details

property	value
status	complete
benchmark	consts1.t2_fixed.smt2
ran by	Akihisa Yamada
cpu timeout	1200 seconds
wallclock timeout	300 seconds
memory limit	137438953472 bytes
execution host	n111.star.cs.uiowa.edu
space	From_T2

run statistics

property	value
solver	Ctrl
configuration	Transition
runtime (wallclock)	1.18620419502 seconds
cpu usage	1.216920681
max memory	1.3201408E7

stage attributes

key	value
output-size	1922
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#(300) f3#(I1) -> f1#(I1) f2#(I2) -> f3#(I2) [101 <= I2] f2#(I3) -> f3#(I3) [1 + I3 <= 100] f1#(I4) -> f2#(-1 + I4) R = f5(x1) -> f4(x1) f4(I0) -> f1(300) f3(I1) -> f1(I1) f2(I2) -> f3(I2) [101 <= I2] f2(I3) -> f3(I3) [1 + I3 <= 100] 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#(300) 2) f3#(I1) -> f1#(I1) 3) f2#(I2) -> f3#(I2) [101 <= I2] 4) f2#(I3) -> f3#(I3) [1 + I3 <= 100] 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) [101 <= I2] f2#(I3) -> f3#(I3) [1 + I3 <= 100] f1#(I4) -> f2#(-1 + I4) R = f5(x1) -> f4(x1) f4(I0) -> f1(300) f3(I1) -> f1(I1) f2(I2) -> f3(I2) [101 <= I2] f2(I3) -> f3(I3) [1 + I3 <= 100] f1(I4) -> f2(-1 + I4) We use the extended value criterion with the projection function NU: NU[f2#(x0)] = x0 NU[f1#(x0)] = x0 - 1 NU[f3#(x0)] = x0 - 1 This gives the following inequalities: ==> I1 - 1 >= I1 - 1 101 <= I2 ==> I2 > I2 - 1 with I2 >= 0 1 + I3 <= 100 ==> I3 >= I3 - 1 ==> I4 - 1 >= (-1 + I4) We remove all the strictly oriented dependency pairs. DP problem for innermost termination. P = f3#(I1) -> f1#(I1) f2#(I3) -> f3#(I3) [1 + I3 <= 100] f1#(I4) -> f2#(-1 + I4) R = f5(x1) -> f4(x1) f4(I0) -> f1(300) f3(I1) -> f1(I1) f2(I2) -> f3(I2) [101 <= I2] f2(I3) -> f3(I3) [1 + I3 <= 100] f1(I4) -> f2(-1 + I4) popout

output may be truncated. 'popout' for the full output.

job log

popout

actions

all output return to Integer Transition Systems