details

property	value
status	complete
benchmark	Convert.jar-obl-9.smt2
ran by	Akihisa Yamada
cpu timeout	1200 seconds
wallclock timeout	300 seconds
memory limit	137438953472 bytes
execution host	n140.star.cs.uiowa.edu
space	From_AProVE_2014

run statistics

property	value
solver	Ctrl
configuration	Transition
runtime (wallclock)	1.02179 seconds
cpu usage	1.04255
user time	0.567659
system time	0.474894
max virtual memory	720288.0
max residence set size	8388.0

stage attributes

key	value
starexec-result	MAYBE

output

MAYBE DP problem for innermost termination. P = init#(x1, x2) -> f3#(rnd1, rnd2) f5#(I0, I1) -> f5#(I0 - 1, I1 + 1) [0 <= I1 - 1 /\ 0 <= I0 - 1] f3#(I2, I3) -> f5#(I4, 1) [0 <= I2 - 1 /\ -1 <= I4 - 1 /\ -1 <= I3 - 1] f4#(I5, I6) -> f4#(I7, I8) [I7 + 2 <= I5 /\ y1 <= 0 /\ 2 <= I5 - 1 /\ 0 <= I7 - 1] f4#(I9, I10) -> f4#(I11, I12) [I11 <= I9 /\ 0 <= I13 - 1 /\ 0 <= I9 - 1 /\ 0 <= I11 - 1] f4#(I14, I15) -> f4#(I16, I17) [I16 + 2 <= I14 /\ I18 <= 0 /\ 1 <= I14 - 1 /\ -1 <= I16 - 1] f2#(I19, I20) -> f4#(I21, I22) [I21 <= I19 /\ 0 <= I23 - 1 /\ -1 <= I19 - 1 /\ -1 <= I21 - 1] f3#(I24, I25) -> f2#(I26, I27) [-1 <= I26 - 1 /\ 0 <= I24 - 1] f1#(I28, I29) -> f2#(I30, I31) [-1 <= I30 - 1 /\ -1 <= I28 - 1 /\ I30 <= I28] R = init(x1, x2) -> f3(rnd1, rnd2) f5(I0, I1) -> f5(I0 - 1, I1 + 1) [0 <= I1 - 1 /\ 0 <= I0 - 1] f3(I2, I3) -> f5(I4, 1) [0 <= I2 - 1 /\ -1 <= I4 - 1 /\ -1 <= I3 - 1] f4(I5, I6) -> f4(I7, I8) [I7 + 2 <= I5 /\ y1 <= 0 /\ 2 <= I5 - 1 /\ 0 <= I7 - 1] f4(I9, I10) -> f4(I11, I12) [I11 <= I9 /\ 0 <= I13 - 1 /\ 0 <= I9 - 1 /\ 0 <= I11 - 1] f4(I14, I15) -> f4(I16, I17) [I16 + 2 <= I14 /\ I18 <= 0 /\ 1 <= I14 - 1 /\ -1 <= I16 - 1] f2(I19, I20) -> f4(I21, I22) [I21 <= I19 /\ 0 <= I23 - 1 /\ -1 <= I19 - 1 /\ -1 <= I21 - 1] f3(I24, I25) -> f2(I26, I27) [-1 <= I26 - 1 /\ 0 <= I24 - 1] f1(I28, I29) -> f2(I30, I31) [-1 <= I30 - 1 /\ -1 <= I28 - 1 /\ I30 <= I28] The dependency graph for this problem is: 0 -> 2, 7 1 -> 1 2 -> 1 3 -> 3, 4, 5 4 -> 3, 4, 5 5 -> 3, 4, 5 6 -> 3, 4, 5 7 -> 6 8 -> 6 Where: 0) init#(x1, x2) -> f3#(rnd1, rnd2) 1) f5#(I0, I1) -> f5#(I0 - 1, I1 + 1) [0 <= I1 - 1 /\ 0 <= I0 - 1] 2) f3#(I2, I3) -> f5#(I4, 1) [0 <= I2 - 1 /\ -1 <= I4 - 1 /\ -1 <= I3 - 1] 3) f4#(I5, I6) -> f4#(I7, I8) [I7 + 2 <= I5 /\ y1 <= 0 /\ 2 <= I5 - 1 /\ 0 <= I7 - 1] 4) f4#(I9, I10) -> f4#(I11, I12) [I11 <= I9 /\ 0 <= I13 - 1 /\ 0 <= I9 - 1 /\ 0 <= I11 - 1] 5) f4#(I14, I15) -> f4#(I16, I17) [I16 + 2 <= I14 /\ I18 <= 0 /\ 1 <= I14 - 1 /\ -1 <= I16 - 1] 6) f2#(I19, I20) -> f4#(I21, I22) [I21 <= I19 /\ 0 <= I23 - 1 /\ -1 <= I19 - 1 /\ -1 <= I21 - 1] 7) f3#(I24, I25) -> f2#(I26, I27) [-1 <= I26 - 1 /\ 0 <= I24 - 1] 8) f1#(I28, I29) -> f2#(I30, I31) [-1 <= I30 - 1 /\ -1 <= I28 - 1 /\ I30 <= I28] We have the following SCCs. { 1 } { 3, 4, 5 } DP problem for innermost termination. P = f4#(I5, I6) -> f4#(I7, I8) [I7 + 2 <= I5 /\ y1 <= 0 /\ 2 <= I5 - 1 /\ 0 <= I7 - 1] f4#(I9, I10) -> f4#(I11, I12) [I11 <= I9 /\ 0 <= I13 - 1 /\ 0 <= I9 - 1 /\ 0 <= I11 - 1] f4#(I14, I15) -> f4#(I16, I17) [I16 + 2 <= I14 /\ I18 <= 0 /\ 1 <= I14 - 1 /\ -1 <= I16 - 1] R = init(x1, x2) -> f3(rnd1, rnd2) f5(I0, I1) -> f5(I0 - 1, I1 + 1) [0 <= I1 - 1 /\ 0 <= I0 - 1] f3(I2, I3) -> f5(I4, 1) [0 <= I2 - 1 /\ -1 <= I4 - 1 /\ -1 <= I3 - 1] f4(I5, I6) -> f4(I7, I8) [I7 + 2 <= I5 /\ y1 <= 0 /\ 2 <= I5 - 1 /\ 0 <= I7 - 1] f4(I9, I10) -> f4(I11, I12) [I11 <= I9 /\ 0 <= I13 - 1 /\ 0 <= I9 - 1 /\ 0 <= I11 - 1] f4(I14, I15) -> f4(I16, I17) [I16 + 2 <= I14 /\ I18 <= 0 /\ 1 <= I14 - 1 /\ -1 <= I16 - 1] f2(I19, I20) -> f4(I21, I22) [I21 <= I19 /\ 0 <= I23 - 1 /\ -1 <= I19 - 1 /\ -1 <= I21 - 1] f3(I24, I25) -> f2(I26, I27) [-1 <= I26 - 1 /\ 0 <= I24 - 1] f1(I28, I29) -> f2(I30, I31) [-1 <= I30 - 1 /\ -1 <= I28 - 1 /\ I30 <= I28] We use the basic value criterion with the projection function NU: NU[f4#(z1,z2)] = z1 This gives the following inequalities: I7 + 2 <= I5 /\ y1 <= 0 /\ 2 <= I5 - 1 /\ 0 <= I7 - 1 ==> I5 >! I7 I11 <= I9 /\ 0 <= I13 - 1 /\ 0 <= I9 - 1 /\ 0 <= I11 - 1 ==> I9 (>! \union =) I11 I16 + 2 <= I14 /\ I18 <= 0 /\ 1 <= I14 - 1 /\ -1 <= I16 - 1 ==> I14 >! I16 We remove all the strictly oriented dependency pairs. DP problem for innermost termination. P = f4#(I9, I10) -> f4#(I11, I12) [I11 <= I9 /\ 0 <= I13 - 1 /\ 0 <= I9 - 1 /\ 0 <= I11 - 1] R = init(x1, x2) -> f3(rnd1, rnd2) f5(I0, I1) -> f5(I0 - 1, I1 + 1) [0 <= I1 - 1 /\ 0 <= I0 - 1] f3(I2, I3) -> f5(I4, 1) [0 <= I2 - 1 /\ -1 <= I4 - 1 /\ -1 <= I3 - 1] f4(I5, I6) -> f4(I7, I8) [I7 + 2 <= I5 /\ y1 <= 0 /\ 2 <= I5 - 1 /\ 0 <= I7 - 1] f4(I9, I10) -> f4(I11, I12) [I11 <= I9 /\ 0 <= I13 - 1 /\ 0 <= I9 - 1 /\ 0 <= I11 - 1] f4(I14, I15) -> f4(I16, I17) [I16 + 2 <= I14 /\ I18 <= 0 /\ 1 <= I14 - 1 /\ -1 <= I16 - 1] f2(I19, I20) -> f4(I21, I22) [I21 <= I19 /\ 0 <= I23 - 1 /\ -1 <= I19 - 1 /\ -1 <= I21 - 1] f3(I24, I25) -> f2(I26, I27) [-1 <= I26 - 1 /\ 0 <= I24 - 1] f1(I28, I29) -> f2(I30, I31) [-1 <= I30 - 1 /\ -1 <= I28 - 1 /\ I30 <= I28] popout

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