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Integ Trans Syste 27634 pair #381739933
details
property
value
status
complete
benchmark
eric.t2.smt2
ran by
Akihisa Yamada
cpu timeout
1200 seconds
wallclock timeout
300 seconds
memory limit
137438953472 bytes
execution host
n008.star.cs.uiowa.edu
space
From_T2
run statistics
property
value
solver
Ctrl
configuration
Transition
runtime (wallclock)
16.072715044 seconds
cpu usage
17.087597428
max memory
2.8151808E7
stage attributes
key
value
output-size
2573
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 = f4#(x1, x2, x3) -> f3#(x1, x2, x3) f3#(I0, I1, I2) -> f2#(I0, I1, 1 + I0) [0 <= 1 + I0 /\ 1 + I0 <= 1 + I1 /\ 1 <= I1 /\ 1 <= I0 /\ I0 <= I1] f2#(I3, I4, I5) -> f1#(I3, I4, I5) [1 + I5 <= I3] f2#(I6, I7, I8) -> f1#(I6, I7, I8) [1 + I6 <= I8] f1#(I9, I10, I11) -> f2#(I9, I10, 0) [0 <= 0 /\ 0 <= 1 + I10 /\ 1 <= I11 /\ 1 + I10 <= I11] f1#(I12, I13, I14) -> f2#(I12, I13, 1 + I14) [0 <= 1 + I14 /\ 1 + I14 <= 1 + I13 /\ I14 <= I13] R = f4(x1, x2, x3) -> f3(x1, x2, x3) f3(I0, I1, I2) -> f2(I0, I1, 1 + I0) [0 <= 1 + I0 /\ 1 + I0 <= 1 + I1 /\ 1 <= I1 /\ 1 <= I0 /\ I0 <= I1] f2(I3, I4, I5) -> f1(I3, I4, I5) [1 + I5 <= I3] f2(I6, I7, I8) -> f1(I6, I7, I8) [1 + I6 <= I8] f1(I9, I10, I11) -> f2(I9, I10, 0) [0 <= 0 /\ 0 <= 1 + I10 /\ 1 <= I11 /\ 1 + I10 <= I11] f1(I12, I13, I14) -> f2(I12, I13, 1 + I14) [0 <= 1 + I14 /\ 1 + I14 <= 1 + I13 /\ I14 <= I13] The dependency graph for this problem is: 0 -> 1 1 -> 3 2 -> 4, 5 3 -> 4, 5 4 -> 2, 3 5 -> 2, 3 Where: 0) f4#(x1, x2, x3) -> f3#(x1, x2, x3) 1) f3#(I0, I1, I2) -> f2#(I0, I1, 1 + I0) [0 <= 1 + I0 /\ 1 + I0 <= 1 + I1 /\ 1 <= I1 /\ 1 <= I0 /\ I0 <= I1] 2) f2#(I3, I4, I5) -> f1#(I3, I4, I5) [1 + I5 <= I3] 3) f2#(I6, I7, I8) -> f1#(I6, I7, I8) [1 + I6 <= I8] 4) f1#(I9, I10, I11) -> f2#(I9, I10, 0) [0 <= 0 /\ 0 <= 1 + I10 /\ 1 <= I11 /\ 1 + I10 <= I11] 5) f1#(I12, I13, I14) -> f2#(I12, I13, 1 + I14) [0 <= 1 + I14 /\ 1 + I14 <= 1 + I13 /\ I14 <= I13] We have the following SCCs. { 2, 3, 4, 5 } DP problem for innermost termination. P = f2#(I3, I4, I5) -> f1#(I3, I4, I5) [1 + I5 <= I3] f2#(I6, I7, I8) -> f1#(I6, I7, I8) [1 + I6 <= I8] f1#(I9, I10, I11) -> f2#(I9, I10, 0) [0 <= 0 /\ 0 <= 1 + I10 /\ 1 <= I11 /\ 1 + I10 <= I11] f1#(I12, I13, I14) -> f2#(I12, I13, 1 + I14) [0 <= 1 + I14 /\ 1 + I14 <= 1 + I13 /\ I14 <= I13] R = f4(x1, x2, x3) -> f3(x1, x2, x3) f3(I0, I1, I2) -> f2(I0, I1, 1 + I0) [0 <= 1 + I0 /\ 1 + I0 <= 1 + I1 /\ 1 <= I1 /\ 1 <= I0 /\ I0 <= I1] f2(I3, I4, I5) -> f1(I3, I4, I5) [1 + I5 <= I3] f2(I6, I7, I8) -> f1(I6, I7, I8) [1 + I6 <= I8] f1(I9, I10, I11) -> f2(I9, I10, 0) [0 <= 0 /\ 0 <= 1 + I10 /\ 1 <= I11 /\ 1 + I10 <= I11] f1(I12, I13, I14) -> f2(I12, I13, 1 + I14) [0 <= 1 + I14 /\ 1 + I14 <= 1 + I13 /\ I14 <= I13]
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