details

property	value
status	complete
benchmark	Et2-rec.jar-obl-8.smt2
ran by	Akihisa Yamada
cpu timeout	1200 seconds
wallclock timeout	300 seconds
memory limit	137438953472 bytes
execution host	n105.star.cs.uiowa.edu
space	From_AProVE_2014

run statistics

property	value
solver	Ctrl
configuration	Transition
runtime (wallclock)	0.702290058136 seconds
cpu usage	0.746482566
max memory	1.1300864E7

stage attributes

key	value
output-size	3134
starexec-result	MAYBE

output

/export/starexec/sandbox/solver/bin/starexec_run_Transition /export/starexec/sandbox/benchmark/theBenchmark.smt2 /export/starexec/sandbox/output/output_files -------------------------------------------------------------------------------- MAYBE DP problem for innermost termination. P = init#(x1, x2, x3, x4) -> f1#(rnd1, rnd2, rnd3, rnd4) f2#(I0, I1, I2, I3) -> f2#(I4, I5, I2, I3 + 1) [0 <= I1 - 1 /\ 0 <= I2 - 1 /\ -1 <= I3 - 1 /\ I3 <= I2 - 1 /\ I3 + 1 <= I2 /\ -1 <= y1 - 1 /\ I0 - 1 - y1 = I4 /\ I0 - 1 - y1 = I5] f2#(I6, I7, I8, I9) -> f2#(I6 - 1, I6 - 1, I8, I9) [0 <= I7 - 1 /\ I8 <= I9 /\ I6 - 1 <= I6 - 1 /\ -1 <= I8 - 1] f1#(I10, I11, I12, I13) -> f2#(I14, I15, I11, 2) [0 <= I10 - 1 /\ -1 <= I14 - 1 /\ 1 <= I11 - 1 /\ -1 <= I15 - 1] f1#(I16, I17, I18, I19) -> f2#(I20, 0, 1, 1) [1 = I17 /\ -1 <= I20 - 1 /\ 0 <= I16 - 1] f1#(I21, I22, I23, I24) -> f2#(0, 0, 0, 0) [0 = I22 /\ 0 <= I21 - 1] R = init(x1, x2, x3, x4) -> f1(rnd1, rnd2, rnd3, rnd4) f2(I0, I1, I2, I3) -> f2(I4, I5, I2, I3 + 1) [0 <= I1 - 1 /\ 0 <= I2 - 1 /\ -1 <= I3 - 1 /\ I3 <= I2 - 1 /\ I3 + 1 <= I2 /\ -1 <= y1 - 1 /\ I0 - 1 - y1 = I4 /\ I0 - 1 - y1 = I5] f2(I6, I7, I8, I9) -> f2(I6 - 1, I6 - 1, I8, I9) [0 <= I7 - 1 /\ I8 <= I9 /\ I6 - 1 <= I6 - 1 /\ -1 <= I8 - 1] f1(I10, I11, I12, I13) -> f2(I14, I15, I11, 2) [0 <= I10 - 1 /\ -1 <= I14 - 1 /\ 1 <= I11 - 1 /\ -1 <= I15 - 1] f1(I16, I17, I18, I19) -> f2(I20, 0, 1, 1) [1 = I17 /\ -1 <= I20 - 1 /\ 0 <= I16 - 1] f1(I21, I22, I23, I24) -> f2(0, 0, 0, 0) [0 = I22 /\ 0 <= I21 - 1] The dependency graph for this problem is: 0 -> 3, 4, 5 1 -> 1, 2 2 -> 2 3 -> 1, 2 4 -> 5 -> Where: 0) init#(x1, x2, x3, x4) -> f1#(rnd1, rnd2, rnd3, rnd4) 1) f2#(I0, I1, I2, I3) -> f2#(I4, I5, I2, I3 + 1) [0 <= I1 - 1 /\ 0 <= I2 - 1 /\ -1 <= I3 - 1 /\ I3 <= I2 - 1 /\ I3 + 1 <= I2 /\ -1 <= y1 - 1 /\ I0 - 1 - y1 = I4 /\ I0 - 1 - y1 = I5] 2) f2#(I6, I7, I8, I9) -> f2#(I6 - 1, I6 - 1, I8, I9) [0 <= I7 - 1 /\ I8 <= I9 /\ I6 - 1 <= I6 - 1 /\ -1 <= I8 - 1] 3) f1#(I10, I11, I12, I13) -> f2#(I14, I15, I11, 2) [0 <= I10 - 1 /\ -1 <= I14 - 1 /\ 1 <= I11 - 1 /\ -1 <= I15 - 1] 4) f1#(I16, I17, I18, I19) -> f2#(I20, 0, 1, 1) [1 = I17 /\ -1 <= I20 - 1 /\ 0 <= I16 - 1] 5) f1#(I21, I22, I23, I24) -> f2#(0, 0, 0, 0) [0 = I22 /\ 0 <= I21 - 1] We have the following SCCs. { 1 } { 2 } DP problem for innermost termination. P = f2#(I6, I7, I8, I9) -> f2#(I6 - 1, I6 - 1, I8, I9) [0 <= I7 - 1 /\ I8 <= I9 /\ I6 - 1 <= I6 - 1 /\ -1 <= I8 - 1] R = init(x1, x2, x3, x4) -> f1(rnd1, rnd2, rnd3, rnd4) f2(I0, I1, I2, I3) -> f2(I4, I5, I2, I3 + 1) [0 <= I1 - 1 /\ 0 <= I2 - 1 /\ -1 <= I3 - 1 /\ I3 <= I2 - 1 /\ I3 + 1 <= I2 /\ -1 <= y1 - 1 /\ I0 - 1 - y1 = I4 /\ I0 - 1 - y1 = I5] f2(I6, I7, I8, I9) -> f2(I6 - 1, I6 - 1, I8, I9) [0 <= I7 - 1 /\ I8 <= I9 /\ I6 - 1 <= I6 - 1 /\ -1 <= I8 - 1] f1(I10, I11, I12, I13) -> f2(I14, I15, I11, 2) [0 <= I10 - 1 /\ -1 <= I14 - 1 /\ 1 <= I11 - 1 /\ -1 <= I15 - 1] f1(I16, I17, I18, I19) -> f2(I20, 0, 1, 1) [1 = I17 /\ -1 <= I20 - 1 /\ 0 <= I16 - 1] f1(I21, I22, I23, I24) -> f2(0, 0, 0, 0) [0 = I22 /\ 0 <= I21 - 1] popout

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