details

property	value
status	complete
benchmark	GCD.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)	0.781659 seconds
cpu usage	0.782746
user time	0.390881
system time	0.391865
max virtual memory	125064.0
max residence set size	8656.0

stage attributes

key	value
starexec-result	YES

output

0.75/0.78 YES 0.75/0.78 0.75/0.78 DP problem for innermost termination. 0.75/0.78 P = 0.75/0.78 init#(x1, x2) -> f1#(rnd1, rnd2) 0.75/0.78 f3#(I0, I1) -> f2#(I1, I0) [I0 <= I1 - 1] 0.75/0.78 f3#(I2, I3) -> f3#(I2 - I3, I3) [0 <= I3 - 1 /\ 0 <= I2 - 1 /\ I3 <= I2 - 1] 0.75/0.78 f2#(I4, I5) -> f2#(I4, 0) [I4 = I5 /\ 0 <= I4 - 1] 0.75/0.78 f2#(I6, I7) -> f2#(I7, 0) [0 = I6 /\ 0 <= I7 - 1] 0.75/0.78 f2#(I8, I9) -> f3#(I8, I9) [0 <= I8 - 1 /\ I9 <= I8 - 1 /\ 0 <= I9 - 1] 0.75/0.78 f2#(I10, I11) -> f3#(I10, I11) [0 <= I10 - 1 /\ I10 <= I11 - 1 /\ 0 <= I11 - 1] 0.75/0.78 f1#(I12, I13) -> f2#(I14, I15) [0 <= I12 - 1 /\ -1 <= I14 - 1 /\ -1 <= I13 - 1 /\ -1 <= I15 - 1] 0.75/0.78 R = 0.75/0.78 init(x1, x2) -> f1(rnd1, rnd2) 0.75/0.78 f3(I0, I1) -> f2(I1, I0) [I0 <= I1 - 1] 0.75/0.78 f3(I2, I3) -> f3(I2 - I3, I3) [0 <= I3 - 1 /\ 0 <= I2 - 1 /\ I3 <= I2 - 1] 0.75/0.78 f2(I4, I5) -> f2(I4, 0) [I4 = I5 /\ 0 <= I4 - 1] 0.75/0.78 f2(I6, I7) -> f2(I7, 0) [0 = I6 /\ 0 <= I7 - 1] 0.75/0.78 f2(I8, I9) -> f3(I8, I9) [0 <= I8 - 1 /\ I9 <= I8 - 1 /\ 0 <= I9 - 1] 0.75/0.78 f2(I10, I11) -> f3(I10, I11) [0 <= I10 - 1 /\ I10 <= I11 - 1 /\ 0 <= I11 - 1] 0.75/0.78 f1(I12, I13) -> f2(I14, I15) [0 <= I12 - 1 /\ -1 <= I14 - 1 /\ -1 <= I13 - 1 /\ -1 <= I15 - 1] 0.75/0.78 0.75/0.78 The dependency graph for this problem is: 0.75/0.78 0 -> 7 0.75/0.78 1 -> 5 0.75/0.78 2 -> 1, 2 0.75/0.78 3 -> 0.75/0.78 4 -> 0.75/0.78 5 -> 2 0.75/0.78 6 -> 1 0.75/0.78 7 -> 3, 4, 5, 6 0.75/0.78 Where: 0.75/0.78 0) init#(x1, x2) -> f1#(rnd1, rnd2) 0.75/0.78 1) f3#(I0, I1) -> f2#(I1, I0) [I0 <= I1 - 1] 0.75/0.78 2) f3#(I2, I3) -> f3#(I2 - I3, I3) [0 <= I3 - 1 /\ 0 <= I2 - 1 /\ I3 <= I2 - 1] 0.75/0.78 3) f2#(I4, I5) -> f2#(I4, 0) [I4 = I5 /\ 0 <= I4 - 1] 0.75/0.78 4) f2#(I6, I7) -> f2#(I7, 0) [0 = I6 /\ 0 <= I7 - 1] 0.75/0.78 5) f2#(I8, I9) -> f3#(I8, I9) [0 <= I8 - 1 /\ I9 <= I8 - 1 /\ 0 <= I9 - 1] 0.75/0.78 6) f2#(I10, I11) -> f3#(I10, I11) [0 <= I10 - 1 /\ I10 <= I11 - 1 /\ 0 <= I11 - 1] 0.75/0.78 7) f1#(I12, I13) -> f2#(I14, I15) [0 <= I12 - 1 /\ -1 <= I14 - 1 /\ -1 <= I13 - 1 /\ -1 <= I15 - 1] 0.75/0.78 0.75/0.78 We have the following SCCs. 0.75/0.78 { 1, 2, 5 } 0.75/0.78 0.75/0.78 DP problem for innermost termination. 0.75/0.78 P = 0.75/0.78 f3#(I0, I1) -> f2#(I1, I0) [I0 <= I1 - 1] 0.75/0.78 f3#(I2, I3) -> f3#(I2 - I3, I3) [0 <= I3 - 1 /\ 0 <= I2 - 1 /\ I3 <= I2 - 1] 0.75/0.78 f2#(I8, I9) -> f3#(I8, I9) [0 <= I8 - 1 /\ I9 <= I8 - 1 /\ 0 <= I9 - 1] 0.75/0.78 R = 0.75/0.78 init(x1, x2) -> f1(rnd1, rnd2) 0.75/0.78 f3(I0, I1) -> f2(I1, I0) [I0 <= I1 - 1] 0.75/0.78 f3(I2, I3) -> f3(I2 - I3, I3) [0 <= I3 - 1 /\ 0 <= I2 - 1 /\ I3 <= I2 - 1] 0.75/0.78 f2(I4, I5) -> f2(I4, 0) [I4 = I5 /\ 0 <= I4 - 1] 0.75/0.78 f2(I6, I7) -> f2(I7, 0) [0 = I6 /\ 0 <= I7 - 1] 0.75/0.78 f2(I8, I9) -> f3(I8, I9) [0 <= I8 - 1 /\ I9 <= I8 - 1 /\ 0 <= I9 - 1] 0.75/0.78 f2(I10, I11) -> f3(I10, I11) [0 <= I10 - 1 /\ I10 <= I11 - 1 /\ 0 <= I11 - 1] 0.75/0.78 f1(I12, I13) -> f2(I14, I15) [0 <= I12 - 1 /\ -1 <= I14 - 1 /\ -1 <= I13 - 1 /\ -1 <= I15 - 1] 0.75/0.78 0.75/0.78 We use the basic value criterion with the projection function NU: 0.75/0.78 NU[f2#(z1,z2)] = z1 0.75/0.78 NU[f3#(z1,z2)] = z2 0.75/0.78 0.75/0.78 This gives the following inequalities: 0.75/0.78 I0 <= I1 - 1 ==> I1 (>! \union =) I1 0.75/0.78 0 <= I3 - 1 /\ 0 <= I2 - 1 /\ I3 <= I2 - 1 ==> I3 (>! \union =) I3 0.75/0.78 0 <= I8 - 1 /\ I9 <= I8 - 1 /\ 0 <= I9 - 1 ==> I8 >! I9 0.75/0.78 0.75/0.78 We remove all the strictly oriented dependency pairs. 0.75/0.78 0.75/0.78 DP problem for innermost termination. 0.75/0.78 P = 0.75/0.78 f3#(I0, I1) -> f2#(I1, I0) [I0 <= I1 - 1] 0.75/0.78 f3#(I2, I3) -> f3#(I2 - I3, I3) [0 <= I3 - 1 /\ 0 <= I2 - 1 /\ I3 <= I2 - 1] 0.75/0.78 R = 0.75/0.78 init(x1, x2) -> f1(rnd1, rnd2) 0.75/0.78 f3(I0, I1) -> f2(I1, I0) [I0 <= I1 - 1] 0.75/0.78 f3(I2, I3) -> f3(I2 - I3, I3) [0 <= I3 - 1 /\ 0 <= I2 - 1 /\ I3 <= I2 - 1] 0.75/0.78 f2(I4, I5) -> f2(I4, 0) [I4 = I5 /\ 0 <= I4 - 1] 0.75/0.78 f2(I6, I7) -> f2(I7, 0) [0 = I6 /\ 0 <= I7 - 1] 0.75/0.78 f2(I8, I9) -> f3(I8, I9) [0 <= I8 - 1 /\ I9 <= I8 - 1 /\ 0 <= I9 - 1] 0.75/0.78 f2(I10, I11) -> f3(I10, I11) [0 <= I10 - 1 /\ I10 <= I11 - 1 /\ 0 <= I11 - 1] 0.75/0.78 f1(I12, I13) -> f2(I14, I15) [0 <= I12 - 1 /\ -1 <= I14 - 1 /\ -1 <= I13 - 1 /\ -1 <= I15 - 1] 0.75/0.78 0.75/0.78 The dependency graph for this problem is: 0.75/0.78 1 -> 0.75/0.78 2 -> 1, 2 0.75/0.78 Where: 0.75/0.78 1) f3#(I0, I1) -> f2#(I1, I0) [I0 <= I1 - 1] 0.75/0.78 2) f3#(I2, I3) -> f3#(I2 - I3, I3) [0 <= I3 - 1 /\ 0 <= I2 - 1 /\ I3 <= I2 - 1] 0.75/0.78 0.75/0.78 We have the following SCCs. 0.75/0.78 { 2 } 0.75/0.78 0.75/0.78 DP problem for innermost termination. 0.75/0.78 P = 0.75/0.78 f3#(I2, I3) -> f3#(I2 - I3, I3) [0 <= I3 - 1 /\ 0 <= I2 - 1 /\ I3 <= I2 - 1] 0.75/0.78 R = 0.75/0.78 init(x1, x2) -> f1(rnd1, rnd2) 0.75/0.78 f3(I0, I1) -> f2(I1, I0) [I0 <= I1 - 1] popout

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

job log

popout

actions

all output return to Integer_Transition_Systems 2019-03-29 01.54