details

property	value
status	complete
benchmark	Velroyen08-alternatingIncr.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)	2.48489 seconds
cpu usage	2.52231
user time	1.26097
system time	1.26133
max virtual memory	692256.0
max residence set size	8708.0

stage attributes

key	value
starexec-result	MAYBE

output

MAYBE DP problem for innermost termination. P = init#(x1, x2) -> f1#(rnd1, rnd2) f3#(I0, I1) -> f2#(I0 + 3, I2) [I0 - 2 * y1 = 1 /\ 0 <= I0 - 1 /\ I0 - 2 * y1 <= 1 /\ 0 <= I0 - 2 * y1] f2#(I3, I4) -> f3#(I3, I5) [0 <= I3 - 1 /\ I3 - 2 * I6 = 1] f3#(I7, I8) -> f2#(I7 - 1, I9) [0 <= I7 - 1 /\ I7 - 2 * I10 = 0 /\ I7 - 2 * I10 <= 1 /\ 0 <= I7 - 2 * I10] f2#(I11, I12) -> f3#(I11, I13) [I11 - 2 * I14 = 0 /\ 0 <= I11 - 1] f1#(I15, I16) -> f2#(I16, I17) [-1 <= I16 - 1 /\ 0 <= I15 - 1] R = init(x1, x2) -> f1(rnd1, rnd2) f3(I0, I1) -> f2(I0 + 3, I2) [I0 - 2 * y1 = 1 /\ 0 <= I0 - 1 /\ I0 - 2 * y1 <= 1 /\ 0 <= I0 - 2 * y1] f2(I3, I4) -> f3(I3, I5) [0 <= I3 - 1 /\ I3 - 2 * I6 = 1] f3(I7, I8) -> f2(I7 - 1, I9) [0 <= I7 - 1 /\ I7 - 2 * I10 = 0 /\ I7 - 2 * I10 <= 1 /\ 0 <= I7 - 2 * I10] f2(I11, I12) -> f3(I11, I13) [I11 - 2 * I14 = 0 /\ 0 <= I11 - 1] f1(I15, I16) -> f2(I16, I17) [-1 <= I16 - 1 /\ 0 <= I15 - 1] The dependency graph for this problem is: 0 -> 5 1 -> 4 2 -> 1 3 -> 2 4 -> 3 5 -> 2, 4 Where: 0) init#(x1, x2) -> f1#(rnd1, rnd2) 1) f3#(I0, I1) -> f2#(I0 + 3, I2) [I0 - 2 * y1 = 1 /\ 0 <= I0 - 1 /\ I0 - 2 * y1 <= 1 /\ 0 <= I0 - 2 * y1] 2) f2#(I3, I4) -> f3#(I3, I5) [0 <= I3 - 1 /\ I3 - 2 * I6 = 1] 3) f3#(I7, I8) -> f2#(I7 - 1, I9) [0 <= I7 - 1 /\ I7 - 2 * I10 = 0 /\ I7 - 2 * I10 <= 1 /\ 0 <= I7 - 2 * I10] 4) f2#(I11, I12) -> f3#(I11, I13) [I11 - 2 * I14 = 0 /\ 0 <= I11 - 1] 5) f1#(I15, I16) -> f2#(I16, I17) [-1 <= I16 - 1 /\ 0 <= I15 - 1] We have the following SCCs. { 1, 2, 3, 4 } DP problem for innermost termination. P = f3#(I0, I1) -> f2#(I0 + 3, I2) [I0 - 2 * y1 = 1 /\ 0 <= I0 - 1 /\ I0 - 2 * y1 <= 1 /\ 0 <= I0 - 2 * y1] f2#(I3, I4) -> f3#(I3, I5) [0 <= I3 - 1 /\ I3 - 2 * I6 = 1] f3#(I7, I8) -> f2#(I7 - 1, I9) [0 <= I7 - 1 /\ I7 - 2 * I10 = 0 /\ I7 - 2 * I10 <= 1 /\ 0 <= I7 - 2 * I10] f2#(I11, I12) -> f3#(I11, I13) [I11 - 2 * I14 = 0 /\ 0 <= I11 - 1] R = init(x1, x2) -> f1(rnd1, rnd2) f3(I0, I1) -> f2(I0 + 3, I2) [I0 - 2 * y1 = 1 /\ 0 <= I0 - 1 /\ I0 - 2 * y1 <= 1 /\ 0 <= I0 - 2 * y1] f2(I3, I4) -> f3(I3, I5) [0 <= I3 - 1 /\ I3 - 2 * I6 = 1] f3(I7, I8) -> f2(I7 - 1, I9) [0 <= I7 - 1 /\ I7 - 2 * I10 = 0 /\ I7 - 2 * I10 <= 1 /\ 0 <= I7 - 2 * I10] f2(I11, I12) -> f3(I11, I13) [I11 - 2 * I14 = 0 /\ 0 <= I11 - 1] f1(I15, I16) -> f2(I16, I17) [-1 <= I16 - 1 /\ 0 <= I15 - 1] popout

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actions

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