details

property	value
status	complete
benchmark	AProVE12-cyclic-Iterate.jar-obl-9.smt2
ran by	Akihisa Yamada
cpu timeout	1200 seconds
wallclock timeout	300 seconds
memory limit	137438953472 bytes
execution host	n141.star.cs.uiowa.edu
space	From_AProVE_2014

run statistics

property	value
solver	Ctrl
configuration	Transition
runtime (wallclock)	1.56791 seconds
cpu usage	1.59719
user time	0.824303
system time	0.772887
max virtual memory	349976.0
max residence set size	8532.0

stage attributes

key	value
starexec-result	MAYBE

output

MAYBE DP problem for innermost termination. P = init#(x1, x2, x3, x4) -> f1#(rnd1, rnd2, rnd3, rnd4) f3#(I0, I1, I2, I3) -> f3#(1, I4, I5, I6) [I5 <= I1 - 1 /\ 0 <= I1 - 1 /\ I6 <= y1 - 1 /\ -1 <= y1 - 1 /\ I1 = I2] f3#(I7, I8, I9, I10) -> f3#(I11, I8, I12, I13) [0 <= I7 - 1 /\ 0 <= I10 - 1 /\ 0 <= I9 - 1 /\ I7 <= I11 - 1 /\ I7 <= I8 - 1 /\ I12 <= I9 - 1 /\ I13 <= I10 - 1 /\ I7 <= I14 - 1] f2#(I15, I16, I17, I18) -> f3#(0, I17, I17, 0) [0 <= I17 - 1 /\ I16 <= 0] f2#(I19, I20, I21, I22) -> f2#(I19 - 1, I19, 1, I23) [0 <= I20 - 1] f2#(I24, I25, I26, I27) -> f2#(I24 - 1, I24, I28, I29) [0 <= I26 - 1 /\ I26 <= I28 - 1 /\ 0 <= I25 - 1] f1#(I30, I31, I32, I33) -> f2#(I31 - 1, I31, 0, I34) [-1 <= I31 - 1 /\ 0 <= I30 - 1] R = init(x1, x2, x3, x4) -> f1(rnd1, rnd2, rnd3, rnd4) f3(I0, I1, I2, I3) -> f3(1, I4, I5, I6) [I5 <= I1 - 1 /\ 0 <= I1 - 1 /\ I6 <= y1 - 1 /\ -1 <= y1 - 1 /\ I1 = I2] f3(I7, I8, I9, I10) -> f3(I11, I8, I12, I13) [0 <= I7 - 1 /\ 0 <= I10 - 1 /\ 0 <= I9 - 1 /\ I7 <= I11 - 1 /\ I7 <= I8 - 1 /\ I12 <= I9 - 1 /\ I13 <= I10 - 1 /\ I7 <= I14 - 1] f2(I15, I16, I17, I18) -> f3(0, I17, I17, 0) [0 <= I17 - 1 /\ I16 <= 0] f2(I19, I20, I21, I22) -> f2(I19 - 1, I19, 1, I23) [0 <= I20 - 1] f2(I24, I25, I26, I27) -> f2(I24 - 1, I24, I28, I29) [0 <= I26 - 1 /\ I26 <= I28 - 1 /\ 0 <= I25 - 1] f1(I30, I31, I32, I33) -> f2(I31 - 1, I31, 0, I34) [-1 <= I31 - 1 /\ 0 <= I30 - 1] The dependency graph for this problem is: 0 -> 6 1 -> 1, 2 2 -> 1, 2 3 -> 1 4 -> 3, 4, 5 5 -> 3, 4, 5 6 -> 4 Where: 0) init#(x1, x2, x3, x4) -> f1#(rnd1, rnd2, rnd3, rnd4) 1) f3#(I0, I1, I2, I3) -> f3#(1, I4, I5, I6) [I5 <= I1 - 1 /\ 0 <= I1 - 1 /\ I6 <= y1 - 1 /\ -1 <= y1 - 1 /\ I1 = I2] 2) f3#(I7, I8, I9, I10) -> f3#(I11, I8, I12, I13) [0 <= I7 - 1 /\ 0 <= I10 - 1 /\ 0 <= I9 - 1 /\ I7 <= I11 - 1 /\ I7 <= I8 - 1 /\ I12 <= I9 - 1 /\ I13 <= I10 - 1 /\ I7 <= I14 - 1] 3) f2#(I15, I16, I17, I18) -> f3#(0, I17, I17, 0) [0 <= I17 - 1 /\ I16 <= 0] 4) f2#(I19, I20, I21, I22) -> f2#(I19 - 1, I19, 1, I23) [0 <= I20 - 1] 5) f2#(I24, I25, I26, I27) -> f2#(I24 - 1, I24, I28, I29) [0 <= I26 - 1 /\ I26 <= I28 - 1 /\ 0 <= I25 - 1] 6) f1#(I30, I31, I32, I33) -> f2#(I31 - 1, I31, 0, I34) [-1 <= I31 - 1 /\ 0 <= I30 - 1] We have the following SCCs. { 4, 5 } { 1, 2 } DP problem for innermost termination. P = f3#(I0, I1, I2, I3) -> f3#(1, I4, I5, I6) [I5 <= I1 - 1 /\ 0 <= I1 - 1 /\ I6 <= y1 - 1 /\ -1 <= y1 - 1 /\ I1 = I2] f3#(I7, I8, I9, I10) -> f3#(I11, I8, I12, I13) [0 <= I7 - 1 /\ 0 <= I10 - 1 /\ 0 <= I9 - 1 /\ I7 <= I11 - 1 /\ I7 <= I8 - 1 /\ I12 <= I9 - 1 /\ I13 <= I10 - 1 /\ I7 <= I14 - 1] R = init(x1, x2, x3, x4) -> f1(rnd1, rnd2, rnd3, rnd4) f3(I0, I1, I2, I3) -> f3(1, I4, I5, I6) [I5 <= I1 - 1 /\ 0 <= I1 - 1 /\ I6 <= y1 - 1 /\ -1 <= y1 - 1 /\ I1 = I2] f3(I7, I8, I9, I10) -> f3(I11, I8, I12, I13) [0 <= I7 - 1 /\ 0 <= I10 - 1 /\ 0 <= I9 - 1 /\ I7 <= I11 - 1 /\ I7 <= I8 - 1 /\ I12 <= I9 - 1 /\ I13 <= I10 - 1 /\ I7 <= I14 - 1] f2(I15, I16, I17, I18) -> f3(0, I17, I17, 0) [0 <= I17 - 1 /\ I16 <= 0] f2(I19, I20, I21, I22) -> f2(I19 - 1, I19, 1, I23) [0 <= I20 - 1] f2(I24, I25, I26, I27) -> f2(I24 - 1, I24, I28, I29) [0 <= I26 - 1 /\ I26 <= I28 - 1 /\ 0 <= I25 - 1] f1(I30, I31, I32, I33) -> f2(I31 - 1, I31, 0, I34) [-1 <= I31 - 1 /\ 0 <= I30 - 1] We use the basic value criterion with the projection function NU: NU[f3#(z1,z2,z3,z4)] = z3 This gives the following inequalities: I5 <= I1 - 1 /\ 0 <= I1 - 1 /\ I6 <= y1 - 1 /\ -1 <= y1 - 1 /\ I1 = I2 ==> I2 >! I5 0 <= I7 - 1 /\ 0 <= I10 - 1 /\ 0 <= I9 - 1 /\ I7 <= I11 - 1 /\ I7 <= I8 - 1 /\ I12 <= I9 - 1 /\ I13 <= I10 - 1 /\ I7 <= I14 - 1 ==> I9 >! I12 All dependency pairs are strictly oriented, so the entire dependency pair problem may be removed. DP problem for innermost termination. P = f2#(I19, I20, I21, I22) -> f2#(I19 - 1, I19, 1, I23) [0 <= I20 - 1] f2#(I24, I25, I26, I27) -> f2#(I24 - 1, I24, I28, I29) [0 <= I26 - 1 /\ I26 <= I28 - 1 /\ 0 <= I25 - 1] R = init(x1, x2, x3, x4) -> f1(rnd1, rnd2, rnd3, rnd4) f3(I0, I1, I2, I3) -> f3(1, I4, I5, I6) [I5 <= I1 - 1 /\ 0 <= I1 - 1 /\ I6 <= y1 - 1 /\ -1 <= y1 - 1 /\ I1 = I2] f3(I7, I8, I9, I10) -> f3(I11, I8, I12, I13) [0 <= I7 - 1 /\ 0 <= I10 - 1 /\ 0 <= I9 - 1 /\ I7 <= I11 - 1 /\ I7 <= I8 - 1 /\ I12 <= I9 - 1 /\ I13 <= I10 - 1 /\ I7 <= I14 - 1] f2(I15, I16, I17, I18) -> f3(0, I17, I17, 0) [0 <= I17 - 1 /\ I16 <= 0] f2(I19, I20, I21, I22) -> f2(I19 - 1, I19, 1, I23) [0 <= I20 - 1] f2(I24, I25, I26, I27) -> f2(I24 - 1, I24, I28, I29) [0 <= I26 - 1 /\ I26 <= I28 - 1 /\ 0 <= I25 - 1] f1(I30, I31, I32, I33) -> f2(I31 - 1, I31, 0, I34) [-1 <= I31 - 1 /\ 0 <= I30 - 1] popout

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