details

property	value
status	complete
benchmark	FactSum.jar-obl-9.smt2
ran by	Akihisa Yamada
cpu timeout	1200 seconds
wallclock timeout	300 seconds
memory limit	137438953472 bytes
execution host	n096.star.cs.uiowa.edu
space	From_AProVE_2014

run statistics

property	value
solver	Ctrl
configuration	Transition
runtime (wallclock)	0.183984041214 seconds
cpu usage	0.189311763
max memory	8749056.0

stage attributes

key	value
output-size	2073
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) -> f1#(rnd1) f3#(I0) -> f3#(I0 - 1) [I0 - 1 <= I0 - 1 /\ 0 <= I0 - 1] f2#(I1) -> f2#(I1 - 1) [-1 <= I1 - 1] f2#(I2) -> f3#(I2) [-1 <= I2 - 1] f1#(I3) -> f2#(10) R = init(x1) -> f1(rnd1) f3(I0) -> f3(I0 - 1) [I0 - 1 <= I0 - 1 /\ 0 <= I0 - 1] f2(I1) -> f2(I1 - 1) [-1 <= I1 - 1] f2(I2) -> f3(I2) [-1 <= I2 - 1] f1(I3) -> f2(10) The dependency graph for this problem is: 0 -> 4 1 -> 1 2 -> 2, 3 3 -> 1 4 -> 2, 3 Where: 0) init#(x1) -> f1#(rnd1) 1) f3#(I0) -> f3#(I0 - 1) [I0 - 1 <= I0 - 1 /\ 0 <= I0 - 1] 2) f2#(I1) -> f2#(I1 - 1) [-1 <= I1 - 1] 3) f2#(I2) -> f3#(I2) [-1 <= I2 - 1] 4) f1#(I3) -> f2#(10) We have the following SCCs. { 2 } { 1 } DP problem for innermost termination. P = f3#(I0) -> f3#(I0 - 1) [I0 - 1 <= I0 - 1 /\ 0 <= I0 - 1] R = init(x1) -> f1(rnd1) f3(I0) -> f3(I0 - 1) [I0 - 1 <= I0 - 1 /\ 0 <= I0 - 1] f2(I1) -> f2(I1 - 1) [-1 <= I1 - 1] f2(I2) -> f3(I2) [-1 <= I2 - 1] f1(I3) -> f2(10) We use the basic value criterion with the projection function NU: NU[f3#(z1)] = z1 This gives the following inequalities: I0 - 1 <= I0 - 1 /\ 0 <= I0 - 1 ==> I0 >! I0 - 1 All dependency pairs are strictly oriented, so the entire dependency pair problem may be removed. DP problem for innermost termination. P = f2#(I1) -> f2#(I1 - 1) [-1 <= I1 - 1] R = init(x1) -> f1(rnd1) f3(I0) -> f3(I0 - 1) [I0 - 1 <= I0 - 1 /\ 0 <= I0 - 1] f2(I1) -> f2(I1 - 1) [-1 <= I1 - 1] f2(I2) -> f3(I2) [-1 <= I2 - 1] f1(I3) -> f2(10) We use the basic value criterion with the projection function NU: NU[f2#(z1)] = z1 This gives the following inequalities: -1 <= I1 - 1 ==> I1 >! I1 - 1 All dependency pairs are strictly oriented, so the entire dependency pair problem may be removed. popout

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all output return to Integ Trans Syste 27634