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	n143.star.cs.uiowa.edu
space	Mixed_ITRS_2014

run statistics

property	value
solver	Ctrl
configuration	Itrs
runtime (wallclock)	0.50911 seconds
cpu usage	0.519291
user time	0.273922
system time	0.245369
max virtual memory	113188.0
max residence set size	8652.0

stage attributes

key	value
starexec-result	YES

output

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 Integer TRS Innermost