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	n103.star.cs.uiowa.edu
space	From_T2

run statistics

property	value
solver	Ctrl
configuration	Transition
runtime (wallclock)	9.03741 seconds
cpu usage	9.18122
user time	4.89073
system time	4.29049
max virtual memory	493568.0
max residence set size	11000.0

stage attributes

key	value
starexec-result	YES

output

9.12/9.03 YES 9.12/9.03 9.12/9.03 DP problem for innermost termination. 9.12/9.03 P = 9.12/9.03 f7#(x1, x2, x3, x4, x5) -> f6#(x1, x2, x3, x4, x5) 9.12/9.03 f4#(I5, I6, I7, I8, I9) -> f1#(I5, I6, 1, 1 + I8, I9) [0 <= I7 /\ I7 <= 0 /\ 0 <= -1 - I8 + I9] 9.12/9.03 f6#(I10, I11, I12, I13, I14) -> f4#(I10, I11, 0, I13, I14) 9.12/9.03 f3#(I21, I22, I23, I24, I25) -> f4#(I21, I22, 0, I24, -1 + I25) 9.12/9.03 f2#(I26, I27, I28, I29, I30) -> f3#(I26, I27, I28, I29, I30) [I27 = I27] 9.12/9.03 f1#(I31, I32, I33, I34, I35) -> f2#(I31, I32, I33, I34, I35) [0 <= -1 - I34 + I35] 9.12/9.03 R = 9.12/9.03 f7(x1, x2, x3, x4, x5) -> f6(x1, x2, x3, x4, x5) 9.12/9.03 f4(I0, I1, I2, I3, I4) -> f5(rnd1, I1, I2, I3, I4) [rnd1 = rnd1 /\ -1 * I3 + I4 <= 0] 9.12/9.03 f4(I5, I6, I7, I8, I9) -> f1(I5, I6, 1, 1 + I8, I9) [0 <= I7 /\ I7 <= 0 /\ 0 <= -1 - I8 + I9] 9.12/9.03 f6(I10, I11, I12, I13, I14) -> f4(I10, I11, 0, I13, I14) 9.12/9.03 f1(I15, I16, I17, I18, I19) -> f5(I20, I16, I17, I18, I19) [I20 = I20 /\ -1 * I18 + I19 <= 0] 9.12/9.03 f3(I21, I22, I23, I24, I25) -> f4(I21, I22, 0, I24, -1 + I25) 9.12/9.03 f2(I26, I27, I28, I29, I30) -> f3(I26, I27, I28, I29, I30) [I27 = I27] 9.12/9.03 f1(I31, I32, I33, I34, I35) -> f2(I31, I32, I33, I34, I35) [0 <= -1 - I34 + I35] 9.12/9.03 9.12/9.03 The dependency graph for this problem is: 9.12/9.03 0 -> 2 9.12/9.03 1 -> 5 9.12/9.03 2 -> 1 9.12/9.03 3 -> 1 9.12/9.03 4 -> 3 9.12/9.03 5 -> 4 9.12/9.03 Where: 9.12/9.03 0) f7#(x1, x2, x3, x4, x5) -> f6#(x1, x2, x3, x4, x5) 9.12/9.03 1) f4#(I5, I6, I7, I8, I9) -> f1#(I5, I6, 1, 1 + I8, I9) [0 <= I7 /\ I7 <= 0 /\ 0 <= -1 - I8 + I9] 9.12/9.03 2) f6#(I10, I11, I12, I13, I14) -> f4#(I10, I11, 0, I13, I14) 9.12/9.03 3) f3#(I21, I22, I23, I24, I25) -> f4#(I21, I22, 0, I24, -1 + I25) 9.12/9.03 4) f2#(I26, I27, I28, I29, I30) -> f3#(I26, I27, I28, I29, I30) [I27 = I27] 9.12/9.03 5) f1#(I31, I32, I33, I34, I35) -> f2#(I31, I32, I33, I34, I35) [0 <= -1 - I34 + I35] 9.12/9.03 9.12/9.03 We have the following SCCs. 9.12/9.03 { 1, 3, 4, 5 } 9.12/9.03 9.12/9.03 DP problem for innermost termination. 9.12/9.03 P = 9.12/9.03 f4#(I5, I6, I7, I8, I9) -> f1#(I5, I6, 1, 1 + I8, I9) [0 <= I7 /\ I7 <= 0 /\ 0 <= -1 - I8 + I9] 9.12/9.03 f3#(I21, I22, I23, I24, I25) -> f4#(I21, I22, 0, I24, -1 + I25) 9.12/9.03 f2#(I26, I27, I28, I29, I30) -> f3#(I26, I27, I28, I29, I30) [I27 = I27] 9.12/9.03 f1#(I31, I32, I33, I34, I35) -> f2#(I31, I32, I33, I34, I35) [0 <= -1 - I34 + I35] 9.12/9.03 R = 9.12/9.03 f7(x1, x2, x3, x4, x5) -> f6(x1, x2, x3, x4, x5) 9.12/9.03 f4(I0, I1, I2, I3, I4) -> f5(rnd1, I1, I2, I3, I4) [rnd1 = rnd1 /\ -1 * I3 + I4 <= 0] 9.12/9.03 f4(I5, I6, I7, I8, I9) -> f1(I5, I6, 1, 1 + I8, I9) [0 <= I7 /\ I7 <= 0 /\ 0 <= -1 - I8 + I9] 9.12/9.03 f6(I10, I11, I12, I13, I14) -> f4(I10, I11, 0, I13, I14) 9.12/9.03 f1(I15, I16, I17, I18, I19) -> f5(I20, I16, I17, I18, I19) [I20 = I20 /\ -1 * I18 + I19 <= 0] 9.12/9.03 f3(I21, I22, I23, I24, I25) -> f4(I21, I22, 0, I24, -1 + I25) 9.12/9.03 f2(I26, I27, I28, I29, I30) -> f3(I26, I27, I28, I29, I30) [I27 = I27] 9.12/9.03 f1(I31, I32, I33, I34, I35) -> f2(I31, I32, I33, I34, I35) [0 <= -1 - I34 + I35] 9.12/9.03 9.12/9.03 We use the extended value criterion with the projection function NU: 9.12/9.03 NU[f2#(x0,x1,x2,x3,x4)] = -x3 + x4 - 2 9.12/9.03 NU[f3#(x0,x1,x2,x3,x4)] = -x3 + x4 - 2 9.12/9.03 NU[f1#(x0,x1,x2,x3,x4)] = -x3 + x4 - 1 9.12/9.03 NU[f4#(x0,x1,x2,x3,x4)] = -x2 - x3 + x4 - 2 9.12/9.03 9.12/9.03 This gives the following inequalities: 9.12/9.03 0 <= I7 /\ I7 <= 0 /\ 0 <= -1 - I8 + I9 ==> -I7 - I8 + I9 - 2 >= -(1 + I8) + I9 - 1 9.12/9.03 ==> -I24 + I25 - 2 >= -0 - I24 + (-1 + I25) - 2 9.12/9.03 I27 = I27 ==> -I29 + I30 - 2 >= -I29 + I30 - 2 9.12/9.03 0 <= -1 - I34 + I35 ==> -I34 + I35 - 1 > -I34 + I35 - 2 with -I34 + I35 - 1 >= 0 9.12/9.03 9.12/9.03 We remove all the strictly oriented dependency pairs. 9.12/9.03 9.12/9.03 DP problem for innermost termination. 9.12/9.03 P = 9.12/9.03 f4#(I5, I6, I7, I8, I9) -> f1#(I5, I6, 1, 1 + I8, I9) [0 <= I7 /\ I7 <= 0 /\ 0 <= -1 - I8 + I9] 9.12/9.03 f3#(I21, I22, I23, I24, I25) -> f4#(I21, I22, 0, I24, -1 + I25) 9.12/9.03 f2#(I26, I27, I28, I29, I30) -> f3#(I26, I27, I28, I29, I30) [I27 = I27] 9.12/9.03 R = 9.12/9.03 f7(x1, x2, x3, x4, x5) -> f6(x1, x2, x3, x4, x5) 9.12/9.03 f4(I0, I1, I2, I3, I4) -> f5(rnd1, I1, I2, I3, I4) [rnd1 = rnd1 /\ -1 * I3 + I4 <= 0] 9.12/9.03 f4(I5, I6, I7, I8, I9) -> f1(I5, I6, 1, 1 + I8, I9) [0 <= I7 /\ I7 <= 0 /\ 0 <= -1 - I8 + I9] 9.12/9.03 f6(I10, I11, I12, I13, I14) -> f4(I10, I11, 0, I13, I14) 9.12/9.03 f1(I15, I16, I17, I18, I19) -> f5(I20, I16, I17, I18, I19) [I20 = I20 /\ -1 * I18 + I19 <= 0] 9.12/9.03 f3(I21, I22, I23, I24, I25) -> f4(I21, I22, 0, I24, -1 + I25) 9.12/9.03 f2(I26, I27, I28, I29, I30) -> f3(I26, I27, I28, I29, I30) [I27 = I27] 9.12/9.03 f1(I31, I32, I33, I34, I35) -> f2(I31, I32, I33, I34, I35) [0 <= -1 - I34 + I35] 9.12/9.03 9.12/9.03 The dependency graph for this problem is: 9.12/9.03 1 -> 9.12/9.03 3 -> 1 9.12/9.03 4 -> 3 9.12/9.03 Where: 9.12/9.03 1) f4#(I5, I6, I7, I8, I9) -> f1#(I5, I6, 1, 1 + I8, I9) [0 <= I7 /\ I7 <= 0 /\ 0 <= -1 - I8 + I9] 9.12/9.03 3) f3#(I21, I22, I23, I24, I25) -> f4#(I21, I22, 0, I24, -1 + I25) 9.12/9.03 4) f2#(I26, I27, I28, I29, I30) -> f3#(I26, I27, I28, I29, I30) [I27 = I27] 9.12/9.03 9.12/9.03 We have the following SCCs. 9.12/9.03 9.12/12.01 EOF popout

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job log

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actions

all output return to Integer_Transition_Systems 2019-03-29 01.54