Integer_Transition_Systems 2019-03-29 01.54 pair #432273369

details

property	value
status	complete
benchmark	GCD2.jar-obl-8.smt2
ran by	Akihisa Yamada
cpu timeout	1200 seconds
wallclock timeout	300 seconds
memory limit	137438953472 bytes
execution host	n149.star.cs.uiowa.edu
space	From_AProVE_2014

run statistics

property	value
solver	Ctrl
configuration	Transition
runtime (wallclock)	0.726039 seconds
cpu usage	0.740416
user time	0.366624
system time	0.373792
max virtual memory	350648.0
max residence set size	8660.0

stage attributes

key	value
starexec-result	YES

output

0.62/0.72	YES
0.62/0.72	
0.62/0.72	DP problem for innermost termination.
0.62/0.72	P =
0.62/0.72	  init#(x1, x2) -> f1#(rnd1, rnd2) 
0.62/0.72	  f3#(I0, I1) -> f2#(I1, I0) [I0 <= I1 - 1]
0.62/0.72	  f2#(I2, I3) -> f2#(I2, 0) [I2 = I3 /\ 0 <= I2 - 1]
0.62/0.72	  f3#(I4, I5) -> f3#(I4 - I5, I5) [0 <= I5 - 1 /\ 0 <= I4 - 1 /\ I5 <= I4 - 1]
0.62/0.72	  f2#(I6, I7) -> f3#(I6, I7) [0 <= I7 - 1 /\ -1 <= I6 - 1 /\ I7 <= I6 - 1]
0.62/0.72	  f2#(I8, I9) -> f3#(I8, I9) [0 <= I9 - 1 /\ -1 <= I8 - 1 /\ I8 <= I9 - 1]
0.62/0.72	  f1#(I10, I11) -> f2#(I12, I13) [0 <= I10 - 1 /\ -1 <= I12 - 1 /\ -1 <= I11 - 1 /\ -1 <= I13 - 1]
0.62/0.72	R = 
0.62/0.72	  init(x1, x2) -> f1(rnd1, rnd2) 
0.62/0.72	  f3(I0, I1) -> f2(I1, I0) [I0 <= I1 - 1]
0.62/0.72	  f2(I2, I3) -> f2(I2, 0) [I2 = I3 /\ 0 <= I2 - 1]
0.62/0.72	  f3(I4, I5) -> f3(I4 - I5, I5) [0 <= I5 - 1 /\ 0 <= I4 - 1 /\ I5 <= I4 - 1]
0.62/0.72	  f2(I6, I7) -> f3(I6, I7) [0 <= I7 - 1 /\ -1 <= I6 - 1 /\ I7 <= I6 - 1]
0.62/0.72	  f2(I8, I9) -> f3(I8, I9) [0 <= I9 - 1 /\ -1 <= I8 - 1 /\ I8 <= I9 - 1]
0.62/0.72	  f1(I10, I11) -> f2(I12, I13) [0 <= I10 - 1 /\ -1 <= I12 - 1 /\ -1 <= I11 - 1 /\ -1 <= I13 - 1]
0.62/0.72	
0.62/0.72	The dependency graph for this problem is:
0.62/0.72	  0 -> 6
0.62/0.72	  1 -> 4
0.62/0.72	  2 -> 
0.62/0.72	  3 -> 1, 3
0.62/0.72	  4 -> 3
0.62/0.72	  5 -> 1
0.62/0.72	  6 -> 2, 4, 5
0.62/0.72	Where:
0.62/0.72	  0) init#(x1, x2) -> f1#(rnd1, rnd2) 
0.62/0.72	  1) f3#(I0, I1) -> f2#(I1, I0) [I0 <= I1 - 1]
0.62/0.72	  2) f2#(I2, I3) -> f2#(I2, 0) [I2 = I3 /\ 0 <= I2 - 1]
0.62/0.72	  3) f3#(I4, I5) -> f3#(I4 - I5, I5) [0 <= I5 - 1 /\ 0 <= I4 - 1 /\ I5 <= I4 - 1]
0.62/0.72	  4) f2#(I6, I7) -> f3#(I6, I7) [0 <= I7 - 1 /\ -1 <= I6 - 1 /\ I7 <= I6 - 1]
0.62/0.72	  5) f2#(I8, I9) -> f3#(I8, I9) [0 <= I9 - 1 /\ -1 <= I8 - 1 /\ I8 <= I9 - 1]
0.62/0.72	  6) f1#(I10, I11) -> f2#(I12, I13) [0 <= I10 - 1 /\ -1 <= I12 - 1 /\ -1 <= I11 - 1 /\ -1 <= I13 - 1]
0.62/0.72	
0.62/0.72	We have the following SCCs.
0.62/0.72	  { 1, 3, 4 }
0.62/0.72	
0.62/0.72	DP problem for innermost termination.
0.62/0.72	P =
0.62/0.72	  f3#(I0, I1) -> f2#(I1, I0) [I0 <= I1 - 1]
0.62/0.72	  f3#(I4, I5) -> f3#(I4 - I5, I5) [0 <= I5 - 1 /\ 0 <= I4 - 1 /\ I5 <= I4 - 1]
0.62/0.72	  f2#(I6, I7) -> f3#(I6, I7) [0 <= I7 - 1 /\ -1 <= I6 - 1 /\ I7 <= I6 - 1]
0.62/0.72	R = 
0.62/0.72	  init(x1, x2) -> f1(rnd1, rnd2) 
0.62/0.72	  f3(I0, I1) -> f2(I1, I0) [I0 <= I1 - 1]
0.62/0.72	  f2(I2, I3) -> f2(I2, 0) [I2 = I3 /\ 0 <= I2 - 1]
0.62/0.72	  f3(I4, I5) -> f3(I4 - I5, I5) [0 <= I5 - 1 /\ 0 <= I4 - 1 /\ I5 <= I4 - 1]
0.62/0.72	  f2(I6, I7) -> f3(I6, I7) [0 <= I7 - 1 /\ -1 <= I6 - 1 /\ I7 <= I6 - 1]
0.62/0.72	  f2(I8, I9) -> f3(I8, I9) [0 <= I9 - 1 /\ -1 <= I8 - 1 /\ I8 <= I9 - 1]
0.62/0.72	  f1(I10, I11) -> f2(I12, I13) [0 <= I10 - 1 /\ -1 <= I12 - 1 /\ -1 <= I11 - 1 /\ -1 <= I13 - 1]
0.62/0.72	
0.62/0.72	We use the basic value criterion with the projection function NU:
0.62/0.72	  NU[f2#(z1,z2)] = z1
0.62/0.72	  NU[f3#(z1,z2)] = z2
0.62/0.72	
0.62/0.72	This gives the following inequalities:
0.62/0.72	  I0 <= I1 - 1  ==>  I1 (>! \union =) I1
0.62/0.72	  0 <= I5 - 1 /\ 0 <= I4 - 1 /\ I5 <= I4 - 1  ==>  I5 (>! \union =) I5
0.62/0.72	  0 <= I7 - 1 /\ -1 <= I6 - 1 /\ I7 <= I6 - 1  ==>  I6 >! I7
0.62/0.72	
0.62/0.72	We remove all the strictly oriented dependency pairs.
0.62/0.72	
0.62/0.72	DP problem for innermost termination.
0.62/0.72	P =
0.62/0.72	  f3#(I0, I1) -> f2#(I1, I0) [I0 <= I1 - 1]
0.62/0.72	  f3#(I4, I5) -> f3#(I4 - I5, I5) [0 <= I5 - 1 /\ 0 <= I4 - 1 /\ I5 <= I4 - 1]
0.62/0.72	R = 
0.62/0.72	  init(x1, x2) -> f1(rnd1, rnd2) 
0.62/0.72	  f3(I0, I1) -> f2(I1, I0) [I0 <= I1 - 1]
0.62/0.72	  f2(I2, I3) -> f2(I2, 0) [I2 = I3 /\ 0 <= I2 - 1]
0.62/0.72	  f3(I4, I5) -> f3(I4 - I5, I5) [0 <= I5 - 1 /\ 0 <= I4 - 1 /\ I5 <= I4 - 1]
0.62/0.72	  f2(I6, I7) -> f3(I6, I7) [0 <= I7 - 1 /\ -1 <= I6 - 1 /\ I7 <= I6 - 1]
0.62/0.72	  f2(I8, I9) -> f3(I8, I9) [0 <= I9 - 1 /\ -1 <= I8 - 1 /\ I8 <= I9 - 1]
0.62/0.72	  f1(I10, I11) -> f2(I12, I13) [0 <= I10 - 1 /\ -1 <= I12 - 1 /\ -1 <= I11 - 1 /\ -1 <= I13 - 1]
0.62/0.72	
0.62/0.72	The dependency graph for this problem is:
0.62/0.72	  1 -> 
0.62/0.72	  3 -> 1, 3
0.62/0.72	Where:
0.62/0.72	  1) f3#(I0, I1) -> f2#(I1, I0) [I0 <= I1 - 1]
0.62/0.72	  3) f3#(I4, I5) -> f3#(I4 - I5, I5) [0 <= I5 - 1 /\ 0 <= I4 - 1 /\ I5 <= I4 - 1]
0.62/0.72	
0.62/0.72	We have the following SCCs.
0.62/0.72	  { 3 }
0.62/0.72	
0.62/0.72	DP problem for innermost termination.
0.62/0.72	P =
0.62/0.72	  f3#(I4, I5) -> f3#(I4 - I5, I5) [0 <= I5 - 1 /\ 0 <= I4 - 1 /\ I5 <= I4 - 1]
0.62/0.72	R = 
0.62/0.72	  init(x1, x2) -> f1(rnd1, rnd2) 
0.62/0.72	  f3(I0, I1) -> f2(I1, I0) [I0 <= I1 - 1]
0.62/0.72	  f2(I2, I3) -> f2(I2, 0) [I2 = I3 /\ 0 <= I2 - 1]
0.62/0.72	  f3(I4, I5) -> f3(I4 - I5, I5) [0 <= I5 - 1 /\ 0 <= I4 - 1 /\ I5 <= I4 - 1]
0.62/0.72	  f2(I6, I7) -> f3(I6, I7) [0 <= I7 - 1 /\ -1 <= I6 - 1 /\ I7 <= I6 - 1]
0.62/0.72	  f2(I8, I9) -> f3(I8, I9) [0 <= I9 - 1 /\ -1 <= I8 - 1 /\ I8 <= I9 - 1]
0.62/0.72	  f1(I10, I11) -> f2(I12, I13) [0 <= I10 - 1 /\ -1 <= I12 - 1 /\ -1 <= I11 - 1 /\ -1 <= I13 - 1]
0.62/0.72

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