details

property	value
status	complete
benchmark	fuhs-inflasso.t2.smt2
ran by	Akihisa Yamada
cpu timeout	1200 seconds
wallclock timeout	300 seconds
memory limit	137438953472 bytes
execution host	n083.star.cs.uiowa.edu
space	From_T2

run statistics

property	value
solver	Ctrl
configuration	Transition
runtime (wallclock)	7.03988695145 seconds
cpu usage	7.480024194
max memory	2.275328E7

stage attributes

key	value
output-size	4414
starexec-result	YES

output

/export/starexec/sandbox2/solver/bin/starexec_run_Transition /export/starexec/sandbox2/benchmark/theBenchmark.smt2 /export/starexec/sandbox2/output/output_files -------------------------------------------------------------------------------- YES DP problem for innermost termination. P = f5#(x1, x2, x3) -> f4#(x1, x2, x3) f4#(I0, I1, I2) -> f2#(0, I1, I2) f2#(I3, I4, I5) -> f1#(I3, 0, I5) [1 + I3 <= I5] f3#(I6, I7, I8) -> f1#(I6, I7, I8) f1#(I9, I10, I11) -> f3#(I9, 1 + I10, I11) [1 + I10 <= I9] f1#(I12, I13, I14) -> f2#(1 + I12, I13, I14) [I12 <= I13] R = f5(x1, x2, x3) -> f4(x1, x2, x3) f4(I0, I1, I2) -> f2(0, I1, I2) f2(I3, I4, I5) -> f1(I3, 0, I5) [1 + I3 <= I5] f3(I6, I7, I8) -> f1(I6, I7, I8) f1(I9, I10, I11) -> f3(I9, 1 + I10, I11) [1 + I10 <= I9] f1(I12, I13, I14) -> f2(1 + I12, I13, I14) [I12 <= I13] The dependency graph for this problem is: 0 -> 1 1 -> 2 2 -> 4, 5 3 -> 4, 5 4 -> 3 5 -> 2 Where: 0) f5#(x1, x2, x3) -> f4#(x1, x2, x3) 1) f4#(I0, I1, I2) -> f2#(0, I1, I2) 2) f2#(I3, I4, I5) -> f1#(I3, 0, I5) [1 + I3 <= I5] 3) f3#(I6, I7, I8) -> f1#(I6, I7, I8) 4) f1#(I9, I10, I11) -> f3#(I9, 1 + I10, I11) [1 + I10 <= I9] 5) f1#(I12, I13, I14) -> f2#(1 + I12, I13, I14) [I12 <= I13] We have the following SCCs. { 2, 3, 4, 5 } DP problem for innermost termination. P = f2#(I3, I4, I5) -> f1#(I3, 0, I5) [1 + I3 <= I5] f3#(I6, I7, I8) -> f1#(I6, I7, I8) f1#(I9, I10, I11) -> f3#(I9, 1 + I10, I11) [1 + I10 <= I9] f1#(I12, I13, I14) -> f2#(1 + I12, I13, I14) [I12 <= I13] R = f5(x1, x2, x3) -> f4(x1, x2, x3) f4(I0, I1, I2) -> f2(0, I1, I2) f2(I3, I4, I5) -> f1(I3, 0, I5) [1 + I3 <= I5] f3(I6, I7, I8) -> f1(I6, I7, I8) f1(I9, I10, I11) -> f3(I9, 1 + I10, I11) [1 + I10 <= I9] f1(I12, I13, I14) -> f2(1 + I12, I13, I14) [I12 <= I13] We use the extended value criterion with the projection function NU: NU[f3#(x0,x1,x2)] = -x0 + x2 - 2 NU[f1#(x0,x1,x2)] = -x0 + x2 - 2 NU[f2#(x0,x1,x2)] = -x0 + x2 - 1 This gives the following inequalities: 1 + I3 <= I5 ==> -I3 + I5 - 1 > -I3 + I5 - 2 with -I3 + I5 - 1 >= 0 ==> -I6 + I8 - 2 >= -I6 + I8 - 2 1 + I10 <= I9 ==> -I9 + I11 - 2 >= -I9 + I11 - 2 I12 <= I13 ==> -I12 + I14 - 2 >= -(1 + I12) + I14 - 1 We remove all the strictly oriented dependency pairs. DP problem for innermost termination. P = f3#(I6, I7, I8) -> f1#(I6, I7, I8) f1#(I9, I10, I11) -> f3#(I9, 1 + I10, I11) [1 + I10 <= I9] f1#(I12, I13, I14) -> f2#(1 + I12, I13, I14) [I12 <= I13] R = f5(x1, x2, x3) -> f4(x1, x2, x3) f4(I0, I1, I2) -> f2(0, I1, I2) f2(I3, I4, I5) -> f1(I3, 0, I5) [1 + I3 <= I5] f3(I6, I7, I8) -> f1(I6, I7, I8) f1(I9, I10, I11) -> f3(I9, 1 + I10, I11) [1 + I10 <= I9] f1(I12, I13, I14) -> f2(1 + I12, I13, I14) [I12 <= I13] The dependency graph for this problem is: 3 -> 4, 5 4 -> 3 5 -> Where: 3) f3#(I6, I7, I8) -> f1#(I6, I7, I8) 4) f1#(I9, I10, I11) -> f3#(I9, 1 + I10, I11) [1 + I10 <= I9] 5) f1#(I12, I13, I14) -> f2#(1 + I12, I13, I14) [I12 <= I13] We have the following SCCs. { 3, 4 } DP problem for innermost termination. P = f3#(I6, I7, I8) -> f1#(I6, I7, I8) f1#(I9, I10, I11) -> f3#(I9, 1 + I10, I11) [1 + I10 <= I9] R = f5(x1, x2, x3) -> f4(x1, x2, x3) popout

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