details

property	value
status	complete
benchmark	IntRTA.jar-obl-8.smt2
ran by	Akihisa Yamada
cpu timeout	1200 seconds
wallclock timeout	300 seconds
memory limit	137438953472 bytes
execution host	n143.star.cs.uiowa.edu
space	From_AProVE_2014

run statistics

property	value
solver	Ctrl
configuration	Transition
runtime (wallclock)	0.644052 seconds
cpu usage	0.655061
user time	0.348764
system time	0.306297
max virtual memory	123984.0
max residence set size	8568.0

stage attributes

key	value
starexec-result	YES

output

YES DP problem for innermost termination. P = init#(x1, x2, x3, x4) -> f1#(rnd1, rnd2, rnd3, rnd4) f2#(I0, I1, I2, I3) -> f2#(I4, I5, I2, I3 + 1) [I3 + 2 <= I1 /\ I2 + 2 <= I0 /\ 0 <= I5 - 1 /\ 0 <= I4 - 1 /\ 0 <= I1 - 1 /\ 0 <= I0 - 1 /\ I5 - 1 <= I1 /\ I4 <= I0 /\ I3 <= I2 - 1 /\ -1 <= I3 - 1] f1#(I6, I7, I8, I9) -> f2#(I10, I11, I12, I13) [0 <= I11 - 1 /\ 0 <= I10 - 1 /\ 0 <= I6 - 1 /\ -1 <= I13 - 1 /\ -1 <= I7 - 1 /\ -1 <= I12 - 1] R = init(x1, x2, x3, x4) -> f1(rnd1, rnd2, rnd3, rnd4) f2(I0, I1, I2, I3) -> f2(I4, I5, I2, I3 + 1) [I3 + 2 <= I1 /\ I2 + 2 <= I0 /\ 0 <= I5 - 1 /\ 0 <= I4 - 1 /\ 0 <= I1 - 1 /\ 0 <= I0 - 1 /\ I5 - 1 <= I1 /\ I4 <= I0 /\ I3 <= I2 - 1 /\ -1 <= I3 - 1] f1(I6, I7, I8, I9) -> f2(I10, I11, I12, I13) [0 <= I11 - 1 /\ 0 <= I10 - 1 /\ 0 <= I6 - 1 /\ -1 <= I13 - 1 /\ -1 <= I7 - 1 /\ -1 <= I12 - 1] The dependency graph for this problem is: 0 -> 2 1 -> 1 2 -> 1 Where: 0) init#(x1, x2, x3, x4) -> f1#(rnd1, rnd2, rnd3, rnd4) 1) f2#(I0, I1, I2, I3) -> f2#(I4, I5, I2, I3 + 1) [I3 + 2 <= I1 /\ I2 + 2 <= I0 /\ 0 <= I5 - 1 /\ 0 <= I4 - 1 /\ 0 <= I1 - 1 /\ 0 <= I0 - 1 /\ I5 - 1 <= I1 /\ I4 <= I0 /\ I3 <= I2 - 1 /\ -1 <= I3 - 1] 2) f1#(I6, I7, I8, I9) -> f2#(I10, I11, I12, I13) [0 <= I11 - 1 /\ 0 <= I10 - 1 /\ 0 <= I6 - 1 /\ -1 <= I13 - 1 /\ -1 <= I7 - 1 /\ -1 <= I12 - 1] We have the following SCCs. { 1 } DP problem for innermost termination. P = f2#(I0, I1, I2, I3) -> f2#(I4, I5, I2, I3 + 1) [I3 + 2 <= I1 /\ I2 + 2 <= I0 /\ 0 <= I5 - 1 /\ 0 <= I4 - 1 /\ 0 <= I1 - 1 /\ 0 <= I0 - 1 /\ I5 - 1 <= I1 /\ I4 <= I0 /\ I3 <= I2 - 1 /\ -1 <= I3 - 1] R = init(x1, x2, x3, x4) -> f1(rnd1, rnd2, rnd3, rnd4) f2(I0, I1, I2, I3) -> f2(I4, I5, I2, I3 + 1) [I3 + 2 <= I1 /\ I2 + 2 <= I0 /\ 0 <= I5 - 1 /\ 0 <= I4 - 1 /\ 0 <= I1 - 1 /\ 0 <= I0 - 1 /\ I5 - 1 <= I1 /\ I4 <= I0 /\ I3 <= I2 - 1 /\ -1 <= I3 - 1] f1(I6, I7, I8, I9) -> f2(I10, I11, I12, I13) [0 <= I11 - 1 /\ 0 <= I10 - 1 /\ 0 <= I6 - 1 /\ -1 <= I13 - 1 /\ -1 <= I7 - 1 /\ -1 <= I12 - 1] We use the reverse value criterion with the projection function NU: NU[f2#(z1,z2,z3,z4)] = z3 - 1 + -1 * z4 This gives the following inequalities: I3 + 2 <= I1 /\ I2 + 2 <= I0 /\ 0 <= I5 - 1 /\ 0 <= I4 - 1 /\ 0 <= I1 - 1 /\ 0 <= I0 - 1 /\ I5 - 1 <= I1 /\ I4 <= I0 /\ I3 <= I2 - 1 /\ -1 <= I3 - 1 ==> I2 - 1 + -1 * I3 > I2 - 1 + -1 * (I3 + 1) with I2 - 1 + -1 * I3 >= 0 All dependency pairs are strictly oriented, so the entire dependency pair problem may be removed. popout

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