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Integ Trans Syste 27634 pair #381738674
details
property
value
status
complete
benchmark
FibSLR.jar-obl-8.smt2
ran by
Akihisa Yamada
cpu timeout
1200 seconds
wallclock timeout
300 seconds
memory limit
137438953472 bytes
execution host
n021.star.cs.uiowa.edu
space
From_AProVE_2014
run statistics
property
value
solver
Ctrl
configuration
Transition
runtime (wallclock)
0.57145690918 seconds
cpu usage
0.593861961
max memory
1.110016E7
stage attributes
key
value
output-size
2873
starexec-result
MAYBE
output
/export/starexec/sandbox2/solver/bin/starexec_run_Transition /export/starexec/sandbox2/benchmark/theBenchmark.smt2 /export/starexec/sandbox2/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#(I0 - 1, I0, I2, I3 + 1) [I3 <= I2 - 1 /\ -1 <= I3 - 1 /\ -1 <= I2 - 1 /\ 0 <= I1 - 1] f2#(I4, I5, I6, I7) -> f2#(I4 - 1, I4, I6, I7) [-1 <= I6 - 1 /\ I6 <= I7 /\ 0 <= I5 - 1] f1#(I8, I9, I10, I11) -> f2#(I12, I13, I9, 2) [-1 <= y2 - 1 /\ 1 <= I9 - 1 /\ -1 <= y1 - 1 /\ 0 <= I8 - 1 /\ y1 * y2 - 1 = I12 /\ y1 * y2 = I13] f1#(I14, I15, I16, I17) -> f2#(-1, 0, 1, 1) [0 <= I14 - 1 /\ -1 <= I18 - 1 /\ 1 = I15] f1#(I19, I20, I21, I22) -> f2#(-1, 0, 0, 0) [0 = I20 /\ 0 <= I19 - 1] R = init(x1, x2, x3, x4) -> f1(rnd1, rnd2, rnd3, rnd4) f2(I0, I1, I2, I3) -> f2(I0 - 1, I0, I2, I3 + 1) [I3 <= I2 - 1 /\ -1 <= I3 - 1 /\ -1 <= I2 - 1 /\ 0 <= I1 - 1] f2(I4, I5, I6, I7) -> f2(I4 - 1, I4, I6, I7) [-1 <= I6 - 1 /\ I6 <= I7 /\ 0 <= I5 - 1] f1(I8, I9, I10, I11) -> f2(I12, I13, I9, 2) [-1 <= y2 - 1 /\ 1 <= I9 - 1 /\ -1 <= y1 - 1 /\ 0 <= I8 - 1 /\ y1 * y2 - 1 = I12 /\ y1 * y2 = I13] f1(I14, I15, I16, I17) -> f2(-1, 0, 1, 1) [0 <= I14 - 1 /\ -1 <= I18 - 1 /\ 1 = I15] f1(I19, I20, I21, I22) -> f2(-1, 0, 0, 0) [0 = I20 /\ 0 <= I19 - 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#(I0 - 1, I0, I2, I3 + 1) [I3 <= I2 - 1 /\ -1 <= I3 - 1 /\ -1 <= I2 - 1 /\ 0 <= I1 - 1] 2) f2#(I4, I5, I6, I7) -> f2#(I4 - 1, I4, I6, I7) [-1 <= I6 - 1 /\ I6 <= I7 /\ 0 <= I5 - 1] 3) f1#(I8, I9, I10, I11) -> f2#(I12, I13, I9, 2) [-1 <= y2 - 1 /\ 1 <= I9 - 1 /\ -1 <= y1 - 1 /\ 0 <= I8 - 1 /\ y1 * y2 - 1 = I12 /\ y1 * y2 = I13] 4) f1#(I14, I15, I16, I17) -> f2#(-1, 0, 1, 1) [0 <= I14 - 1 /\ -1 <= I18 - 1 /\ 1 = I15] 5) f1#(I19, I20, I21, I22) -> f2#(-1, 0, 0, 0) [0 = I20 /\ 0 <= I19 - 1] We have the following SCCs. { 1 } { 2 } DP problem for innermost termination. P = f2#(I4, I5, I6, I7) -> f2#(I4 - 1, I4, I6, I7) [-1 <= I6 - 1 /\ I6 <= I7 /\ 0 <= I5 - 1] R = init(x1, x2, x3, x4) -> f1(rnd1, rnd2, rnd3, rnd4) f2(I0, I1, I2, I3) -> f2(I0 - 1, I0, I2, I3 + 1) [I3 <= I2 - 1 /\ -1 <= I3 - 1 /\ -1 <= I2 - 1 /\ 0 <= I1 - 1] f2(I4, I5, I6, I7) -> f2(I4 - 1, I4, I6, I7) [-1 <= I6 - 1 /\ I6 <= I7 /\ 0 <= I5 - 1] f1(I8, I9, I10, I11) -> f2(I12, I13, I9, 2) [-1 <= y2 - 1 /\ 1 <= I9 - 1 /\ -1 <= y1 - 1 /\ 0 <= I8 - 1 /\ y1 * y2 - 1 = I12 /\ y1 * y2 = I13] f1(I14, I15, I16, I17) -> f2(-1, 0, 1, 1) [0 <= I14 - 1 /\ -1 <= I18 - 1 /\ 1 = I15] f1(I19, I20, I21, I22) -> f2(-1, 0, 0, 0) [0 = I20 /\ 0 <= I19 - 1]
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