details

property	value
status	complete
benchmark	countup.itrs
ran by	Akihisa Yamada
cpu timeout	1200 seconds
wallclock timeout	300 seconds
memory limit	137438953472 bytes
execution host	n145.star.cs.uiowa.edu
space	Mixed_ITRS_2014

run statistics

property	value
solver	Ctrl
configuration	Itrs
runtime (wallclock)	0.134685 seconds
cpu usage	0.137141
user time	0.08463
system time	0.052511
max virtual memory	113188.0
max residence set size	7776.0

stage attributes

key	value
starexec-result	YES

output

YES DP problem for innermost termination. P = cu#(true, x) -> cu#(x < 100000, x + 1) R = cu(true, x) -> cu(x < 100000, x + 1) This problem is converted using chaining, where edges between chained DPs are removed. DP problem for innermost termination. P = cu#(true, x) -> cu_1#(x) cu_1#(x) -> cu#(x < 100000, x + 1) cu_1#(x) -> cu_1#(x + 1) [x < 100000] R = cu(true, x) -> cu(x < 100000, x + 1) The dependency graph for this problem is: 0 -> 2, 1 1 -> 2 -> 2, 1 Where: 0) cu#(true, x) -> cu_1#(x) 1) cu_1#(x) -> cu#(x < 100000, x + 1) 2) cu_1#(x) -> cu_1#(x + 1) [x < 100000] We have the following SCCs. { 2 } DP problem for innermost termination. P = cu_1#(x) -> cu_1#(x + 1) [x < 100000] R = cu(true, x) -> cu(x < 100000, x + 1) We use the reverse value criterion with the projection function NU: NU[cu_1#(z1)] = 100000 + -1 * z1 This gives the following inequalities: x < 100000 ==> 100000 + -1 * x > 100000 + -1 * (x + 1) with 100000 + -1 * x >= 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