details

property	value
status	complete
benchmark	Exc5.jar-obl-8.smt2
ran by	Akihisa Yamada
cpu timeout	1200 seconds
wallclock timeout	300 seconds
memory limit	137438953472 bytes
execution host	n056.star.cs.uiowa.edu
space	From_AProVE_2014

run statistics

property	value
solver	Ctrl
configuration	Transition
runtime (wallclock)	0.403347 seconds
cpu usage	0.411041
user time	0.217217
system time	0.193824
max virtual memory	113176.0
max residence set size	8000.0

stage attributes

key	value
starexec-result	YES

output

0.00/0.40 YES 0.00/0.40 0.00/0.40 DP problem for innermost termination. 0.00/0.40 P = 0.00/0.40 init#(x1) -> f1#(rnd1) 0.00/0.40 f2#(I0) -> f2#(I0 + 1) [I0 <= 19 /\ 11 <= I0 - 1] 0.00/0.40 f2#(I1) -> f2#(I1 + 1) [I1 <= 19 /\ I1 <= 11] 0.00/0.40 f1#(I2) -> f2#(0) 0.00/0.40 R = 0.00/0.40 init(x1) -> f1(rnd1) 0.00/0.40 f2(I0) -> f2(I0 + 1) [I0 <= 19 /\ 11 <= I0 - 1] 0.00/0.40 f2(I1) -> f2(I1 + 1) [I1 <= 19 /\ I1 <= 11] 0.00/0.40 f1(I2) -> f2(0) 0.00/0.40 0.00/0.40 The dependency graph for this problem is: 0.00/0.40 0 -> 3 0.00/0.40 1 -> 1 0.00/0.40 2 -> 1, 2 0.00/0.40 3 -> 2 0.00/0.40 Where: 0.00/0.40 0) init#(x1) -> f1#(rnd1) 0.00/0.40 1) f2#(I0) -> f2#(I0 + 1) [I0 <= 19 /\ 11 <= I0 - 1] 0.00/0.40 2) f2#(I1) -> f2#(I1 + 1) [I1 <= 19 /\ I1 <= 11] 0.00/0.40 3) f1#(I2) -> f2#(0) 0.00/0.40 0.00/0.40 We have the following SCCs. 0.00/0.40 { 2 } 0.00/0.40 { 1 } 0.00/0.40 0.00/0.40 DP problem for innermost termination. 0.00/0.40 P = 0.00/0.40 f2#(I0) -> f2#(I0 + 1) [I0 <= 19 /\ 11 <= I0 - 1] 0.00/0.40 R = 0.00/0.40 init(x1) -> f1(rnd1) 0.00/0.40 f2(I0) -> f2(I0 + 1) [I0 <= 19 /\ 11 <= I0 - 1] 0.00/0.40 f2(I1) -> f2(I1 + 1) [I1 <= 19 /\ I1 <= 11] 0.00/0.40 f1(I2) -> f2(0) 0.00/0.40 0.00/0.40 We use the reverse value criterion with the projection function NU: 0.00/0.40 NU[f2#(z1)] = 19 + -1 * z1 0.00/0.40 0.00/0.40 This gives the following inequalities: 0.00/0.40 I0 <= 19 /\ 11 <= I0 - 1 ==> 19 + -1 * I0 > 19 + -1 * (I0 + 1) with 19 + -1 * I0 >= 0 0.00/0.40 0.00/0.40 All dependency pairs are strictly oriented, so the entire dependency pair problem may be removed. 0.00/0.40 0.00/0.40 DP problem for innermost termination. 0.00/0.40 P = 0.00/0.40 f2#(I1) -> f2#(I1 + 1) [I1 <= 19 /\ I1 <= 11] 0.00/0.40 R = 0.00/0.40 init(x1) -> f1(rnd1) 0.00/0.40 f2(I0) -> f2(I0 + 1) [I0 <= 19 /\ 11 <= I0 - 1] 0.00/0.40 f2(I1) -> f2(I1 + 1) [I1 <= 19 /\ I1 <= 11] 0.00/0.40 f1(I2) -> f2(0) 0.00/0.40 0.00/0.40 We use the reverse value criterion with the projection function NU: 0.00/0.40 NU[f2#(z1)] = 11 + -1 * z1 0.00/0.40 0.00/0.40 This gives the following inequalities: 0.00/0.40 I1 <= 19 /\ I1 <= 11 ==> 11 + -1 * I1 > 11 + -1 * (I1 + 1) with 11 + -1 * I1 >= 0 0.00/0.40 0.00/0.40 All dependency pairs are strictly oriented, so the entire dependency pair problem may be removed. 0.00/3.38 EOF popout

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all output return to Integer_Transition_Systems 2019-03-29 01.54