Integer_Transition_Systems 2019-03-29 01.54 pair #432274890

details

property	value
status	complete
benchmark	acqrel-succeed2.t2.smt2
ran by	Akihisa Yamada
cpu timeout	1200 seconds
wallclock timeout	300 seconds
memory limit	137438953472 bytes
execution host	n003.star.cs.uiowa.edu
space	From_T2

run statistics

property	value
solver	Ctrl
configuration	Transition
runtime (wallclock)	6.30927 seconds
cpu usage	6.40613
user time	3.25005
system time	3.15608
max virtual memory	684948.0
max residence set size	8616.0

stage attributes

key	value
starexec-result	MAYBE

output

6.34/6.30	MAYBE
6.34/6.30	
6.34/6.30	DP problem for innermost termination.
6.34/6.30	P =
6.34/6.30	  f5#(x1, x2, x3, x4) -> f4#(x1, x2, x3, x4) 
6.34/6.30	  f4#(I0, I1, I2, I3) -> f2#(0, 0, rnd3, rnd4) [rnd4 = rnd4 /\ rnd3 = rnd3]
6.34/6.30	  f2#(I4, I5, I6, I7) -> f1#(0, I5, I6, I7) [I6 <= 0 /\ y1 = 1]
6.34/6.30	  f3#(I8, I9, I10, I11) -> f1#(I8, I9, I10, I11) 
6.34/6.30	  f1#(I12, I13, I14, I15) -> f3#(I12, I13, I14, -1 + I15) [1 <= I15]
6.34/6.30	  f1#(I16, I17, I18, I19) -> f2#(I16, 0, I20, I19) [I19 <= 0 /\ I21 = 0 /\ I20 = I20]
6.34/6.30	R = 
6.34/6.30	  f5(x1, x2, x3, x4) -> f4(x1, x2, x3, x4) 
6.34/6.30	  f4(I0, I1, I2, I3) -> f2(0, 0, rnd3, rnd4) [rnd4 = rnd4 /\ rnd3 = rnd3]
6.34/6.30	  f2(I4, I5, I6, I7) -> f1(0, I5, I6, I7) [I6 <= 0 /\ y1 = 1]
6.34/6.30	  f3(I8, I9, I10, I11) -> f1(I8, I9, I10, I11) 
6.34/6.30	  f1(I12, I13, I14, I15) -> f3(I12, I13, I14, -1 + I15) [1 <= I15]
6.34/6.30	  f1(I16, I17, I18, I19) -> f2(I16, 0, I20, I19) [I19 <= 0 /\ I21 = 0 /\ I20 = I20]
6.34/6.30	
6.34/6.30	The dependency graph for this problem is:
6.34/6.30	  0 -> 1
6.34/6.30	  1 -> 2
6.34/6.30	  2 -> 4, 5
6.34/6.30	  3 -> 4, 5
6.34/6.30	  4 -> 3
6.34/6.30	  5 -> 2
6.34/6.30	Where:
6.34/6.30	  0) f5#(x1, x2, x3, x4) -> f4#(x1, x2, x3, x4) 
6.34/6.30	  1) f4#(I0, I1, I2, I3) -> f2#(0, 0, rnd3, rnd4) [rnd4 = rnd4 /\ rnd3 = rnd3]
6.34/6.30	  2) f2#(I4, I5, I6, I7) -> f1#(0, I5, I6, I7) [I6 <= 0 /\ y1 = 1]
6.34/6.30	  3) f3#(I8, I9, I10, I11) -> f1#(I8, I9, I10, I11) 
6.34/6.30	  4) f1#(I12, I13, I14, I15) -> f3#(I12, I13, I14, -1 + I15) [1 <= I15]
6.34/6.30	  5) f1#(I16, I17, I18, I19) -> f2#(I16, 0, I20, I19) [I19 <= 0 /\ I21 = 0 /\ I20 = I20]
6.34/6.30	
6.34/6.30	We have the following SCCs.
6.34/6.30	  { 2, 3, 4, 5 }
6.34/6.30	
6.34/6.30	DP problem for innermost termination.
6.34/6.30	P =
6.34/6.30	  f2#(I4, I5, I6, I7) -> f1#(0, I5, I6, I7) [I6 <= 0 /\ y1 = 1]
6.34/6.30	  f3#(I8, I9, I10, I11) -> f1#(I8, I9, I10, I11) 
6.34/6.30	  f1#(I12, I13, I14, I15) -> f3#(I12, I13, I14, -1 + I15) [1 <= I15]
6.34/6.30	  f1#(I16, I17, I18, I19) -> f2#(I16, 0, I20, I19) [I19 <= 0 /\ I21 = 0 /\ I20 = I20]
6.34/6.30	R = 
6.34/6.30	  f5(x1, x2, x3, x4) -> f4(x1, x2, x3, x4) 
6.34/6.30	  f4(I0, I1, I2, I3) -> f2(0, 0, rnd3, rnd4) [rnd4 = rnd4 /\ rnd3 = rnd3]
6.34/6.30	  f2(I4, I5, I6, I7) -> f1(0, I5, I6, I7) [I6 <= 0 /\ y1 = 1]
6.34/6.30	  f3(I8, I9, I10, I11) -> f1(I8, I9, I10, I11) 
6.34/6.30	  f1(I12, I13, I14, I15) -> f3(I12, I13, I14, -1 + I15) [1 <= I15]
6.34/6.30	  f1(I16, I17, I18, I19) -> f2(I16, 0, I20, I19) [I19 <= 0 /\ I21 = 0 /\ I20 = I20]
6.34/6.30	
6.34/6.30	We use the basic value criterion with the projection function NU:
6.34/6.30	  NU[f3#(z1,z2,z3,z4)] = z4
6.34/6.30	  NU[f1#(z1,z2,z3,z4)] = z4
6.34/6.30	  NU[f2#(z1,z2,z3,z4)] = z4
6.34/6.30	
6.34/6.30	This gives the following inequalities:
6.34/6.30	  I6 <= 0 /\ y1 = 1  ==>  I7 (>! \union =) I7
6.34/6.30	    ==>  I11 (>! \union =) I11
6.34/6.30	  1 <= I15  ==>  I15 >! -1 + I15
6.34/6.30	  I19 <= 0 /\ I21 = 0 /\ I20 = I20  ==>  I19 (>! \union =) I19
6.34/6.30	
6.34/6.30	We remove all the strictly oriented dependency pairs.
6.34/6.30	
6.34/6.30	DP problem for innermost termination.
6.34/6.30	P =
6.34/6.30	  f2#(I4, I5, I6, I7) -> f1#(0, I5, I6, I7) [I6 <= 0 /\ y1 = 1]
6.34/6.30	  f3#(I8, I9, I10, I11) -> f1#(I8, I9, I10, I11) 
6.34/6.30	  f1#(I16, I17, I18, I19) -> f2#(I16, 0, I20, I19) [I19 <= 0 /\ I21 = 0 /\ I20 = I20]
6.34/6.30	R = 
6.34/6.30	  f5(x1, x2, x3, x4) -> f4(x1, x2, x3, x4) 
6.34/6.30	  f4(I0, I1, I2, I3) -> f2(0, 0, rnd3, rnd4) [rnd4 = rnd4 /\ rnd3 = rnd3]
6.34/6.30	  f2(I4, I5, I6, I7) -> f1(0, I5, I6, I7) [I6 <= 0 /\ y1 = 1]
6.34/6.30	  f3(I8, I9, I10, I11) -> f1(I8, I9, I10, I11) 
6.34/6.30	  f1(I12, I13, I14, I15) -> f3(I12, I13, I14, -1 + I15) [1 <= I15]
6.34/6.30	  f1(I16, I17, I18, I19) -> f2(I16, 0, I20, I19) [I19 <= 0 /\ I21 = 0 /\ I20 = I20]
6.34/6.30	
6.34/6.30	The dependency graph for this problem is:
6.34/6.30	  2 -> 5
6.34/6.30	  3 -> 5
6.34/6.30	  5 -> 2
6.34/6.30	Where:
6.34/6.30	  2) f2#(I4, I5, I6, I7) -> f1#(0, I5, I6, I7) [I6 <= 0 /\ y1 = 1]
6.34/6.30	  3) f3#(I8, I9, I10, I11) -> f1#(I8, I9, I10, I11) 
6.34/6.30	  5) f1#(I16, I17, I18, I19) -> f2#(I16, 0, I20, I19) [I19 <= 0 /\ I21 = 0 /\ I20 = I20]
6.34/6.30	
6.34/6.30	We have the following SCCs.
6.34/6.30	  { 2, 5 }
6.34/6.30	
6.34/6.30	DP problem for innermost termination.
6.34/6.30	P =
6.34/6.30	  f2#(I4, I5, I6, I7) -> f1#(0, I5, I6, I7) [I6 <= 0 /\ y1 = 1]
6.34/6.30	  f1#(I16, I17, I18, I19) -> f2#(I16, 0, I20, I19) [I19 <= 0 /\ I21 = 0 /\ I20 = I20]
6.34/6.30	R = 
6.34/6.30	  f5(x1, x2, x3, x4) -> f4(x1, x2, x3, x4) 
6.34/6.30	  f4(I0, I1, I2, I3) -> f2(0, 0, rnd3, rnd4) [rnd4 = rnd4 /\ rnd3 = rnd3]
6.34/6.30	  f2(I4, I5, I6, I7) -> f1(0, I5, I6, I7) [I6 <= 0 /\ y1 = 1]
6.34/6.30	  f3(I8, I9, I10, I11) -> f1(I8, I9, I10, I11) 
6.34/6.30	  f1(I12, I13, I14, I15) -> f3(I12, I13, I14, -1 + I15) [1 <= I15]
6.34/6.30	  f1(I16, I17, I18, I19) -> f2(I16, 0, I20, I19) [I19 <= 0 /\ I21 = 0 /\ I20 = I20]
6.34/6.30

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