details

property	value
status	complete
benchmark	increase3.itrs
ran by	Akihisa Yamada
cpu timeout	1200 seconds
wallclock timeout	300 seconds
memory limit	137438953472 bytes
execution host	n110.star.cs.uiowa.edu
space	Mixed_ITRS_2014

run statistics

property	value
solver	Ctrl
configuration	Itrs
runtime (wallclock)	0.460204839706 seconds
cpu usage	0.467322112
max memory	1.024E7

stage attributes

key	value
output-size	1377
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 && x > z] f#(I0, I1, I2) -> f#(I0, I1 + 1, I2) [I0 > I1 && I0 > I2] R = f(x, y, z) -> f(x, y, z + 1) [x > y && x > z] f(I0, I1, I2) -> f(I0, I1 + 1, I2) [I0 > I1 && I0 > I2] We use the reverse value criterion with the projection function NU: NU[f#(z1,z2,z3)] = z1 + -1 * z3 This gives the following inequalities: x > y && x > z ==> x + -1 * z > x + -1 * (z + 1) with x + -1 * z >= 0 I0 > I1 && I0 > I2 ==> I0 + -1 * I2 >= I0 + -1 * I2 We remove all the strictly oriented dependency pairs. DP problem for innermost termination. P = f#(I0, I1, I2) -> f#(I0, I1 + 1, I2) [I0 > I1 && I0 > I2] R = f(x, y, z) -> f(x, y, z + 1) [x > y && x > z] f(I0, I1, I2) -> f(I0, I1 + 1, I2) [I0 > I1 && I0 > I2] We use the reverse value criterion with the projection function NU: NU[f#(z1,z2,z3)] = z1 + -1 * z2 This gives the following inequalities: I0 > I1 && I0 > I2 ==> I0 + -1 * I1 > I0 + -1 * (I1 + 1) with I0 + -1 * I1 >= 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