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Integ Trans Syste 27634 pair #381739298
details
property
value
status
complete
benchmark
ex08_rec.jar-obl-8.smt2
ran by
Akihisa Yamada
cpu timeout
1200 seconds
wallclock timeout
300 seconds
memory limit
137438953472 bytes
execution host
n023.star.cs.uiowa.edu
space
From_AProVE_2014
run statistics
property
value
solver
Ctrl
configuration
Transition
runtime (wallclock)
8.00261282921 seconds
cpu usage
8.535844257
max memory
2.5862144E7
stage attributes
key
value
output-size
2982
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) -> f1#(rnd1, rnd2, rnd3) f3#(I0, I1, I2) -> f2#(1, 1, I1 + 1) [1 = I2 /\ 1 <= I1 - 1 /\ I0 <= 1] f2#(I3, I4, I5) -> f2#(0, 0, 9) [10 = I5 /\ I3 <= 1] f2#(I6, I7, I8) -> f2#(1, 1, 2) [1 = I8 /\ I6 <= 1] f3#(I9, I10, I11) -> f2#(0, 0, I10 - 1) [0 = I11 /\ 1 <= I10 - 1 /\ I9 <= 1] f2#(I12, I13, I14) -> f3#(I12, I14, I13) [10 <= I14 - 1] f2#(I15, I16, I17) -> f3#(I15, I17, I16) [1 <= I17 - 1 /\ I17 <= 9] f1#(I18, I19, I20) -> f2#(0, 0, 10 * I19) [-1 <= I19 - 1 /\ 0 <= I18 - 1] R = init(x1, x2, x3) -> f1(rnd1, rnd2, rnd3) f3(I0, I1, I2) -> f2(1, 1, I1 + 1) [1 = I2 /\ 1 <= I1 - 1 /\ I0 <= 1] f2(I3, I4, I5) -> f2(0, 0, 9) [10 = I5 /\ I3 <= 1] f2(I6, I7, I8) -> f2(1, 1, 2) [1 = I8 /\ I6 <= 1] f3(I9, I10, I11) -> f2(0, 0, I10 - 1) [0 = I11 /\ 1 <= I10 - 1 /\ I9 <= 1] f2(I12, I13, I14) -> f3(I12, I14, I13) [10 <= I14 - 1] f2(I15, I16, I17) -> f3(I15, I17, I16) [1 <= I17 - 1 /\ I17 <= 9] f1(I18, I19, I20) -> f2(0, 0, 10 * I19) [-1 <= I19 - 1 /\ 0 <= I18 - 1] The dependency graph for this problem is: 0 -> 7 1 -> 2, 5, 6 2 -> 6 3 -> 6 4 -> 2, 3, 5, 6 5 -> 1, 4 6 -> 1, 4 7 -> 2, 5 Where: 0) init#(x1, x2, x3) -> f1#(rnd1, rnd2, rnd3) 1) f3#(I0, I1, I2) -> f2#(1, 1, I1 + 1) [1 = I2 /\ 1 <= I1 - 1 /\ I0 <= 1] 2) f2#(I3, I4, I5) -> f2#(0, 0, 9) [10 = I5 /\ I3 <= 1] 3) f2#(I6, I7, I8) -> f2#(1, 1, 2) [1 = I8 /\ I6 <= 1] 4) f3#(I9, I10, I11) -> f2#(0, 0, I10 - 1) [0 = I11 /\ 1 <= I10 - 1 /\ I9 <= 1] 5) f2#(I12, I13, I14) -> f3#(I12, I14, I13) [10 <= I14 - 1] 6) f2#(I15, I16, I17) -> f3#(I15, I17, I16) [1 <= I17 - 1 /\ I17 <= 9] 7) f1#(I18, I19, I20) -> f2#(0, 0, 10 * I19) [-1 <= I19 - 1 /\ 0 <= I18 - 1] We have the following SCCs. { 1, 2, 3, 4, 5, 6 } DP problem for innermost termination. P = f3#(I0, I1, I2) -> f2#(1, 1, I1 + 1) [1 = I2 /\ 1 <= I1 - 1 /\ I0 <= 1] f2#(I3, I4, I5) -> f2#(0, 0, 9) [10 = I5 /\ I3 <= 1] f2#(I6, I7, I8) -> f2#(1, 1, 2) [1 = I8 /\ I6 <= 1] f3#(I9, I10, I11) -> f2#(0, 0, I10 - 1) [0 = I11 /\ 1 <= I10 - 1 /\ I9 <= 1] f2#(I12, I13, I14) -> f3#(I12, I14, I13) [10 <= I14 - 1] f2#(I15, I16, I17) -> f3#(I15, I17, I16) [1 <= I17 - 1 /\ I17 <= 9] R = init(x1, x2, x3) -> f1(rnd1, rnd2, rnd3) f3(I0, I1, I2) -> f2(1, 1, I1 + 1) [1 = I2 /\ 1 <= I1 - 1 /\ I0 <= 1] f2(I3, I4, I5) -> f2(0, 0, 9) [10 = I5 /\ I3 <= 1] f2(I6, I7, I8) -> f2(1, 1, 2) [1 = I8 /\ I6 <= 1] f3(I9, I10, I11) -> f2(0, 0, I10 - 1) [0 = I11 /\ 1 <= I10 - 1 /\ I9 <= 1] f2(I12, I13, I14) -> f3(I12, I14, I13) [10 <= I14 - 1] f2(I15, I16, I17) -> f3(I15, I17, I16) [1 <= I17 - 1 /\ I17 <= 9] f1(I18, I19, I20) -> f2(0, 0, 10 * I19) [-1 <= I19 - 1 /\ 0 <= I18 - 1]
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