ITS pair #487099687

details

property	value
status	complete
benchmark	costa09-example_1.jar-obl-10.smt2
ran by	Akihisa Yamada
cpu timeout	1200 seconds
wallclock timeout	300 seconds
memory limit	137438953472 bytes
execution host	n151.star.cs.uiowa.edu
space	From_AProVE_2014

run statistics

property	value
solver	Ctrl
configuration	Transition
runtime (wallclock)	1.21877 seconds
cpu usage	1.24162
user time	0.659282
system time	0.582341
max virtual memory	350552.0
max residence set size	8568.0

stage attributes

key	value
starexec-result	YES

output

YES

DP problem for innermost termination.
P =
  init#(x1, x2, x3) -> f1#(rnd1, rnd2, rnd3) 
  f4#(I0, I1, I2) -> f4#(I0 + I1, I1 + 3, I1 + 3) [I1 = I2 /\ I1 <= 1000 /\ -1 <= I0 - 1 /\ -1 <= I1 - 1]
  f3#(I3, I4, I5) -> f4#(0, 0, 0) [I4 = I5 /\ 1000 <= I4 - 1]
  f3#(I6, I7, I8) -> f3#(I6 + I7, I7 + 2, I7 + 2) [I7 = I8 /\ I7 <= 1000 /\ -1 <= I6 - 1 /\ -1 <= I7 - 1]
  f2#(I9, I10, I11) -> f3#(0, 0, 0) [I10 = I11 /\ 1000 <= I10 - 1]
  f2#(I12, I13, I14) -> f2#(I12 + I13, I13 + 1, I13 + 1) [I13 = I14 /\ I13 <= 1000 /\ -1 <= I12 - 1 /\ -1 <= I13 - 1]
  f1#(I15, I16, I17) -> f2#(0, 0, 0) 
R = 
  init(x1, x2, x3) -> f1(rnd1, rnd2, rnd3) 
  f4(I0, I1, I2) -> f4(I0 + I1, I1 + 3, I1 + 3) [I1 = I2 /\ I1 <= 1000 /\ -1 <= I0 - 1 /\ -1 <= I1 - 1]
  f3(I3, I4, I5) -> f4(0, 0, 0) [I4 = I5 /\ 1000 <= I4 - 1]
  f3(I6, I7, I8) -> f3(I6 + I7, I7 + 2, I7 + 2) [I7 = I8 /\ I7 <= 1000 /\ -1 <= I6 - 1 /\ -1 <= I7 - 1]
  f2(I9, I10, I11) -> f3(0, 0, 0) [I10 = I11 /\ 1000 <= I10 - 1]
  f2(I12, I13, I14) -> f2(I12 + I13, I13 + 1, I13 + 1) [I13 = I14 /\ I13 <= 1000 /\ -1 <= I12 - 1 /\ -1 <= I13 - 1]
  f1(I15, I16, I17) -> f2(0, 0, 0) 

The dependency graph for this problem is:
  0 -> 6
  1 -> 1
  2 -> 1
  3 -> 2, 3
  4 -> 3
  5 -> 4, 5
  6 -> 5
Where:
  0) init#(x1, x2, x3) -> f1#(rnd1, rnd2, rnd3) 
  1) f4#(I0, I1, I2) -> f4#(I0 + I1, I1 + 3, I1 + 3) [I1 = I2 /\ I1 <= 1000 /\ -1 <= I0 - 1 /\ -1 <= I1 - 1]
  2) f3#(I3, I4, I5) -> f4#(0, 0, 0) [I4 = I5 /\ 1000 <= I4 - 1]
  3) f3#(I6, I7, I8) -> f3#(I6 + I7, I7 + 2, I7 + 2) [I7 = I8 /\ I7 <= 1000 /\ -1 <= I6 - 1 /\ -1 <= I7 - 1]
  4) f2#(I9, I10, I11) -> f3#(0, 0, 0) [I10 = I11 /\ 1000 <= I10 - 1]
  5) f2#(I12, I13, I14) -> f2#(I12 + I13, I13 + 1, I13 + 1) [I13 = I14 /\ I13 <= 1000 /\ -1 <= I12 - 1 /\ -1 <= I13 - 1]
  6) f1#(I15, I16, I17) -> f2#(0, 0, 0) 

We have the following SCCs.
  { 5 }
  { 3 }
  { 1 }

DP problem for innermost termination.
P =
  f4#(I0, I1, I2) -> f4#(I0 + I1, I1 + 3, I1 + 3) [I1 = I2 /\ I1 <= 1000 /\ -1 <= I0 - 1 /\ -1 <= I1 - 1]
R = 
  init(x1, x2, x3) -> f1(rnd1, rnd2, rnd3) 
  f4(I0, I1, I2) -> f4(I0 + I1, I1 + 3, I1 + 3) [I1 = I2 /\ I1 <= 1000 /\ -1 <= I0 - 1 /\ -1 <= I1 - 1]
  f3(I3, I4, I5) -> f4(0, 0, 0) [I4 = I5 /\ 1000 <= I4 - 1]
  f3(I6, I7, I8) -> f3(I6 + I7, I7 + 2, I7 + 2) [I7 = I8 /\ I7 <= 1000 /\ -1 <= I6 - 1 /\ -1 <= I7 - 1]
  f2(I9, I10, I11) -> f3(0, 0, 0) [I10 = I11 /\ 1000 <= I10 - 1]
  f2(I12, I13, I14) -> f2(I12 + I13, I13 + 1, I13 + 1) [I13 = I14 /\ I13 <= 1000 /\ -1 <= I12 - 1 /\ -1 <= I13 - 1]
  f1(I15, I16, I17) -> f2(0, 0, 0) 

We use the reverse value criterion with the projection function NU:
  NU[f4#(z1,z2,z3)] = 1000 + -1 * z2

This gives the following inequalities:
  I1 = I2 /\ I1 <= 1000 /\ -1 <= I0 - 1 /\ -1 <= I1 - 1  ==>  1000 + -1 * I1 > 1000 + -1 * (I1 + 3)  with  1000 + -1 * I1 >= 0

All dependency pairs are strictly oriented, so the entire dependency pair problem may be removed.

DP problem for innermost termination.
P =
  f3#(I6, I7, I8) -> f3#(I6 + I7, I7 + 2, I7 + 2) [I7 = I8 /\ I7 <= 1000 /\ -1 <= I6 - 1 /\ -1 <= I7 - 1]
R = 
  init(x1, x2, x3) -> f1(rnd1, rnd2, rnd3) 
  f4(I0, I1, I2) -> f4(I0 + I1, I1 + 3, I1 + 3) [I1 = I2 /\ I1 <= 1000 /\ -1 <= I0 - 1 /\ -1 <= I1 - 1]
  f3(I3, I4, I5) -> f4(0, 0, 0) [I4 = I5 /\ 1000 <= I4 - 1]
  f3(I6, I7, I8) -> f3(I6 + I7, I7 + 2, I7 + 2) [I7 = I8 /\ I7 <= 1000 /\ -1 <= I6 - 1 /\ -1 <= I7 - 1]
  f2(I9, I10, I11) -> f3(0, 0, 0) [I10 = I11 /\ 1000 <= I10 - 1]
  f2(I12, I13, I14) -> f2(I12 + I13, I13 + 1, I13 + 1) [I13 = I14 /\ I13 <= 1000 /\ -1 <= I12 - 1 /\ -1 <= I13 - 1]
  f1(I15, I16, I17) -> f2(0, 0, 0) 

We use the reverse value criterion with the projection function NU:
  NU[f3#(z1,z2,z3)] = 1000 + -1 * z2

This gives the following inequalities:
  I7 = I8 /\ I7 <= 1000 /\ -1 <= I6 - 1 /\ -1 <= I7 - 1  ==>  1000 + -1 * I7 > 1000 + -1 * (I7 + 2)  with  1000 + -1 * I7 >= 0

All dependency pairs are strictly oriented, so the entire dependency pair problem may be removed.

DP problem for innermost termination.
P =
  f2#(I12, I13, I14) -> f2#(I12 + I13, I13 + 1, I13 + 1) [I13 = I14 /\ I13 <= 1000 /\ -1 <= I12 - 1 /\ -1 <= I13 - 1]
R = 
  init(x1, x2, x3) -> f1(rnd1, rnd2, rnd3) 
  f4(I0, I1, I2) -> f4(I0 + I1, I1 + 3, I1 + 3) [I1 = I2 /\ I1 <= 1000 /\ -1 <= I0 - 1 /\ -1 <= I1 - 1]
  f3(I3, I4, I5) -> f4(0, 0, 0) [I4 = I5 /\ 1000 <= I4 - 1]
  f3(I6, I7, I8) -> f3(I6 + I7, I7 + 2, I7 + 2) [I7 = I8 /\ I7 <= 1000 /\ -1 <= I6 - 1 /\ -1 <= I7 - 1]
  f2(I9, I10, I11) -> f3(0, 0, 0) [I10 = I11 /\ 1000 <= I10 - 1]
  f2(I12, I13, I14) -> f2(I12 + I13, I13 + 1, I13 + 1) [I13 = I14 /\ I13 <= 1000 /\ -1 <= I12 - 1 /\ -1 <= I13 - 1]
  f1(I15, I16, I17) -> f2(0, 0, 0) 

We use the reverse value criterion with the projection function NU:
  NU[f2#(z1,z2,z3)] = 1000 + -1 * z2

This gives the following inequalities:
  I13 = I14 /\ I13 <= 1000 /\ -1 <= I12 - 1 /\ -1 <= I13 - 1  ==>  1000 + -1 * I13 > 1000 + -1 * (I13 + 1)  with  1000 + -1 * I13 >= 0

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