details

property	value
status	complete
benchmark	nested2.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)	25.6569 seconds
cpu usage	26.0563
user time	12.9959
system time	13.0604
max virtual memory	727584.0
max residence set size	13864.0

stage attributes

key	value
starexec-result	YES

output

25.96/25.65 YES 25.96/25.65 25.96/25.65 DP problem for innermost termination. 25.96/25.65 P = 25.96/25.65 f6#(x1, x2, x3, x4, x5, x6, x7) -> f1#(x1, x2, x3, x4, x5, x6, x7) 25.96/25.65 f5#(I0, I1, I2, I3, I4, I5, I6) -> f3#(I0, I1, I2, I3, I4, I5, I6) 25.96/25.65 f3#(I7, I8, I9, I10, I11, I12, I13) -> f5#(I7, -1 + I8, I9, rnd4, -1 + I11, I12, rnd7) [1 <= rnd7 /\ -1 + rnd7 <= -1 + I11 /\ -1 + I11 <= -1 + rnd7 /\ -1 + rnd4 <= -1 + I8 /\ -1 + I8 <= -1 + rnd4 /\ 1 <= I11 /\ rnd7 = rnd7 /\ rnd4 = rnd4] 25.96/25.65 f3#(I14, I15, I16, I17, I18, I19, I20) -> f2#(I14, I15, I16, I17, I18, I19, I20) [I18 <= 0 /\ I18 <= 0] 25.96/25.65 f4#(I21, I22, I23, I24, I25, I26, I27) -> f3#(I21, -1 + I22, rnd3, I24, -1 + I25, rnd6, I27) [1 <= rnd6 /\ 1 <= rnd3 /\ -1 + rnd6 <= -1 + I25 /\ -1 + I25 <= -1 + rnd6 /\ -1 + rnd3 <= -1 + I22 /\ -1 + I22 <= -1 + rnd3 /\ 1 <= I25 /\ rnd6 = rnd6 /\ rnd3 = rnd3] 25.96/25.65 f2#(I28, I29, I30, I31, I32, I33, I34) -> f3#(I28, I29, I30, I31, I28, I33, I34) [1 <= I29 /\ 1 <= I28 /\ 1 <= I29] 25.96/25.65 f1#(I35, I36, I37, I38, I39, I40, I41) -> f2#(I35, rnd2, I37, I38, rnd5, I40, I41) [rnd5 = rnd5 /\ rnd2 = rnd2] 25.96/25.65 R = 25.96/25.65 f6(x1, x2, x3, x4, x5, x6, x7) -> f1(x1, x2, x3, x4, x5, x6, x7) 25.96/25.65 f5(I0, I1, I2, I3, I4, I5, I6) -> f3(I0, I1, I2, I3, I4, I5, I6) 25.96/25.65 f3(I7, I8, I9, I10, I11, I12, I13) -> f5(I7, -1 + I8, I9, rnd4, -1 + I11, I12, rnd7) [1 <= rnd7 /\ -1 + rnd7 <= -1 + I11 /\ -1 + I11 <= -1 + rnd7 /\ -1 + rnd4 <= -1 + I8 /\ -1 + I8 <= -1 + rnd4 /\ 1 <= I11 /\ rnd7 = rnd7 /\ rnd4 = rnd4] 25.96/25.65 f3(I14, I15, I16, I17, I18, I19, I20) -> f2(I14, I15, I16, I17, I18, I19, I20) [I18 <= 0 /\ I18 <= 0] 25.96/25.65 f4(I21, I22, I23, I24, I25, I26, I27) -> f3(I21, -1 + I22, rnd3, I24, -1 + I25, rnd6, I27) [1 <= rnd6 /\ 1 <= rnd3 /\ -1 + rnd6 <= -1 + I25 /\ -1 + I25 <= -1 + rnd6 /\ -1 + rnd3 <= -1 + I22 /\ -1 + I22 <= -1 + rnd3 /\ 1 <= I25 /\ rnd6 = rnd6 /\ rnd3 = rnd3] 25.96/25.65 f2(I28, I29, I30, I31, I32, I33, I34) -> f3(I28, I29, I30, I31, I28, I33, I34) [1 <= I29 /\ 1 <= I28 /\ 1 <= I29] 25.96/25.65 f1(I35, I36, I37, I38, I39, I40, I41) -> f2(I35, rnd2, I37, I38, rnd5, I40, I41) [rnd5 = rnd5 /\ rnd2 = rnd2] 25.96/25.65 25.96/25.65 The dependency graph for this problem is: 25.96/25.65 0 -> 6 25.96/25.65 1 -> 2, 3 25.96/25.65 2 -> 1 25.96/25.65 3 -> 5 25.96/25.65 4 -> 2, 3 25.96/25.65 5 -> 2 25.96/25.65 6 -> 5 25.96/25.65 Where: 25.96/25.65 0) f6#(x1, x2, x3, x4, x5, x6, x7) -> f1#(x1, x2, x3, x4, x5, x6, x7) 25.96/25.65 1) f5#(I0, I1, I2, I3, I4, I5, I6) -> f3#(I0, I1, I2, I3, I4, I5, I6) 25.96/25.65 2) f3#(I7, I8, I9, I10, I11, I12, I13) -> f5#(I7, -1 + I8, I9, rnd4, -1 + I11, I12, rnd7) [1 <= rnd7 /\ -1 + rnd7 <= -1 + I11 /\ -1 + I11 <= -1 + rnd7 /\ -1 + rnd4 <= -1 + I8 /\ -1 + I8 <= -1 + rnd4 /\ 1 <= I11 /\ rnd7 = rnd7 /\ rnd4 = rnd4] 25.96/25.65 3) f3#(I14, I15, I16, I17, I18, I19, I20) -> f2#(I14, I15, I16, I17, I18, I19, I20) [I18 <= 0 /\ I18 <= 0] 25.96/25.65 4) f4#(I21, I22, I23, I24, I25, I26, I27) -> f3#(I21, -1 + I22, rnd3, I24, -1 + I25, rnd6, I27) [1 <= rnd6 /\ 1 <= rnd3 /\ -1 + rnd6 <= -1 + I25 /\ -1 + I25 <= -1 + rnd6 /\ -1 + rnd3 <= -1 + I22 /\ -1 + I22 <= -1 + rnd3 /\ 1 <= I25 /\ rnd6 = rnd6 /\ rnd3 = rnd3] 25.96/25.65 5) f2#(I28, I29, I30, I31, I32, I33, I34) -> f3#(I28, I29, I30, I31, I28, I33, I34) [1 <= I29 /\ 1 <= I28 /\ 1 <= I29] 25.96/25.65 6) f1#(I35, I36, I37, I38, I39, I40, I41) -> f2#(I35, rnd2, I37, I38, rnd5, I40, I41) [rnd5 = rnd5 /\ rnd2 = rnd2] 25.96/25.65 25.96/25.65 We have the following SCCs. 25.96/25.65 { 1, 2, 3, 5 } 25.96/25.65 25.96/25.65 DP problem for innermost termination. 25.96/25.65 P = 25.96/25.65 f5#(I0, I1, I2, I3, I4, I5, I6) -> f3#(I0, I1, I2, I3, I4, I5, I6) 25.96/25.65 f3#(I7, I8, I9, I10, I11, I12, I13) -> f5#(I7, -1 + I8, I9, rnd4, -1 + I11, I12, rnd7) [1 <= rnd7 /\ -1 + rnd7 <= -1 + I11 /\ -1 + I11 <= -1 + rnd7 /\ -1 + rnd4 <= -1 + I8 /\ -1 + I8 <= -1 + rnd4 /\ 1 <= I11 /\ rnd7 = rnd7 /\ rnd4 = rnd4] 25.96/25.65 f3#(I14, I15, I16, I17, I18, I19, I20) -> f2#(I14, I15, I16, I17, I18, I19, I20) [I18 <= 0 /\ I18 <= 0] 25.96/25.65 f2#(I28, I29, I30, I31, I32, I33, I34) -> f3#(I28, I29, I30, I31, I28, I33, I34) [1 <= I29 /\ 1 <= I28 /\ 1 <= I29] 25.96/25.65 R = 25.96/25.65 f6(x1, x2, x3, x4, x5, x6, x7) -> f1(x1, x2, x3, x4, x5, x6, x7) 25.96/25.65 f5(I0, I1, I2, I3, I4, I5, I6) -> f3(I0, I1, I2, I3, I4, I5, I6) 25.96/25.65 f3(I7, I8, I9, I10, I11, I12, I13) -> f5(I7, -1 + I8, I9, rnd4, -1 + I11, I12, rnd7) [1 <= rnd7 /\ -1 + rnd7 <= -1 + I11 /\ -1 + I11 <= -1 + rnd7 /\ -1 + rnd4 <= -1 + I8 /\ -1 + I8 <= -1 + rnd4 /\ 1 <= I11 /\ rnd7 = rnd7 /\ rnd4 = rnd4] 25.96/25.65 f3(I14, I15, I16, I17, I18, I19, I20) -> f2(I14, I15, I16, I17, I18, I19, I20) [I18 <= 0 /\ I18 <= 0] 25.96/25.65 f4(I21, I22, I23, I24, I25, I26, I27) -> f3(I21, -1 + I22, rnd3, I24, -1 + I25, rnd6, I27) [1 <= rnd6 /\ 1 <= rnd3 /\ -1 + rnd6 <= -1 + I25 /\ -1 + I25 <= -1 + rnd6 /\ -1 + rnd3 <= -1 + I22 /\ -1 + I22 <= -1 + rnd3 /\ 1 <= I25 /\ rnd6 = rnd6 /\ rnd3 = rnd3] 25.96/25.65 f2(I28, I29, I30, I31, I32, I33, I34) -> f3(I28, I29, I30, I31, I28, I33, I34) [1 <= I29 /\ 1 <= I28 /\ 1 <= I29] 25.96/25.65 f1(I35, I36, I37, I38, I39, I40, I41) -> f2(I35, rnd2, I37, I38, rnd5, I40, I41) [rnd5 = rnd5 /\ rnd2 = rnd2] 25.96/25.65 25.96/25.65 We use the extended value criterion with the projection function NU: 25.96/25.65 NU[f2#(x0,x1,x2,x3,x4,x5,x6)] = x1 - 1 25.96/25.65 NU[f3#(x0,x1,x2,x3,x4,x5,x6)] = x1 - x4 - 1 25.96/25.65 NU[f5#(x0,x1,x2,x3,x4,x5,x6)] = x1 - x4 - 1 25.96/25.65 25.96/25.65 This gives the following inequalities: 25.96/25.65 ==> I1 - I4 - 1 >= I1 - I4 - 1 25.96/25.65 1 <= rnd7 /\ -1 + rnd7 <= -1 + I11 /\ -1 + I11 <= -1 + rnd7 /\ -1 + rnd4 <= -1 + I8 /\ -1 + I8 <= -1 + rnd4 /\ 1 <= I11 /\ rnd7 = rnd7 /\ rnd4 = rnd4 ==> I8 - I11 - 1 >= (-1 + I8) - (-1 + I11) - 1 25.96/25.65 I18 <= 0 /\ I18 <= 0 ==> I15 - I18 - 1 >= I15 - 1 25.96/25.65 1 <= I29 /\ 1 <= I28 /\ 1 <= I29 ==> I29 - 1 > I29 - I28 - 1 with I29 - 1 >= 0 25.96/25.65 25.96/25.65 We remove all the strictly oriented dependency pairs. 25.96/25.65 25.96/25.65 DP problem for innermost termination. 25.96/25.65 P = 25.96/25.65 f5#(I0, I1, I2, I3, I4, I5, I6) -> f3#(I0, I1, I2, I3, I4, I5, I6) 25.96/25.65 f3#(I7, I8, I9, I10, I11, I12, I13) -> f5#(I7, -1 + I8, I9, rnd4, -1 + I11, I12, rnd7) [1 <= rnd7 /\ -1 + rnd7 <= -1 + I11 /\ -1 + I11 <= -1 + rnd7 /\ -1 + rnd4 <= -1 + I8 /\ -1 + I8 <= -1 + rnd4 /\ 1 <= I11 /\ rnd7 = rnd7 /\ rnd4 = rnd4] 25.96/25.65 f3#(I14, I15, I16, I17, I18, I19, I20) -> f2#(I14, I15, I16, I17, I18, I19, I20) [I18 <= 0 /\ I18 <= 0] 25.96/25.65 R = 25.96/25.65 f6(x1, x2, x3, x4, x5, x6, x7) -> f1(x1, x2, x3, x4, x5, x6, x7) 25.96/25.65 f5(I0, I1, I2, I3, I4, I5, I6) -> f3(I0, I1, I2, I3, I4, I5, I6) 25.96/25.65 f3(I7, I8, I9, I10, I11, I12, I13) -> f5(I7, -1 + I8, I9, rnd4, -1 + I11, I12, rnd7) [1 <= rnd7 /\ -1 + rnd7 <= -1 + I11 /\ -1 + I11 <= -1 + rnd7 /\ -1 + rnd4 <= -1 + I8 /\ -1 + I8 <= -1 + rnd4 /\ 1 <= I11 /\ rnd7 = rnd7 /\ rnd4 = rnd4] 25.96/25.65 f3(I14, I15, I16, I17, I18, I19, I20) -> f2(I14, I15, I16, I17, I18, I19, I20) [I18 <= 0 /\ I18 <= 0] 25.96/25.65 f4(I21, I22, I23, I24, I25, I26, I27) -> f3(I21, -1 + I22, rnd3, I24, -1 + I25, rnd6, I27) [1 <= rnd6 /\ 1 <= rnd3 /\ -1 + rnd6 <= -1 + I25 /\ -1 + I25 <= -1 + rnd6 /\ -1 + rnd3 <= -1 + I22 /\ -1 + I22 <= -1 + rnd3 /\ 1 <= I25 /\ rnd6 = rnd6 /\ rnd3 = rnd3] 25.96/25.65 f2(I28, I29, I30, I31, I32, I33, I34) -> f3(I28, I29, I30, I31, I28, I33, I34) [1 <= I29 /\ 1 <= I28 /\ 1 <= I29] 25.96/25.65 f1(I35, I36, I37, I38, I39, I40, I41) -> f2(I35, rnd2, I37, I38, rnd5, I40, I41) [rnd5 = rnd5 /\ rnd2 = rnd2] 25.96/25.65 25.96/25.65 The dependency graph for this problem is: 25.96/25.65 1 -> 2, 3 25.96/25.65 2 -> 1 25.96/25.65 3 -> 25.96/25.65 Where: 25.96/25.65 1) f5#(I0, I1, I2, I3, I4, I5, I6) -> f3#(I0, I1, I2, I3, I4, I5, I6) 25.96/25.65 2) f3#(I7, I8, I9, I10, I11, I12, I13) -> f5#(I7, -1 + I8, I9, rnd4, -1 + I11, I12, rnd7) [1 <= rnd7 /\ -1 + rnd7 <= -1 + I11 /\ -1 + I11 <= -1 + rnd7 /\ -1 + rnd4 <= -1 + I8 /\ -1 + I8 <= -1 + rnd4 /\ 1 <= I11 /\ rnd7 = rnd7 /\ rnd4 = rnd4] 25.96/25.65 3) f3#(I14, I15, I16, I17, I18, I19, I20) -> f2#(I14, I15, I16, I17, I18, I19, I20) [I18 <= 0 /\ I18 <= 0] 25.96/25.65 25.96/25.65 We have the following SCCs. 25.96/25.65 { 1, 2 } 25.96/25.65 25.96/25.65 DP problem for innermost termination. 25.96/25.65 P = 25.96/25.65 f5#(I0, I1, I2, I3, I4, I5, I6) -> f3#(I0, I1, I2, I3, I4, I5, I6) 25.96/25.65 f3#(I7, I8, I9, I10, I11, I12, I13) -> f5#(I7, -1 + I8, I9, rnd4, -1 + I11, I12, rnd7) [1 <= rnd7 /\ -1 + rnd7 <= -1 + I11 /\ -1 + I11 <= -1 + rnd7 /\ -1 + rnd4 <= -1 + I8 /\ -1 + I8 <= -1 + rnd4 /\ 1 <= I11 /\ rnd7 = rnd7 /\ rnd4 = rnd4] 25.96/25.65 R = 25.96/25.65 f6(x1, x2, x3, x4, x5, x6, x7) -> f1(x1, x2, x3, x4, x5, x6, x7) popout

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all output return to Integer_Transition_Systems 2019-03-29 01.54