details

property	value
status	complete
benchmark	eratosthenes_small.itrs
ran by	Akihisa Yamada
cpu timeout	1200 seconds
wallclock timeout	300 seconds
memory limit	137438953472 bytes
execution host	n146.star.cs.uiowa.edu
space	Mixed_ITRS_2014

run statistics

property	value
solver	Ctrl
configuration	Itrs
runtime (wallclock)	0.532814 seconds
cpu usage	0.540205
user time	0.304082
system time	0.236123
max virtual memory	123856.0
max residence set size	8652.0

stage attributes

key	value
starexec-result	MAYBE

output

MAYBE DP problem for innermost termination. P = isdiv#(x, y) -> isdiv#(x, y - x) [y >= x && x > 0] if#(false, I3, I4, zs) -> filter#(I3, zs) if#(true, I5, I6, B0) -> filter#(I5, B0) filter#(I7, cons(I8, B1)) -> if#(isdiv(I7, I8), I7, I8, B1) filter#(I7, cons(I8, B1)) -> isdiv#(I7, I8) sieve#(cons(I10, ys)) -> sieve#(filter(I10, ys)) sieve#(cons(I10, ys)) -> filter#(I10, ys) nats#(I11, I12) -> nats#(I11 + 1, I12) [I12 >= I11] primes#(I15) -> sieve#(nats(2, I15)) primes#(I15) -> nats#(2, I15) R = isdiv(x, y) -> isdiv(x, y - x) [y >= x && x > 0] isdiv(I0, I1) -> false [I0 > I1 && I1 > 0] isdiv(I2, 0) -> true [I2 > 0] if(false, I3, I4, zs) -> cons(I4, filter(I3, zs)) if(true, I5, I6, B0) -> filter(I5, B0) filter(I7, cons(I8, B1)) -> if(isdiv(I7, I8), I7, I8, B1) filter(I9, nil) -> nil sieve(cons(I10, ys)) -> cons(I10, sieve(filter(I10, ys))) sieve(nil) -> nil nats(I11, I12) -> cons(I11, nats(I11 + 1, I12)) [I12 >= I11] nats(I13, I14) -> nil [I13 > I14] primes(I15) -> sieve(nats(2, I15)) The dependency graph for this problem is: 0 -> 0 1 -> 3, 4 2 -> 3, 4 3 -> 1, 2 4 -> 0 5 -> 5, 6 6 -> 3, 4 7 -> 7 8 -> 5, 6 9 -> 7 Where: 0) isdiv#(x, y) -> isdiv#(x, y - x) [y >= x && x > 0] 1) if#(false, I3, I4, zs) -> filter#(I3, zs) 2) if#(true, I5, I6, B0) -> filter#(I5, B0) 3) filter#(I7, cons(I8, B1)) -> if#(isdiv(I7, I8), I7, I8, B1) 4) filter#(I7, cons(I8, B1)) -> isdiv#(I7, I8) 5) sieve#(cons(I10, ys)) -> sieve#(filter(I10, ys)) 6) sieve#(cons(I10, ys)) -> filter#(I10, ys) 7) nats#(I11, I12) -> nats#(I11 + 1, I12) [I12 >= I11] 8) primes#(I15) -> sieve#(nats(2, I15)) 9) primes#(I15) -> nats#(2, I15) We have the following SCCs. { 5 } { 1, 2, 3 } { 0 } { 7 } DP problem for innermost termination. P = nats#(I11, I12) -> nats#(I11 + 1, I12) [I12 >= I11] R = isdiv(x, y) -> isdiv(x, y - x) [y >= x && x > 0] isdiv(I0, I1) -> false [I0 > I1 && I1 > 0] isdiv(I2, 0) -> true [I2 > 0] if(false, I3, I4, zs) -> cons(I4, filter(I3, zs)) if(true, I5, I6, B0) -> filter(I5, B0) filter(I7, cons(I8, B1)) -> if(isdiv(I7, I8), I7, I8, B1) filter(I9, nil) -> nil sieve(cons(I10, ys)) -> cons(I10, sieve(filter(I10, ys))) sieve(nil) -> nil nats(I11, I12) -> cons(I11, nats(I11 + 1, I12)) [I12 >= I11] nats(I13, I14) -> nil [I13 > I14] primes(I15) -> sieve(nats(2, I15)) We use the reverse value criterion with the projection function NU: NU[nats#(z1,z2)] = z2 + -1 * z1 This gives the following inequalities: I12 >= I11 ==> I12 + -1 * I11 > I12 + -1 * (I11 + 1) with I12 + -1 * I11 >= 0 All dependency pairs are strictly oriented, so the entire dependency pair problem may be removed. DP problem for innermost termination. P = isdiv#(x, y) -> isdiv#(x, y - x) [y >= x && x > 0] R = isdiv(x, y) -> isdiv(x, y - x) [y >= x && x > 0] isdiv(I0, I1) -> false [I0 > I1 && I1 > 0] isdiv(I2, 0) -> true [I2 > 0] if(false, I3, I4, zs) -> cons(I4, filter(I3, zs)) if(true, I5, I6, B0) -> filter(I5, B0) filter(I7, cons(I8, B1)) -> if(isdiv(I7, I8), I7, I8, B1) filter(I9, nil) -> nil sieve(cons(I10, ys)) -> cons(I10, sieve(filter(I10, ys))) sieve(nil) -> nil nats(I11, I12) -> cons(I11, nats(I11 + 1, I12)) [I12 >= I11] nats(I13, I14) -> nil [I13 > I14] primes(I15) -> sieve(nats(2, I15)) We use the reverse value criterion with the projection function NU: popout

output may be truncated. 'popout' for the full output.

job log

popout

actions

all output return to Integer TRS Innermost