Spaces
Explore
Communities
Statistics
Reports
Cluster
Status
Help
ITS pair #487100194
details
property
value
status
complete
benchmark
Break.jar-obl-8.smt2
ran by
Akihisa Yamada
cpu timeout
1200 seconds
wallclock timeout
300 seconds
memory limit
137438953472 bytes
execution host
n141.star.cs.uiowa.edu
space
From_AProVE_2014
run statistics
property
value
solver
Ctrl
configuration
Transition
runtime (wallclock)
0.188508 seconds
cpu usage
0.192211
user time
0.098797
system time
0.093414
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 = init#(x1) -> f1#(rnd1) f2#(I0) -> f2#(I0 + 1) [I0 <= 10] f1#(I1) -> f2#(0) R = init(x1) -> f1(rnd1) f2(I0) -> f2(I0 + 1) [I0 <= 10] f1(I1) -> f2(0) The dependency graph for this problem is: 0 -> 2 1 -> 1 2 -> 1 Where: 0) init#(x1) -> f1#(rnd1) 1) f2#(I0) -> f2#(I0 + 1) [I0 <= 10] 2) f1#(I1) -> f2#(0) We have the following SCCs. { 1 } DP problem for innermost termination. P = f2#(I0) -> f2#(I0 + 1) [I0 <= 10] R = init(x1) -> f1(rnd1) f2(I0) -> f2(I0 + 1) [I0 <= 10] f1(I1) -> f2(0) We use the reverse value criterion with the projection function NU: NU[f2#(z1)] = 10 + -1 * z1 This gives the following inequalities: I0 <= 10 ==> 10 + -1 * I0 > 10 + -1 * (I0 + 1) with 10 + -1 * I0 >= 0 All dependency pairs are strictly oriented, so the entire dependency pair problem may be removed.
popout
output may be truncated. 'popout' for the full output.
job log
popout
actions
all output
return to ITS