details

property	value
status	complete
benchmark	poly3.itrs
ran by	Akihisa Yamada
cpu timeout	1200 seconds
wallclock timeout	300 seconds
memory limit	137438953472 bytes
execution host	n093.star.cs.uiowa.edu
space	Mixed_ITRS_2014

run statistics

property	value
solver	Ctrl
configuration	Itrs
runtime (wallclock)	0.937294960022 seconds
cpu usage	0.943051754
max memory	1.513472E7

stage attributes

key	value
output-size	1154
starexec-result	MAYBE

output

/export/starexec/sandbox2/solver/bin/starexec_run_Itrs /export/starexec/sandbox2/benchmark/theBenchmark.itrs /export/starexec/sandbox2/output/output_files -------------------------------------------------------------------------------- MAYBE DP problem for innermost termination. P = eval#(x, y, z) -> eval#(x, y - 1, z + y) [x >= 0 && z * z * z >= y] eval#(I0, I1, I2) -> eval#(I0 - 1, I1 - 1, I2) [I0 >= 0 && I2 * I2 * I2 >= I1] R = eval(x, y, z) -> eval(x, y - 1, z + y) [x >= 0 && z * z * z >= y] eval(I0, I1, I2) -> eval(I0 - 1, I1 - 1, I2) [I0 >= 0 && I2 * I2 * I2 >= I1] We use the reverse value criterion with the projection function NU: NU[eval#(z1,z2,z3)] = z1 This gives the following inequalities: x >= 0 && z * z * z >= y ==> x >= x I0 >= 0 && I2 * I2 * I2 >= I1 ==> I0 > I0 - 1 with I0 >= 0 We remove all the strictly oriented dependency pairs. DP problem for innermost termination. P = eval#(x, y, z) -> eval#(x, y - 1, z + y) [x >= 0 && z * z * z >= y] R = eval(x, y, z) -> eval(x, y - 1, z + y) [x >= 0 && z * z * z >= y] eval(I0, I1, I2) -> eval(I0 - 1, I1 - 1, I2) [I0 >= 0 && I2 * I2 * I2 >= I1] popout

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

job log

popout

actions

all output return to Integ TRS Inner 43557