details

property	value
status	complete
benchmark	simple_test.t2.smt2
ran by	Akihisa Yamada
cpu timeout	1200 seconds
wallclock timeout	300 seconds
memory limit	137438953472 bytes
execution host	n105.star.cs.uiowa.edu
space	From_T2

run statistics

property	value
solver	Ctrl
configuration	Transition
runtime (wallclock)	5.50961780548 seconds
cpu usage	5.802763084
max memory	2.0406272E7

stage attributes

key	value
output-size	10371
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 = f10#(x1, x2, x3) -> f9#(x1, x2, x3) f9#(I0, I1, I2) -> f7#(0, I1, I2) f8#(I3, I4, I5) -> f7#(I3, I4, I5) f7#(I6, I7, I8) -> f8#(I6, -1 + I7, I8) [1 + I8 <= I7] f7#(I9, I10, I11) -> f5#(I9, I10, I11) [I10 <= I11] f6#(I12, I13, I14) -> f5#(I12, I13, I14) f5#(I15, I16, I17) -> f6#(1 + I15, I16, I17) [1 + I15 <= I17] f5#(I18, I19, I20) -> f3#(I18, I19, I20) [I20 <= I18] f4#(I21, I22, I23) -> f3#(I21, I22, I23) f3#(I24, I25, I26) -> f4#(I24, -1 + I25, I26) [1 <= I25] f3#(I27, I28, I29) -> f1#(I27, I28, I29) [I28 <= 0] f2#(I30, I31, I32) -> f1#(I30, I31, I32) f1#(I33, I34, I35) -> f2#(-1 + I33, I34, I35) [1 <= I33] R = f10(x1, x2, x3) -> f9(x1, x2, x3) f9(I0, I1, I2) -> f7(0, I1, I2) f8(I3, I4, I5) -> f7(I3, I4, I5) f7(I6, I7, I8) -> f8(I6, -1 + I7, I8) [1 + I8 <= I7] f7(I9, I10, I11) -> f5(I9, I10, I11) [I10 <= I11] f6(I12, I13, I14) -> f5(I12, I13, I14) f5(I15, I16, I17) -> f6(1 + I15, I16, I17) [1 + I15 <= I17] f5(I18, I19, I20) -> f3(I18, I19, I20) [I20 <= I18] f4(I21, I22, I23) -> f3(I21, I22, I23) f3(I24, I25, I26) -> f4(I24, -1 + I25, I26) [1 <= I25] f3(I27, I28, I29) -> f1(I27, I28, I29) [I28 <= 0] f2(I30, I31, I32) -> f1(I30, I31, I32) f1(I33, I34, I35) -> f2(-1 + I33, I34, I35) [1 <= I33] The dependency graph for this problem is: 0 -> 1 1 -> 3, 4 2 -> 3, 4 3 -> 2 4 -> 6, 7 5 -> 6, 7 6 -> 5 7 -> 9, 10 8 -> 9, 10 9 -> 8 10 -> 12 11 -> 12 12 -> 11 Where: 0) f10#(x1, x2, x3) -> f9#(x1, x2, x3) 1) f9#(I0, I1, I2) -> f7#(0, I1, I2) 2) f8#(I3, I4, I5) -> f7#(I3, I4, I5) 3) f7#(I6, I7, I8) -> f8#(I6, -1 + I7, I8) [1 + I8 <= I7] 4) f7#(I9, I10, I11) -> f5#(I9, I10, I11) [I10 <= I11] 5) f6#(I12, I13, I14) -> f5#(I12, I13, I14) 6) f5#(I15, I16, I17) -> f6#(1 + I15, I16, I17) [1 + I15 <= I17] 7) f5#(I18, I19, I20) -> f3#(I18, I19, I20) [I20 <= I18] 8) f4#(I21, I22, I23) -> f3#(I21, I22, I23) 9) f3#(I24, I25, I26) -> f4#(I24, -1 + I25, I26) [1 <= I25] 10) f3#(I27, I28, I29) -> f1#(I27, I28, I29) [I28 <= 0] 11) f2#(I30, I31, I32) -> f1#(I30, I31, I32) 12) f1#(I33, I34, I35) -> f2#(-1 + I33, I34, I35) [1 <= I33] We have the following SCCs. { 2, 3 } { 5, 6 } { 8, 9 } { 11, 12 } DP problem for innermost termination. P = f2#(I30, I31, I32) -> f1#(I30, I31, I32) f1#(I33, I34, I35) -> f2#(-1 + I33, I34, I35) [1 <= I33] R = f10(x1, x2, x3) -> f9(x1, x2, x3) f9(I0, I1, I2) -> f7(0, I1, I2) f8(I3, I4, I5) -> f7(I3, I4, I5) f7(I6, I7, I8) -> f8(I6, -1 + I7, I8) [1 + I8 <= I7] f7(I9, I10, I11) -> f5(I9, I10, I11) [I10 <= I11] f6(I12, I13, I14) -> f5(I12, I13, I14) f5(I15, I16, I17) -> f6(1 + I15, I16, I17) [1 + I15 <= I17] f5(I18, I19, I20) -> f3(I18, I19, I20) [I20 <= I18] f4(I21, I22, I23) -> f3(I21, I22, I23) f3(I24, I25, I26) -> f4(I24, -1 + I25, I26) [1 <= I25] f3(I27, I28, I29) -> f1(I27, I28, I29) [I28 <= 0] f2(I30, I31, I32) -> f1(I30, I31, I32) f1(I33, I34, I35) -> f2(-1 + I33, I34, I35) [1 <= I33] We use the basic value criterion with the projection function NU: NU[f1#(z1,z2,z3)] = z1 NU[f2#(z1,z2,z3)] = z1 This gives the following inequalities: ==> I30 (>! \union =) I30 1 <= I33 ==> I33 >! -1 + I33 popout

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

job log

popout

actions

all output return to Integer Transition Systems