details

property	value
status	complete
benchmark	increase2.itrs
ran by	Akihisa Yamada
cpu timeout	1200 seconds
wallclock timeout	300 seconds
memory limit	137438953472 bytes
execution host	n102.star.cs.uiowa.edu
space	Mixed_ITRS_2014

run statistics

property	value
solver	Ctrl
configuration	Itrs
runtime (wallclock)	0.264089822769 seconds
cpu usage	0.274331111
max memory	9580544.0

stage attributes

key	value
output-size	920
starexec-result	YES

output

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