Integer_Transition_Systems 2019-03-29 01.54 pair #432273543

details

property	value
status	complete
benchmark	Fibonacci.jar-obl-8.smt2
ran by	Akihisa Yamada
cpu timeout	1200 seconds
wallclock timeout	300 seconds
memory limit	137438953472 bytes
execution host	n176.star.cs.uiowa.edu
space	From_AProVE_2014

run statistics

property	value
solver	Ctrl
configuration	Transition
runtime (wallclock)	0.824288 seconds
cpu usage	0.840942
user time	0.448385
system time	0.392557
max virtual memory	125640.0
max residence set size	8496.0

stage attributes

key	value
starexec-result	YES

output

0.77/0.82	YES
0.77/0.82	
0.77/0.82	DP problem for innermost termination.
0.77/0.82	P =
0.77/0.82	  init#(x1, x2) -> f1#(rnd1, rnd2) 
0.77/0.82	  f3#(I0, I1) -> f2#(I1, I2) [I1 <= I0 - 1 /\ 1 <= I0 - 1]
0.77/0.82	  f4#(I3, I4) -> f3#(I3, I3 - 2) [0 <= I3 - 1]
0.77/0.82	  f2#(I5, I6) -> f3#(I5, I5 - 2) [I5 - 1 <= I5 - 1 /\ 1 <= I5 - 1]
0.77/0.82	  f2#(I7, I8) -> f3#(2, 0) [2 = I7]
0.77/0.82	  f2#(I9, I10) -> f2#(I9 - 1, I11) [I9 - 1 <= I9 - 1 /\ 1 <= I9 - 1]
0.77/0.82	  f1#(I12, I13) -> f2#(I13, I14) [-1 <= I13 - 1 /\ 0 <= I12 - 1]
0.77/0.82	R = 
0.77/0.82	  init(x1, x2) -> f1(rnd1, rnd2) 
0.77/0.82	  f3(I0, I1) -> f2(I1, I2) [I1 <= I0 - 1 /\ 1 <= I0 - 1]
0.77/0.82	  f4(I3, I4) -> f3(I3, I3 - 2) [0 <= I3 - 1]
0.77/0.82	  f2(I5, I6) -> f3(I5, I5 - 2) [I5 - 1 <= I5 - 1 /\ 1 <= I5 - 1]
0.77/0.82	  f2(I7, I8) -> f3(2, 0) [2 = I7]
0.77/0.82	  f2(I9, I10) -> f2(I9 - 1, I11) [I9 - 1 <= I9 - 1 /\ 1 <= I9 - 1]
0.77/0.82	  f1(I12, I13) -> f2(I13, I14) [-1 <= I13 - 1 /\ 0 <= I12 - 1]
0.77/0.82	
0.77/0.82	The dependency graph for this problem is:
0.77/0.82	  0 -> 6
0.77/0.82	  1 -> 3, 4, 5
0.77/0.82	  2 -> 1
0.77/0.82	  3 -> 1
0.77/0.82	  4 -> 1
0.77/0.82	  5 -> 3, 4, 5
0.77/0.82	  6 -> 3, 4, 5
0.77/0.82	Where:
0.77/0.82	  0) init#(x1, x2) -> f1#(rnd1, rnd2) 
0.77/0.82	  1) f3#(I0, I1) -> f2#(I1, I2) [I1 <= I0 - 1 /\ 1 <= I0 - 1]
0.77/0.82	  2) f4#(I3, I4) -> f3#(I3, I3 - 2) [0 <= I3 - 1]
0.77/0.82	  3) f2#(I5, I6) -> f3#(I5, I5 - 2) [I5 - 1 <= I5 - 1 /\ 1 <= I5 - 1]
0.77/0.82	  4) f2#(I7, I8) -> f3#(2, 0) [2 = I7]
0.77/0.82	  5) f2#(I9, I10) -> f2#(I9 - 1, I11) [I9 - 1 <= I9 - 1 /\ 1 <= I9 - 1]
0.77/0.82	  6) f1#(I12, I13) -> f2#(I13, I14) [-1 <= I13 - 1 /\ 0 <= I12 - 1]
0.77/0.82	
0.77/0.82	We have the following SCCs.
0.77/0.82	  { 1, 3, 4, 5 }
0.77/0.82	
0.77/0.82	DP problem for innermost termination.
0.77/0.82	P =
0.77/0.82	  f3#(I0, I1) -> f2#(I1, I2) [I1 <= I0 - 1 /\ 1 <= I0 - 1]
0.77/0.82	  f2#(I5, I6) -> f3#(I5, I5 - 2) [I5 - 1 <= I5 - 1 /\ 1 <= I5 - 1]
0.77/0.82	  f2#(I7, I8) -> f3#(2, 0) [2 = I7]
0.77/0.82	  f2#(I9, I10) -> f2#(I9 - 1, I11) [I9 - 1 <= I9 - 1 /\ 1 <= I9 - 1]
0.77/0.82	R = 
0.77/0.82	  init(x1, x2) -> f1(rnd1, rnd2) 
0.77/0.82	  f3(I0, I1) -> f2(I1, I2) [I1 <= I0 - 1 /\ 1 <= I0 - 1]
0.77/0.82	  f4(I3, I4) -> f3(I3, I3 - 2) [0 <= I3 - 1]
0.77/0.82	  f2(I5, I6) -> f3(I5, I5 - 2) [I5 - 1 <= I5 - 1 /\ 1 <= I5 - 1]
0.77/0.82	  f2(I7, I8) -> f3(2, 0) [2 = I7]
0.77/0.82	  f2(I9, I10) -> f2(I9 - 1, I11) [I9 - 1 <= I9 - 1 /\ 1 <= I9 - 1]
0.77/0.82	  f1(I12, I13) -> f2(I13, I14) [-1 <= I13 - 1 /\ 0 <= I12 - 1]
0.77/0.82	
0.77/0.82	We use the basic value criterion with the projection function NU:
0.77/0.82	  NU[f2#(z1,z2)] = z1
0.77/0.82	  NU[f3#(z1,z2)] = z2
0.77/0.82	
0.77/0.82	This gives the following inequalities:
0.77/0.82	  I1 <= I0 - 1 /\ 1 <= I0 - 1  ==>  I1 (>! \union =) I1
0.77/0.82	  I5 - 1 <= I5 - 1 /\ 1 <= I5 - 1  ==>  I5 >! I5 - 2
0.77/0.82	  2 = I7  ==>  I7 >! 0
0.77/0.82	  I9 - 1 <= I9 - 1 /\ 1 <= I9 - 1  ==>  I9 >! I9 - 1
0.77/0.82	
0.77/0.82	We remove all the strictly oriented dependency pairs.
0.77/0.82	
0.77/0.82	DP problem for innermost termination.
0.77/0.82	P =
0.77/0.82	  f3#(I0, I1) -> f2#(I1, I2) [I1 <= I0 - 1 /\ 1 <= I0 - 1]
0.77/0.82	R = 
0.77/0.82	  init(x1, x2) -> f1(rnd1, rnd2) 
0.77/0.82	  f3(I0, I1) -> f2(I1, I2) [I1 <= I0 - 1 /\ 1 <= I0 - 1]
0.77/0.82	  f4(I3, I4) -> f3(I3, I3 - 2) [0 <= I3 - 1]
0.77/0.82	  f2(I5, I6) -> f3(I5, I5 - 2) [I5 - 1 <= I5 - 1 /\ 1 <= I5 - 1]
0.77/0.82	  f2(I7, I8) -> f3(2, 0) [2 = I7]
0.77/0.82	  f2(I9, I10) -> f2(I9 - 1, I11) [I9 - 1 <= I9 - 1 /\ 1 <= I9 - 1]
0.77/0.82	  f1(I12, I13) -> f2(I13, I14) [-1 <= I13 - 1 /\ 0 <= I12 - 1]
0.77/0.82	
0.77/0.82	The dependency graph for this problem is:
0.77/0.82	  1 -> 
0.77/0.82	Where:
0.77/0.82	  1) f3#(I0, I1) -> f2#(I1, I2) [I1 <= I0 - 1 /\ 1 <= I0 - 1]
0.77/0.82	
0.77/0.82	We have the following SCCs.
0.77/0.82	  
0.77/3.80	EOF

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