details

property	value
status	complete
benchmark	11.itrs
ran by	Akihisa Yamada
cpu timeout	1200 seconds
wallclock timeout	300 seconds
memory limit	137438953472 bytes
execution host	n009.star.cs.uiowa.edu
space	Mixed_ITRS_2014

run statistics

property	value
solver	Ctrl
configuration	Itrs
runtime (wallclock)	0.150391101837 seconds
cpu usage	0.155076627
max memory	8876032.0

stage attributes

key	value
output-size	1334
starexec-result	YES

output

/export/starexec/sandbox2/solver/bin/starexec_run_Itrs /export/starexec/sandbox2/benchmark/theBenchmark.itrs /export/starexec/sandbox2/output/output_files -------------------------------------------------------------------------------- YES DP problem for innermost termination. P = eval#(x, y) -> eval#(x - 1, y) [x > 0 && not(x = 0) && x <= y] eval#(I0, I1) -> eval#(I1, I1) [I0 > 0 && not(I0 = 0) && I0 > I1] R = eval(x, y) -> eval(x - 1, y) [x > 0 && not(x = 0) && x <= y] eval(I0, I1) -> eval(I1, I1) [I0 > 0 && not(I0 = 0) && I0 > I1] The dependency graph for this problem is: 0 -> 0 1 -> 0 Where: 0) eval#(x, y) -> eval#(x - 1, y) [x > 0 && not(x = 0) && x <= y] 1) eval#(I0, I1) -> eval#(I1, I1) [I0 > 0 && not(I0 = 0) && I0 > I1] We have the following SCCs. { 0 } DP problem for innermost termination. P = eval#(x, y) -> eval#(x - 1, y) [x > 0 && not(x = 0) && x <= y] R = eval(x, y) -> eval(x - 1, y) [x > 0 && not(x = 0) && x <= y] eval(I0, I1) -> eval(I1, I1) [I0 > 0 && not(I0 = 0) && I0 > I1] We use the reverse value criterion with the projection function NU: NU[eval#(z1,z2)] = z1 This gives the following inequalities: x > 0 && not(x = 0) && x <= y ==> x > x - 1 with x >= 0 All dependency pairs are strictly oriented, so the entire dependency pair problem may be removed. popout

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job log

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actions

all output return to Integ TRS Inner 43557