details

property	value
status	complete
benchmark	Nested.jar-obl-8.smt2
ran by	Akihisa Yamada
cpu timeout	1200 seconds
wallclock timeout	300 seconds
memory limit	137438953472 bytes
execution host	n038.star.cs.uiowa.edu
space	From_AProVE_2014

run statistics

property	value
solver	Ctrl
configuration	Transition
runtime (wallclock)	1.99352 seconds
cpu usage	2.0375
user time	1.08257
system time	0.954927
max virtual memory	691036.0
max residence set size	8700.0

stage attributes

key	value
starexec-result	YES

output

1.93/1.99 YES 1.93/1.99 1.93/1.99 DP problem for innermost termination. 1.93/1.99 P = 1.93/1.99 init#(x1, x2) -> f1#(rnd1, rnd2) 1.93/1.99 f3#(I0, I1) -> f3#(I0, I1 + 1) [I1 <= 11] 1.93/1.99 f3#(I2, I3) -> f2#(I2 + 1, I4) [11 <= I3 - 1] 1.93/1.99 f2#(I5, I6) -> f3#(I5, 3) [I5 <= 9] 1.93/1.99 f1#(I7, I8) -> f2#(0, I9) 1.93/1.99 R = 1.93/1.99 init(x1, x2) -> f1(rnd1, rnd2) 1.93/1.99 f3(I0, I1) -> f3(I0, I1 + 1) [I1 <= 11] 1.93/1.99 f3(I2, I3) -> f2(I2 + 1, I4) [11 <= I3 - 1] 1.93/1.99 f2(I5, I6) -> f3(I5, 3) [I5 <= 9] 1.93/1.99 f1(I7, I8) -> f2(0, I9) 1.93/1.99 1.93/1.99 The dependency graph for this problem is: 1.93/1.99 0 -> 4 1.93/1.99 1 -> 1, 2 1.93/1.99 2 -> 3 1.93/1.99 3 -> 1 1.93/1.99 4 -> 3 1.93/1.99 Where: 1.93/1.99 0) init#(x1, x2) -> f1#(rnd1, rnd2) 1.93/1.99 1) f3#(I0, I1) -> f3#(I0, I1 + 1) [I1 <= 11] 1.93/1.99 2) f3#(I2, I3) -> f2#(I2 + 1, I4) [11 <= I3 - 1] 1.93/1.99 3) f2#(I5, I6) -> f3#(I5, 3) [I5 <= 9] 1.93/1.99 4) f1#(I7, I8) -> f2#(0, I9) 1.93/1.99 1.93/1.99 We have the following SCCs. 1.93/1.99 { 1, 2, 3 } 1.93/1.99 1.93/1.99 DP problem for innermost termination. 1.93/1.99 P = 1.93/1.99 f3#(I0, I1) -> f3#(I0, I1 + 1) [I1 <= 11] 1.93/1.99 f3#(I2, I3) -> f2#(I2 + 1, I4) [11 <= I3 - 1] 1.93/1.99 f2#(I5, I6) -> f3#(I5, 3) [I5 <= 9] 1.93/1.99 R = 1.93/1.99 init(x1, x2) -> f1(rnd1, rnd2) 1.93/1.99 f3(I0, I1) -> f3(I0, I1 + 1) [I1 <= 11] 1.93/1.99 f3(I2, I3) -> f2(I2 + 1, I4) [11 <= I3 - 1] 1.93/1.99 f2(I5, I6) -> f3(I5, 3) [I5 <= 9] 1.93/1.99 f1(I7, I8) -> f2(0, I9) 1.93/1.99 1.93/1.99 We use the reverse value criterion with the projection function NU: 1.93/1.99 NU[f2#(z1,z2)] = 9 + -1 * z1 1.93/1.99 NU[f3#(z1,z2)] = 9 + -1 * (z1 + 1) 1.93/1.99 1.93/1.99 This gives the following inequalities: 1.93/1.99 I1 <= 11 ==> 9 + -1 * (I0 + 1) >= 9 + -1 * (I0 + 1) 1.93/1.99 11 <= I3 - 1 ==> 9 + -1 * (I2 + 1) >= 9 + -1 * (I2 + 1) 1.93/1.99 I5 <= 9 ==> 9 + -1 * I5 > 9 + -1 * (I5 + 1) with 9 + -1 * I5 >= 0 1.93/1.99 1.93/1.99 We remove all the strictly oriented dependency pairs. 1.93/1.99 1.93/1.99 DP problem for innermost termination. 1.93/1.99 P = 1.93/1.99 f3#(I0, I1) -> f3#(I0, I1 + 1) [I1 <= 11] 1.93/1.99 f3#(I2, I3) -> f2#(I2 + 1, I4) [11 <= I3 - 1] 1.93/1.99 R = 1.93/1.99 init(x1, x2) -> f1(rnd1, rnd2) 1.93/1.99 f3(I0, I1) -> f3(I0, I1 + 1) [I1 <= 11] 1.93/1.99 f3(I2, I3) -> f2(I2 + 1, I4) [11 <= I3 - 1] 1.93/1.99 f2(I5, I6) -> f3(I5, 3) [I5 <= 9] 1.93/1.99 f1(I7, I8) -> f2(0, I9) 1.93/1.99 1.93/1.99 The dependency graph for this problem is: 1.93/1.99 1 -> 1, 2 1.93/1.99 2 -> 1.93/1.99 Where: 1.93/1.99 1) f3#(I0, I1) -> f3#(I0, I1 + 1) [I1 <= 11] 1.93/1.99 2) f3#(I2, I3) -> f2#(I2 + 1, I4) [11 <= I3 - 1] 1.93/1.99 1.93/1.99 We have the following SCCs. 1.93/1.99 { 1 } 1.93/1.99 1.93/1.99 DP problem for innermost termination. 1.93/1.99 P = 1.93/1.99 f3#(I0, I1) -> f3#(I0, I1 + 1) [I1 <= 11] 1.93/1.99 R = 1.93/1.99 init(x1, x2) -> f1(rnd1, rnd2) 1.93/1.99 f3(I0, I1) -> f3(I0, I1 + 1) [I1 <= 11] 1.93/1.99 f3(I2, I3) -> f2(I2 + 1, I4) [11 <= I3 - 1] 1.93/1.99 f2(I5, I6) -> f3(I5, 3) [I5 <= 9] 1.93/1.99 f1(I7, I8) -> f2(0, I9) 1.93/1.99 1.93/1.99 We use the reverse value criterion with the projection function NU: 1.93/1.99 NU[f3#(z1,z2)] = 11 + -1 * z2 1.93/1.99 1.93/1.99 This gives the following inequalities: 1.93/1.99 I1 <= 11 ==> 11 + -1 * I1 > 11 + -1 * (I1 + 1) with 11 + -1 * I1 >= 0 1.93/1.99 1.93/1.99 All dependency pairs are strictly oriented, so the entire dependency pair problem may be removed. 1.93/4.97 EOF popout

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all output return to Integer_Transition_Systems 2019-03-29 01.54