details

property	value
status	complete
benchmark	TerminatorRec04.jar-obl-9.smt2
ran by	Akihisa Yamada
cpu timeout	1200 seconds
wallclock timeout	300 seconds
memory limit	137438953472 bytes
execution host	n142.star.cs.uiowa.edu
space	From_AProVE_2014

run statistics

property	value
solver	Ctrl
configuration	Transition
runtime (wallclock)	0.507107 seconds
cpu usage	0.517385
user time	0.276298
system time	0.241087
max virtual memory	113188.0
max residence set size	8388.0

stage attributes

key	value
starexec-result	MAYBE

output

MAYBE DP problem for innermost termination. P = init#(x1, x2) -> f1#(rnd1, rnd2) f3#(I0, I1) -> f3#(I0 - 1, I2) [I0 - 1 <= I0 - 1 /\ 0 <= I0 - 1] f2#(I3, I4) -> f2#(I3 + 1, I5) [0 <= I3 - 1] f2#(I6, I7) -> f3#(I6, I8) [0 <= I6 - 1] f1#(I9, I10) -> f2#(I10, I11) [-1 <= I10 - 1 /\ 0 <= I9 - 1] R = init(x1, x2) -> f1(rnd1, rnd2) f3(I0, I1) -> f3(I0 - 1, I2) [I0 - 1 <= I0 - 1 /\ 0 <= I0 - 1] f2(I3, I4) -> f2(I3 + 1, I5) [0 <= I3 - 1] f2(I6, I7) -> f3(I6, I8) [0 <= I6 - 1] f1(I9, I10) -> f2(I10, I11) [-1 <= I10 - 1 /\ 0 <= I9 - 1] The dependency graph for this problem is: 0 -> 4 1 -> 1 2 -> 2, 3 3 -> 1 4 -> 2, 3 Where: 0) init#(x1, x2) -> f1#(rnd1, rnd2) 1) f3#(I0, I1) -> f3#(I0 - 1, I2) [I0 - 1 <= I0 - 1 /\ 0 <= I0 - 1] 2) f2#(I3, I4) -> f2#(I3 + 1, I5) [0 <= I3 - 1] 3) f2#(I6, I7) -> f3#(I6, I8) [0 <= I6 - 1] 4) f1#(I9, I10) -> f2#(I10, I11) [-1 <= I10 - 1 /\ 0 <= I9 - 1] We have the following SCCs. { 2 } { 1 } DP problem for innermost termination. P = f3#(I0, I1) -> f3#(I0 - 1, I2) [I0 - 1 <= I0 - 1 /\ 0 <= I0 - 1] R = init(x1, x2) -> f1(rnd1, rnd2) f3(I0, I1) -> f3(I0 - 1, I2) [I0 - 1 <= I0 - 1 /\ 0 <= I0 - 1] f2(I3, I4) -> f2(I3 + 1, I5) [0 <= I3 - 1] f2(I6, I7) -> f3(I6, I8) [0 <= I6 - 1] f1(I9, I10) -> f2(I10, I11) [-1 <= I10 - 1 /\ 0 <= I9 - 1] We use the basic value criterion with the projection function NU: NU[f3#(z1,z2)] = z1 This gives the following inequalities: I0 - 1 <= I0 - 1 /\ 0 <= I0 - 1 ==> I0 >! I0 - 1 All dependency pairs are strictly oriented, so the entire dependency pair problem may be removed. DP problem for innermost termination. P = f2#(I3, I4) -> f2#(I3 + 1, I5) [0 <= I3 - 1] R = init(x1, x2) -> f1(rnd1, rnd2) f3(I0, I1) -> f3(I0 - 1, I2) [I0 - 1 <= I0 - 1 /\ 0 <= I0 - 1] f2(I3, I4) -> f2(I3 + 1, I5) [0 <= I3 - 1] f2(I6, I7) -> f3(I6, I8) [0 <= I6 - 1] f1(I9, I10) -> f2(I10, I11) [-1 <= I10 - 1 /\ 0 <= I9 - 1] popout

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