Integer_Transition_Systems 2019-03-29 01.54 pair #432273900

details

property	value
status	complete
benchmark	ex08_rec.jar-obl-8.smt2
ran by	Akihisa Yamada
cpu timeout	1200 seconds
wallclock timeout	300 seconds
memory limit	137438953472 bytes
execution host	n062.star.cs.uiowa.edu
space	From_AProVE_2014

run statistics

property	value
solver	Ctrl
configuration	Transition
runtime (wallclock)	11.0096 seconds
cpu usage	11.2048
user time	5.46229
system time	5.74246
max virtual memory	718572.0
max residence set size	12112.0

stage attributes

key	value
starexec-result	MAYBE

output

11.17/11.00	MAYBE
11.17/11.00	
11.17/11.00	DP problem for innermost termination.
11.17/11.00	P =
11.17/11.00	  init#(x1, x2, x3) -> f1#(rnd1, rnd2, rnd3) 
11.17/11.00	  f3#(I0, I1, I2) -> f2#(1, 1, I1 + 1) [1 = I2 /\ 1 <= I1 - 1 /\ I0 <= 1]
11.17/11.00	  f2#(I3, I4, I5) -> f2#(0, 0, 9) [10 = I5 /\ I3 <= 1]
11.17/11.00	  f2#(I6, I7, I8) -> f2#(1, 1, 2) [1 = I8 /\ I6 <= 1]
11.17/11.00	  f3#(I9, I10, I11) -> f2#(0, 0, I10 - 1) [0 = I11 /\ 1 <= I10 - 1 /\ I9 <= 1]
11.17/11.00	  f2#(I12, I13, I14) -> f3#(I12, I14, I13) [10 <= I14 - 1]
11.17/11.00	  f2#(I15, I16, I17) -> f3#(I15, I17, I16) [1 <= I17 - 1 /\ I17 <= 9]
11.17/11.00	  f1#(I18, I19, I20) -> f2#(0, 0, 10 * I19) [-1 <= I19 - 1 /\ 0 <= I18 - 1]
11.17/11.00	R = 
11.17/11.00	  init(x1, x2, x3) -> f1(rnd1, rnd2, rnd3) 
11.17/11.00	  f3(I0, I1, I2) -> f2(1, 1, I1 + 1) [1 = I2 /\ 1 <= I1 - 1 /\ I0 <= 1]
11.17/11.00	  f2(I3, I4, I5) -> f2(0, 0, 9) [10 = I5 /\ I3 <= 1]
11.17/11.00	  f2(I6, I7, I8) -> f2(1, 1, 2) [1 = I8 /\ I6 <= 1]
11.17/11.00	  f3(I9, I10, I11) -> f2(0, 0, I10 - 1) [0 = I11 /\ 1 <= I10 - 1 /\ I9 <= 1]
11.17/11.00	  f2(I12, I13, I14) -> f3(I12, I14, I13) [10 <= I14 - 1]
11.17/11.00	  f2(I15, I16, I17) -> f3(I15, I17, I16) [1 <= I17 - 1 /\ I17 <= 9]
11.17/11.00	  f1(I18, I19, I20) -> f2(0, 0, 10 * I19) [-1 <= I19 - 1 /\ 0 <= I18 - 1]
11.17/11.00	
11.17/11.00	The dependency graph for this problem is:
11.17/11.00	  0 -> 7
11.17/11.00	  1 -> 2, 5, 6
11.17/11.00	  2 -> 6
11.17/11.00	  3 -> 6
11.17/11.00	  4 -> 2, 3, 5, 6
11.17/11.00	  5 -> 1, 4
11.17/11.00	  6 -> 1, 4
11.17/11.00	  7 -> 2, 5
11.17/11.00	Where:
11.17/11.00	  0) init#(x1, x2, x3) -> f1#(rnd1, rnd2, rnd3) 
11.17/11.00	  1) f3#(I0, I1, I2) -> f2#(1, 1, I1 + 1) [1 = I2 /\ 1 <= I1 - 1 /\ I0 <= 1]
11.17/11.00	  2) f2#(I3, I4, I5) -> f2#(0, 0, 9) [10 = I5 /\ I3 <= 1]
11.17/11.00	  3) f2#(I6, I7, I8) -> f2#(1, 1, 2) [1 = I8 /\ I6 <= 1]
11.17/11.00	  4) f3#(I9, I10, I11) -> f2#(0, 0, I10 - 1) [0 = I11 /\ 1 <= I10 - 1 /\ I9 <= 1]
11.17/11.00	  5) f2#(I12, I13, I14) -> f3#(I12, I14, I13) [10 <= I14 - 1]
11.17/11.00	  6) f2#(I15, I16, I17) -> f3#(I15, I17, I16) [1 <= I17 - 1 /\ I17 <= 9]
11.17/11.00	  7) f1#(I18, I19, I20) -> f2#(0, 0, 10 * I19) [-1 <= I19 - 1 /\ 0 <= I18 - 1]
11.17/11.00	
11.17/11.00	We have the following SCCs.
11.17/11.00	  { 1, 2, 3, 4, 5, 6 }
11.17/11.00	
11.17/11.00	DP problem for innermost termination.
11.17/11.00	P =
11.17/11.00	  f3#(I0, I1, I2) -> f2#(1, 1, I1 + 1) [1 = I2 /\ 1 <= I1 - 1 /\ I0 <= 1]
11.17/11.00	  f2#(I3, I4, I5) -> f2#(0, 0, 9) [10 = I5 /\ I3 <= 1]
11.17/11.00	  f2#(I6, I7, I8) -> f2#(1, 1, 2) [1 = I8 /\ I6 <= 1]
11.17/11.00	  f3#(I9, I10, I11) -> f2#(0, 0, I10 - 1) [0 = I11 /\ 1 <= I10 - 1 /\ I9 <= 1]
11.17/11.00	  f2#(I12, I13, I14) -> f3#(I12, I14, I13) [10 <= I14 - 1]
11.17/11.00	  f2#(I15, I16, I17) -> f3#(I15, I17, I16) [1 <= I17 - 1 /\ I17 <= 9]
11.17/11.00	R = 
11.17/11.00	  init(x1, x2, x3) -> f1(rnd1, rnd2, rnd3) 
11.17/11.00	  f3(I0, I1, I2) -> f2(1, 1, I1 + 1) [1 = I2 /\ 1 <= I1 - 1 /\ I0 <= 1]
11.17/11.00	  f2(I3, I4, I5) -> f2(0, 0, 9) [10 = I5 /\ I3 <= 1]
11.17/11.00	  f2(I6, I7, I8) -> f2(1, 1, 2) [1 = I8 /\ I6 <= 1]
11.17/11.00	  f3(I9, I10, I11) -> f2(0, 0, I10 - 1) [0 = I11 /\ 1 <= I10 - 1 /\ I9 <= 1]
11.17/11.00	  f2(I12, I13, I14) -> f3(I12, I14, I13) [10 <= I14 - 1]
11.17/11.00	  f2(I15, I16, I17) -> f3(I15, I17, I16) [1 <= I17 - 1 /\ I17 <= 9]
11.17/11.00	  f1(I18, I19, I20) -> f2(0, 0, 10 * I19) [-1 <= I19 - 1 /\ 0 <= I18 - 1]
11.17/11.00	
11.17/13.98	EOF

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