details

property	value
status	complete
benchmark	PastaB13.jar-obl-8.smt2
ran by	Akihisa Yamada
cpu timeout	1200 seconds
wallclock timeout	300 seconds
memory limit	137438953472 bytes
execution host	n072.star.cs.uiowa.edu
space	From_AProVE_2014

run statistics

property	value
solver	Ctrl
configuration	Transition
runtime (wallclock)	4.4523 seconds
cpu usage	4.52563
user time	2.36782
system time	2.15781
max virtual memory	728184.0
max residence set size	8740.0

stage attributes

key	value
starexec-result	YES

output

4.44/4.45 YES 4.44/4.45 4.44/4.45 DP problem for innermost termination. 4.44/4.45 P = 4.44/4.45 init#(x1, x2, x3) -> f1#(rnd1, rnd2, rnd3) 4.44/4.45 f3#(I0, I1, I2) -> f2#(I0 - 1, I1, I2) [I2 <= I0 - 1 /\ I1 <= I2] 4.44/4.45 f3#(I3, I4, I5) -> f2#(I3, I4, I5) [I3 <= I5 /\ I4 <= I5] 4.44/4.45 f3#(I6, I7, I8) -> f2#(I6, I7 - 1, I8) [I8 <= I7 - 1] 4.44/4.45 f2#(I9, I10, I11) -> f3#(I9, I10, I11) [I11 <= I9 - 1 /\ I10 <= I11] 4.44/4.45 f2#(I12, I13, I14) -> f3#(I12, I13, I14) [I14 <= I13 - 1] 4.44/4.45 f1#(I15, I16, I17) -> f2#(I18, I19, I20) [0 <= I15 - 1 /\ -1 <= I18 - 1 /\ -1 <= I20 - 1 /\ -1 <= I16 - 1 /\ -1 <= I19 - 1] 4.44/4.45 R = 4.44/4.45 init(x1, x2, x3) -> f1(rnd1, rnd2, rnd3) 4.44/4.45 f3(I0, I1, I2) -> f2(I0 - 1, I1, I2) [I2 <= I0 - 1 /\ I1 <= I2] 4.44/4.45 f3(I3, I4, I5) -> f2(I3, I4, I5) [I3 <= I5 /\ I4 <= I5] 4.44/4.45 f3(I6, I7, I8) -> f2(I6, I7 - 1, I8) [I8 <= I7 - 1] 4.44/4.45 f2(I9, I10, I11) -> f3(I9, I10, I11) [I11 <= I9 - 1 /\ I10 <= I11] 4.44/4.45 f2(I12, I13, I14) -> f3(I12, I13, I14) [I14 <= I13 - 1] 4.44/4.45 f1(I15, I16, I17) -> f2(I18, I19, I20) [0 <= I15 - 1 /\ -1 <= I18 - 1 /\ -1 <= I20 - 1 /\ -1 <= I16 - 1 /\ -1 <= I19 - 1] 4.44/4.45 4.44/4.45 The dependency graph for this problem is: 4.44/4.45 0 -> 6 4.44/4.45 1 -> 4 4.44/4.45 2 -> 4.44/4.45 3 -> 4, 5 4.44/4.45 4 -> 1 4.44/4.45 5 -> 3 4.44/4.45 6 -> 4, 5 4.44/4.45 Where: 4.44/4.45 0) init#(x1, x2, x3) -> f1#(rnd1, rnd2, rnd3) 4.44/4.45 1) f3#(I0, I1, I2) -> f2#(I0 - 1, I1, I2) [I2 <= I0 - 1 /\ I1 <= I2] 4.44/4.45 2) f3#(I3, I4, I5) -> f2#(I3, I4, I5) [I3 <= I5 /\ I4 <= I5] 4.44/4.45 3) f3#(I6, I7, I8) -> f2#(I6, I7 - 1, I8) [I8 <= I7 - 1] 4.44/4.45 4) f2#(I9, I10, I11) -> f3#(I9, I10, I11) [I11 <= I9 - 1 /\ I10 <= I11] 4.44/4.45 5) f2#(I12, I13, I14) -> f3#(I12, I13, I14) [I14 <= I13 - 1] 4.44/4.45 6) f1#(I15, I16, I17) -> f2#(I18, I19, I20) [0 <= I15 - 1 /\ -1 <= I18 - 1 /\ -1 <= I20 - 1 /\ -1 <= I16 - 1 /\ -1 <= I19 - 1] 4.44/4.45 4.44/4.45 We have the following SCCs. 4.44/4.45 { 3, 5 } 4.44/4.45 { 1, 4 } 4.44/4.45 4.44/4.45 DP problem for innermost termination. 4.44/4.45 P = 4.44/4.45 f3#(I0, I1, I2) -> f2#(I0 - 1, I1, I2) [I2 <= I0 - 1 /\ I1 <= I2] 4.44/4.45 f2#(I9, I10, I11) -> f3#(I9, I10, I11) [I11 <= I9 - 1 /\ I10 <= I11] 4.44/4.45 R = 4.44/4.45 init(x1, x2, x3) -> f1(rnd1, rnd2, rnd3) 4.44/4.45 f3(I0, I1, I2) -> f2(I0 - 1, I1, I2) [I2 <= I0 - 1 /\ I1 <= I2] 4.44/4.45 f3(I3, I4, I5) -> f2(I3, I4, I5) [I3 <= I5 /\ I4 <= I5] 4.44/4.45 f3(I6, I7, I8) -> f2(I6, I7 - 1, I8) [I8 <= I7 - 1] 4.44/4.45 f2(I9, I10, I11) -> f3(I9, I10, I11) [I11 <= I9 - 1 /\ I10 <= I11] 4.44/4.45 f2(I12, I13, I14) -> f3(I12, I13, I14) [I14 <= I13 - 1] 4.44/4.45 f1(I15, I16, I17) -> f2(I18, I19, I20) [0 <= I15 - 1 /\ -1 <= I18 - 1 /\ -1 <= I20 - 1 /\ -1 <= I16 - 1 /\ -1 <= I19 - 1] 4.44/4.45 4.44/4.45 We use the reverse value criterion with the projection function NU: 4.44/4.45 NU[f2#(z1,z2,z3)] = z1 - 1 + -1 * z3 4.44/4.45 NU[f3#(z1,z2,z3)] = z1 - 1 - 1 + -1 * z3 4.44/4.45 4.44/4.45 This gives the following inequalities: 4.44/4.45 I2 <= I0 - 1 /\ I1 <= I2 ==> I0 - 1 - 1 + -1 * I2 >= I0 - 1 - 1 + -1 * I2 4.44/4.45 I11 <= I9 - 1 /\ I10 <= I11 ==> I9 - 1 + -1 * I11 > I9 - 1 - 1 + -1 * I11 with I9 - 1 + -1 * I11 >= 0 4.44/4.45 4.44/4.45 We remove all the strictly oriented dependency pairs. 4.44/4.45 4.44/4.45 DP problem for innermost termination. 4.44/4.45 P = 4.44/4.45 f3#(I0, I1, I2) -> f2#(I0 - 1, I1, I2) [I2 <= I0 - 1 /\ I1 <= I2] 4.44/4.45 R = 4.44/4.45 init(x1, x2, x3) -> f1(rnd1, rnd2, rnd3) 4.44/4.45 f3(I0, I1, I2) -> f2(I0 - 1, I1, I2) [I2 <= I0 - 1 /\ I1 <= I2] 4.44/4.45 f3(I3, I4, I5) -> f2(I3, I4, I5) [I3 <= I5 /\ I4 <= I5] 4.44/4.45 f3(I6, I7, I8) -> f2(I6, I7 - 1, I8) [I8 <= I7 - 1] 4.44/4.45 f2(I9, I10, I11) -> f3(I9, I10, I11) [I11 <= I9 - 1 /\ I10 <= I11] 4.44/4.45 f2(I12, I13, I14) -> f3(I12, I13, I14) [I14 <= I13 - 1] 4.44/4.45 f1(I15, I16, I17) -> f2(I18, I19, I20) [0 <= I15 - 1 /\ -1 <= I18 - 1 /\ -1 <= I20 - 1 /\ -1 <= I16 - 1 /\ -1 <= I19 - 1] 4.44/4.45 4.44/4.45 The dependency graph for this problem is: 4.44/4.45 1 -> 4.44/4.45 Where: 4.44/4.45 1) f3#(I0, I1, I2) -> f2#(I0 - 1, I1, I2) [I2 <= I0 - 1 /\ I1 <= I2] 4.44/4.45 4.44/4.45 We have the following SCCs. 4.44/4.45 4.44/4.45 4.44/4.45 DP problem for innermost termination. 4.44/4.45 P = 4.44/4.45 f3#(I6, I7, I8) -> f2#(I6, I7 - 1, I8) [I8 <= I7 - 1] 4.44/4.45 f2#(I12, I13, I14) -> f3#(I12, I13, I14) [I14 <= I13 - 1] 4.44/4.45 R = 4.44/4.45 init(x1, x2, x3) -> f1(rnd1, rnd2, rnd3) 4.44/4.45 f3(I0, I1, I2) -> f2(I0 - 1, I1, I2) [I2 <= I0 - 1 /\ I1 <= I2] 4.44/4.45 f3(I3, I4, I5) -> f2(I3, I4, I5) [I3 <= I5 /\ I4 <= I5] 4.44/4.45 f3(I6, I7, I8) -> f2(I6, I7 - 1, I8) [I8 <= I7 - 1] 4.44/4.45 f2(I9, I10, I11) -> f3(I9, I10, I11) [I11 <= I9 - 1 /\ I10 <= I11] 4.44/4.45 f2(I12, I13, I14) -> f3(I12, I13, I14) [I14 <= I13 - 1] 4.44/4.45 f1(I15, I16, I17) -> f2(I18, I19, I20) [0 <= I15 - 1 /\ -1 <= I18 - 1 /\ -1 <= I20 - 1 /\ -1 <= I16 - 1 /\ -1 <= I19 - 1] 4.44/4.45 4.44/4.45 We use the reverse value criterion with the projection function NU: 4.44/4.45 NU[f2#(z1,z2,z3)] = z2 - 1 + -1 * z3 4.44/4.45 NU[f3#(z1,z2,z3)] = z2 - 1 - 1 + -1 * z3 popout

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

job log

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actions

all output return to Integer_Transition_Systems 2019-03-29 01.54