details

property	value
status	complete
benchmark	Ex09.itrs
ran by	Akihisa Yamada
cpu timeout	1200 seconds
wallclock timeout	300 seconds
memory limit	137438953472 bytes
execution host	n149.star.cs.uiowa.edu
space	Mixed_ITRS_2014

run statistics

property	value
solver	Ctrl
configuration	Itrs
runtime (wallclock)	0.220926 seconds
cpu usage	0.224983
user time	0.107123
system time	0.11786
max virtual memory	113188.0
max residence set size	7764.0

stage attributes

key	value
starexec-result	MAYBE

output

MAYBE DP problem for innermost termination. P = round#(x) -> if#(x % 2 = 0, x, x + 1) minusNat#(true, I2, y) -> minus#(I2, round(I2)) minusNat#(true, I2, y) -> round#(I2) minus#(I3, I4) -> minusNat#(I4 >= 0 && I3 = I4 + 1, I3, I4) R = if(false, u, v) -> v if(true, I0, I1) -> I0 round(x) -> if(x % 2 = 0, x, x + 1) minusNat(true, I2, y) -> minus(I2, round(I2)) minus(I3, I4) -> minusNat(I4 >= 0 && I3 = I4 + 1, I3, I4) This problem is converted using chaining, where edges between chained DPs are removed. DP problem for innermost termination. P = round#(x) -> if#(x % 2 = 0, x, x + 1) minusNat#(true, I2, y) -> minus#(I2, round(I2)) minusNat#(true, I2, y) -> round#(I2) minus#(I3, I4) -> minusNat#(I4 >= 0 && I3 = I4 + 1, I3, I4) minus#(I3, I4) -> minus#(I3, round(I3)) [I4 >= 0 && I3 = I4 + 1] minus#(I3, I4) -> round#(I3) [I4 >= 0 && I3 = I4 + 1] R = if(false, u, v) -> v if(true, I0, I1) -> I0 round(x) -> if(x % 2 = 0, x, x + 1) minusNat(true, I2, y) -> minus(I2, round(I2)) minus(I3, I4) -> minusNat(I4 >= 0 && I3 = I4 + 1, I3, I4) The dependency graph for this problem is: 0 -> 1 -> 5, 4, 3 2 -> 0 3 -> 4 -> 5, 4, 3 5 -> 0 Where: 0) round#(x) -> if#(x % 2 = 0, x, x + 1) 1) minusNat#(true, I2, y) -> minus#(I2, round(I2)) 2) minusNat#(true, I2, y) -> round#(I2) 3) minus#(I3, I4) -> minusNat#(I4 >= 0 && I3 = I4 + 1, I3, I4) 4) minus#(I3, I4) -> minus#(I3, round(I3)) [I4 >= 0 && I3 = I4 + 1] 5) minus#(I3, I4) -> round#(I3) [I4 >= 0 && I3 = I4 + 1] We have the following SCCs. { 4 } DP problem for innermost termination. P = minus#(I3, I4) -> minus#(I3, round(I3)) [I4 >= 0 && I3 = I4 + 1] R = if(false, u, v) -> v if(true, I0, I1) -> I0 round(x) -> if(x % 2 = 0, x, x + 1) minusNat(true, I2, y) -> minus(I2, round(I2)) minus(I3, I4) -> minusNat(I4 >= 0 && I3 = I4 + 1, I3, I4) popout

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

job log

popout

actions

all output return to Integer TRS Innermost