details

property	value
status	complete
benchmark	ex03_rec.jar-obl-8.smt2
ran by	Akihisa Yamada
cpu timeout	1200 seconds
wallclock timeout	300 seconds
memory limit	137438953472 bytes
execution host	n151.star.cs.uiowa.edu
space	From_AProVE_2014

run statistics

property	value
solver	Ctrl
configuration	Transition
runtime (wallclock)	0.691836 seconds
cpu usage	0.704329
user time	0.332155
system time	0.372174
max virtual memory	607464.0
max residence set size	8364.0

stage attributes

key	value
starexec-result	MAYBE

output

MAYBE DP problem for innermost termination. P = init#(x1, x2) -> f1#(rnd1, rnd2) f2#(I0, I1) -> f2#(I0 + 1, I2) [-5 <= I0 - 1 /\ I0 <= 0 /\ I0 <= -1] f2#(I3, I4) -> f2#(I3 + 1, I5) [I3 <= -5 - 1 /\ I3 <= 0 /\ I3 <= -1] f2#(I6, I7) -> f2#(-5, I8) [-5 = I6] f1#(I9, I10) -> f2#(-1 * I10, I11) [0 <= I9 - 1 /\ -1 <= I10 - 1 /\ -1 * I10 <= 0] R = init(x1, x2) -> f1(rnd1, rnd2) f2(I0, I1) -> f2(I0 + 1, I2) [-5 <= I0 - 1 /\ I0 <= 0 /\ I0 <= -1] f2(I3, I4) -> f2(I3 + 1, I5) [I3 <= -5 - 1 /\ I3 <= 0 /\ I3 <= -1] f2(I6, I7) -> f2(-5, I8) [-5 = I6] f1(I9, I10) -> f2(-1 * I10, I11) [0 <= I9 - 1 /\ -1 <= I10 - 1 /\ -1 * I10 <= 0] The dependency graph for this problem is: 0 -> 4 1 -> 1 2 -> 2, 3 3 -> 3 4 -> 1, 2, 3 Where: 0) init#(x1, x2) -> f1#(rnd1, rnd2) 1) f2#(I0, I1) -> f2#(I0 + 1, I2) [-5 <= I0 - 1 /\ I0 <= 0 /\ I0 <= -1] 2) f2#(I3, I4) -> f2#(I3 + 1, I5) [I3 <= -5 - 1 /\ I3 <= 0 /\ I3 <= -1] 3) f2#(I6, I7) -> f2#(-5, I8) [-5 = I6] 4) f1#(I9, I10) -> f2#(-1 * I10, I11) [0 <= I9 - 1 /\ -1 <= I10 - 1 /\ -1 * I10 <= 0] We have the following SCCs. { 2 } { 3 } { 1 } DP problem for innermost termination. P = f2#(I0, I1) -> f2#(I0 + 1, I2) [-5 <= I0 - 1 /\ I0 <= 0 /\ I0 <= -1] R = init(x1, x2) -> f1(rnd1, rnd2) f2(I0, I1) -> f2(I0 + 1, I2) [-5 <= I0 - 1 /\ I0 <= 0 /\ I0 <= -1] f2(I3, I4) -> f2(I3 + 1, I5) [I3 <= -5 - 1 /\ I3 <= 0 /\ I3 <= -1] f2(I6, I7) -> f2(-5, I8) [-5 = I6] f1(I9, I10) -> f2(-1 * I10, I11) [0 <= I9 - 1 /\ -1 <= I10 - 1 /\ -1 * I10 <= 0] We use the reverse value criterion with the projection function NU: NU[f2#(z1,z2)] = 0 + -1 * z1 This gives the following inequalities: -5 <= I0 - 1 /\ I0 <= 0 /\ I0 <= -1 ==> 0 + -1 * I0 > 0 + -1 * (I0 + 1) with 0 + -1 * I0 >= 0 All dependency pairs are strictly oriented, so the entire dependency pair problem may be removed. DP problem for innermost termination. P = f2#(I6, I7) -> f2#(-5, I8) [-5 = I6] R = init(x1, x2) -> f1(rnd1, rnd2) f2(I0, I1) -> f2(I0 + 1, I2) [-5 <= I0 - 1 /\ I0 <= 0 /\ I0 <= -1] f2(I3, I4) -> f2(I3 + 1, I5) [I3 <= -5 - 1 /\ I3 <= 0 /\ I3 <= -1] f2(I6, I7) -> f2(-5, I8) [-5 = I6] f1(I9, I10) -> f2(-1 * I10, I11) [0 <= I9 - 1 /\ -1 <= I10 - 1 /\ -1 * I10 <= 0] popout

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