details

property	value
status	complete
benchmark	AProVEMath.jar-obl-8.smt2
ran by	Akihisa Yamada
cpu timeout	1200 seconds
wallclock timeout	300 seconds
memory limit	137438953472 bytes
execution host	n139.star.cs.uiowa.edu
space	From_AProVE_2014

run statistics

property	value
solver	Ctrl
configuration	Transition
runtime (wallclock)	1.20628 seconds
cpu usage	1.21063
user time	0.646382
system time	0.564248
max virtual memory	721552.0
max residence set size	8572.0

stage attributes

key	value
starexec-result	YES

output

YES DP problem for innermost termination. P = init#(x1, x2) -> f1#(rnd1, rnd2) f3#(I0, I1) -> f2#(I2, I3) [I0 - 2 * y1 = 1 /\ 0 <= I0 - 1 /\ I2 <= I0 - 1 /\ 0 <= I0 - 2 * y1 /\ I0 - 2 * y1 <= 1 /\ I0 - 2 * I2 <= 1 /\ 0 <= I0 - 2 * I2] f2#(I4, I5) -> f3#(I4, I6) [I4 - 2 * I7 = 1 /\ y2 <= I4 - 1 /\ 0 <= I4 - 1] f3#(I8, I9) -> f2#(I10, I11) [I8 - 2 * I12 = 0 /\ 0 <= I8 - 1 /\ I10 <= I8 - 1 /\ 0 <= I8 - 2 * I12 /\ I8 - 2 * I12 <= 1 /\ I8 - 2 * I10 <= 1 /\ 0 <= I8 - 2 * I10] f2#(I13, I14) -> f3#(I13, I15) [I13 - 2 * I16 = 0 /\ I17 <= I13 - 1 /\ 0 <= I13 - 1] f1#(I18, I19) -> f2#(I20, I21) [-1 <= I19 - 1 /\ 1 <= I20 - 1 /\ 2 <= I22 - 1 /\ 0 <= I18 - 1] f1#(I23, I24) -> f2#(I25, I26) [-1 <= I24 - 1 /\ 1 <= I25 - 1 /\ I27 <= 1 /\ -1 <= I27 - 1 /\ 0 <= I23 - 1] R = init(x1, x2) -> f1(rnd1, rnd2) f3(I0, I1) -> f2(I2, I3) [I0 - 2 * y1 = 1 /\ 0 <= I0 - 1 /\ I2 <= I0 - 1 /\ 0 <= I0 - 2 * y1 /\ I0 - 2 * y1 <= 1 /\ I0 - 2 * I2 <= 1 /\ 0 <= I0 - 2 * I2] f2(I4, I5) -> f3(I4, I6) [I4 - 2 * I7 = 1 /\ y2 <= I4 - 1 /\ 0 <= I4 - 1] f3(I8, I9) -> f2(I10, I11) [I8 - 2 * I12 = 0 /\ 0 <= I8 - 1 /\ I10 <= I8 - 1 /\ 0 <= I8 - 2 * I12 /\ I8 - 2 * I12 <= 1 /\ I8 - 2 * I10 <= 1 /\ 0 <= I8 - 2 * I10] f2(I13, I14) -> f3(I13, I15) [I13 - 2 * I16 = 0 /\ I17 <= I13 - 1 /\ 0 <= I13 - 1] f1(I18, I19) -> f2(I20, I21) [-1 <= I19 - 1 /\ 1 <= I20 - 1 /\ 2 <= I22 - 1 /\ 0 <= I18 - 1] f1(I23, I24) -> f2(I25, I26) [-1 <= I24 - 1 /\ 1 <= I25 - 1 /\ I27 <= 1 /\ -1 <= I27 - 1 /\ 0 <= I23 - 1] The dependency graph for this problem is: 0 -> 5, 6 1 -> 2, 4 2 -> 1 3 -> 2, 4 4 -> 3 5 -> 2, 4 6 -> 2, 4 Where: 0) init#(x1, x2) -> f1#(rnd1, rnd2) 1) f3#(I0, I1) -> f2#(I2, I3) [I0 - 2 * y1 = 1 /\ 0 <= I0 - 1 /\ I2 <= I0 - 1 /\ 0 <= I0 - 2 * y1 /\ I0 - 2 * y1 <= 1 /\ I0 - 2 * I2 <= 1 /\ 0 <= I0 - 2 * I2] 2) f2#(I4, I5) -> f3#(I4, I6) [I4 - 2 * I7 = 1 /\ y2 <= I4 - 1 /\ 0 <= I4 - 1] 3) f3#(I8, I9) -> f2#(I10, I11) [I8 - 2 * I12 = 0 /\ 0 <= I8 - 1 /\ I10 <= I8 - 1 /\ 0 <= I8 - 2 * I12 /\ I8 - 2 * I12 <= 1 /\ I8 - 2 * I10 <= 1 /\ 0 <= I8 - 2 * I10] 4) f2#(I13, I14) -> f3#(I13, I15) [I13 - 2 * I16 = 0 /\ I17 <= I13 - 1 /\ 0 <= I13 - 1] 5) f1#(I18, I19) -> f2#(I20, I21) [-1 <= I19 - 1 /\ 1 <= I20 - 1 /\ 2 <= I22 - 1 /\ 0 <= I18 - 1] 6) f1#(I23, I24) -> f2#(I25, I26) [-1 <= I24 - 1 /\ 1 <= I25 - 1 /\ I27 <= 1 /\ -1 <= I27 - 1 /\ 0 <= I23 - 1] We have the following SCCs. { 1, 2, 3, 4 } DP problem for innermost termination. P = f3#(I0, I1) -> f2#(I2, I3) [I0 - 2 * y1 = 1 /\ 0 <= I0 - 1 /\ I2 <= I0 - 1 /\ 0 <= I0 - 2 * y1 /\ I0 - 2 * y1 <= 1 /\ I0 - 2 * I2 <= 1 /\ 0 <= I0 - 2 * I2] f2#(I4, I5) -> f3#(I4, I6) [I4 - 2 * I7 = 1 /\ y2 <= I4 - 1 /\ 0 <= I4 - 1] f3#(I8, I9) -> f2#(I10, I11) [I8 - 2 * I12 = 0 /\ 0 <= I8 - 1 /\ I10 <= I8 - 1 /\ 0 <= I8 - 2 * I12 /\ I8 - 2 * I12 <= 1 /\ I8 - 2 * I10 <= 1 /\ 0 <= I8 - 2 * I10] f2#(I13, I14) -> f3#(I13, I15) [I13 - 2 * I16 = 0 /\ I17 <= I13 - 1 /\ 0 <= I13 - 1] R = init(x1, x2) -> f1(rnd1, rnd2) f3(I0, I1) -> f2(I2, I3) [I0 - 2 * y1 = 1 /\ 0 <= I0 - 1 /\ I2 <= I0 - 1 /\ 0 <= I0 - 2 * y1 /\ I0 - 2 * y1 <= 1 /\ I0 - 2 * I2 <= 1 /\ 0 <= I0 - 2 * I2] f2(I4, I5) -> f3(I4, I6) [I4 - 2 * I7 = 1 /\ y2 <= I4 - 1 /\ 0 <= I4 - 1] f3(I8, I9) -> f2(I10, I11) [I8 - 2 * I12 = 0 /\ 0 <= I8 - 1 /\ I10 <= I8 - 1 /\ 0 <= I8 - 2 * I12 /\ I8 - 2 * I12 <= 1 /\ I8 - 2 * I10 <= 1 /\ 0 <= I8 - 2 * I10] f2(I13, I14) -> f3(I13, I15) [I13 - 2 * I16 = 0 /\ I17 <= I13 - 1 /\ 0 <= I13 - 1] f1(I18, I19) -> f2(I20, I21) [-1 <= I19 - 1 /\ 1 <= I20 - 1 /\ 2 <= I22 - 1 /\ 0 <= I18 - 1] f1(I23, I24) -> f2(I25, I26) [-1 <= I24 - 1 /\ 1 <= I25 - 1 /\ I27 <= 1 /\ -1 <= I27 - 1 /\ 0 <= I23 - 1] We use the basic value criterion with the projection function NU: NU[f2#(z1,z2)] = z1 NU[f3#(z1,z2)] = z1 This gives the following inequalities: I0 - 2 * y1 = 1 /\ 0 <= I0 - 1 /\ I2 <= I0 - 1 /\ 0 <= I0 - 2 * y1 /\ I0 - 2 * y1 <= 1 /\ I0 - 2 * I2 <= 1 /\ 0 <= I0 - 2 * I2 ==> I0 >! I2 I4 - 2 * I7 = 1 /\ y2 <= I4 - 1 /\ 0 <= I4 - 1 ==> I4 (>! \union =) I4 I8 - 2 * I12 = 0 /\ 0 <= I8 - 1 /\ I10 <= I8 - 1 /\ 0 <= I8 - 2 * I12 /\ I8 - 2 * I12 <= 1 /\ I8 - 2 * I10 <= 1 /\ 0 <= I8 - 2 * I10 ==> I8 >! I10 I13 - 2 * I16 = 0 /\ I17 <= I13 - 1 /\ 0 <= I13 - 1 ==> I13 (>! \union =) I13 We remove all the strictly oriented dependency pairs. DP problem for innermost termination. P = f2#(I4, I5) -> f3#(I4, I6) [I4 - 2 * I7 = 1 /\ y2 <= I4 - 1 /\ 0 <= I4 - 1] f2#(I13, I14) -> f3#(I13, I15) [I13 - 2 * I16 = 0 /\ I17 <= I13 - 1 /\ 0 <= I13 - 1] R = init(x1, x2) -> f1(rnd1, rnd2) f3(I0, I1) -> f2(I2, I3) [I0 - 2 * y1 = 1 /\ 0 <= I0 - 1 /\ I2 <= I0 - 1 /\ 0 <= I0 - 2 * y1 /\ I0 - 2 * y1 <= 1 /\ I0 - 2 * I2 <= 1 /\ 0 <= I0 - 2 * I2] f2(I4, I5) -> f3(I4, I6) [I4 - 2 * I7 = 1 /\ y2 <= I4 - 1 /\ 0 <= I4 - 1] f3(I8, I9) -> f2(I10, I11) [I8 - 2 * I12 = 0 /\ 0 <= I8 - 1 /\ I10 <= I8 - 1 /\ 0 <= I8 - 2 * I12 /\ I8 - 2 * I12 <= 1 /\ I8 - 2 * I10 <= 1 /\ 0 <= I8 - 2 * I10] f2(I13, I14) -> f3(I13, I15) [I13 - 2 * I16 = 0 /\ I17 <= I13 - 1 /\ 0 <= I13 - 1] f1(I18, I19) -> f2(I20, I21) [-1 <= I19 - 1 /\ 1 <= I20 - 1 /\ 2 <= I22 - 1 /\ 0 <= I18 - 1] f1(I23, I24) -> f2(I25, I26) [-1 <= I24 - 1 /\ 1 <= I25 - 1 /\ I27 <= 1 /\ -1 <= I27 - 1 /\ 0 <= I23 - 1] The dependency graph for this problem is: 2 -> 4 -> Where: 2) f2#(I4, I5) -> f3#(I4, I6) [I4 - 2 * I7 = 1 /\ y2 <= I4 - 1 /\ 0 <= I4 - 1] 4) f2#(I13, I14) -> f3#(I13, I15) [I13 - 2 * I16 = 0 /\ I17 <= I13 - 1 /\ 0 <= I13 - 1] We have the following SCCs. popout

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job log

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actions

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