1.08/1.10	MAYBE
1.08/1.10	
1.08/1.10	DP problem for innermost termination.
1.08/1.10	P =
1.08/1.10	  isdiv#(x, y) -> isdiv#(0 - x, y) [0 > x]
1.08/1.10	  isdiv#(I0, I1) -> isdiv#(I0, 0 - I1) [0 > I1]
1.08/1.10	  isdiv#(I2, I3) -> isdiv#(I2, 0 - I2 + I3) [I3 >= I2 && I2 > 0]
1.08/1.10	  if_2#(false, I7, I8, zs) -> filter#(I7, zs) 
1.08/1.10	  filter#(I9, cons(I10, B0)) -> if_2#(isdiv(I9, I10), I9, I10, B0) 
1.08/1.10	  filter#(I9, cons(I10, B0)) -> isdiv#(I9, I10) 
1.08/1.10	  if_1#(true, I11, I12, B1) -> filter#(I11, B1) 
1.08/1.10	  filter#(I13, cons(I14, B2)) -> if_1#(isdiv(I13, I14), I13, I14, B2) 
1.08/1.10	  filter#(I13, cons(I14, B2)) -> isdiv#(I13, I14) 
1.08/1.10	  sieve#(cons(I16, ys)) -> sieve#(filter(I16, ys)) 
1.08/1.10	  sieve#(cons(I16, ys)) -> filter#(I16, ys) 
1.08/1.10	  nats#(I17, I18) -> nats#(I17 + 1, I18) [I18 > I17]
1.08/1.10	  primes#(I23) -> sieve#(nats(2, I23)) 
1.08/1.10	  primes#(I23) -> nats#(2, I23) 
1.08/1.10	  mem#(I24, cons(I25, B3)) -> mem#(I24, B3) [I24 > I25]
1.08/1.10	  mem#(I26, cons(I27, B4)) -> mem#(I26, B4) [I27 > I26]
1.08/1.10	  isprime#(I31) -> mem#(I31, primes(I31)) 
1.08/1.10	  isprime#(I31) -> primes#(I31) 
1.08/1.10	R = 
1.08/1.10	  isdiv(x, y) -> isdiv(0 - x, y) [0 > x]
1.08/1.10	  isdiv(I0, I1) -> isdiv(I0, 0 - I1) [0 > I1]
1.08/1.10	  isdiv(I2, I3) -> isdiv(I2, 0 - I2 + I3) [I3 >= I2 && I2 > 0]
1.08/1.10	  isdiv(I4, I5) -> false [I4 > I5 && I5 > 0]
1.08/1.10	  isdiv(I6, 0) -> true [I6 > 0]
1.08/1.10	  if_2(false, I7, I8, zs) -> cons(I7, filter(I7, zs)) 
1.08/1.10	  filter(I9, cons(I10, B0)) -> if_2(isdiv(I9, I10), I9, I10, B0) 
1.08/1.10	  if_1(true, I11, I12, B1) -> filter(I11, B1) 
1.08/1.10	  filter(I13, cons(I14, B2)) -> if_1(isdiv(I13, I14), I13, I14, B2) 
1.08/1.10	  filter(I15, nil) -> nil 
1.08/1.10	  sieve(cons(I16, ys)) -> cons(I16, sieve(filter(I16, ys))) 
1.08/1.10	  sieve(nil) -> nil 
1.08/1.10	  nats(I17, I18) -> cons(I17, nats(I17 + 1, I18)) [I18 > I17]
1.08/1.10	  nats(I19, I20) -> cons(I19, nil) [I19 = I20]
1.08/1.10	  nats(I21, I22) -> nil [I21 > I22]
1.08/1.10	  primes(I23) -> sieve(nats(2, I23)) 
1.08/1.10	  mem(I24, cons(I25, B3)) -> mem(I24, B3) [I24 > I25]
1.08/1.10	  mem(I26, cons(I27, B4)) -> mem(I26, B4) [I27 > I26]
1.08/1.10	  mem(I28, cons(I29, B5)) -> true [I28 = I29]
1.08/1.10	  mem(I30, nil) -> false 
1.08/1.10	  isprime(I31) -> mem(I31, primes(I31)) 
1.08/1.10	
1.08/1.10	The dependency graph for this problem is:
1.08/1.10	  0 -> 1, 2
1.08/1.10	  1 -> 0, 2
1.08/1.10	  2 -> 2
1.08/1.10	  3 -> 4, 5, 7, 8
1.08/1.10	  4 -> 3
1.08/1.10	  5 -> 0, 1, 2
1.08/1.10	  6 -> 4, 5, 7, 8
1.08/1.10	  7 -> 6
1.08/1.10	  8 -> 0, 1, 2
1.08/1.10	  9 -> 9, 10
1.08/1.10	  10 -> 4, 5, 7, 8
1.08/1.10	  11 -> 11
1.08/1.10	  12 -> 9, 10
1.08/1.10	  13 -> 11
1.08/1.10	  14 -> 14, 15
1.08/1.10	  15 -> 14, 15
1.08/1.10	  16 -> 14, 15
1.08/1.10	  17 -> 12, 13
1.08/1.10	Where:
1.08/1.10	  0) isdiv#(x, y) -> isdiv#(0 - x, y) [0 > x]
1.08/1.10	  1) isdiv#(I0, I1) -> isdiv#(I0, 0 - I1) [0 > I1]
1.08/1.10	  2) isdiv#(I2, I3) -> isdiv#(I2, 0 - I2 + I3) [I3 >= I2 && I2 > 0]
1.08/1.10	  3) if_2#(false, I7, I8, zs) -> filter#(I7, zs) 
1.08/1.10	  4) filter#(I9, cons(I10, B0)) -> if_2#(isdiv(I9, I10), I9, I10, B0) 
1.08/1.10	  5) filter#(I9, cons(I10, B0)) -> isdiv#(I9, I10) 
1.08/1.10	  6) if_1#(true, I11, I12, B1) -> filter#(I11, B1) 
1.08/1.10	  7) filter#(I13, cons(I14, B2)) -> if_1#(isdiv(I13, I14), I13, I14, B2) 
1.08/1.10	  8) filter#(I13, cons(I14, B2)) -> isdiv#(I13, I14) 
1.08/1.10	  9) sieve#(cons(I16, ys)) -> sieve#(filter(I16, ys)) 
1.08/1.10	  10) sieve#(cons(I16, ys)) -> filter#(I16, ys) 
1.08/1.10	  11) nats#(I17, I18) -> nats#(I17 + 1, I18) [I18 > I17]
1.08/1.10	  12) primes#(I23) -> sieve#(nats(2, I23)) 
1.08/1.10	  13) primes#(I23) -> nats#(2, I23) 
1.08/1.10	  14) mem#(I24, cons(I25, B3)) -> mem#(I24, B3) [I24 > I25]
1.08/1.10	  15) mem#(I26, cons(I27, B4)) -> mem#(I26, B4) [I27 > I26]
1.08/1.10	  16) isprime#(I31) -> mem#(I31, primes(I31)) 
1.08/1.10	  17) isprime#(I31) -> primes#(I31) 
1.08/1.10	
1.08/1.10	We have the following SCCs.
1.08/1.10	  { 14, 15 }
1.08/1.10	  { 11 }
1.08/1.10	  { 9 }
1.08/1.10	  { 3, 4, 6, 7 }
1.08/1.10	  { 0, 1 }
1.08/1.10	  { 2 }
1.08/1.10	
1.08/1.10	DP problem for innermost termination.
1.08/1.10	P =
1.08/1.10	  isdiv#(I2, I3) -> isdiv#(I2, 0 - I2 + I3) [I3 >= I2 && I2 > 0]
1.08/1.10	R = 
1.08/1.10	  isdiv(x, y) -> isdiv(0 - x, y) [0 > x]
1.08/1.10	  isdiv(I0, I1) -> isdiv(I0, 0 - I1) [0 > I1]
1.08/1.10	  isdiv(I2, I3) -> isdiv(I2, 0 - I2 + I3) [I3 >= I2 && I2 > 0]
1.08/1.10	  isdiv(I4, I5) -> false [I4 > I5 && I5 > 0]
1.08/1.10	  isdiv(I6, 0) -> true [I6 > 0]
1.08/1.10	  if_2(false, I7, I8, zs) -> cons(I7, filter(I7, zs)) 
1.08/1.10	  filter(I9, cons(I10, B0)) -> if_2(isdiv(I9, I10), I9, I10, B0) 
1.08/1.10	  if_1(true, I11, I12, B1) -> filter(I11, B1) 
1.08/1.10	  filter(I13, cons(I14, B2)) -> if_1(isdiv(I13, I14), I13, I14, B2) 
1.08/1.10	  filter(I15, nil) -> nil 
1.08/1.10	  sieve(cons(I16, ys)) -> cons(I16, sieve(filter(I16, ys))) 
1.08/1.10	  sieve(nil) -> nil 
1.08/1.10	  nats(I17, I18) -> cons(I17, nats(I17 + 1, I18)) [I18 > I17]
1.08/1.10	  nats(I19, I20) -> cons(I19, nil) [I19 = I20]
1.08/1.10	  nats(I21, I22) -> nil [I21 > I22]
1.08/1.10	  primes(I23) -> sieve(nats(2, I23)) 
1.08/1.10	  mem(I24, cons(I25, B3)) -> mem(I24, B3) [I24 > I25]
1.08/1.10	  mem(I26, cons(I27, B4)) -> mem(I26, B4) [I27 > I26]
1.08/1.10	  mem(I28, cons(I29, B5)) -> true [I28 = I29]
1.08/1.10	  mem(I30, nil) -> false 
1.08/1.10	  isprime(I31) -> mem(I31, primes(I31)) 
1.08/1.10	
1.08/1.10	We use the reverse value criterion with the projection function NU:
1.08/1.10	  NU[isdiv#(z1,z2)] = z2 + -1 * z1
1.08/1.10	
1.08/1.10	This gives the following inequalities:
1.08/1.10	  I3 >= I2 && I2 > 0  ==>  I3 + -1 * I2 > 0 - I2 + I3 + -1 * I2  with  I3 + -1 * I2 >= 0
1.08/1.10	
1.08/1.10	All dependency pairs are strictly oriented, so the entire dependency pair problem may be removed.
1.08/1.10	
1.08/1.10	DP problem for innermost termination.
1.08/1.10	P =
1.08/1.10	  isdiv#(x, y) -> isdiv#(0 - x, y) [0 > x]
1.08/1.10	  isdiv#(I0, I1) -> isdiv#(I0, 0 - I1) [0 > I1]
1.08/1.10	R = 
1.08/1.10	  isdiv(x, y) -> isdiv(0 - x, y) [0 > x]
1.08/1.10	  isdiv(I0, I1) -> isdiv(I0, 0 - I1) [0 > I1]
1.08/1.10	  isdiv(I2, I3) -> isdiv(I2, 0 - I2 + I3) [I3 >= I2 && I2 > 0]
1.08/1.10	  isdiv(I4, I5) -> false [I4 > I5 && I5 > 0]
1.08/1.10	  isdiv(I6, 0) -> true [I6 > 0]
1.08/1.10	  if_2(false, I7, I8, zs) -> cons(I7, filter(I7, zs)) 
1.08/1.10	  filter(I9, cons(I10, B0)) -> if_2(isdiv(I9, I10), I9, I10, B0) 
1.08/1.10	  if_1(true, I11, I12, B1) -> filter(I11, B1) 
1.08/1.10	  filter(I13, cons(I14, B2)) -> if_1(isdiv(I13, I14), I13, I14, B2) 
1.08/1.10	  filter(I15, nil) -> nil 
1.08/1.10	  sieve(cons(I16, ys)) -> cons(I16, sieve(filter(I16, ys))) 
1.08/1.10	  sieve(nil) -> nil 
1.08/1.10	  nats(I17, I18) -> cons(I17, nats(I17 + 1, I18)) [I18 > I17]
1.08/1.10	  nats(I19, I20) -> cons(I19, nil) [I19 = I20]
1.08/1.10	  nats(I21, I22) -> nil [I21 > I22]
1.08/1.10	  primes(I23) -> sieve(nats(2, I23)) 
1.08/1.10	  mem(I24, cons(I25, B3)) -> mem(I24, B3) [I24 > I25]
1.08/1.10	  mem(I26, cons(I27, B4)) -> mem(I26, B4) [I27 > I26]
1.08/1.10	  mem(I28, cons(I29, B5)) -> true [I28 = I29]
1.08/1.10	  mem(I30, nil) -> false 
1.08/1.10	  isprime(I31) -> mem(I31, primes(I31)) 
1.08/1.10	
1.08/1.10	We use the reverse value criterion with the projection function NU:
1.08/1.10	  NU[isdiv#(z1,z2)] = 0 + -1 * z1
1.08/1.10	
1.08/1.10	This gives the following inequalities:
1.08/1.10	  0 > x  ==>  0 + -1 * x > 0 + -1 * (0 - x)  with  0 + -1 * x >= 0
1.08/1.10	  0 > I1  ==>  0 + -1 * I0 >= 0 + -1 * I0
1.08/1.10	
1.08/1.10	We remove all the strictly oriented dependency pairs.
1.08/1.10	
1.08/1.10	DP problem for innermost termination.
1.08/1.10	P =
1.08/1.10	  isdiv#(I0, I1) -> isdiv#(I0, 0 - I1) [0 > I1]
1.08/1.10	R = 
1.08/1.10	  isdiv(x, y) -> isdiv(0 - x, y) [0 > x]
1.08/1.10	  isdiv(I0, I1) -> isdiv(I0, 0 - I1) [0 > I1]
1.08/1.10	  isdiv(I2, I3) -> isdiv(I2, 0 - I2 + I3) [I3 >= I2 && I2 > 0]
1.08/1.10	  isdiv(I4, I5) -> false [I4 > I5 && I5 > 0]
1.08/1.10	  isdiv(I6, 0) -> true [I6 > 0]
1.08/1.10	  if_2(false, I7, I8, zs) -> cons(I7, filter(I7, zs)) 
1.08/1.10	  filter(I9, cons(I10, B0)) -> if_2(isdiv(I9, I10), I9, I10, B0) 
1.08/1.10	  if_1(true, I11, I12, B1) -> filter(I11, B1) 
1.08/1.10	  filter(I13, cons(I14, B2)) -> if_1(isdiv(I13, I14), I13, I14, B2) 
1.08/1.10	  filter(I15, nil) -> nil 
1.08/1.10	  sieve(cons(I16, ys)) -> cons(I16, sieve(filter(I16, ys))) 
1.08/1.10	  sieve(nil) -> nil 
1.08/1.10	  nats(I17, I18) -> cons(I17, nats(I17 + 1, I18)) [I18 > I17]
1.08/1.10	  nats(I19, I20) -> cons(I19, nil) [I19 = I20]
1.08/1.10	  nats(I21, I22) -> nil [I21 > I22]
1.08/1.10	  primes(I23) -> sieve(nats(2, I23)) 
1.08/1.10	  mem(I24, cons(I25, B3)) -> mem(I24, B3) [I24 > I25]
1.08/1.10	  mem(I26, cons(I27, B4)) -> mem(I26, B4) [I27 > I26]
1.08/1.10	  mem(I28, cons(I29, B5)) -> true [I28 = I29]
1.08/1.10	  mem(I30, nil) -> false 
1.08/1.10	  isprime(I31) -> mem(I31, primes(I31)) 
1.08/1.10	
1.08/1.10	The dependency graph for this problem is:
1.08/1.10	  1 -> 
1.08/1.10	Where:
1.08/1.10	  1) isdiv#(I0, I1) -> isdiv#(I0, 0 - I1) [0 > I1]
1.08/1.10	
1.08/1.10	We have the following SCCs.
1.08/1.10	  
1.08/1.10	
1.08/1.10	DP problem for innermost termination.
1.08/1.10	P =
1.08/1.10	  if_2#(false, I7, I8, zs) -> filter#(I7, zs) 
1.08/1.10	  filter#(I9, cons(I10, B0)) -> if_2#(isdiv(I9, I10), I9, I10, B0) 
1.08/1.10	  if_1#(true, I11, I12, B1) -> filter#(I11, B1) 
1.08/1.10	  filter#(I13, cons(I14, B2)) -> if_1#(isdiv(I13, I14), I13, I14, B2) 
1.08/1.10	R = 
1.08/1.10	  isdiv(x, y) -> isdiv(0 - x, y) [0 > x]
1.08/1.10	  isdiv(I0, I1) -> isdiv(I0, 0 - I1) [0 > I1]
1.08/1.10	  isdiv(I2, I3) -> isdiv(I2, 0 - I2 + I3) [I3 >= I2 && I2 > 0]
1.08/1.10	  isdiv(I4, I5) -> false [I4 > I5 && I5 > 0]
1.08/1.10	  isdiv(I6, 0) -> true [I6 > 0]
1.08/1.10	  if_2(false, I7, I8, zs) -> cons(I7, filter(I7, zs)) 
1.08/1.10	  filter(I9, cons(I10, B0)) -> if_2(isdiv(I9, I10), I9, I10, B0) 
1.08/1.10	  if_1(true, I11, I12, B1) -> filter(I11, B1) 
1.08/1.10	  filter(I13, cons(I14, B2)) -> if_1(isdiv(I13, I14), I13, I14, B2) 
1.08/1.10	  filter(I15, nil) -> nil 
1.08/1.10	  sieve(cons(I16, ys)) -> cons(I16, sieve(filter(I16, ys))) 
1.08/1.10	  sieve(nil) -> nil 
1.08/1.10	  nats(I17, I18) -> cons(I17, nats(I17 + 1, I18)) [I18 > I17]
1.08/1.10	  nats(I19, I20) -> cons(I19, nil) [I19 = I20]
1.08/1.10	  nats(I21, I22) -> nil [I21 > I22]
1.08/1.10	  primes(I23) -> sieve(nats(2, I23)) 
1.08/1.10	  mem(I24, cons(I25, B3)) -> mem(I24, B3) [I24 > I25]
1.08/1.10	  mem(I26, cons(I27, B4)) -> mem(I26, B4) [I27 > I26]
1.08/1.10	  mem(I28, cons(I29, B5)) -> true [I28 = I29]
1.08/1.10	  mem(I30, nil) -> false 
1.08/1.10	  isprime(I31) -> mem(I31, primes(I31)) 
1.08/1.10	
1.08/1.10	We use the subterm criterion with the projection function NU:
1.08/1.10	  NU[if_2#(x0,x1,x2,x3)] = x3
1.08/1.10	  NU[filter#(x0,x1)] = x1
1.08/1.10	  NU[if_1#(x0,x1,x2,x3)] = x3
1.08/1.10	
1.08/1.10	This gives the following inequalities:
1.08/1.10	  zs = zs
1.08/1.10	  cons(I10, B0) |> B0
1.08/1.10	  B1 = B1
1.08/1.10	  cons(I14, B2) |> B2
1.08/1.10	
1.08/1.10	We remove all the strictly oriented dependency pairs.
1.08/1.10	
1.08/1.10	DP problem for innermost termination.
1.08/1.10	P =
1.08/1.10	  if_2#(false, I7, I8, zs) -> filter#(I7, zs) 
1.08/1.10	  if_1#(true, I11, I12, B1) -> filter#(I11, B1) 
1.08/1.10	R = 
1.08/1.10	  isdiv(x, y) -> isdiv(0 - x, y) [0 > x]
1.08/1.10	  isdiv(I0, I1) -> isdiv(I0, 0 - I1) [0 > I1]
1.08/1.10	  isdiv(I2, I3) -> isdiv(I2, 0 - I2 + I3) [I3 >= I2 && I2 > 0]
1.08/1.10	  isdiv(I4, I5) -> false [I4 > I5 && I5 > 0]
1.08/1.10	  isdiv(I6, 0) -> true [I6 > 0]
1.08/1.10	  if_2(false, I7, I8, zs) -> cons(I7, filter(I7, zs)) 
1.08/1.10	  filter(I9, cons(I10, B0)) -> if_2(isdiv(I9, I10), I9, I10, B0) 
1.08/1.10	  if_1(true, I11, I12, B1) -> filter(I11, B1) 
1.08/1.10	  filter(I13, cons(I14, B2)) -> if_1(isdiv(I13, I14), I13, I14, B2) 
1.08/1.10	  filter(I15, nil) -> nil 
1.08/1.10	  sieve(cons(I16, ys)) -> cons(I16, sieve(filter(I16, ys))) 
1.08/1.10	  sieve(nil) -> nil 
1.08/1.10	  nats(I17, I18) -> cons(I17, nats(I17 + 1, I18)) [I18 > I17]
1.08/1.10	  nats(I19, I20) -> cons(I19, nil) [I19 = I20]
1.08/1.10	  nats(I21, I22) -> nil [I21 > I22]
1.08/1.10	  primes(I23) -> sieve(nats(2, I23)) 
1.08/1.10	  mem(I24, cons(I25, B3)) -> mem(I24, B3) [I24 > I25]
1.08/1.10	  mem(I26, cons(I27, B4)) -> mem(I26, B4) [I27 > I26]
1.08/1.10	  mem(I28, cons(I29, B5)) -> true [I28 = I29]
1.08/1.10	  mem(I30, nil) -> false 
1.08/1.10	  isprime(I31) -> mem(I31, primes(I31)) 
1.08/1.10	
1.08/1.10	The dependency graph for this problem is:
1.08/1.10	  3 -> 
1.08/1.10	  6 -> 
1.08/1.10	Where:
1.08/1.10	  3) if_2#(false, I7, I8, zs) -> filter#(I7, zs) 
1.08/1.10	  6) if_1#(true, I11, I12, B1) -> filter#(I11, B1) 
1.08/1.10	
1.08/1.10	We have the following SCCs.
1.08/1.10	  
1.08/1.10	
1.08/1.10	DP problem for innermost termination.
1.08/1.10	P =
1.08/1.10	  sieve#(cons(I16, ys)) -> sieve#(filter(I16, ys)) 
1.08/1.10	R = 
1.08/1.10	  isdiv(x, y) -> isdiv(0 - x, y) [0 > x]
1.08/1.10	  isdiv(I0, I1) -> isdiv(I0, 0 - I1) [0 > I1]
1.08/1.10	  isdiv(I2, I3) -> isdiv(I2, 0 - I2 + I3) [I3 >= I2 && I2 > 0]
1.08/1.10	  isdiv(I4, I5) -> false [I4 > I5 && I5 > 0]
1.08/1.10	  isdiv(I6, 0) -> true [I6 > 0]
1.08/1.10	  if_2(false, I7, I8, zs) -> cons(I7, filter(I7, zs)) 
1.08/1.10	  filter(I9, cons(I10, B0)) -> if_2(isdiv(I9, I10), I9, I10, B0) 
1.08/1.10	  if_1(true, I11, I12, B1) -> filter(I11, B1) 
1.08/1.10	  filter(I13, cons(I14, B2)) -> if_1(isdiv(I13, I14), I13, I14, B2) 
1.08/1.10	  filter(I15, nil) -> nil 
1.08/1.10	  sieve(cons(I16, ys)) -> cons(I16, sieve(filter(I16, ys))) 
1.08/1.10	  sieve(nil) -> nil 
1.08/1.10	  nats(I17, I18) -> cons(I17, nats(I17 + 1, I18)) [I18 > I17]
1.08/1.10	  nats(I19, I20) -> cons(I19, nil) [I19 = I20]
1.08/1.10	  nats(I21, I22) -> nil [I21 > I22]
1.08/1.10	  primes(I23) -> sieve(nats(2, I23)) 
1.08/1.10	  mem(I24, cons(I25, B3)) -> mem(I24, B3) [I24 > I25]
1.08/1.10	  mem(I26, cons(I27, B4)) -> mem(I26, B4) [I27 > I26]
1.08/1.10	  mem(I28, cons(I29, B5)) -> true [I28 = I29]
1.08/1.10	  mem(I30, nil) -> false 
1.08/1.10	  isprime(I31) -> mem(I31, primes(I31)) 
1.08/1.10	
1.08/1.10	EOF