details

property	value
status	complete
benchmark	22.itrs
ran by	Akihisa Yamada
cpu timeout	1200 seconds
wallclock timeout	300 seconds
memory limit	137438953472 bytes
execution host	n119.star.cs.uiowa.edu
space	Mixed_ITRS_2014

run statistics

property	value
solver	Ctrl
configuration	Itrs
runtime (wallclock)	2.03369 seconds
cpu usage	2.0722
user time	1.02517
system time	1.04703
max virtual memory	356800.0
max residence set size	8664.0

stage attributes

key	value
starexec-result	YES

output

2.02/2.03 YES 2.02/2.03 2.02/2.03 DP problem for innermost termination. 2.02/2.03 P = 2.02/2.03 eval_2#(x, y) -> eval_1#(x - 1, y) [x > 0 && 0 >= y] 2.02/2.03 eval_2#(I0, I1) -> eval_2#(I0, I1 - 1) [I0 > 0 && I1 > 0] 2.02/2.03 eval_1#(I2, I3) -> eval_2#(I2, I3) [I2 > 0] 2.02/2.03 R = 2.02/2.03 eval_2(x, y) -> eval_1(x - 1, y) [x > 0 && 0 >= y] 2.02/2.03 eval_2(I0, I1) -> eval_2(I0, I1 - 1) [I0 > 0 && I1 > 0] 2.02/2.03 eval_1(I2, I3) -> eval_2(I2, I3) [I2 > 0] 2.02/2.03 2.02/2.03 We use the reverse value criterion with the projection function NU: 2.02/2.03 NU[eval_1#(z1,z2)] = z1 2.02/2.03 NU[eval_2#(z1,z2)] = z1 - 1 + -1 * 0 2.02/2.03 2.02/2.03 This gives the following inequalities: 2.02/2.03 x > 0 && 0 >= y ==> x - 1 + -1 * 0 >= x - 1 2.02/2.03 I0 > 0 && I1 > 0 ==> I0 - 1 + -1 * 0 >= I0 - 1 + -1 * 0 2.02/2.03 I2 > 0 ==> I2 > I2 - 1 + -1 * 0 with I2 >= 0 2.02/2.03 2.02/2.03 We remove all the strictly oriented dependency pairs. 2.02/2.03 2.02/2.03 DP problem for innermost termination. 2.02/2.03 P = 2.02/2.03 eval_2#(x, y) -> eval_1#(x - 1, y) [x > 0 && 0 >= y] 2.02/2.03 eval_2#(I0, I1) -> eval_2#(I0, I1 - 1) [I0 > 0 && I1 > 0] 2.02/2.03 R = 2.02/2.03 eval_2(x, y) -> eval_1(x - 1, y) [x > 0 && 0 >= y] 2.02/2.03 eval_2(I0, I1) -> eval_2(I0, I1 - 1) [I0 > 0 && I1 > 0] 2.02/2.03 eval_1(I2, I3) -> eval_2(I2, I3) [I2 > 0] 2.02/2.03 2.02/2.03 The dependency graph for this problem is: 2.02/2.03 0 -> 2.02/2.03 1 -> 0, 1 2.02/2.03 Where: 2.02/2.03 0) eval_2#(x, y) -> eval_1#(x - 1, y) [x > 0 && 0 >= y] 2.02/2.03 1) eval_2#(I0, I1) -> eval_2#(I0, I1 - 1) [I0 > 0 && I1 > 0] 2.02/2.03 2.02/2.03 We have the following SCCs. 2.02/2.03 { 1 } 2.02/2.03 2.02/2.03 DP problem for innermost termination. 2.02/2.03 P = 2.02/2.03 eval_2#(I0, I1) -> eval_2#(I0, I1 - 1) [I0 > 0 && I1 > 0] 2.02/2.03 R = 2.02/2.03 eval_2(x, y) -> eval_1(x - 1, y) [x > 0 && 0 >= y] 2.02/2.03 eval_2(I0, I1) -> eval_2(I0, I1 - 1) [I0 > 0 && I1 > 0] 2.02/2.03 eval_1(I2, I3) -> eval_2(I2, I3) [I2 > 0] 2.02/2.03 2.02/2.03 We use the reverse value criterion with the projection function NU: 2.02/2.03 NU[eval_2#(z1,z2)] = z2 2.02/2.03 2.02/2.03 This gives the following inequalities: 2.02/2.03 I0 > 0 && I1 > 0 ==> I1 > I1 - 1 with I1 >= 0 2.02/2.03 2.02/2.03 All dependency pairs are strictly oriented, so the entire dependency pair problem may be removed. 2.02/5.01 EOF popout

output may be truncated. 'popout' for the full output.

job log

popout

actions

all output return to Integer_TRS_Innermost 2019-03-21 04.53