details

property	value
status	complete
benchmark	nested2.t2_fixed.smt2
ran by	Akihisa Yamada
cpu timeout	1200 seconds
wallclock timeout	300 seconds
memory limit	137438953472 bytes
execution host	n062.star.cs.uiowa.edu
space	From_T2

run statistics

property	value
solver	Ctrl
configuration	Transition
runtime (wallclock)	19.5344 seconds
cpu usage	19.4662
user time	10.0509
system time	9.41531
max virtual memory	753932.0
max residence set size	10108.0

stage attributes

key	value
starexec-result	YES

output

19.38/19.53 YES 19.38/19.53 19.38/19.53 DP problem for innermost termination. 19.38/19.53 P = 19.38/19.53 f6#(x1, x2, x3, x4, x5, x6) -> f1#(x1, x2, x3, x4, x5, x6) 19.38/19.53 f5#(I0, I1, I2, I3, I4, I5) -> f3#(I0, I1, I2, I3, I4, I5) 19.38/19.53 f3#(I6, I7, I8, I9, I10, I11) -> f5#(-1 + I6, I7, rnd3, -1 + I9, I10, rnd6) [1 <= rnd6 /\ -1 + rnd6 <= -1 + I9 /\ -1 + I9 <= -1 + rnd6 /\ -1 + rnd3 <= -1 + I6 /\ -1 + I6 <= -1 + rnd3 /\ 1 <= I9 /\ rnd6 = rnd6 /\ rnd3 = rnd3] 19.38/19.53 f3#(I12, I13, I14, I15, I16, I17) -> f2#(I12, I13, I14, I15, I16, I17) [I15 <= 0 /\ I15 <= 0] 19.38/19.53 f4#(I18, I19, I20, I21, I22, I23) -> f3#(-1 + I18, rnd2, I20, -1 + I21, rnd5, I23) [1 <= rnd5 /\ 1 <= rnd2 /\ -1 + rnd5 <= -1 + I21 /\ -1 + I21 <= -1 + rnd5 /\ -1 + rnd2 <= -1 + I18 /\ -1 + I18 <= -1 + rnd2 /\ 1 <= I21 /\ rnd5 = rnd5 /\ rnd2 = rnd2] 19.38/19.53 f2#(I24, I25, I26, I27, I28, I29) -> f3#(I24, I25, I26, 5000, I28, I29) [1 <= I24 /\ 1 <= 5000 /\ 1 <= I24] 19.38/19.53 f1#(I30, I31, I32, I33, I34, I35) -> f2#(rnd1, I31, I32, rnd4, I34, I35) [rnd4 = rnd4 /\ rnd1 = rnd1] 19.38/19.53 R = 19.38/19.53 f6(x1, x2, x3, x4, x5, x6) -> f1(x1, x2, x3, x4, x5, x6) 19.38/19.53 f5(I0, I1, I2, I3, I4, I5) -> f3(I0, I1, I2, I3, I4, I5) 19.38/19.53 f3(I6, I7, I8, I9, I10, I11) -> f5(-1 + I6, I7, rnd3, -1 + I9, I10, rnd6) [1 <= rnd6 /\ -1 + rnd6 <= -1 + I9 /\ -1 + I9 <= -1 + rnd6 /\ -1 + rnd3 <= -1 + I6 /\ -1 + I6 <= -1 + rnd3 /\ 1 <= I9 /\ rnd6 = rnd6 /\ rnd3 = rnd3] 19.38/19.53 f3(I12, I13, I14, I15, I16, I17) -> f2(I12, I13, I14, I15, I16, I17) [I15 <= 0 /\ I15 <= 0] 19.38/19.53 f4(I18, I19, I20, I21, I22, I23) -> f3(-1 + I18, rnd2, I20, -1 + I21, rnd5, I23) [1 <= rnd5 /\ 1 <= rnd2 /\ -1 + rnd5 <= -1 + I21 /\ -1 + I21 <= -1 + rnd5 /\ -1 + rnd2 <= -1 + I18 /\ -1 + I18 <= -1 + rnd2 /\ 1 <= I21 /\ rnd5 = rnd5 /\ rnd2 = rnd2] 19.38/19.53 f2(I24, I25, I26, I27, I28, I29) -> f3(I24, I25, I26, 5000, I28, I29) [1 <= I24 /\ 1 <= 5000 /\ 1 <= I24] 19.38/19.53 f1(I30, I31, I32, I33, I34, I35) -> f2(rnd1, I31, I32, rnd4, I34, I35) [rnd4 = rnd4 /\ rnd1 = rnd1] 19.38/19.53 19.38/19.53 The dependency graph for this problem is: 19.38/19.53 0 -> 6 19.38/19.53 1 -> 2, 3 19.38/19.53 2 -> 1 19.38/19.53 3 -> 5 19.38/19.53 4 -> 2, 3 19.38/19.53 5 -> 2 19.38/19.53 6 -> 5 19.38/19.53 Where: 19.38/19.53 0) f6#(x1, x2, x3, x4, x5, x6) -> f1#(x1, x2, x3, x4, x5, x6) 19.38/19.53 1) f5#(I0, I1, I2, I3, I4, I5) -> f3#(I0, I1, I2, I3, I4, I5) 19.38/19.53 2) f3#(I6, I7, I8, I9, I10, I11) -> f5#(-1 + I6, I7, rnd3, -1 + I9, I10, rnd6) [1 <= rnd6 /\ -1 + rnd6 <= -1 + I9 /\ -1 + I9 <= -1 + rnd6 /\ -1 + rnd3 <= -1 + I6 /\ -1 + I6 <= -1 + rnd3 /\ 1 <= I9 /\ rnd6 = rnd6 /\ rnd3 = rnd3] 19.38/19.53 3) f3#(I12, I13, I14, I15, I16, I17) -> f2#(I12, I13, I14, I15, I16, I17) [I15 <= 0 /\ I15 <= 0] 19.38/19.53 4) f4#(I18, I19, I20, I21, I22, I23) -> f3#(-1 + I18, rnd2, I20, -1 + I21, rnd5, I23) [1 <= rnd5 /\ 1 <= rnd2 /\ -1 + rnd5 <= -1 + I21 /\ -1 + I21 <= -1 + rnd5 /\ -1 + rnd2 <= -1 + I18 /\ -1 + I18 <= -1 + rnd2 /\ 1 <= I21 /\ rnd5 = rnd5 /\ rnd2 = rnd2] 19.38/19.53 5) f2#(I24, I25, I26, I27, I28, I29) -> f3#(I24, I25, I26, 5000, I28, I29) [1 <= I24 /\ 1 <= 5000 /\ 1 <= I24] 19.38/19.53 6) f1#(I30, I31, I32, I33, I34, I35) -> f2#(rnd1, I31, I32, rnd4, I34, I35) [rnd4 = rnd4 /\ rnd1 = rnd1] 19.38/19.53 19.38/19.53 We have the following SCCs. 19.38/19.53 { 1, 2, 3, 5 } 19.38/19.53 19.38/19.53 DP problem for innermost termination. 19.38/19.53 P = 19.38/19.53 f5#(I0, I1, I2, I3, I4, I5) -> f3#(I0, I1, I2, I3, I4, I5) 19.38/19.53 f3#(I6, I7, I8, I9, I10, I11) -> f5#(-1 + I6, I7, rnd3, -1 + I9, I10, rnd6) [1 <= rnd6 /\ -1 + rnd6 <= -1 + I9 /\ -1 + I9 <= -1 + rnd6 /\ -1 + rnd3 <= -1 + I6 /\ -1 + I6 <= -1 + rnd3 /\ 1 <= I9 /\ rnd6 = rnd6 /\ rnd3 = rnd3] 19.38/19.53 f3#(I12, I13, I14, I15, I16, I17) -> f2#(I12, I13, I14, I15, I16, I17) [I15 <= 0 /\ I15 <= 0] 19.38/19.53 f2#(I24, I25, I26, I27, I28, I29) -> f3#(I24, I25, I26, 5000, I28, I29) [1 <= I24 /\ 1 <= 5000 /\ 1 <= I24] 19.38/19.53 R = 19.38/19.53 f6(x1, x2, x3, x4, x5, x6) -> f1(x1, x2, x3, x4, x5, x6) 19.38/19.53 f5(I0, I1, I2, I3, I4, I5) -> f3(I0, I1, I2, I3, I4, I5) 19.38/19.53 f3(I6, I7, I8, I9, I10, I11) -> f5(-1 + I6, I7, rnd3, -1 + I9, I10, rnd6) [1 <= rnd6 /\ -1 + rnd6 <= -1 + I9 /\ -1 + I9 <= -1 + rnd6 /\ -1 + rnd3 <= -1 + I6 /\ -1 + I6 <= -1 + rnd3 /\ 1 <= I9 /\ rnd6 = rnd6 /\ rnd3 = rnd3] 19.38/19.53 f3(I12, I13, I14, I15, I16, I17) -> f2(I12, I13, I14, I15, I16, I17) [I15 <= 0 /\ I15 <= 0] 19.38/19.53 f4(I18, I19, I20, I21, I22, I23) -> f3(-1 + I18, rnd2, I20, -1 + I21, rnd5, I23) [1 <= rnd5 /\ 1 <= rnd2 /\ -1 + rnd5 <= -1 + I21 /\ -1 + I21 <= -1 + rnd5 /\ -1 + rnd2 <= -1 + I18 /\ -1 + I18 <= -1 + rnd2 /\ 1 <= I21 /\ rnd5 = rnd5 /\ rnd2 = rnd2] 19.38/19.53 f2(I24, I25, I26, I27, I28, I29) -> f3(I24, I25, I26, 5000, I28, I29) [1 <= I24 /\ 1 <= 5000 /\ 1 <= I24] 19.38/19.53 f1(I30, I31, I32, I33, I34, I35) -> f2(rnd1, I31, I32, rnd4, I34, I35) [rnd4 = rnd4 /\ rnd1 = rnd1] 19.38/19.53 19.38/19.53 We use the extended value criterion with the projection function NU: 19.38/19.53 NU[f2#(x0,x1,x2,x3,x4,x5)] = x0 - 1 19.38/19.53 NU[f3#(x0,x1,x2,x3,x4,x5)] = x0 - x3 - 1 19.38/19.53 NU[f5#(x0,x1,x2,x3,x4,x5)] = x0 - x3 - 1 19.38/19.53 19.38/19.53 This gives the following inequalities: 19.38/19.53 ==> I0 - I3 - 1 >= I0 - I3 - 1 19.38/19.53 1 <= rnd6 /\ -1 + rnd6 <= -1 + I9 /\ -1 + I9 <= -1 + rnd6 /\ -1 + rnd3 <= -1 + I6 /\ -1 + I6 <= -1 + rnd3 /\ 1 <= I9 /\ rnd6 = rnd6 /\ rnd3 = rnd3 ==> I6 - I9 - 1 >= (-1 + I6) - (-1 + I9) - 1 19.38/19.53 I15 <= 0 /\ I15 <= 0 ==> I12 - I15 - 1 >= I12 - 1 19.38/19.53 1 <= I24 /\ 1 <= 5000 /\ 1 <= I24 ==> I24 - 1 > I24 - 5000 - 1 with I24 - 1 >= 0 19.38/19.53 19.38/19.53 We remove all the strictly oriented dependency pairs. 19.38/19.53 19.38/19.53 DP problem for innermost termination. 19.38/19.53 P = 19.38/19.53 f5#(I0, I1, I2, I3, I4, I5) -> f3#(I0, I1, I2, I3, I4, I5) 19.38/19.53 f3#(I6, I7, I8, I9, I10, I11) -> f5#(-1 + I6, I7, rnd3, -1 + I9, I10, rnd6) [1 <= rnd6 /\ -1 + rnd6 <= -1 + I9 /\ -1 + I9 <= -1 + rnd6 /\ -1 + rnd3 <= -1 + I6 /\ -1 + I6 <= -1 + rnd3 /\ 1 <= I9 /\ rnd6 = rnd6 /\ rnd3 = rnd3] 19.38/19.53 f3#(I12, I13, I14, I15, I16, I17) -> f2#(I12, I13, I14, I15, I16, I17) [I15 <= 0 /\ I15 <= 0] 19.38/19.53 R = 19.38/19.53 f6(x1, x2, x3, x4, x5, x6) -> f1(x1, x2, x3, x4, x5, x6) 19.38/19.53 f5(I0, I1, I2, I3, I4, I5) -> f3(I0, I1, I2, I3, I4, I5) 19.38/19.53 f3(I6, I7, I8, I9, I10, I11) -> f5(-1 + I6, I7, rnd3, -1 + I9, I10, rnd6) [1 <= rnd6 /\ -1 + rnd6 <= -1 + I9 /\ -1 + I9 <= -1 + rnd6 /\ -1 + rnd3 <= -1 + I6 /\ -1 + I6 <= -1 + rnd3 /\ 1 <= I9 /\ rnd6 = rnd6 /\ rnd3 = rnd3] 19.38/19.53 f3(I12, I13, I14, I15, I16, I17) -> f2(I12, I13, I14, I15, I16, I17) [I15 <= 0 /\ I15 <= 0] 19.38/19.53 f4(I18, I19, I20, I21, I22, I23) -> f3(-1 + I18, rnd2, I20, -1 + I21, rnd5, I23) [1 <= rnd5 /\ 1 <= rnd2 /\ -1 + rnd5 <= -1 + I21 /\ -1 + I21 <= -1 + rnd5 /\ -1 + rnd2 <= -1 + I18 /\ -1 + I18 <= -1 + rnd2 /\ 1 <= I21 /\ rnd5 = rnd5 /\ rnd2 = rnd2] 19.38/19.53 f2(I24, I25, I26, I27, I28, I29) -> f3(I24, I25, I26, 5000, I28, I29) [1 <= I24 /\ 1 <= 5000 /\ 1 <= I24] 19.38/19.53 f1(I30, I31, I32, I33, I34, I35) -> f2(rnd1, I31, I32, rnd4, I34, I35) [rnd4 = rnd4 /\ rnd1 = rnd1] 19.38/19.53 19.38/19.53 The dependency graph for this problem is: 19.38/19.53 1 -> 2, 3 19.38/19.53 2 -> 1 19.38/19.53 3 -> 19.38/19.53 Where: 19.38/19.53 1) f5#(I0, I1, I2, I3, I4, I5) -> f3#(I0, I1, I2, I3, I4, I5) 19.38/19.53 2) f3#(I6, I7, I8, I9, I10, I11) -> f5#(-1 + I6, I7, rnd3, -1 + I9, I10, rnd6) [1 <= rnd6 /\ -1 + rnd6 <= -1 + I9 /\ -1 + I9 <= -1 + rnd6 /\ -1 + rnd3 <= -1 + I6 /\ -1 + I6 <= -1 + rnd3 /\ 1 <= I9 /\ rnd6 = rnd6 /\ rnd3 = rnd3] 19.38/19.53 3) f3#(I12, I13, I14, I15, I16, I17) -> f2#(I12, I13, I14, I15, I16, I17) [I15 <= 0 /\ I15 <= 0] 19.38/19.53 19.38/19.53 We have the following SCCs. 19.38/19.53 { 1, 2 } 19.38/19.53 19.38/19.53 DP problem for innermost termination. 19.38/19.53 P = 19.38/19.53 f5#(I0, I1, I2, I3, I4, I5) -> f3#(I0, I1, I2, I3, I4, I5) 19.38/19.53 f3#(I6, I7, I8, I9, I10, I11) -> f5#(-1 + I6, I7, rnd3, -1 + I9, I10, rnd6) [1 <= rnd6 /\ -1 + rnd6 <= -1 + I9 /\ -1 + I9 <= -1 + rnd6 /\ -1 + rnd3 <= -1 + I6 /\ -1 + I6 <= -1 + rnd3 /\ 1 <= I9 /\ rnd6 = rnd6 /\ rnd3 = rnd3] 19.38/19.53 R = 19.38/19.53 f6(x1, x2, x3, x4, x5, x6) -> f1(x1, x2, x3, x4, x5, x6) popout

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

job log

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actions

all output return to Integer_Transition_Systems 2019-03-29 01.54