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	n009.star.cs.uiowa.edu
space	Mixed_ITRS_2014

run statistics

property	value
solver	Ctrl
configuration	Itrs
runtime (wallclock)	0.501551 seconds
cpu usage	0.512173
user time	0.27843
system time	0.233743
max virtual memory	113176.0
max residence set size	8644.0

stage attributes

key	value
starexec-result	YES

output

0.00/0.50 YES 0.00/0.50 0.00/0.50 DP problem for innermost termination. 0.00/0.50 P = 0.00/0.50 f#(x, y, z) -> f#(x, y, z + 1) [x > y && x > z] 0.00/0.50 f#(I0, I1, I2) -> f#(I0, I1 + 1, I2) [I0 > I1 && I0 > I2] 0.00/0.50 R = 0.00/0.50 f(x, y, z) -> f(x, y, z + 1) [x > y && x > z] 0.00/0.50 f(I0, I1, I2) -> f(I0, I1 + 1, I2) [I0 > I1 && I0 > I2] 0.00/0.50 0.00/0.50 We use the reverse value criterion with the projection function NU: 0.00/0.50 NU[f#(z1,z2,z3)] = z1 + -1 * z3 0.00/0.50 0.00/0.50 This gives the following inequalities: 0.00/0.50 x > y && x > z ==> x + -1 * z > x + -1 * (z + 1) with x + -1 * z >= 0 0.00/0.50 I0 > I1 && I0 > I2 ==> I0 + -1 * I2 >= I0 + -1 * I2 0.00/0.50 0.00/0.50 We remove all the strictly oriented dependency pairs. 0.00/0.50 0.00/0.50 DP problem for innermost termination. 0.00/0.50 P = 0.00/0.50 f#(I0, I1, I2) -> f#(I0, I1 + 1, I2) [I0 > I1 && I0 > I2] 0.00/0.50 R = 0.00/0.50 f(x, y, z) -> f(x, y, z + 1) [x > y && x > z] 0.00/0.50 f(I0, I1, I2) -> f(I0, I1 + 1, I2) [I0 > I1 && I0 > I2] 0.00/0.50 0.00/0.50 We use the reverse value criterion with the projection function NU: 0.00/0.50 NU[f#(z1,z2,z3)] = z1 + -1 * z2 0.00/0.50 0.00/0.50 This gives the following inequalities: 0.00/0.50 I0 > I1 && I0 > I2 ==> I0 + -1 * I1 > I0 + -1 * (I1 + 1) with I0 + -1 * I1 >= 0 0.00/0.50 0.00/0.50 All dependency pairs are strictly oriented, so the entire dependency pair problem may be removed. 0.00/3.48 EOF popout

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job log

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actions

all output return to Integer_TRS_Innermost 2019-03-21 04.53