details

property	value
status	complete
benchmark	s2.t2.smt2
ran by	Akihisa Yamada
cpu timeout	1200 seconds
wallclock timeout	300 seconds
memory limit	137438953472 bytes
execution host	n127.star.cs.uiowa.edu
space	From_T2

run statistics

property	value
solver	Ctrl
configuration	Transition
runtime (wallclock)	6.7644 seconds
cpu usage	6.87481
user time	3.41501
system time	3.4598
max virtual memory	239392.0
max residence set size	10200.0

stage attributes

key	value
starexec-result	YES

output

6.81/6.76 YES 6.81/6.76 6.81/6.76 DP problem for innermost termination. 6.81/6.76 P = 6.81/6.76 f5#(x1, x2, x3, x4, x5) -> f1#(x1, x2, x3, x4, x5) 6.81/6.76 f4#(I0, I1, I2, I3, I4) -> f2#(I0, I1, I2, I3, I4) 6.81/6.76 f2#(I5, I6, I7, I8, I9) -> f4#(I5, 1 + I6, I7, I8, I9) [1 + I7 <= I5 /\ 1 + I7 <= 1 + I6 /\ 1 + I6 <= 1 + I7 /\ 1 + I6 <= I5] 6.81/6.76 f1#(I15, I16, I17, I18, I19) -> f2#(I15, rnd2, I17, I18, I19) [y1 = 0 /\ 0 <= y1 /\ y1 <= 0 /\ 0 <= y1 /\ y1 <= 0 /\ 1 + y1 <= I15 /\ y2 = 1 + y1 /\ 1 <= y2 /\ y2 <= 1 /\ 1 <= y2 /\ y2 <= 1 /\ 1 + y2 <= I15 /\ rnd2 = 1 + y2 /\ 2 <= rnd2 /\ rnd2 <= 2] 6.81/6.76 R = 6.81/6.76 f5(x1, x2, x3, x4, x5) -> f1(x1, x2, x3, x4, x5) 6.81/6.76 f4(I0, I1, I2, I3, I4) -> f2(I0, I1, I2, I3, I4) 6.81/6.76 f2(I5, I6, I7, I8, I9) -> f4(I5, 1 + I6, I7, I8, I9) [1 + I7 <= I5 /\ 1 + I7 <= 1 + I6 /\ 1 + I6 <= 1 + I7 /\ 1 + I6 <= I5] 6.81/6.76 f2(I10, I11, I12, I13, I14) -> f3(I10, I11, I12, I14, I14) [I10 <= I11] 6.81/6.76 f1(I15, I16, I17, I18, I19) -> f2(I15, rnd2, I17, I18, I19) [y1 = 0 /\ 0 <= y1 /\ y1 <= 0 /\ 0 <= y1 /\ y1 <= 0 /\ 1 + y1 <= I15 /\ y2 = 1 + y1 /\ 1 <= y2 /\ y2 <= 1 /\ 1 <= y2 /\ y2 <= 1 /\ 1 + y2 <= I15 /\ rnd2 = 1 + y2 /\ 2 <= rnd2 /\ rnd2 <= 2] 6.81/6.76 6.81/6.76 The dependency graph for this problem is: 6.81/6.76 0 -> 3 6.81/6.76 1 -> 2 6.81/6.76 2 -> 1 6.81/6.76 3 -> 2 6.81/6.76 Where: 6.81/6.76 0) f5#(x1, x2, x3, x4, x5) -> f1#(x1, x2, x3, x4, x5) 6.81/6.76 1) f4#(I0, I1, I2, I3, I4) -> f2#(I0, I1, I2, I3, I4) 6.81/6.76 2) f2#(I5, I6, I7, I8, I9) -> f4#(I5, 1 + I6, I7, I8, I9) [1 + I7 <= I5 /\ 1 + I7 <= 1 + I6 /\ 1 + I6 <= 1 + I7 /\ 1 + I6 <= I5] 6.81/6.76 3) f1#(I15, I16, I17, I18, I19) -> f2#(I15, rnd2, I17, I18, I19) [y1 = 0 /\ 0 <= y1 /\ y1 <= 0 /\ 0 <= y1 /\ y1 <= 0 /\ 1 + y1 <= I15 /\ y2 = 1 + y1 /\ 1 <= y2 /\ y2 <= 1 /\ 1 <= y2 /\ y2 <= 1 /\ 1 + y2 <= I15 /\ rnd2 = 1 + y2 /\ 2 <= rnd2 /\ rnd2 <= 2] 6.81/6.76 6.81/6.76 We have the following SCCs. 6.81/6.76 { 1, 2 } 6.81/6.76 6.81/6.76 DP problem for innermost termination. 6.81/6.76 P = 6.81/6.76 f4#(I0, I1, I2, I3, I4) -> f2#(I0, I1, I2, I3, I4) 6.81/6.76 f2#(I5, I6, I7, I8, I9) -> f4#(I5, 1 + I6, I7, I8, I9) [1 + I7 <= I5 /\ 1 + I7 <= 1 + I6 /\ 1 + I6 <= 1 + I7 /\ 1 + I6 <= I5] 6.81/6.76 R = 6.81/6.76 f5(x1, x2, x3, x4, x5) -> f1(x1, x2, x3, x4, x5) 6.81/6.76 f4(I0, I1, I2, I3, I4) -> f2(I0, I1, I2, I3, I4) 6.81/6.76 f2(I5, I6, I7, I8, I9) -> f4(I5, 1 + I6, I7, I8, I9) [1 + I7 <= I5 /\ 1 + I7 <= 1 + I6 /\ 1 + I6 <= 1 + I7 /\ 1 + I6 <= I5] 6.81/6.76 f2(I10, I11, I12, I13, I14) -> f3(I10, I11, I12, I14, I14) [I10 <= I11] 6.81/6.76 f1(I15, I16, I17, I18, I19) -> f2(I15, rnd2, I17, I18, I19) [y1 = 0 /\ 0 <= y1 /\ y1 <= 0 /\ 0 <= y1 /\ y1 <= 0 /\ 1 + y1 <= I15 /\ y2 = 1 + y1 /\ 1 <= y2 /\ y2 <= 1 /\ 1 <= y2 /\ y2 <= 1 /\ 1 + y2 <= I15 /\ rnd2 = 1 + y2 /\ 2 <= rnd2 /\ rnd2 <= 2] 6.81/6.76 6.81/6.76 We use the reverse value criterion with the projection function NU: 6.81/6.76 NU[f2#(z1,z2,z3,z4,z5)] = 1 + z3 + -1 * (1 + z2) 6.81/6.76 NU[f4#(z1,z2,z3,z4,z5)] = 1 + z3 + -1 * (1 + z2) 6.81/6.76 6.81/6.76 This gives the following inequalities: 6.81/6.76 ==> 1 + I2 + -1 * (1 + I1) >= 1 + I2 + -1 * (1 + I1) 6.81/6.76 1 + I7 <= I5 /\ 1 + I7 <= 1 + I6 /\ 1 + I6 <= 1 + I7 /\ 1 + I6 <= I5 ==> 1 + I7 + -1 * (1 + I6) > 1 + I7 + -1 * (1 + (1 + I6)) with 1 + I7 + -1 * (1 + I6) >= 0 6.81/6.76 6.81/6.76 We remove all the strictly oriented dependency pairs. 6.81/6.76 6.81/6.76 DP problem for innermost termination. 6.81/6.76 P = 6.81/6.76 f4#(I0, I1, I2, I3, I4) -> f2#(I0, I1, I2, I3, I4) 6.81/6.76 R = 6.81/6.76 f5(x1, x2, x3, x4, x5) -> f1(x1, x2, x3, x4, x5) 6.81/6.76 f4(I0, I1, I2, I3, I4) -> f2(I0, I1, I2, I3, I4) 6.81/6.76 f2(I5, I6, I7, I8, I9) -> f4(I5, 1 + I6, I7, I8, I9) [1 + I7 <= I5 /\ 1 + I7 <= 1 + I6 /\ 1 + I6 <= 1 + I7 /\ 1 + I6 <= I5] 6.81/6.76 f2(I10, I11, I12, I13, I14) -> f3(I10, I11, I12, I14, I14) [I10 <= I11] 6.81/6.76 f1(I15, I16, I17, I18, I19) -> f2(I15, rnd2, I17, I18, I19) [y1 = 0 /\ 0 <= y1 /\ y1 <= 0 /\ 0 <= y1 /\ y1 <= 0 /\ 1 + y1 <= I15 /\ y2 = 1 + y1 /\ 1 <= y2 /\ y2 <= 1 /\ 1 <= y2 /\ y2 <= 1 /\ 1 + y2 <= I15 /\ rnd2 = 1 + y2 /\ 2 <= rnd2 /\ rnd2 <= 2] 6.81/6.76 6.81/6.76 The dependency graph for this problem is: 6.81/6.76 1 -> 6.81/6.76 Where: 6.81/6.76 1) f4#(I0, I1, I2, I3, I4) -> f2#(I0, I1, I2, I3, I4) 6.81/6.76 6.81/6.76 We have the following SCCs. 6.81/6.76 6.81/9.73 EOF popout

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all output return to Integer_Transition_Systems 2019-03-29 01.54