details

property	value
status	complete
benchmark	A09.itrs
ran by	Akihisa Yamada
cpu timeout	1200 seconds
wallclock timeout	300 seconds
memory limit	137438953472 bytes
execution host	n139.star.cs.uiowa.edu
space	Mixed_ITRS_2014

run statistics

property	value
solver	Ctrl
configuration	Itrs
runtime (wallclock)	0.835984 seconds
cpu usage	0.494851
user time	0.240514
system time	0.254337
max virtual memory	231076.0
max residence set size	8656.0

stage attributes

key	value
starexec-result	MAYBE

output

MAYBE DP problem for innermost termination. P = round#(x) -> if#(x % 2 = 0, x, x + 1) fNat#(true, I2, y) -> f#(I2 > y, I2, round(y + 1)) fNat#(true, I2, y) -> round#(y + 1) f#(true, I3, I4) -> fNat#(I3 >= 0 && I4 >= 0, I3, I4) R = if(false, u, v) -> v if(true, I0, I1) -> I0 round(x) -> if(x % 2 = 0, x, x + 1) fNat(true, I2, y) -> f(I2 > y, I2, round(y + 1)) f(true, I3, I4) -> fNat(I3 >= 0 && I4 >= 0, 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) fNat#(true, I2, y) -> f#(I2 > y, I2, round(y + 1)) fNat#(true, I2, y) -> round#(y + 1) f#(true, I3, I4) -> f_1#(I3, I4) f_1#(I3, I4) -> fNat#(I3 >= 0 && I4 >= 0, I3, I4) f_1#(I3, I4) -> round#(I4 + 1) [I3 >= 0 && I4 >= 0] f_1#(I3, I4) -> f#(I3 > I4, I3, round(I4 + 1)) [I3 >= 0 && I4 >= 0] f_1#(I3, I4) -> f_1#(I3, round(I4 + 1)) [I3 >= 0 && I4 >= 0, I3 > I4] fNat#(true, I2, y) -> f_1#(I2, round(y + 1)) [I2 > y] R = if(false, u, v) -> v if(true, I0, I1) -> I0 round(x) -> if(x % 2 = 0, x, x + 1) fNat(true, I2, y) -> f(I2 > y, I2, round(y + 1)) f(true, I3, I4) -> fNat(I3 >= 0 && I4 >= 0, I3, I4) The dependency graph for this problem is: 0 -> 1 -> 2 -> 0 3 -> 7, 6, 5, 4 4 -> 5 -> 0 6 -> 7 -> 7, 4, 5, 6 8 -> 4, 5, 6, 7 Where: 0) round#(x) -> if#(x % 2 = 0, x, x + 1) 1) fNat#(true, I2, y) -> f#(I2 > y, I2, round(y + 1)) 2) fNat#(true, I2, y) -> round#(y + 1) 3) f#(true, I3, I4) -> f_1#(I3, I4) 4) f_1#(I3, I4) -> fNat#(I3 >= 0 && I4 >= 0, I3, I4) 5) f_1#(I3, I4) -> round#(I4 + 1) [I3 >= 0 && I4 >= 0] 6) f_1#(I3, I4) -> f#(I3 > I4, I3, round(I4 + 1)) [I3 >= 0 && I4 >= 0] 7) f_1#(I3, I4) -> f_1#(I3, round(I4 + 1)) [I3 >= 0 && I4 >= 0, I3 > I4] 8) fNat#(true, I2, y) -> f_1#(I2, round(y + 1)) [I2 > y] We have the following SCCs. { 7 } DP problem for innermost termination. P = f_1#(I3, I4) -> f_1#(I3, round(I4 + 1)) [I3 >= 0 && I4 >= 0, I3 > I4] R = if(false, u, v) -> v if(true, I0, I1) -> I0 round(x) -> if(x % 2 = 0, x, x + 1) fNat(true, I2, y) -> f(I2 > y, I2, round(y + 1)) f(true, I3, I4) -> fNat(I3 >= 0 && I4 >= 0, I3, I4) popout

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

job log

popout

actions

all output return to Integer TRS Innermost