details

property	value
status	complete
benchmark	c.03.itrs
ran by	Akihisa Yamada
cpu timeout	1200 seconds
wallclock timeout	300 seconds
memory limit	137438953472 bytes
execution host	n047.star.cs.uiowa.edu
space	Mixed_ITRS_2014

run statistics

property	value
solver	Ctrl
configuration	Itrs
runtime (wallclock)	0.462646961212 seconds
cpu usage	0.482589956
max memory	1.038336E7

stage attributes

key	value
output-size	1416
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, z) -> eval#(x, y, z + 1) [y > x && x >= z] eval#(I0, I1, I2) -> eval#(I0 + 1, I1, I2) [I1 > I0 && I2 > I0] R = eval(x, y, z) -> eval(x, y, z + 1) [y > x && x >= z] eval(I0, I1, I2) -> eval(I0 + 1, I1, I2) [I1 > I0 && I2 > I0] We use the reverse value criterion with the projection function NU: NU[eval#(z1,z2,z3)] = z2 + -1 * z1 This gives the following inequalities: y > x && x >= z ==> y + -1 * x >= y + -1 * x I1 > I0 && I2 > I0 ==> I1 + -1 * I0 > I1 + -1 * (I0 + 1) with I1 + -1 * I0 >= 0 We remove all the strictly oriented dependency pairs. DP problem for innermost termination. P = eval#(x, y, z) -> eval#(x, y, z + 1) [y > x && x >= z] R = eval(x, y, z) -> eval(x, y, z + 1) [y > x && x >= z] eval(I0, I1, I2) -> eval(I0 + 1, I1, I2) [I1 > I0 && I2 > I0] We use the reverse value criterion with the projection function NU: NU[eval#(z1,z2,z3)] = z1 + -1 * z3 This gives the following inequalities: y > x && x >= z ==> x + -1 * z > x + -1 * (z + 1) with x + -1 * z >= 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