details

property	value
status	complete
benchmark	dsa_test10.t2.smt2
ran by	Akihisa Yamada
cpu timeout	1200 seconds
wallclock timeout	300 seconds
memory limit	137438953472 bytes
execution host	n070.star.cs.uiowa.edu
space	From_T2

run statistics

property	value
solver	Ctrl
configuration	Transition
runtime (wallclock)	12.2491 seconds
cpu usage	12.1441
user time	6.19006
system time	5.95401
max virtual memory	720596.0
max residence set size	9796.0

stage attributes

key	value
starexec-result	YES

output

12.09/12.24 YES 12.09/12.24 12.09/12.24 DP problem for innermost termination. 12.09/12.24 P = 12.09/12.24 f7#(x1, x2, x3, x4, x5, x6) -> f6#(x1, x2, x3, x4, x5, x6) 12.09/12.24 f6#(I0, I1, I2, I3, I4, I5) -> f1#(I0, 0, I2, rnd4, rnd5, I5) [rnd5 = rnd5 /\ rnd4 = rnd4] 12.09/12.24 f2#(I6, I7, I8, I9, I10, I11) -> f1#(I6, 1 + I7, I8, I9, I10, I11) [1 + I7 <= I6] 12.09/12.24 f2#(I12, I13, I14, I15, I16, I17) -> f5#(I12, I13, 0, I15, I16, rnd6) [rnd6 = rnd6 /\ I12 <= I13] 12.09/12.24 f5#(I18, I19, I20, I21, I22, I23) -> f3#(I18, I19, I20, I21, I22, I23) 12.09/12.24 f3#(I24, I25, I26, I27, I28, I29) -> f5#(I24, I25, 1 + I26, I27, I28, I29) [1 + I26 <= I24] 12.09/12.24 f1#(I36, I37, I38, I39, I40, I41) -> f2#(I36, I37, I38, I39, I40, I41) 12.09/12.24 R = 12.09/12.24 f7(x1, x2, x3, x4, x5, x6) -> f6(x1, x2, x3, x4, x5, x6) 12.09/12.24 f6(I0, I1, I2, I3, I4, I5) -> f1(I0, 0, I2, rnd4, rnd5, I5) [rnd5 = rnd5 /\ rnd4 = rnd4] 12.09/12.24 f2(I6, I7, I8, I9, I10, I11) -> f1(I6, 1 + I7, I8, I9, I10, I11) [1 + I7 <= I6] 12.09/12.24 f2(I12, I13, I14, I15, I16, I17) -> f5(I12, I13, 0, I15, I16, rnd6) [rnd6 = rnd6 /\ I12 <= I13] 12.09/12.24 f5(I18, I19, I20, I21, I22, I23) -> f3(I18, I19, I20, I21, I22, I23) 12.09/12.24 f3(I24, I25, I26, I27, I28, I29) -> f5(I24, I25, 1 + I26, I27, I28, I29) [1 + I26 <= I24] 12.09/12.24 f3(I30, I31, I32, I33, I34, I35) -> f4(I30, I31, I32, I33, I34, I35) [I30 <= I32] 12.09/12.24 f1(I36, I37, I38, I39, I40, I41) -> f2(I36, I37, I38, I39, I40, I41) 12.09/12.24 12.09/12.24 The dependency graph for this problem is: 12.09/12.24 0 -> 1 12.09/12.24 1 -> 6 12.09/12.24 2 -> 6 12.09/12.24 3 -> 4 12.09/12.24 4 -> 5 12.09/12.24 5 -> 4 12.09/12.24 6 -> 2, 3 12.09/12.24 Where: 12.09/12.24 0) f7#(x1, x2, x3, x4, x5, x6) -> f6#(x1, x2, x3, x4, x5, x6) 12.09/12.24 1) f6#(I0, I1, I2, I3, I4, I5) -> f1#(I0, 0, I2, rnd4, rnd5, I5) [rnd5 = rnd5 /\ rnd4 = rnd4] 12.09/12.24 2) f2#(I6, I7, I8, I9, I10, I11) -> f1#(I6, 1 + I7, I8, I9, I10, I11) [1 + I7 <= I6] 12.09/12.24 3) f2#(I12, I13, I14, I15, I16, I17) -> f5#(I12, I13, 0, I15, I16, rnd6) [rnd6 = rnd6 /\ I12 <= I13] 12.09/12.24 4) f5#(I18, I19, I20, I21, I22, I23) -> f3#(I18, I19, I20, I21, I22, I23) 12.09/12.24 5) f3#(I24, I25, I26, I27, I28, I29) -> f5#(I24, I25, 1 + I26, I27, I28, I29) [1 + I26 <= I24] 12.09/12.24 6) f1#(I36, I37, I38, I39, I40, I41) -> f2#(I36, I37, I38, I39, I40, I41) 12.09/12.24 12.09/12.24 We have the following SCCs. 12.09/12.24 { 2, 6 } 12.09/12.24 { 4, 5 } 12.09/12.24 12.09/12.24 DP problem for innermost termination. 12.09/12.24 P = 12.09/12.24 f5#(I18, I19, I20, I21, I22, I23) -> f3#(I18, I19, I20, I21, I22, I23) 12.09/12.24 f3#(I24, I25, I26, I27, I28, I29) -> f5#(I24, I25, 1 + I26, I27, I28, I29) [1 + I26 <= I24] 12.09/12.24 R = 12.09/12.24 f7(x1, x2, x3, x4, x5, x6) -> f6(x1, x2, x3, x4, x5, x6) 12.09/12.24 f6(I0, I1, I2, I3, I4, I5) -> f1(I0, 0, I2, rnd4, rnd5, I5) [rnd5 = rnd5 /\ rnd4 = rnd4] 12.09/12.24 f2(I6, I7, I8, I9, I10, I11) -> f1(I6, 1 + I7, I8, I9, I10, I11) [1 + I7 <= I6] 12.09/12.24 f2(I12, I13, I14, I15, I16, I17) -> f5(I12, I13, 0, I15, I16, rnd6) [rnd6 = rnd6 /\ I12 <= I13] 12.09/12.24 f5(I18, I19, I20, I21, I22, I23) -> f3(I18, I19, I20, I21, I22, I23) 12.09/12.24 f3(I24, I25, I26, I27, I28, I29) -> f5(I24, I25, 1 + I26, I27, I28, I29) [1 + I26 <= I24] 12.09/12.24 f3(I30, I31, I32, I33, I34, I35) -> f4(I30, I31, I32, I33, I34, I35) [I30 <= I32] 12.09/12.24 f1(I36, I37, I38, I39, I40, I41) -> f2(I36, I37, I38, I39, I40, I41) 12.09/12.24 12.09/12.24 We use the reverse value criterion with the projection function NU: 12.09/12.24 NU[f3#(z1,z2,z3,z4,z5,z6)] = z1 + -1 * (1 + z3) 12.09/12.24 NU[f5#(z1,z2,z3,z4,z5,z6)] = z1 + -1 * (1 + z3) 12.09/12.24 12.09/12.24 This gives the following inequalities: 12.09/12.24 ==> I18 + -1 * (1 + I20) >= I18 + -1 * (1 + I20) 12.09/12.24 1 + I26 <= I24 ==> I24 + -1 * (1 + I26) > I24 + -1 * (1 + (1 + I26)) with I24 + -1 * (1 + I26) >= 0 12.09/12.24 12.09/12.24 We remove all the strictly oriented dependency pairs. 12.09/12.24 12.09/12.24 DP problem for innermost termination. 12.09/12.24 P = 12.09/12.24 f5#(I18, I19, I20, I21, I22, I23) -> f3#(I18, I19, I20, I21, I22, I23) 12.09/12.24 R = 12.09/12.24 f7(x1, x2, x3, x4, x5, x6) -> f6(x1, x2, x3, x4, x5, x6) 12.09/12.24 f6(I0, I1, I2, I3, I4, I5) -> f1(I0, 0, I2, rnd4, rnd5, I5) [rnd5 = rnd5 /\ rnd4 = rnd4] 12.09/12.24 f2(I6, I7, I8, I9, I10, I11) -> f1(I6, 1 + I7, I8, I9, I10, I11) [1 + I7 <= I6] 12.09/12.24 f2(I12, I13, I14, I15, I16, I17) -> f5(I12, I13, 0, I15, I16, rnd6) [rnd6 = rnd6 /\ I12 <= I13] 12.09/12.24 f5(I18, I19, I20, I21, I22, I23) -> f3(I18, I19, I20, I21, I22, I23) 12.09/12.24 f3(I24, I25, I26, I27, I28, I29) -> f5(I24, I25, 1 + I26, I27, I28, I29) [1 + I26 <= I24] 12.09/12.24 f3(I30, I31, I32, I33, I34, I35) -> f4(I30, I31, I32, I33, I34, I35) [I30 <= I32] 12.09/12.24 f1(I36, I37, I38, I39, I40, I41) -> f2(I36, I37, I38, I39, I40, I41) 12.09/12.24 12.09/12.24 The dependency graph for this problem is: 12.09/12.24 4 -> 12.09/12.24 Where: 12.09/12.24 4) f5#(I18, I19, I20, I21, I22, I23) -> f3#(I18, I19, I20, I21, I22, I23) 12.09/12.24 12.09/12.24 We have the following SCCs. 12.09/12.24 12.09/12.24 12.09/12.24 DP problem for innermost termination. 12.09/12.24 P = 12.09/12.24 f2#(I6, I7, I8, I9, I10, I11) -> f1#(I6, 1 + I7, I8, I9, I10, I11) [1 + I7 <= I6] 12.09/12.24 f1#(I36, I37, I38, I39, I40, I41) -> f2#(I36, I37, I38, I39, I40, I41) 12.09/12.24 R = 12.09/12.24 f7(x1, x2, x3, x4, x5, x6) -> f6(x1, x2, x3, x4, x5, x6) 12.09/12.24 f6(I0, I1, I2, I3, I4, I5) -> f1(I0, 0, I2, rnd4, rnd5, I5) [rnd5 = rnd5 /\ rnd4 = rnd4] 12.09/12.24 f2(I6, I7, I8, I9, I10, I11) -> f1(I6, 1 + I7, I8, I9, I10, I11) [1 + I7 <= I6] 12.09/12.24 f2(I12, I13, I14, I15, I16, I17) -> f5(I12, I13, 0, I15, I16, rnd6) [rnd6 = rnd6 /\ I12 <= I13] 12.09/12.24 f5(I18, I19, I20, I21, I22, I23) -> f3(I18, I19, I20, I21, I22, I23) 12.09/12.24 f3(I24, I25, I26, I27, I28, I29) -> f5(I24, I25, 1 + I26, I27, I28, I29) [1 + I26 <= I24] 12.09/12.24 f3(I30, I31, I32, I33, I34, I35) -> f4(I30, I31, I32, I33, I34, I35) [I30 <= I32] 12.09/12.24 f1(I36, I37, I38, I39, I40, I41) -> f2(I36, I37, I38, I39, I40, I41) 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