details

property	value
status	complete
benchmark	small01.t2_fixed.smt2
ran by	Akihisa Yamada
cpu timeout	1200 seconds
wallclock timeout	300 seconds
memory limit	137438953472 bytes
execution host	n073.star.cs.uiowa.edu
space	From_T2

run statistics

property	value
solver	Ctrl
configuration	Transition
runtime (wallclock)	1.15633106232 seconds
cpu usage	1.227583915
max memory	1.294336E7

stage attributes

key	value
output-size	1952
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 = f5#(x1) -> f4#(x1) f4#(I0) -> f1#(I0) [1 <= I0] f3#(I1) -> f1#(I1) f2#(I2) -> f3#(I2) [1 + I2 <= 0] f2#(I3) -> f3#(I3) [1 <= I3] f1#(I4) -> f2#(-1 + I4) R = f5(x1) -> f4(x1) f4(I0) -> f1(I0) [1 <= I0] f3(I1) -> f1(I1) f2(I2) -> f3(I2) [1 + I2 <= 0] f2(I3) -> f3(I3) [1 <= I3] f1(I4) -> f2(-1 + I4) The dependency graph for this problem is: 0 -> 1 1 -> 5 2 -> 5 3 -> 2 4 -> 2 5 -> 3, 4 Where: 0) f5#(x1) -> f4#(x1) 1) f4#(I0) -> f1#(I0) [1 <= I0] 2) f3#(I1) -> f1#(I1) 3) f2#(I2) -> f3#(I2) [1 + I2 <= 0] 4) f2#(I3) -> f3#(I3) [1 <= I3] 5) f1#(I4) -> f2#(-1 + I4) We have the following SCCs. { 2, 3, 4, 5 } DP problem for innermost termination. P = f3#(I1) -> f1#(I1) f2#(I2) -> f3#(I2) [1 + I2 <= 0] f2#(I3) -> f3#(I3) [1 <= I3] f1#(I4) -> f2#(-1 + I4) R = f5(x1) -> f4(x1) f4(I0) -> f1(I0) [1 <= I0] f3(I1) -> f1(I1) f2(I2) -> f3(I2) [1 + I2 <= 0] f2(I3) -> f3(I3) [1 <= I3] f1(I4) -> f2(-1 + I4) We use the extended value criterion with the projection function NU: NU[f2#(x0)] = x0 - 1 NU[f1#(x0)] = x0 - 2 NU[f3#(x0)] = x0 - 2 This gives the following inequalities: ==> I1 - 2 >= I1 - 2 1 + I2 <= 0 ==> I2 - 1 >= I2 - 2 1 <= I3 ==> I3 - 1 > I3 - 2 with I3 - 1 >= 0 ==> I4 - 2 >= (-1 + I4) - 1 We remove all the strictly oriented dependency pairs. DP problem for innermost termination. P = f3#(I1) -> f1#(I1) f2#(I2) -> f3#(I2) [1 + I2 <= 0] f1#(I4) -> f2#(-1 + I4) R = f5(x1) -> f4(x1) f4(I0) -> f1(I0) [1 <= I0] f3(I1) -> f1(I1) f2(I2) -> f3(I2) [1 + I2 <= 0] f2(I3) -> f3(I3) [1 <= I3] f1(I4) -> f2(-1 + I4) popout

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job log

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actions

all output return to Integer Transition Systems