Spaces
Explore
Communities
Statistics
Reports
Cluster
Status
Help
ITS pair #487099846
details
property
value
status
complete
benchmark
PastaC5.jar-obl-8.smt2
ran by
Akihisa Yamada
cpu timeout
1200 seconds
wallclock timeout
300 seconds
memory limit
137438953472 bytes
execution host
n140.star.cs.uiowa.edu
space
From_AProVE_2014
run statistics
property
value
solver
Ctrl
configuration
Transition
runtime (wallclock)
0.396885 seconds
cpu usage
0.405363
user time
0.234289
system time
0.171074
max virtual memory
113188.0
max residence set size
8684.0
stage attributes
key
value
starexec-result
YES
output
YES DP problem for innermost termination. P = init#(x1, x2) -> f1#(rnd1, rnd2) f2#(I0, I1) -> f2#(I0 - I1, I1) [I1 <= I0 - 1 /\ 0 <= I1 - 1 /\ 0 <= I0 - 1] f2#(I2, I3) -> f2#(I2, I3 - I2) [I2 <= I3 - 1 /\ 0 <= I2 - 1 /\ 0 <= I3 - 1] f1#(I4, I5) -> f2#(I6, I7) [0 <= I4 - 1 /\ 0 <= I7 - 1 /\ 0 <= I6 - 1 /\ -1 <= I5 - 1] R = init(x1, x2) -> f1(rnd1, rnd2) f2(I0, I1) -> f2(I0 - I1, I1) [I1 <= I0 - 1 /\ 0 <= I1 - 1 /\ 0 <= I0 - 1] f2(I2, I3) -> f2(I2, I3 - I2) [I2 <= I3 - 1 /\ 0 <= I2 - 1 /\ 0 <= I3 - 1] f1(I4, I5) -> f2(I6, I7) [0 <= I4 - 1 /\ 0 <= I7 - 1 /\ 0 <= I6 - 1 /\ -1 <= I5 - 1] The dependency graph for this problem is: 0 -> 3 1 -> 1, 2 2 -> 1, 2 3 -> 1, 2 Where: 0) init#(x1, x2) -> f1#(rnd1, rnd2) 1) f2#(I0, I1) -> f2#(I0 - I1, I1) [I1 <= I0 - 1 /\ 0 <= I1 - 1 /\ 0 <= I0 - 1] 2) f2#(I2, I3) -> f2#(I2, I3 - I2) [I2 <= I3 - 1 /\ 0 <= I2 - 1 /\ 0 <= I3 - 1] 3) f1#(I4, I5) -> f2#(I6, I7) [0 <= I4 - 1 /\ 0 <= I7 - 1 /\ 0 <= I6 - 1 /\ -1 <= I5 - 1] We have the following SCCs. { 1, 2 } DP problem for innermost termination. P = f2#(I0, I1) -> f2#(I0 - I1, I1) [I1 <= I0 - 1 /\ 0 <= I1 - 1 /\ 0 <= I0 - 1] f2#(I2, I3) -> f2#(I2, I3 - I2) [I2 <= I3 - 1 /\ 0 <= I2 - 1 /\ 0 <= I3 - 1] R = init(x1, x2) -> f1(rnd1, rnd2) f2(I0, I1) -> f2(I0 - I1, I1) [I1 <= I0 - 1 /\ 0 <= I1 - 1 /\ 0 <= I0 - 1] f2(I2, I3) -> f2(I2, I3 - I2) [I2 <= I3 - 1 /\ 0 <= I2 - 1 /\ 0 <= I3 - 1] f1(I4, I5) -> f2(I6, I7) [0 <= I4 - 1 /\ 0 <= I7 - 1 /\ 0 <= I6 - 1 /\ -1 <= I5 - 1] We use the basic value criterion with the projection function NU: NU[f2#(z1,z2)] = z2 This gives the following inequalities: I1 <= I0 - 1 /\ 0 <= I1 - 1 /\ 0 <= I0 - 1 ==> I1 (>! \union =) I1 I2 <= I3 - 1 /\ 0 <= I2 - 1 /\ 0 <= I3 - 1 ==> I3 >! I3 - I2 We remove all the strictly oriented dependency pairs. DP problem for innermost termination. P = f2#(I0, I1) -> f2#(I0 - I1, I1) [I1 <= I0 - 1 /\ 0 <= I1 - 1 /\ 0 <= I0 - 1] R = init(x1, x2) -> f1(rnd1, rnd2) f2(I0, I1) -> f2(I0 - I1, I1) [I1 <= I0 - 1 /\ 0 <= I1 - 1 /\ 0 <= I0 - 1] f2(I2, I3) -> f2(I2, I3 - I2) [I2 <= I3 - 1 /\ 0 <= I2 - 1 /\ 0 <= I3 - 1] f1(I4, I5) -> f2(I6, I7) [0 <= I4 - 1 /\ 0 <= I7 - 1 /\ 0 <= I6 - 1 /\ -1 <= I5 - 1] We use the basic value criterion with the projection function NU: NU[f2#(z1,z2)] = z1 This gives the following inequalities: I1 <= I0 - 1 /\ 0 <= I1 - 1 /\ 0 <= I0 - 1 ==> I0 >! I0 - I1 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