Integer_Transition_Systems 2019-03-29 01.54 pair #432275454

details

property	value
status	complete
benchmark	ex7.t2.smt2
ran by	Akihisa Yamada
cpu timeout	1200 seconds
wallclock timeout	300 seconds
memory limit	137438953472 bytes
execution host	n167.star.cs.uiowa.edu
space	From_T2

run statistics

property	value
solver	Ctrl
configuration	Transition
runtime (wallclock)	6.32318 seconds
cpu usage	6.43201
user time	3.19332
system time	3.23869
max virtual memory	726108.0
max residence set size	8768.0

stage attributes

key	value
starexec-result	YES

output

6.30/6.32	YES
6.30/6.32	
6.30/6.32	DP problem for innermost termination.
6.30/6.32	P =
6.30/6.32	  f5#(x1, x2, x3, x4, x5, x6) -> f4#(x1, x2, x3, x4, x5, x6) 
6.30/6.32	  f4#(I0, I1, I2, I3, I4, I5) -> f3#(I0, 0, I0, rnd4, rnd5, I5) [rnd4 = rnd5 /\ rnd5 = rnd5]
6.30/6.32	  f3#(I6, I7, I8, I9, I10, I11) -> f1#(I6, I7, I8, I9, I10, I11) 
6.30/6.32	  f1#(I12, I13, I14, I15, I16, I17) -> f3#(I12, 1 + I13, I14, I15, I16, rnd6) [rnd6 = rnd6 /\ 1 + I13 <= I14]
6.30/6.32	R = 
6.30/6.32	  f5(x1, x2, x3, x4, x5, x6) -> f4(x1, x2, x3, x4, x5, x6) 
6.30/6.32	  f4(I0, I1, I2, I3, I4, I5) -> f3(I0, 0, I0, rnd4, rnd5, I5) [rnd4 = rnd5 /\ rnd5 = rnd5]
6.30/6.32	  f3(I6, I7, I8, I9, I10, I11) -> f1(I6, I7, I8, I9, I10, I11) 
6.30/6.32	  f1(I12, I13, I14, I15, I16, I17) -> f3(I12, 1 + I13, I14, I15, I16, rnd6) [rnd6 = rnd6 /\ 1 + I13 <= I14]
6.30/6.32	  f1(I18, I19, I20, I21, I22, I23) -> f2(I18, I19, I20, I21, I22, I23) [I20 <= I19]
6.30/6.32	
6.30/6.32	The dependency graph for this problem is:
6.30/6.32	  0 -> 1
6.30/6.32	  1 -> 2
6.30/6.32	  2 -> 3
6.30/6.32	  3 -> 2
6.30/6.32	Where:
6.30/6.32	  0) f5#(x1, x2, x3, x4, x5, x6) -> f4#(x1, x2, x3, x4, x5, x6) 
6.30/6.32	  1) f4#(I0, I1, I2, I3, I4, I5) -> f3#(I0, 0, I0, rnd4, rnd5, I5) [rnd4 = rnd5 /\ rnd5 = rnd5]
6.30/6.32	  2) f3#(I6, I7, I8, I9, I10, I11) -> f1#(I6, I7, I8, I9, I10, I11) 
6.30/6.32	  3) f1#(I12, I13, I14, I15, I16, I17) -> f3#(I12, 1 + I13, I14, I15, I16, rnd6) [rnd6 = rnd6 /\ 1 + I13 <= I14]
6.30/6.32	
6.30/6.32	We have the following SCCs.
6.30/6.32	  { 2, 3 }
6.30/6.32	
6.30/6.32	DP problem for innermost termination.
6.30/6.32	P =
6.30/6.32	  f3#(I6, I7, I8, I9, I10, I11) -> f1#(I6, I7, I8, I9, I10, I11) 
6.30/6.32	  f1#(I12, I13, I14, I15, I16, I17) -> f3#(I12, 1 + I13, I14, I15, I16, rnd6) [rnd6 = rnd6 /\ 1 + I13 <= I14]
6.30/6.32	R = 
6.30/6.32	  f5(x1, x2, x3, x4, x5, x6) -> f4(x1, x2, x3, x4, x5, x6) 
6.30/6.32	  f4(I0, I1, I2, I3, I4, I5) -> f3(I0, 0, I0, rnd4, rnd5, I5) [rnd4 = rnd5 /\ rnd5 = rnd5]
6.30/6.32	  f3(I6, I7, I8, I9, I10, I11) -> f1(I6, I7, I8, I9, I10, I11) 
6.30/6.32	  f1(I12, I13, I14, I15, I16, I17) -> f3(I12, 1 + I13, I14, I15, I16, rnd6) [rnd6 = rnd6 /\ 1 + I13 <= I14]
6.30/6.32	  f1(I18, I19, I20, I21, I22, I23) -> f2(I18, I19, I20, I21, I22, I23) [I20 <= I19]
6.30/6.32	
6.30/6.32	We use the reverse value criterion with the projection function NU:
6.30/6.32	  NU[f1#(z1,z2,z3,z4,z5,z6)] = z3 + -1 * (1 + z2)
6.30/6.32	  NU[f3#(z1,z2,z3,z4,z5,z6)] = z3 + -1 * (1 + z2)
6.30/6.32	
6.30/6.32	This gives the following inequalities:
6.30/6.32	    ==>  I8 + -1 * (1 + I7) >= I8 + -1 * (1 + I7)
6.30/6.32	  rnd6 = rnd6 /\ 1 + I13 <= I14  ==>  I14 + -1 * (1 + I13) > I14 + -1 * (1 + (1 + I13))  with  I14 + -1 * (1 + I13) >= 0
6.30/6.32	
6.30/6.32	We remove all the strictly oriented dependency pairs.
6.30/6.32	
6.30/6.32	DP problem for innermost termination.
6.30/6.32	P =
6.30/6.32	  f3#(I6, I7, I8, I9, I10, I11) -> f1#(I6, I7, I8, I9, I10, I11) 
6.30/6.32	R = 
6.30/6.32	  f5(x1, x2, x3, x4, x5, x6) -> f4(x1, x2, x3, x4, x5, x6) 
6.30/6.32	  f4(I0, I1, I2, I3, I4, I5) -> f3(I0, 0, I0, rnd4, rnd5, I5) [rnd4 = rnd5 /\ rnd5 = rnd5]
6.30/6.32	  f3(I6, I7, I8, I9, I10, I11) -> f1(I6, I7, I8, I9, I10, I11) 
6.30/6.32	  f1(I12, I13, I14, I15, I16, I17) -> f3(I12, 1 + I13, I14, I15, I16, rnd6) [rnd6 = rnd6 /\ 1 + I13 <= I14]
6.30/6.32	  f1(I18, I19, I20, I21, I22, I23) -> f2(I18, I19, I20, I21, I22, I23) [I20 <= I19]
6.30/6.32	
6.30/6.32	The dependency graph for this problem is:
6.30/6.32	  2 -> 
6.30/6.32	Where:
6.30/6.32	  2) f3#(I6, I7, I8, I9, I10, I11) -> f1#(I6, I7, I8, I9, I10, I11) 
6.30/6.32	
6.30/6.32	We have the following SCCs.
6.30/6.32	  
6.40/9.29	EOF

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