details

property	value
status	complete
benchmark	p-7.t2.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_T2

run statistics

property	value
solver	Ctrl
configuration	Transition
runtime (wallclock)	9.14921 seconds
cpu usage	9.29942
user time	4.84155
system time	4.45787
max virtual memory	723304.0
max residence set size	11004.0

stage attributes

key	value
starexec-result	YES

output

YES DP problem for innermost termination. P = f7#(x1, x2, x3, x4, x5) -> f6#(x1, x2, x3, x4, x5) f4#(I5, I6, I7, I8, I9) -> f1#(I5, I6, 1, 1 + I8, I9) [0 <= I7 /\ I7 <= 0 /\ 0 <= -1 - I8 + I9] f6#(I10, I11, I12, I13, I14) -> f4#(I10, I11, 0, I13, I14) f3#(I21, I22, I23, I24, I25) -> f4#(I21, I22, 0, I24, -1 + I25) f2#(I26, I27, I28, I29, I30) -> f3#(I26, I27, I28, I29, I30) [I27 = I27] f1#(I31, I32, I33, I34, I35) -> f2#(I31, I32, I33, I34, I35) [0 <= -1 - I34 + I35] R = f7(x1, x2, x3, x4, x5) -> f6(x1, x2, x3, x4, x5) f4(I0, I1, I2, I3, I4) -> f5(rnd1, I1, I2, I3, I4) [rnd1 = rnd1 /\ -1 * I3 + I4 <= 0] f4(I5, I6, I7, I8, I9) -> f1(I5, I6, 1, 1 + I8, I9) [0 <= I7 /\ I7 <= 0 /\ 0 <= -1 - I8 + I9] f6(I10, I11, I12, I13, I14) -> f4(I10, I11, 0, I13, I14) f1(I15, I16, I17, I18, I19) -> f5(I20, I16, I17, I18, I19) [I20 = I20 /\ -1 * I18 + I19 <= 0] f3(I21, I22, I23, I24, I25) -> f4(I21, I22, 0, I24, -1 + I25) f2(I26, I27, I28, I29, I30) -> f3(I26, I27, I28, I29, I30) [I27 = I27] f1(I31, I32, I33, I34, I35) -> f2(I31, I32, I33, I34, I35) [0 <= -1 - I34 + I35] The dependency graph for this problem is: 0 -> 2 1 -> 5 2 -> 1 3 -> 1 4 -> 3 5 -> 4 Where: 0) f7#(x1, x2, x3, x4, x5) -> f6#(x1, x2, x3, x4, x5) 1) f4#(I5, I6, I7, I8, I9) -> f1#(I5, I6, 1, 1 + I8, I9) [0 <= I7 /\ I7 <= 0 /\ 0 <= -1 - I8 + I9] 2) f6#(I10, I11, I12, I13, I14) -> f4#(I10, I11, 0, I13, I14) 3) f3#(I21, I22, I23, I24, I25) -> f4#(I21, I22, 0, I24, -1 + I25) 4) f2#(I26, I27, I28, I29, I30) -> f3#(I26, I27, I28, I29, I30) [I27 = I27] 5) f1#(I31, I32, I33, I34, I35) -> f2#(I31, I32, I33, I34, I35) [0 <= -1 - I34 + I35] We have the following SCCs. { 1, 3, 4, 5 } DP problem for innermost termination. P = f4#(I5, I6, I7, I8, I9) -> f1#(I5, I6, 1, 1 + I8, I9) [0 <= I7 /\ I7 <= 0 /\ 0 <= -1 - I8 + I9] f3#(I21, I22, I23, I24, I25) -> f4#(I21, I22, 0, I24, -1 + I25) f2#(I26, I27, I28, I29, I30) -> f3#(I26, I27, I28, I29, I30) [I27 = I27] f1#(I31, I32, I33, I34, I35) -> f2#(I31, I32, I33, I34, I35) [0 <= -1 - I34 + I35] R = f7(x1, x2, x3, x4, x5) -> f6(x1, x2, x3, x4, x5) f4(I0, I1, I2, I3, I4) -> f5(rnd1, I1, I2, I3, I4) [rnd1 = rnd1 /\ -1 * I3 + I4 <= 0] f4(I5, I6, I7, I8, I9) -> f1(I5, I6, 1, 1 + I8, I9) [0 <= I7 /\ I7 <= 0 /\ 0 <= -1 - I8 + I9] f6(I10, I11, I12, I13, I14) -> f4(I10, I11, 0, I13, I14) f1(I15, I16, I17, I18, I19) -> f5(I20, I16, I17, I18, I19) [I20 = I20 /\ -1 * I18 + I19 <= 0] f3(I21, I22, I23, I24, I25) -> f4(I21, I22, 0, I24, -1 + I25) f2(I26, I27, I28, I29, I30) -> f3(I26, I27, I28, I29, I30) [I27 = I27] f1(I31, I32, I33, I34, I35) -> f2(I31, I32, I33, I34, I35) [0 <= -1 - I34 + I35] We use the extended value criterion with the projection function NU: NU[f2#(x0,x1,x2,x3,x4)] = -x3 + x4 - 2 NU[f3#(x0,x1,x2,x3,x4)] = -x3 + x4 - 2 NU[f1#(x0,x1,x2,x3,x4)] = -x3 + x4 - 1 NU[f4#(x0,x1,x2,x3,x4)] = -x2 - x3 + x4 - 2 This gives the following inequalities: 0 <= I7 /\ I7 <= 0 /\ 0 <= -1 - I8 + I9 ==> -I7 - I8 + I9 - 2 >= -(1 + I8) + I9 - 1 ==> -I24 + I25 - 2 >= -0 - I24 + (-1 + I25) - 2 I27 = I27 ==> -I29 + I30 - 2 >= -I29 + I30 - 2 0 <= -1 - I34 + I35 ==> -I34 + I35 - 1 > -I34 + I35 - 2 with -I34 + I35 - 1 >= 0 We remove all the strictly oriented dependency pairs. DP problem for innermost termination. P = f4#(I5, I6, I7, I8, I9) -> f1#(I5, I6, 1, 1 + I8, I9) [0 <= I7 /\ I7 <= 0 /\ 0 <= -1 - I8 + I9] f3#(I21, I22, I23, I24, I25) -> f4#(I21, I22, 0, I24, -1 + I25) f2#(I26, I27, I28, I29, I30) -> f3#(I26, I27, I28, I29, I30) [I27 = I27] R = f7(x1, x2, x3, x4, x5) -> f6(x1, x2, x3, x4, x5) f4(I0, I1, I2, I3, I4) -> f5(rnd1, I1, I2, I3, I4) [rnd1 = rnd1 /\ -1 * I3 + I4 <= 0] f4(I5, I6, I7, I8, I9) -> f1(I5, I6, 1, 1 + I8, I9) [0 <= I7 /\ I7 <= 0 /\ 0 <= -1 - I8 + I9] f6(I10, I11, I12, I13, I14) -> f4(I10, I11, 0, I13, I14) f1(I15, I16, I17, I18, I19) -> f5(I20, I16, I17, I18, I19) [I20 = I20 /\ -1 * I18 + I19 <= 0] f3(I21, I22, I23, I24, I25) -> f4(I21, I22, 0, I24, -1 + I25) f2(I26, I27, I28, I29, I30) -> f3(I26, I27, I28, I29, I30) [I27 = I27] f1(I31, I32, I33, I34, I35) -> f2(I31, I32, I33, I34, I35) [0 <= -1 - I34 + I35] The dependency graph for this problem is: 1 -> 3 -> 1 4 -> 3 Where: 1) f4#(I5, I6, I7, I8, I9) -> f1#(I5, I6, 1, 1 + I8, I9) [0 <= I7 /\ I7 <= 0 /\ 0 <= -1 - I8 + I9] 3) f3#(I21, I22, I23, I24, I25) -> f4#(I21, I22, 0, I24, -1 + I25) 4) f2#(I26, I27, I28, I29, I30) -> f3#(I26, I27, I28, I29, I30) [I27 = I27] We have the following SCCs. popout

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job log

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actions

all output return to ITS