details

property	value
status	complete
benchmark	eric3.t2.smt2
ran by	Akihisa Yamada
cpu timeout	1200 seconds
wallclock timeout	300 seconds
memory limit	137438953472 bytes
execution host	n143.star.cs.uiowa.edu
space	From_T2

run statistics

property	value
solver	Ctrl
configuration	Transition
runtime (wallclock)	22.4951 seconds
cpu usage	22.837
user time	11.4601
system time	11.3769
max virtual memory	719612.0
max residence set size	12028.0

stage attributes

key	value
starexec-result	MAYBE

output

MAYBE DP problem for innermost termination. P = f7#(x1, x2, x3, x4, x5, x6) -> f5#(x1, x2, x3, x4, x5, x6) f5#(I6, I7, I8, I9, I10, I11) -> f2#(I6, I7, I8, I8, 2, 2) f2#(I12, I13, I14, I15, I16, I17) -> f3#(I12, 1, I14, I15, I16, I17) [2 <= I17] f2#(I18, I19, I20, I21, I22, I23) -> f3#(I18, 0, I20, I21, I22, I23) [I23 <= 1] f3#(I24, I25, I26, I27, I28, I29) -> f1#(I24, I25, I26, rnd4, I28, I29) [1 <= rnd4 /\ rnd4 = rnd4] f3#(I30, I31, I32, I33, I34, I35) -> f4#(1, I31, I32, I33, I34, I35) f3#(I36, I37, I38, I39, I40, I41) -> f4#(0, I37, I38, I39, I40, I41) [I37 <= 0] f1#(I42, I43, I44, I45, I46, I47) -> f2#(I42, I43, I44, I45, I46, -1 + I47) [I45 <= 0] f1#(I48, I49, I50, I51, I52, I53) -> f2#(I48, I49, I50, I51, I52, 1 + I53) [1 <= I51] R = f7(x1, x2, x3, x4, x5, x6) -> f5(x1, x2, x3, x4, x5, x6) f4(I0, I1, I2, I3, I4, I5) -> f6(I0, I1, I2, I3, I4, I5) [I0 <= 0] f5(I6, I7, I8, I9, I10, I11) -> f2(I6, I7, I8, I8, 2, 2) f2(I12, I13, I14, I15, I16, I17) -> f3(I12, 1, I14, I15, I16, I17) [2 <= I17] f2(I18, I19, I20, I21, I22, I23) -> f3(I18, 0, I20, I21, I22, I23) [I23 <= 1] f3(I24, I25, I26, I27, I28, I29) -> f1(I24, I25, I26, rnd4, I28, I29) [1 <= rnd4 /\ rnd4 = rnd4] f3(I30, I31, I32, I33, I34, I35) -> f4(1, I31, I32, I33, I34, I35) f3(I36, I37, I38, I39, I40, I41) -> f4(0, I37, I38, I39, I40, I41) [I37 <= 0] f1(I42, I43, I44, I45, I46, I47) -> f2(I42, I43, I44, I45, I46, -1 + I47) [I45 <= 0] f1(I48, I49, I50, I51, I52, I53) -> f2(I48, I49, I50, I51, I52, 1 + I53) [1 <= I51] The dependency graph for this problem is: 0 -> 1 1 -> 2 2 -> 4, 5 3 -> 4, 5, 6 4 -> 8 5 -> 6 -> 7 -> 2, 3 8 -> 2, 3 Where: 0) f7#(x1, x2, x3, x4, x5, x6) -> f5#(x1, x2, x3, x4, x5, x6) 1) f5#(I6, I7, I8, I9, I10, I11) -> f2#(I6, I7, I8, I8, 2, 2) 2) f2#(I12, I13, I14, I15, I16, I17) -> f3#(I12, 1, I14, I15, I16, I17) [2 <= I17] 3) f2#(I18, I19, I20, I21, I22, I23) -> f3#(I18, 0, I20, I21, I22, I23) [I23 <= 1] 4) f3#(I24, I25, I26, I27, I28, I29) -> f1#(I24, I25, I26, rnd4, I28, I29) [1 <= rnd4 /\ rnd4 = rnd4] 5) f3#(I30, I31, I32, I33, I34, I35) -> f4#(1, I31, I32, I33, I34, I35) 6) f3#(I36, I37, I38, I39, I40, I41) -> f4#(0, I37, I38, I39, I40, I41) [I37 <= 0] 7) f1#(I42, I43, I44, I45, I46, I47) -> f2#(I42, I43, I44, I45, I46, -1 + I47) [I45 <= 0] 8) f1#(I48, I49, I50, I51, I52, I53) -> f2#(I48, I49, I50, I51, I52, 1 + I53) [1 <= I51] We have the following SCCs. { 2, 3, 4, 8 } DP problem for innermost termination. P = f2#(I12, I13, I14, I15, I16, I17) -> f3#(I12, 1, I14, I15, I16, I17) [2 <= I17] f2#(I18, I19, I20, I21, I22, I23) -> f3#(I18, 0, I20, I21, I22, I23) [I23 <= 1] f3#(I24, I25, I26, I27, I28, I29) -> f1#(I24, I25, I26, rnd4, I28, I29) [1 <= rnd4 /\ rnd4 = rnd4] f1#(I48, I49, I50, I51, I52, I53) -> f2#(I48, I49, I50, I51, I52, 1 + I53) [1 <= I51] R = f7(x1, x2, x3, x4, x5, x6) -> f5(x1, x2, x3, x4, x5, x6) f4(I0, I1, I2, I3, I4, I5) -> f6(I0, I1, I2, I3, I4, I5) [I0 <= 0] f5(I6, I7, I8, I9, I10, I11) -> f2(I6, I7, I8, I8, 2, 2) f2(I12, I13, I14, I15, I16, I17) -> f3(I12, 1, I14, I15, I16, I17) [2 <= I17] f2(I18, I19, I20, I21, I22, I23) -> f3(I18, 0, I20, I21, I22, I23) [I23 <= 1] f3(I24, I25, I26, I27, I28, I29) -> f1(I24, I25, I26, rnd4, I28, I29) [1 <= rnd4 /\ rnd4 = rnd4] f3(I30, I31, I32, I33, I34, I35) -> f4(1, I31, I32, I33, I34, I35) f3(I36, I37, I38, I39, I40, I41) -> f4(0, I37, I38, I39, I40, I41) [I37 <= 0] f1(I42, I43, I44, I45, I46, I47) -> f2(I42, I43, I44, I45, I46, -1 + I47) [I45 <= 0] f1(I48, I49, I50, I51, I52, I53) -> f2(I48, I49, I50, I51, I52, 1 + I53) [1 <= I51] We use the extended value criterion with the projection function NU: NU[f1#(x0,x1,x2,x3,x4,x5)] = -x5 NU[f3#(x0,x1,x2,x3,x4,x5)] = -x5 NU[f2#(x0,x1,x2,x3,x4,x5)] = -x5 + 1 This gives the following inequalities: 2 <= I17 ==> -I17 + 1 >= -I17 I23 <= 1 ==> -I23 + 1 > -I23 with -I23 + 1 >= 0 1 <= rnd4 /\ rnd4 = rnd4 ==> -I29 >= -I29 1 <= I51 ==> -I53 >= -(1 + I53) + 1 We remove all the strictly oriented dependency pairs. DP problem for innermost termination. P = f2#(I12, I13, I14, I15, I16, I17) -> f3#(I12, 1, I14, I15, I16, I17) [2 <= I17] f3#(I24, I25, I26, I27, I28, I29) -> f1#(I24, I25, I26, rnd4, I28, I29) [1 <= rnd4 /\ rnd4 = rnd4] f1#(I48, I49, I50, I51, I52, I53) -> f2#(I48, I49, I50, I51, I52, 1 + I53) [1 <= I51] R = f7(x1, x2, x3, x4, x5, x6) -> f5(x1, x2, x3, x4, x5, x6) f4(I0, I1, I2, I3, I4, I5) -> f6(I0, I1, I2, I3, I4, I5) [I0 <= 0] f5(I6, I7, I8, I9, I10, I11) -> f2(I6, I7, I8, I8, 2, 2) f2(I12, I13, I14, I15, I16, I17) -> f3(I12, 1, I14, I15, I16, I17) [2 <= I17] f2(I18, I19, I20, I21, I22, I23) -> f3(I18, 0, I20, I21, I22, I23) [I23 <= 1] f3(I24, I25, I26, I27, I28, I29) -> f1(I24, I25, I26, rnd4, I28, I29) [1 <= rnd4 /\ rnd4 = rnd4] f3(I30, I31, I32, I33, I34, I35) -> f4(1, I31, I32, I33, I34, I35) f3(I36, I37, I38, I39, I40, I41) -> f4(0, I37, I38, I39, I40, I41) [I37 <= 0] f1(I42, I43, I44, I45, I46, I47) -> f2(I42, I43, I44, I45, I46, -1 + I47) [I45 <= 0] f1(I48, I49, I50, I51, I52, I53) -> f2(I48, I49, I50, I51, I52, 1 + I53) [1 <= I51] popout

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

job log

popout

actions

all output return to ITS