details

property	value
status	complete
benchmark	Test10.jar-obl-9.smt2
ran by	Akihisa Yamada
cpu timeout	1200 seconds
wallclock timeout	300 seconds
memory limit	137438953472 bytes
execution host	n147.star.cs.uiowa.edu
space	From_AProVE_2014

run statistics

property	value
solver	Ctrl
configuration	Transition
runtime (wallclock)	1.6335 seconds
cpu usage	3.20309
user time	1.68724
system time	1.51584
max virtual memory	350684.0
max residence set size	8528.0

stage attributes

key	value
starexec-result	YES

output

YES DP problem for innermost termination. P = init#(x1, x2, x3) -> f1#(rnd1, rnd2, rnd3) f4#(I0, I1, I2) -> f4#(I0 - 1, I3, I4) [I0 - 1 <= I0 - 1 /\ 0 <= I0 - 1] f3#(I5, I6, I7) -> f3#(I5, I6 + 1, I6 + 1) [I6 = I7 /\ 0 <= I6 - 1 /\ I6 <= 99] f3#(I8, I9, I10) -> f4#(I9, I11, I12) [I9 = I10 /\ 0 <= I9 - 1 /\ I9 <= 99] f3#(I13, I14, I15) -> f4#(I14, I16, I17) [I14 = I15 /\ I14 <= 99 /\ 0 <= I14 - 1] f3#(I18, I19, I20) -> f2#(I18 - 1, I21, I22) [I19 = I20 /\ 99 <= I19 - 1 /\ 0 <= I18 - 1 /\ I18 - 1 <= I18 - 1] f2#(I23, I24, I25) -> f3#(I23, I23, I23) f1#(I26, I27, I28) -> f2#(I27, I29, I30) [-1 <= I27 - 1 /\ 0 <= I26 - 1] R = init(x1, x2, x3) -> f1(rnd1, rnd2, rnd3) f4(I0, I1, I2) -> f4(I0 - 1, I3, I4) [I0 - 1 <= I0 - 1 /\ 0 <= I0 - 1] f3(I5, I6, I7) -> f3(I5, I6 + 1, I6 + 1) [I6 = I7 /\ 0 <= I6 - 1 /\ I6 <= 99] f3(I8, I9, I10) -> f4(I9, I11, I12) [I9 = I10 /\ 0 <= I9 - 1 /\ I9 <= 99] f3(I13, I14, I15) -> f4(I14, I16, I17) [I14 = I15 /\ I14 <= 99 /\ 0 <= I14 - 1] f3(I18, I19, I20) -> f2(I18 - 1, I21, I22) [I19 = I20 /\ 99 <= I19 - 1 /\ 0 <= I18 - 1 /\ I18 - 1 <= I18 - 1] f2(I23, I24, I25) -> f3(I23, I23, I23) f1(I26, I27, I28) -> f2(I27, I29, I30) [-1 <= I27 - 1 /\ 0 <= I26 - 1] The dependency graph for this problem is: 0 -> 7 1 -> 1 2 -> 2, 3, 4, 5 3 -> 1 4 -> 1 5 -> 6 6 -> 2, 3, 4, 5 7 -> 6 Where: 0) init#(x1, x2, x3) -> f1#(rnd1, rnd2, rnd3) 1) f4#(I0, I1, I2) -> f4#(I0 - 1, I3, I4) [I0 - 1 <= I0 - 1 /\ 0 <= I0 - 1] 2) f3#(I5, I6, I7) -> f3#(I5, I6 + 1, I6 + 1) [I6 = I7 /\ 0 <= I6 - 1 /\ I6 <= 99] 3) f3#(I8, I9, I10) -> f4#(I9, I11, I12) [I9 = I10 /\ 0 <= I9 - 1 /\ I9 <= 99] 4) f3#(I13, I14, I15) -> f4#(I14, I16, I17) [I14 = I15 /\ I14 <= 99 /\ 0 <= I14 - 1] 5) f3#(I18, I19, I20) -> f2#(I18 - 1, I21, I22) [I19 = I20 /\ 99 <= I19 - 1 /\ 0 <= I18 - 1 /\ I18 - 1 <= I18 - 1] 6) f2#(I23, I24, I25) -> f3#(I23, I23, I23) 7) f1#(I26, I27, I28) -> f2#(I27, I29, I30) [-1 <= I27 - 1 /\ 0 <= I26 - 1] We have the following SCCs. { 2, 5, 6 } { 1 } DP problem for innermost termination. P = f4#(I0, I1, I2) -> f4#(I0 - 1, I3, I4) [I0 - 1 <= I0 - 1 /\ 0 <= I0 - 1] R = init(x1, x2, x3) -> f1(rnd1, rnd2, rnd3) f4(I0, I1, I2) -> f4(I0 - 1, I3, I4) [I0 - 1 <= I0 - 1 /\ 0 <= I0 - 1] f3(I5, I6, I7) -> f3(I5, I6 + 1, I6 + 1) [I6 = I7 /\ 0 <= I6 - 1 /\ I6 <= 99] f3(I8, I9, I10) -> f4(I9, I11, I12) [I9 = I10 /\ 0 <= I9 - 1 /\ I9 <= 99] f3(I13, I14, I15) -> f4(I14, I16, I17) [I14 = I15 /\ I14 <= 99 /\ 0 <= I14 - 1] f3(I18, I19, I20) -> f2(I18 - 1, I21, I22) [I19 = I20 /\ 99 <= I19 - 1 /\ 0 <= I18 - 1 /\ I18 - 1 <= I18 - 1] f2(I23, I24, I25) -> f3(I23, I23, I23) f1(I26, I27, I28) -> f2(I27, I29, I30) [-1 <= I27 - 1 /\ 0 <= I26 - 1] We use the basic value criterion with the projection function NU: NU[f4#(z1,z2,z3)] = z1 This gives the following inequalities: I0 - 1 <= I0 - 1 /\ 0 <= I0 - 1 ==> I0 >! I0 - 1 All dependency pairs are strictly oriented, so the entire dependency pair problem may be removed. DP problem for innermost termination. P = f3#(I5, I6, I7) -> f3#(I5, I6 + 1, I6 + 1) [I6 = I7 /\ 0 <= I6 - 1 /\ I6 <= 99] f3#(I18, I19, I20) -> f2#(I18 - 1, I21, I22) [I19 = I20 /\ 99 <= I19 - 1 /\ 0 <= I18 - 1 /\ I18 - 1 <= I18 - 1] f2#(I23, I24, I25) -> f3#(I23, I23, I23) R = init(x1, x2, x3) -> f1(rnd1, rnd2, rnd3) f4(I0, I1, I2) -> f4(I0 - 1, I3, I4) [I0 - 1 <= I0 - 1 /\ 0 <= I0 - 1] f3(I5, I6, I7) -> f3(I5, I6 + 1, I6 + 1) [I6 = I7 /\ 0 <= I6 - 1 /\ I6 <= 99] f3(I8, I9, I10) -> f4(I9, I11, I12) [I9 = I10 /\ 0 <= I9 - 1 /\ I9 <= 99] f3(I13, I14, I15) -> f4(I14, I16, I17) [I14 = I15 /\ I14 <= 99 /\ 0 <= I14 - 1] f3(I18, I19, I20) -> f2(I18 - 1, I21, I22) [I19 = I20 /\ 99 <= I19 - 1 /\ 0 <= I18 - 1 /\ I18 - 1 <= I18 - 1] f2(I23, I24, I25) -> f3(I23, I23, I23) f1(I26, I27, I28) -> f2(I27, I29, I30) [-1 <= I27 - 1 /\ 0 <= I26 - 1] We use the basic value criterion with the projection function NU: NU[f2#(z1,z2,z3)] = z1 NU[f3#(z1,z2,z3)] = z1 This gives the following inequalities: I6 = I7 /\ 0 <= I6 - 1 /\ I6 <= 99 ==> I5 (>! \union =) I5 I19 = I20 /\ 99 <= I19 - 1 /\ 0 <= I18 - 1 /\ I18 - 1 <= I18 - 1 ==> I18 >! I18 - 1 ==> I23 (>! \union =) I23 We remove all the strictly oriented dependency pairs. DP problem for innermost termination. P = f3#(I5, I6, I7) -> f3#(I5, I6 + 1, I6 + 1) [I6 = I7 /\ 0 <= I6 - 1 /\ I6 <= 99] f2#(I23, I24, I25) -> f3#(I23, I23, I23) R = init(x1, x2, x3) -> f1(rnd1, rnd2, rnd3) f4(I0, I1, I2) -> f4(I0 - 1, I3, I4) [I0 - 1 <= I0 - 1 /\ 0 <= I0 - 1] f3(I5, I6, I7) -> f3(I5, I6 + 1, I6 + 1) [I6 = I7 /\ 0 <= I6 - 1 /\ I6 <= 99] popout

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