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Integ Trans Syste 27634 pair #381738536
details
property
value
status
complete
benchmark
Et1-rec.jar-obl-8.smt2
ran by
Akihisa Yamada
cpu timeout
1200 seconds
wallclock timeout
300 seconds
memory limit
137438953472 bytes
execution host
n090.star.cs.uiowa.edu
space
From_AProVE_2014
run statistics
property
value
solver
Ctrl
configuration
Transition
runtime (wallclock)
0.483822107315 seconds
cpu usage
0.506558446
max memory
1.0584064E7
stage attributes
key
value
output-size
2458
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) -> f1#(rnd1, rnd2, rnd3) f2#(I0, I1, I2) -> f2#(I0 + I1, I1 + 1, I0 + I1) [I0 = I2 /\ I0 <= I1 + 1 - 1 /\ I1 <= I1 + 1 - 1 /\ I0 <= I1 - 1] f1#(I3, I4, I5) -> f2#(I6, I7, I8) [-1 <= y1 - 1 /\ 1 <= I4 - 1 /\ -1 <= y2 - 1 /\ 0 <= I3 - 1 /\ 0 - y2 = I6 /\ 0 - y1 = I7 /\ 0 - y2 = I8] f1#(I9, I10, I11) -> f2#(0, I12, 0) [0 <= I9 - 1 /\ -1 <= I13 - 1 /\ 1 = I10 /\ 0 - I13 = I12] f1#(I14, I15, I16) -> f2#(0, 0, 0) [0 = I15 /\ 0 <= I14 - 1] R = init(x1, x2, x3) -> f1(rnd1, rnd2, rnd3) f2(I0, I1, I2) -> f2(I0 + I1, I1 + 1, I0 + I1) [I0 = I2 /\ I0 <= I1 + 1 - 1 /\ I1 <= I1 + 1 - 1 /\ I0 <= I1 - 1] f1(I3, I4, I5) -> f2(I6, I7, I8) [-1 <= y1 - 1 /\ 1 <= I4 - 1 /\ -1 <= y2 - 1 /\ 0 <= I3 - 1 /\ 0 - y2 = I6 /\ 0 - y1 = I7 /\ 0 - y2 = I8] f1(I9, I10, I11) -> f2(0, I12, 0) [0 <= I9 - 1 /\ -1 <= I13 - 1 /\ 1 = I10 /\ 0 - I13 = I12] f1(I14, I15, I16) -> f2(0, 0, 0) [0 = I15 /\ 0 <= I14 - 1] The dependency graph for this problem is: 0 -> 2, 3, 4 1 -> 1 2 -> 1 3 -> 4 -> Where: 0) init#(x1, x2, x3) -> f1#(rnd1, rnd2, rnd3) 1) f2#(I0, I1, I2) -> f2#(I0 + I1, I1 + 1, I0 + I1) [I0 = I2 /\ I0 <= I1 + 1 - 1 /\ I1 <= I1 + 1 - 1 /\ I0 <= I1 - 1] 2) f1#(I3, I4, I5) -> f2#(I6, I7, I8) [-1 <= y1 - 1 /\ 1 <= I4 - 1 /\ -1 <= y2 - 1 /\ 0 <= I3 - 1 /\ 0 - y2 = I6 /\ 0 - y1 = I7 /\ 0 - y2 = I8] 3) f1#(I9, I10, I11) -> f2#(0, I12, 0) [0 <= I9 - 1 /\ -1 <= I13 - 1 /\ 1 = I10 /\ 0 - I13 = I12] 4) f1#(I14, I15, I16) -> f2#(0, 0, 0) [0 = I15 /\ 0 <= I14 - 1] We have the following SCCs. { 1 } DP problem for innermost termination. P = f2#(I0, I1, I2) -> f2#(I0 + I1, I1 + 1, I0 + I1) [I0 = I2 /\ I0 <= I1 + 1 - 1 /\ I1 <= I1 + 1 - 1 /\ I0 <= I1 - 1] R = init(x1, x2, x3) -> f1(rnd1, rnd2, rnd3) f2(I0, I1, I2) -> f2(I0 + I1, I1 + 1, I0 + I1) [I0 = I2 /\ I0 <= I1 + 1 - 1 /\ I1 <= I1 + 1 - 1 /\ I0 <= I1 - 1] f1(I3, I4, I5) -> f2(I6, I7, I8) [-1 <= y1 - 1 /\ 1 <= I4 - 1 /\ -1 <= y2 - 1 /\ 0 <= I3 - 1 /\ 0 - y2 = I6 /\ 0 - y1 = I7 /\ 0 - y2 = I8] f1(I9, I10, I11) -> f2(0, I12, 0) [0 <= I9 - 1 /\ -1 <= I13 - 1 /\ 1 = I10 /\ 0 - I13 = I12] f1(I14, I15, I16) -> f2(0, 0, 0) [0 = I15 /\ 0 <= I14 - 1]
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