details

property	value
status	complete
benchmark	Fibonacci.jar-obl-8.smt2
ran by	Akihisa Yamada
cpu timeout	1200 seconds
wallclock timeout	300 seconds
memory limit	137438953472 bytes
execution host	n041.star.cs.uiowa.edu
space	From_AProVE_2014

run statistics

property	value
solver	Ctrl
configuration	Transition
runtime (wallclock)	0.720185041428 seconds
cpu usage	0.757075472
max memory	1.189888E7

stage attributes

key	value
output-size	3275
starexec-result	YES

output

/export/starexec/sandbox/solver/bin/starexec_run_Transition /export/starexec/sandbox/benchmark/theBenchmark.smt2 /export/starexec/sandbox/output/output_files -------------------------------------------------------------------------------- YES DP problem for innermost termination. P = init#(x1, x2) -> f1#(rnd1, rnd2) f3#(I0, I1) -> f2#(I1, I2) [I1 <= I0 - 1 /\ 1 <= I0 - 1] f4#(I3, I4) -> f3#(I3, I3 - 2) [0 <= I3 - 1] f2#(I5, I6) -> f3#(I5, I5 - 2) [I5 - 1 <= I5 - 1 /\ 1 <= I5 - 1] f2#(I7, I8) -> f3#(2, 0) [2 = I7] f2#(I9, I10) -> f2#(I9 - 1, I11) [I9 - 1 <= I9 - 1 /\ 1 <= I9 - 1] f1#(I12, I13) -> f2#(I13, I14) [-1 <= I13 - 1 /\ 0 <= I12 - 1] R = init(x1, x2) -> f1(rnd1, rnd2) f3(I0, I1) -> f2(I1, I2) [I1 <= I0 - 1 /\ 1 <= I0 - 1] f4(I3, I4) -> f3(I3, I3 - 2) [0 <= I3 - 1] f2(I5, I6) -> f3(I5, I5 - 2) [I5 - 1 <= I5 - 1 /\ 1 <= I5 - 1] f2(I7, I8) -> f3(2, 0) [2 = I7] f2(I9, I10) -> f2(I9 - 1, I11) [I9 - 1 <= I9 - 1 /\ 1 <= I9 - 1] f1(I12, I13) -> f2(I13, I14) [-1 <= I13 - 1 /\ 0 <= I12 - 1] The dependency graph for this problem is: 0 -> 6 1 -> 3, 4, 5 2 -> 1 3 -> 1 4 -> 1 5 -> 3, 4, 5 6 -> 3, 4, 5 Where: 0) init#(x1, x2) -> f1#(rnd1, rnd2) 1) f3#(I0, I1) -> f2#(I1, I2) [I1 <= I0 - 1 /\ 1 <= I0 - 1] 2) f4#(I3, I4) -> f3#(I3, I3 - 2) [0 <= I3 - 1] 3) f2#(I5, I6) -> f3#(I5, I5 - 2) [I5 - 1 <= I5 - 1 /\ 1 <= I5 - 1] 4) f2#(I7, I8) -> f3#(2, 0) [2 = I7] 5) f2#(I9, I10) -> f2#(I9 - 1, I11) [I9 - 1 <= I9 - 1 /\ 1 <= I9 - 1] 6) f1#(I12, I13) -> f2#(I13, I14) [-1 <= I13 - 1 /\ 0 <= I12 - 1] We have the following SCCs. { 1, 3, 4, 5 } DP problem for innermost termination. P = f3#(I0, I1) -> f2#(I1, I2) [I1 <= I0 - 1 /\ 1 <= I0 - 1] f2#(I5, I6) -> f3#(I5, I5 - 2) [I5 - 1 <= I5 - 1 /\ 1 <= I5 - 1] f2#(I7, I8) -> f3#(2, 0) [2 = I7] f2#(I9, I10) -> f2#(I9 - 1, I11) [I9 - 1 <= I9 - 1 /\ 1 <= I9 - 1] R = init(x1, x2) -> f1(rnd1, rnd2) f3(I0, I1) -> f2(I1, I2) [I1 <= I0 - 1 /\ 1 <= I0 - 1] f4(I3, I4) -> f3(I3, I3 - 2) [0 <= I3 - 1] f2(I5, I6) -> f3(I5, I5 - 2) [I5 - 1 <= I5 - 1 /\ 1 <= I5 - 1] f2(I7, I8) -> f3(2, 0) [2 = I7] f2(I9, I10) -> f2(I9 - 1, I11) [I9 - 1 <= I9 - 1 /\ 1 <= I9 - 1] f1(I12, I13) -> f2(I13, I14) [-1 <= I13 - 1 /\ 0 <= I12 - 1] We use the basic value criterion with the projection function NU: NU[f2#(z1,z2)] = z1 NU[f3#(z1,z2)] = z2 This gives the following inequalities: I1 <= I0 - 1 /\ 1 <= I0 - 1 ==> I1 (>! \union =) I1 I5 - 1 <= I5 - 1 /\ 1 <= I5 - 1 ==> I5 >! I5 - 2 2 = I7 ==> I7 >! 0 I9 - 1 <= I9 - 1 /\ 1 <= I9 - 1 ==> I9 >! I9 - 1 We remove all the strictly oriented dependency pairs. DP problem for innermost termination. P = f3#(I0, I1) -> f2#(I1, I2) [I1 <= I0 - 1 /\ 1 <= I0 - 1] R = init(x1, x2) -> f1(rnd1, rnd2) f3(I0, I1) -> f2(I1, I2) [I1 <= I0 - 1 /\ 1 <= I0 - 1] f4(I3, I4) -> f3(I3, I3 - 2) [0 <= I3 - 1] f2(I5, I6) -> f3(I5, I5 - 2) [I5 - 1 <= I5 - 1 /\ 1 <= I5 - 1] f2(I7, I8) -> f3(2, 0) [2 = I7] f2(I9, I10) -> f2(I9 - 1, I11) [I9 - 1 <= I9 - 1 /\ 1 <= I9 - 1] f1(I12, I13) -> f2(I13, I14) [-1 <= I13 - 1 /\ 0 <= I12 - 1] The dependency graph for this problem is: 1 -> Where: 1) f3#(I0, I1) -> f2#(I1, I2) [I1 <= I0 - 1 /\ 1 <= I0 - 1] We have the following SCCs. popout

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job log

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actions

all output return to Integ Trans Syste 27634