details

property	value
status	complete
benchmark	TerminatorRec02.jar-obl-8.smt2
ran by	Akihisa Yamada
cpu timeout	1200 seconds
wallclock timeout	300 seconds
memory limit	137438953472 bytes
execution host	n150.star.cs.uiowa.edu
space	From_AProVE_2014

run statistics

property	value
solver	Ctrl
configuration	Transition
runtime (wallclock)	0.303835 seconds
cpu usage	0.302428
user time	0.170618
system time	0.13181
max virtual memory	113188.0
max residence set size	8400.0

stage attributes

key	value
starexec-result	YES

output

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

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