsplit(m, st_2), in_3), st_3, m)
if_2(st_1, in_2, st_2, in_3, st_3, m, false) → if_4(st_1, in_2, st_2, in_3, st_3, m, empty(fstsplit(m, app(map_f(two, head(in_2)), st_2))))
if_4(st_1, in_2, st_2, in_3, st_3, m, false) → ring(st_1, tail(in_2), sndsplit(m, app(map_f(two, head(in_2)), st_2)), cons(fstsplit(m, app(map_f(two, head(in_2)), st_2)), in_3), st_3, m)
ring(st_1, in_2, st_2, in_3, st_3, m) → if_5(st_1, in_2, st_2, in_3, st_3, m, empty(map_f(two, head(in_2))))
if_5(st_1, in_2, st_2, in_3, st_3, m, true) → ring(st_1, tail(in_2), st_2, in_3, st_3, m)
ring(st_1, in_2, st_2, in_3, st_3, m) → if_6(st_1, in_2, st_2, in_3, st_3, m, leq(m, length(st_3)))
if_6(st_1, in_2, st_2, in_3, st_3, m, true) → if_7(st_1, in_2, st_2, in_3, st_3, m, empty(fstsplit(m, st_3)))
if_7(st_1, in_2, st_2, in_3, st_3, m, false) → ring(st_1, in_2, st_2, in_3, sndsplit(m, st_3), m)
if_6(st_1, in_2, st_2, in_3, st_3, m, false) → if_8(st_1, in_2, st_2, in_3, st_3, m, empty(fstsplit(m, app(map_f(three, head(in_3)), st_3))))
if_8(st_1, in_2, st_2, in_3, st_3, m, false) → ring(st_1, in_2, st_2, tail(in_3), sndsplit(m, app(map_f(three, head(in_3)), st_3)), m)
ring(st_1, in_2, st_2, in_3, st_3, m) → if_9(st_1, in_2, st_2, in_3, st_3, m, empty(map_f(three, head(in_3))))
if_9(st_1, in_2, st_2, in_3, st_3, m, true) → ring(st_1, in_2, st_2, tail(in_3), st_3, 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))
head(cons(x0, x1))
tail(cons(x0, x1))
ring(x0, x1, x2, x3, x4, x5)
if_1(x0, x1, x2, x3, x4, x5, false)
if_2(x0, x1, x2, x3, x4, x5, true)
if_3(x0, x1, x2, x3, x4, x5, false)
if_2(x0, x1, x2, x3, x4, x5, false)
if_4(x0, x1, x2, x3, x4, x5, false)
if_5(x0, x1, x2, x3, x4, x5, true)
if_6(x0, x1, x2, x3, x4, x5, true)
if_7(x0, x1, x2, x3, x4, x5, false)
if_6(x0, x1, x2, x3, x4, x5, false)
if_8(x0, x1, x2, x3, x4, x5, false)
if_9(x0, x1, x2, x3, x4, x5, true)

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(x2)
s(x1)  =  s(x1)

Lexicographic Path Order [LPO].
Precedence:
s1 > LEQ1

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))
head(cons(h, t)) → h
tail(cons(h, t)) → t
ring(st_1, in_2, st_2, in_3, st_3, m) → if_1(st_1, in_2, st_2, in_3, st_3, m, empty(fstsplit(m, st_1)))
if_1(st_1, in_2, st_2, in_3, st_3, m, false) → ring(sndsplit(m, st_1), cons(fstsplit(m, st_1), in_2), st_2, in_3, st_3, m)
ring(st_1, in_2, st_2, in_3, st_3, m) → if_2(st_1, in_2, st_2, in_3, st_3, m, leq(m, length(st_2)))
if_2(st_1, in_2, st_2, in_3, st_3, m, true) → if_3(st_1, in_2, st_2, in_3, st_3, m, empty(fstsplit(m, st_2)))
if_3(st_1, in_2, st_2, in_3, st_3, m, false) → ring(st_1, in_2, sndsplit(m, st_2), cons(fstsplit(m, st_2), in_3), st_3, m)
if_2(st_1, in_2, st_2, in_3, st_3, m, false) → if_4(st_1, in_2, st_2, in_3, st_3, m, empty(fstsplit(m, app(map_f(two, head(in_2)), st_2))))
if_4(st_1, in_2, st_2, in_3, st_3, m, false) → ring(st_1, tail(in_2), sndsplit(m, app(map_f(two, head(in_2)), st_2)), cons(fstsplit(m, app(map_f(two, head(in_2)), st_2)), in_3), st_3, m)
ring(st_1, in_2, st_2, in_3, st_3, m) → if_5(st_1, in_2, st_2, in_3, st_3, m, empty(map_f(two, head(in_2))))
if_5(st_1, in_2, st_2, in_3, st_3, m, true) → ring(st_1, tail(in_2), st_2, in_3, st_3, m)
ring(st_1, in_2, st_2, in_3, st_3, m) → if_6(st_1, in_2, st_2, in_3, st_3, m, leq(m, length(st_3)))
if_6(st_1, in_2, st_2, in_3, st_3, m, true) → if_7(st_1, in_2, st_2, in_3, st_3, m, empty(fstsplit(m, st_3)))
if_7(st_1, in_2, st_2, in_3, st_3, m, false) → ring(st_1, in_2, st_2, in_3, sndsplit(m, st_3), m)
if_6(st_1, in_2, st_2, in_3, st_3, m, false) → if_8(st_1, in_2, st_2, in_3, st_3, m, empty(fstsplit(m, app(map_f(three, head(in_3)), st_3))))
if_8(st_1, in_2, st_2, in_3, st_3, m, false) → ring(st_1, in_2, st_2, tail(in_3), sndsplit(m, app(map_f(three, head(in_3)), st_3)), m)
ring(st_1, in_2, st_2, in_3, st_3, m) → if_9(st_1, in_2, st_2, in_3, st_3, m, empty(map_f(three, head(in_3))))
if_9(st_1, in_2, st_2, in_3, st_3, m, true) → ring(st_1, in_2, st_2, tail(in_3), st_3, 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))
head(cons(x0, x1))
tail(cons(x0, x1))
ring(x0, x1, x2, x3, x4, x5)
if_1(x0, x1, x2, x3, x4, x5, false)
if_2(x0, x1, x2, x3, x4, x5, true)
if_3(x0, x1, x2, x3, x4, x5, false)
if_2(x0, x1, x2, x3, x4, x5, false)
if_4(x0, x1, x2, x3, x4, x5, false)
if_5(x0, x1, x2, x3, x4, x5, true)
if_6(x0, x1, x2, x3, x4, x5, true)
if_7(x0, x1, x2, x3, x4, x5, false)
if_6(x0, x1, x2, x3, x4, x5, false)
if_8(x0, x1, x2, x3, x4, x5, false)
if_9(x0, x1, x2, x3, x4, x5, true)

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))
head(cons(h, t)) → h
tail(cons(h, t)) → t
ring(st_1, in_2, st_2, in_3, st_3, m) → if_1(st_1, in_2, st_2, in_3, st_3, m, empty(fstsplit(m, st_1)))
if_1(st_1, in_2, st_2, in_3, st_3, m, false) → ring(sndsplit(m, st_1), cons(fstsplit(m, st_1), in_2), st_2, in_3, st_3, m)
ring(st_1, in_2, st_2, in_3, st_3, m) → if_2(st_1, in_2, st_2, in_3, st_3, m, leq(m, length(st_2)))
if_2(st_1, in_2, st_2, in_3, st_3, m, true) → if_3(st_1, in_2, st_2, in_3, st_3, m, empty(fstsplit(m, st_2)))
if_3(st_1, in_2, st_2, in_3, st_3, m, false) → ring(st_1, in_2, sndsplit(m, st_2), cons(fstsplit(m, st_2), in_3), st_3, m)
if_2(st_1, in_2, st_2, in_3, st_3, m, false) → if_4(st_1, in_2, st_2, in_3, st_3, m, empty(fstsplit(m, app(map_f(two, head(in_2)), st_2))))
if_4(st_1, in_2, st_2, in_3, st_3, m, false) → ring(st_1, tail(in_2), sndsplit(m, app(map_f(two, head(in_2)), st_2)), cons(fstsplit(m, app(map_f(two, head(in_2)), st_2)), in_3), st_3, m)
ring(st_1, in_2, st_2, in_3, st_3, m) → if_5(st_1, in_2, st_2, in_3, st_3, m, empty(map_f(two, head(in_2))))
if_5(st_1, in_2, st_2, in_3, st_3, m, true) → ring(st_1, tail(in_2), st_2, in_3, st_3, m)
ring(st_1, in_2, st_2, in_3, st_3, m) → if_6(st_1, in_2, st_2, in_3, st_3, m, leq(m, length(st_3)))
if_6(st_1, in_2, st_2, in_3, st_3, m, true) → if_7(st_1, in_2, st_2, in_3, st_3, m, empty(fstsplit(m, st_3)))
if_7(st_1, in_2, st_2, in_3, st_3, m, false) → ring(st_1, in_2, st_2, in_3, sndsplit(m, st_3), m)
if_6(st_1, in_2, st_2, in_3, st_3, m, false) → if_8(st_1, in_2, st_2, in_3, st_3, m, empty(fstsplit(m, app(map_f(three, head(in_3)), st_3))))
if_8(st_1, in_2, st_2, in_3, st_3, m, false) → ring(st_1, in_2, st_2, tail(in_3), sndsplit(m, app(map_f(three, head(in_3)), st_3)), m)
ring(st_1, in_2, st_2, in_3, st_3, m) → if_9(st_1, in_2, st_2, in_3, st_3, m, empty(map_f(three, head(in_3))))
if_9(st_1, in_2, st_2, in_3, st_3, m, true) → ring(st_1, in_2, st_2, tail(in_3), st_3, 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))
head(cons(x0, x1))
tail(cons(x0, x1))
ring(x0, x1, x2, x3, x4, x5)
if_1(x0, x1, x2, x3, x4, x5, false)
if_2(x0, x1, x2, x3, x4, x5, true)
if_3(x0, x1, x2, x3, x4, x5, false)
if_2(x0, x1, x2, x3, x4, x5, false)
if_4(x0, x1, x2, x3, x4, x5, false)
if_5(x0, x1, x2, x3, x4, x5, true)
if_6(x0, x1, x2, x3, x4, x5, true)
if_7(x0, x1, x2, x3, x4, x5, false)
if_6(x0, x1, x2, x3, x4, x5, false)
if_8(x0, x1, x2, x3, x4, x5, false)
if_9(x0, x1, x2, x3, x4, x5, true)

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

Lexicographic Path Order [LPO].
Precedence:
s1 > SNDSPLIT2

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))
head(cons(h, t)) → h
tail(cons(h, t)) → t
ring(st_1, in_2, st_2, in_3, st_3, m) → if_1(st_1, in_2, st_2, in_3, st_3, m, empty(fstsplit(m, st_1)))
if_1(st_1, in_2, st_2, in_3, st_3, m, false) → ring(sndsplit(m, st_1), cons(fstsplit(m, st_1), in_2), st_2, in_3, st_3, m)
ring(st_1, in_2, st_2, in_3, st_3, m) → if_2(st_1, in_2, st_2, in_3, st_3, m, leq(m, length(st_2)))
if_2(st_1, in_2, st_2, in_3, st_3, m, true) → if_3(st_1, in_2, st_2, in_3, st_3, m, empty(fstsplit(m, st_2)))
if_3(st_1, in_2, st_2, in_3, st_3, m, false) → ring(st_1, in_2, sndsplit(m, st_2), cons(fstsplit(m, st_2), in_3), st_3, m)
if_2(st_1, in_2, st_2, in_3, st_3, m, false) → if_4(st_1, in_2, st_2, in_3, st_3, m, empty(fstsplit(m, app(map_f(two, head(in_2)), st_2))))
if_4(st_1, in_2, st_2, in_3, st_3, m, false) → ring(st_1, tail(in_2), sndsplit(m, app(map_f(two, head(in_2)), st_2)), cons(fstsplit(m, app(map_f(two, head(in_2)), st_2)), in_3), st_3, m)
ring(st_1, in_2, st_2, in_3, st_3, m) → if_5(st_1, in_2, st_2, in_3, st_3, m, empty(map_f(two, head(in_2))))
if_5(st_1, in_2, st_2, in_3, st_3, m, true) → ring(st_1, tail(in_2), st_2, in_3, st_3, m)
ring(st_1, in_2, st_2, in_3, st_3, m) → if_6(st_1, in_2, st_2, in_3, st_3, m, leq(m, length(st_3)))
if_6(st_1, in_2, st_2, in_3, st_3, m, true) → if_7(st_1, in_2, st_2, in_3, st_3, m, empty(fstsplit(m, st_3)))
if_7(st_1, in_2, st_2, in_3, st_3, m, false) → ring(st_1, in_2, st_2, in_3, sndsplit(m, st_3), m)
if_6(st_1, in_2, st_2, in_3, st_3, m, false) → if_8(st_1, in_2, st_2, in_3, st_3, m, empty(fstsplit(m, app(map_f(three, head(in_3)), st_3))))
if_8(st_1, in_2, st_2, in_3, st_3, m, false) → ring(st_1, in_2, st_2, tail(in_3), sndsplit(m, app(map_f(three, head(in_3)), st_3)), m)
ring(st_1, in_2, st_2, in_3, st_3, m) → if_9(st_1, in_2, st_2, in_3, st_3, m, empty(map_f(three, head(in_3))))
if_9(st_1, in_2, st_2, in_3, st_3, m, true) → ring(st_1, in_2, st_2, tail(in_3), st_3, 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))
head(cons(x0, x1))
tail(cons(x0, x1))
ring(x0, x1, x2, x3, x4, x5)
if_1(x0, x1, x2, x3, x4, x5, false)
if_2(x0, x1, x2, x3, x4, x5, true)
if_3(x0, x1, x2, x3, x4, x5, false)
if_2(x0, x1, x2, x3, x4, x5, false)
if_4(x0, x1, x2, x3, x4, x5, false)
if_5(x0, x1, x2, x3, x4, x5, true)
if_6(x0, x1, x2, x3, x4, x5, true)
if_7(x0, x1, x2, x3, x4, x5, false)
if_6(x0, x1, x2, x3, x4, x5, false)
if_8(x0, x1, x2, x3, x4, x5, false)
if_9(x0, x1, x2, x3, x4, x5, true)

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))
head(cons(h, t)) → h
tail(cons(h, t)) → t
ring(st_1, in_2, st_2, in_3, st_3, m) → if_1(st_1, in_2, st_2, in_3, st_3, m, empty(fstsplit(m, st_1)))
if_1(st_1, in_2, st_2, in_3, st_3, m, false) → ring(sndsplit(m, st_1), cons(fstsplit(m, st_1), in_2), st_2, in_3, st_3, m)
ring(st_1, in_2, st_2, in_3, st_3, m) → if_2(st_1, in_2, st_2, in_3, st_3, m, leq(m, length(st_2)))
if_2(st_1, in_2, st_2, in_3, st_3, m, true) → if_3(st_1, in_2, st_2, in_3, st_3, m, empty(fstsplit(m, st_2)))
if_3(st_1, in_2, st_2, in_3, st_3, m, false) → ring(st_1, in_2, sndsplit(m, st_2), cons(fstsplit(m, st_2), in_3), st_3, m)
if_2(st_1, in_2, st_2, in_3, st_3, m, false) → if_4(st_1, in_2, st_2, in_3, st_3, m, empty(fstsplit(m, app(map_f(two, head(in_2)), st_2))))
if_4(st_1, in_2, st_2, in_3, st_3, m, false) → ring(st_1, tail(in_2), sndsplit(m, app(map_f(two, head(in_2)), st_2)), cons(fstsplit(m, app(map_f(two, head(in_2)), st_2)), in_3), st_3, m)
ring(st_1, in_2, st_2, in_3, st_3, m) → if_5(st_1, in_2, st_2, in_3, st_3, m, empty(map_f(two, head(in_2))))
if_5(st_1, in_2, st_2, in_3, st_3, m, true) → ring(st_1, tail(in_2), st_2, in_3, st_3, m)
ring(st_1, in_2, st_2, in_3, st_3, m) → if_6(st_1, in_2, st_2, in_3, st_3, m, leq(m, length(st_3)))
if_6(st_1, in_2, st_2, in_3, st_3, m, true) → if_7(st_1, in_2, st_2, in_3, st_3, m, empty(fstsplit(m, st_3)))
if_7(st_1, in_2, st_2, in_3, st_3, m, false) → ring(st_1, in_2, st_2, in_3, sndsplit(m, st_3), m)
if_6(st_1, in_2, st_2, in_3, st_3, m, false) → if_8(st_1, in_2, st_2, in_3, st_3, m, empty(fstsplit(m, app(map_f(three, head(in_3)), st_3))))
if_8(st_1, in_2, st_2, in_3, st_3, m, false) → ring(st_1, in_2, st_2, tail(in_3), sndsplit(m, app(map_f(three, head(in_3)), st_3)), m)
ring(st_1, in_2, st_2, in_3, st_3, m) → if_9(st_1, in_2, st_2, in_3, st_3, m, empty(map_f(three, head(in_3))))
if_9(st_1, in_2, st_2, in_3, st_3, m, true) → ring(st_1, in_2, st_2, tail(in_3), st_3, 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))
head(cons(x0, x1))
tail(cons(x0, x1))
ring(x0, x1, x2, x3, x4, x5)
if_1(x0, x1, x2, x3, x4, x5, false)
if_2(x0, x1, x2, x3, x4, x5, true)
if_3(x0, x1, x2, x3, x4, x5, false)
if_2(x0, x1, x2, x3, x4, x5, false)
if_4(x0, x1, x2, x3, x4, x5, false)
if_5(x0, x1, x2, x3, x4, x5, true)
if_6(x0, x1, x2, x3, x4, x5, true)
if_7(x0, x1, x2, x3, x4, x5, false)
if_6(x0, x1, x2, x3, x4, x5, false)
if_8(x0, x1, x2, x3, x4, x5, false)
if_9(x0, x1, x2, x3, x4, x5, true)

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

Lexicographic Path Order [LPO].
Precedence:
s1 > FSTSPLIT2

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))
head(cons(h, t)) → h
tail(cons(h, t)) → t
ring(st_1, in_2, st_2, in_3, st_3, m) → if_1(st_1, in_2, st_2, in_3, st_3, m, empty(fstsplit(m, st_1)))
if_1(st_1, in_2, st_2, in_3, st_3, m, false) → ring(sndsplit(m, st_1), cons(fstsplit(m, st_1), in_2), st_2, in_3, st_3, m)
ring(st_1, in_2, st_2, in_3, st_3, m) → if_2(st_1, in_2, st_2, in_3, st_3, m, leq(m, length(st_2)))
if_2(st_1, in_2, st_2, in_3, st_3, m, true) → if_3(st_1, in_2, st_2, in_3, st_3, m, empty(fstsplit(m, st_2)))
if_3(st_1, in_2, st_2, in_3, st_3, m, false) → ring(st_1, in_2, sndsplit(m, st_2), cons(fstsplit(m, st_2), in_3), st_3, m)
if_2(st_1, in_2, st_2, in_3, st_3, m, false) → if_4(st_1, in_2, st_2, in_3, st_3, m, empty(fstsplit(m, app(map_f(two, head(in_2)), st_2))))
if_4(st_1, in_2, st_2, in_3, st_3, m, false) → ring(st_1, tail(in_2), sndsplit(m, app(map_f(two, head(in_2)), st_2)), cons(fstsplit(m, app(map_f(two, head(in_2)), st_2)), in_3), st_3, m)
ring(st_1, in_2, st_2, in_3, st_3, m) → if_5(st_1, in_2, st_2, in_3, st_3, m, empty(map_f(two, head(in_2))))
if_5(st_1, in_2, st_2, in_3, st_3, m, true) → ring(st_1, tail(in_2), st_2, in_3, st_3, m)
ring(st_1, in_2, st_2, in_3, st_3, m) → if_6(st_1, in_2, st_2, in_3, st_3, m, leq(m, length(st_3)))
if_6(st_1, in_2, st_2, in_3, st_3, m, true) → if_7(st_1, in_2, st_2, in_3, st_3, m, empty(fstsplit(m, st_3)))
if_7(st_1, in_2, st_2, in_3, st_3, m, false) → ring(st_1, in_2, st_2, in_3, sndsplit(m, st_3), m)
if_6(st_1, in_2, st_2, in_3, st_3, m, false) → if_8(st_1, in_2, st_2, in_3, st_3, m, empty(fstsplit(m, app(map_f(three, head(in_3)), st_3))))
if_8(st_1, in_2, st_2, in_3, st_3, m, false) → ring(st_1, in_2, st_2, tail(in_3), sndsplit(m, app(map_f(three, head(in_3)), st_3)), m)
ring(st_1, in_2, st_2, in_3, st_3, m) → if_9(st_1, in_2, st_2, in_3, st_3, m, empty(map_f(three, head(in_3))))
if_9(st_1, in_2, st_2, in_3, st_3, m, true) → ring(st_1, in_2, st_2, tail(in_3), st_3, 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))
head(cons(x0, x1))
tail(cons(x0, x1))
ring(x0, x1, x2, x3, x4, x5)
if_1(x0, x1, x2, x3, x4, x5, false)
if_2(x0, x1, x2, x3, x4, x5, true)
if_3(x0, x1, x2, x3, x4, x5, false)
if_2(x0, x1, x2, x3, x4, x5, false)
if_4(x0, x1, x2, x3, x4, x5, false)
if_5(x0, x1, x2, x3, x4, x5, true)
if_6(x0, x1, x2, x3, x4, x5, true)
if_7(x0, x1, x2, x3, x4, x5, false)
if_6(x0, x1, x2, x3, x4, x5, false)
if_8(x0, x1, x2, x3, x4, x5, false)
if_9(x0, x1, x2, x3, x4, x5, true)

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:

RING(st_1, in_2, st_2, in_3, st_3, m) → IF_1(st_1, in_2, st_2, in_3, st_3, m, empty(fstsplit(m, st_1)))
IF_1(st_1, in_2, st_2, in_3, st_3, m, false) → RING(sndsplit(m, st_1), cons(fstsplit(m, st_1), in_2), st_2, in_3, st_3, m)
RING(st_1, in_2, st_2, in_3, st_3, m) → IF_2(st_1, in_2, st_2, in_3, st_3, m, leq(m, length(st_2)))
IF_2(st_1, in_2, st_2, in_3, st_3, m, true) → IF_3(st_1, in_2, st_2, in_3, st_3, m, empty(fstsplit(m, st_2)))
IF_3(st_1, in_2, st_2, in_3, st_3, m, false) → RING(st_1, in_2, sndsplit(m, st_2), cons(fstsplit(m, st_2), in_3), st_3, m)
RING(st_1, in_2, st_2, in_3, st_3, m) → IF_5(st_1, in_2, st_2, in_3, st_3, m, empty(map_f(two, head(in_2))))
IF_5(st_1, in_2, st_2, in_3, st_3, m, true) → RING(st_1, tail(in_2), st_2, in_3, st_3, m)
RING(st_1, in_2, st_2, in_3, st_3, m) → IF_6(st_1, in_2, st_2, in_3, st_3, m, leq(m, length(st_3)))
IF_6(st_1, in_2, st_2, in_3, st_3, m, true) → IF_7(st_1, in_2, st_2, in_3, st_3, m, empty(fstsplit(m, st_3)))
IF_7(st_1, in_2, st_2, in_3, st_3, m, false) → RING(st_1, in_2, st_2, in_3, sndsplit(m, st_3), m)
RING(st_1, in_2, st_2, in_3, st_3, m) → IF_9(st_1, in_2, st_2, in_3, st_3, m, empty(map_f(three, head(in_3))))
IF_9(st_1, in_2, st_2, in_3, st_3, m, true) → RING(st_1, in_2, st_2, tail(in_3), st_3, m)
IF_6(st_1, in_2, st_2, in_3, st_3, m, false) → IF_8(st_1, in_2, st_2, in_3, st_3, m, empty(fstsplit(m, app(map_f(three, head(in_3)), st_3))))
IF_8(st_1, in_2, st_2, in_3, st_3, m, false) → RING(st_1, in_2, st_2, tail(in_3), sndsplit(m, app(map_f(three, head(in_3)), st_3)), m)
IF_2(st_1, in_2, st_2, in_3, st_3, m, false) → IF_4(st_1, in_2, st_2, in_3, st_3, m, empty(fstsplit(m, app(map_f(two, head(in_2)), st_2))))
IF_4(st_1, in_2, st_2, in_3, st_3, m, false) → RING(st_1, tail(in_2), sndsplit(m, app(map_f(two, head(in_2)), st_2)), cons(fstsplit(m, app(map_f(two, head(in_2)), st_2)), in_3), st_3, 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))
head(cons(h, t)) → h
tail(cons(h, t)) → t
ring(st_1, in_2, st_2, in_3, st_3, m) → if_1(st_1, in_2, st_2, in_3, st_3, m, empty(fstsplit(m, st_1)))
if_1(st_1, in_2, st_2, in_3, st_3, m, false) → ring(sndsplit(m, st_1), cons(fstsplit(m, st_1), in_2), st_2, in_3, st_3, m)
ring(st_1, in_2, st_2, in_3, st_3, m) → if_2(st_1, in_2, st_2, in_3, st_3, m, leq(m, length(st_2)))
if_2(st_1, in_2, st_2, in_3, st_3, m, true) → if_3(st_1, in_2, st_2, in_3, st_3, m, empty(fstsplit(m, st_2)))
if_3(st_1, in_2, st_2, in_3, st_3, m, false) → ring(st_1, in_2, sndsplit(m, st_2), cons(fstsplit(m, st_2), in_3), st_3, m)
if_2(st_1, in_2, st_2, in_3, st_3, m, false) → if_4(st_1, in_2, st_2, in_3, st_3, m, empty(fstsplit(m, app(map_f(two, head(in_2)), st_2))))
if_4(st_1, in_2, st_2, in_3, st_3, m, false) → ring(st_1, tail(in_2), sndsplit(m, app(map_f(two, head(in_2)), st_2)), cons(fstsplit(m, app(map_f(two, head(in_2)), st_2)), in_3), st_3, m)
ring(st_1, in_2, st_2, in_3, st_3, m) → if_5(st_1, in_2, st_2, in_3, st_3, m, empty(map_f(two, head(in_2))))
if_5(st_1, in_2, st_2, in_3, st_3, m, true) → ring(st_1, tail(in_2), st_2, in_3, st_3, m)
ring(st_1, in_2, st_2, in_3, st_3, m) → if_6(st_1, in_2, st_2, in_3, st_3, m, leq(m, length(st_3)))
if_6(st_1, in_2, st_2, in_3, st_3, m, true) → if_7(st_1, in_2, st_2, in_3, st_3, m, empty(fstsplit(m, st_3)))
if_7(st_1, in_2, st_2, in_3, st_3, m, false) → ring(st_1, in_2, st_2, in_3, sndsplit(m, st_3), m)
if_6(st_1, in_2, st_2, in_3, st_3, m, false) → if_8(st_1, in_2, st_2, in_3, st_3, m, empty(fstsplit(m, app(map_f(three, head(in_3)), st_3))))
if_8(st_1, in_2, st_2, in_3, st_3, m, false) → ring(st_1, in_2, st_2, tail(in_3), sndsplit(m, app(map_f(three, head(in_3)), st_3)), m)
ring(st_1, in_2, st_2, in_3, st_3, m) → if_9(st_1, in_2, st_2, in_3, st_3, m, empty(map_f(three, head(in_3))))
if_9(st_1, in_2, st_2, in_3, st_3, m, true) → ring(st_1, in_2, st_2, tail(in_3), st_3, 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))
head(cons(x0, x1))
tail(cons(x0, x1))
ring(x0, x1, x2, x3, x4, x5)
if_1(x0, x1, x2, x3, x4, x5, false)
if_2(x0, x1, x2, x3, x4, x5, true)
if_3(x0, x1, x2, x3, x4, x5, false)
if_2(x0, x1, x2, x3, x4, x5, false)
if_4(x0, x1, x2, x3, x4, x5, false)
if_5(x0, x1, x2, x3, x4, x5, true)
if_6(x0, x1, x2, x3, x4, x5, true)
if_7(x0, x1, x2, x3, x4, x5, false)
if_6(x0, x1, x2, x3, x4, x5, false)
if_8(x0, x1, x2, x3, x4, x5, false)
if_9(x0, x1, x2, x3, x4, x5, true)

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