details

property	value
status	complete
benchmark	costa09-example_1.jar-obl-10.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)	1.23042583466 seconds
cpu usage	1.31460623
max memory	1.325056E7

stage attributes

key	value
output-size	5288
starexec-result	YES

output

/export/starexec/sandbox/solver/bin/starexec_run_Transition /export/starexec/sandbox/benchmark/theBenchmark.smt2 /export/starexec/sandbox/output/output_files -------------------------------------------------------------------------------- YES DP problem for innermost termination. P = init#(x1, x2, x3) -> f1#(rnd1, rnd2, rnd3) f4#(I0, I1, I2) -> f4#(I0 + I1, I1 + 3, I1 + 3) [I1 = I2 /\ I1 <= 1000 /\ -1 <= I0 - 1 /\ -1 <= I1 - 1] f3#(I3, I4, I5) -> f4#(0, 0, 0) [I4 = I5 /\ 1000 <= I4 - 1] f3#(I6, I7, I8) -> f3#(I6 + I7, I7 + 2, I7 + 2) [I7 = I8 /\ I7 <= 1000 /\ -1 <= I6 - 1 /\ -1 <= I7 - 1] f2#(I9, I10, I11) -> f3#(0, 0, 0) [I10 = I11 /\ 1000 <= I10 - 1] f2#(I12, I13, I14) -> f2#(I12 + I13, I13 + 1, I13 + 1) [I13 = I14 /\ I13 <= 1000 /\ -1 <= I12 - 1 /\ -1 <= I13 - 1] f1#(I15, I16, I17) -> f2#(0, 0, 0) R = init(x1, x2, x3) -> f1(rnd1, rnd2, rnd3) f4(I0, I1, I2) -> f4(I0 + I1, I1 + 3, I1 + 3) [I1 = I2 /\ I1 <= 1000 /\ -1 <= I0 - 1 /\ -1 <= I1 - 1] f3(I3, I4, I5) -> f4(0, 0, 0) [I4 = I5 /\ 1000 <= I4 - 1] f3(I6, I7, I8) -> f3(I6 + I7, I7 + 2, I7 + 2) [I7 = I8 /\ I7 <= 1000 /\ -1 <= I6 - 1 /\ -1 <= I7 - 1] f2(I9, I10, I11) -> f3(0, 0, 0) [I10 = I11 /\ 1000 <= I10 - 1] f2(I12, I13, I14) -> f2(I12 + I13, I13 + 1, I13 + 1) [I13 = I14 /\ I13 <= 1000 /\ -1 <= I12 - 1 /\ -1 <= I13 - 1] f1(I15, I16, I17) -> f2(0, 0, 0) The dependency graph for this problem is: 0 -> 6 1 -> 1 2 -> 1 3 -> 2, 3 4 -> 3 5 -> 4, 5 6 -> 5 Where: 0) init#(x1, x2, x3) -> f1#(rnd1, rnd2, rnd3) 1) f4#(I0, I1, I2) -> f4#(I0 + I1, I1 + 3, I1 + 3) [I1 = I2 /\ I1 <= 1000 /\ -1 <= I0 - 1 /\ -1 <= I1 - 1] 2) f3#(I3, I4, I5) -> f4#(0, 0, 0) [I4 = I5 /\ 1000 <= I4 - 1] 3) f3#(I6, I7, I8) -> f3#(I6 + I7, I7 + 2, I7 + 2) [I7 = I8 /\ I7 <= 1000 /\ -1 <= I6 - 1 /\ -1 <= I7 - 1] 4) f2#(I9, I10, I11) -> f3#(0, 0, 0) [I10 = I11 /\ 1000 <= I10 - 1] 5) f2#(I12, I13, I14) -> f2#(I12 + I13, I13 + 1, I13 + 1) [I13 = I14 /\ I13 <= 1000 /\ -1 <= I12 - 1 /\ -1 <= I13 - 1] 6) f1#(I15, I16, I17) -> f2#(0, 0, 0) We have the following SCCs. { 5 } { 3 } { 1 } DP problem for innermost termination. P = f4#(I0, I1, I2) -> f4#(I0 + I1, I1 + 3, I1 + 3) [I1 = I2 /\ I1 <= 1000 /\ -1 <= I0 - 1 /\ -1 <= I1 - 1] R = init(x1, x2, x3) -> f1(rnd1, rnd2, rnd3) f4(I0, I1, I2) -> f4(I0 + I1, I1 + 3, I1 + 3) [I1 = I2 /\ I1 <= 1000 /\ -1 <= I0 - 1 /\ -1 <= I1 - 1] f3(I3, I4, I5) -> f4(0, 0, 0) [I4 = I5 /\ 1000 <= I4 - 1] f3(I6, I7, I8) -> f3(I6 + I7, I7 + 2, I7 + 2) [I7 = I8 /\ I7 <= 1000 /\ -1 <= I6 - 1 /\ -1 <= I7 - 1] f2(I9, I10, I11) -> f3(0, 0, 0) [I10 = I11 /\ 1000 <= I10 - 1] f2(I12, I13, I14) -> f2(I12 + I13, I13 + 1, I13 + 1) [I13 = I14 /\ I13 <= 1000 /\ -1 <= I12 - 1 /\ -1 <= I13 - 1] f1(I15, I16, I17) -> f2(0, 0, 0) We use the reverse value criterion with the projection function NU: NU[f4#(z1,z2,z3)] = 1000 + -1 * z2 This gives the following inequalities: I1 = I2 /\ I1 <= 1000 /\ -1 <= I0 - 1 /\ -1 <= I1 - 1 ==> 1000 + -1 * I1 > 1000 + -1 * (I1 + 3) with 1000 + -1 * I1 >= 0 All dependency pairs are strictly oriented, so the entire dependency pair problem may be removed. DP problem for innermost termination. P = f3#(I6, I7, I8) -> f3#(I6 + I7, I7 + 2, I7 + 2) [I7 = I8 /\ I7 <= 1000 /\ -1 <= I6 - 1 /\ -1 <= I7 - 1] R = init(x1, x2, x3) -> f1(rnd1, rnd2, rnd3) f4(I0, I1, I2) -> f4(I0 + I1, I1 + 3, I1 + 3) [I1 = I2 /\ I1 <= 1000 /\ -1 <= I0 - 1 /\ -1 <= I1 - 1] f3(I3, I4, I5) -> f4(0, 0, 0) [I4 = I5 /\ 1000 <= I4 - 1] f3(I6, I7, I8) -> f3(I6 + I7, I7 + 2, I7 + 2) [I7 = I8 /\ I7 <= 1000 /\ -1 <= I6 - 1 /\ -1 <= I7 - 1] f2(I9, I10, I11) -> f3(0, 0, 0) [I10 = I11 /\ 1000 <= I10 - 1] f2(I12, I13, I14) -> f2(I12 + I13, I13 + 1, I13 + 1) [I13 = I14 /\ I13 <= 1000 /\ -1 <= I12 - 1 /\ -1 <= I13 - 1] f1(I15, I16, I17) -> f2(0, 0, 0) We use the reverse value criterion with the projection function NU: NU[f3#(z1,z2,z3)] = 1000 + -1 * z2 This gives the following inequalities: I7 = I8 /\ I7 <= 1000 /\ -1 <= I6 - 1 /\ -1 <= I7 - 1 ==> 1000 + -1 * I7 > 1000 + -1 * (I7 + 2) with 1000 + -1 * I7 >= 0 All dependency pairs are strictly oriented, so the entire dependency pair problem may be removed. DP problem for innermost termination. P = f2#(I12, I13, I14) -> f2#(I12 + I13, I13 + 1, I13 + 1) [I13 = I14 /\ I13 <= 1000 /\ -1 <= I12 - 1 /\ -1 <= I13 - 1] R = init(x1, x2, x3) -> f1(rnd1, rnd2, rnd3) f4(I0, I1, I2) -> f4(I0 + I1, I1 + 3, I1 + 3) [I1 = I2 /\ I1 <= 1000 /\ -1 <= I0 - 1 /\ -1 <= I1 - 1] f3(I3, I4, I5) -> f4(0, 0, 0) [I4 = I5 /\ 1000 <= I4 - 1] f3(I6, I7, I8) -> f3(I6 + I7, I7 + 2, I7 + 2) [I7 = I8 /\ I7 <= 1000 /\ -1 <= I6 - 1 /\ -1 <= I7 - 1] f2(I9, I10, I11) -> f3(0, 0, 0) [I10 = I11 /\ 1000 <= I10 - 1] f2(I12, I13, I14) -> f2(I12 + I13, I13 + 1, I13 + 1) [I13 = I14 /\ I13 <= 1000 /\ -1 <= I12 - 1 /\ -1 <= I13 - 1] f1(I15, I16, I17) -> f2(0, 0, 0) popout

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