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	n146.star.cs.uiowa.edu
space	From_T2

run statistics

property	value
solver	Ctrl
configuration	Transition
runtime (wallclock)	5.29763 seconds
cpu usage	5.03761
user time	2.7014
system time	2.33621
max virtual memory	720476.0
max residence set size	8660.0

stage attributes

key	value
starexec-result	YES

output

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 We remove all the strictly oriented dependency pairs. DP problem for innermost termination. P = f2#(I30, I31, I32) -> f1#(I30, I31, I32) R = popout

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

job log

popout

actions

all output return to ITS