ITS pair #487099567

details
property	value
status	complete
benchmark	ListDuplicate.jar-obl-9.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)	0.829639 seconds
cpu usage	0.837706
user time	0.451805
system time	0.385901
max virtual memory	570864.0
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) -> f3#(rnd1, rnd2, rnd3) 
  f5#(I0, I1, I2) -> f5#(I3, 0, I4) [1 = I1 /\ 0 <= I4 - 1 /\ 0 <= I3 - 1 /\ 0 <= I2 - 1 /\ 0 <= I0 - 1 /\ I4 <= I2 /\ I4 <= I0 /\ I3 <= I2 /\ I3 <= I0]
  f5#(I5, I6, I7) -> f5#(I8, 1, I9) [0 = I6 /\ -1 <= I9 - 1 /\ -1 <= I8 - 1 /\ 0 <= I7 - 1 /\ 0 <= I5 - 1 /\ I9 + 1 <= I7 /\ I9 + 1 <= I5 /\ I8 + 1 <= I7 /\ I8 + 1 <= I5]
  f2#(I10, I11, I12) -> f5#(I13, 1, I14) [I13 <= I11 /\ 0 <= y1 - 1 /\ I14 <= I11 /\ 0 <= I10 - 1 /\ -1 <= I11 - 1 /\ -1 <= I13 - 1 /\ -1 <= I14 - 1]
  f4#(I15, I16, I17) -> f4#(I15 - 1, I18, I19) [0 <= I15 - 1]
  f3#(I20, I21, I22) -> f4#(I23, I24, I25) [0 <= I20 - 1 /\ -1 <= I23 - 1 /\ -1 <= I21 - 1]
  f3#(I26, I27, I28) -> f2#(I29, I30, I31) [-1 <= I30 - 1 /\ 0 <= I29 - 1 /\ 0 <= I26 - 1 /\ I29 <= I26]
  f1#(I32, I33, I34) -> f2#(I35, I36, I37) [-1 <= I36 - 1 /\ 0 <= I35 - 1 /\ -1 <= I33 - 1 /\ 0 <= I32 - 1 /\ I36 <= I33 /\ I35 - 1 <= I33 /\ I35 <= I32]
R = 
  init(x1, x2, x3) -> f3(rnd1, rnd2, rnd3) 
  f5(I0, I1, I2) -> f5(I3, 0, I4) [1 = I1 /\ 0 <= I4 - 1 /\ 0 <= I3 - 1 /\ 0 <= I2 - 1 /\ 0 <= I0 - 1 /\ I4 <= I2 /\ I4 <= I0 /\ I3 <= I2 /\ I3 <= I0]
  f5(I5, I6, I7) -> f5(I8, 1, I9) [0 = I6 /\ -1 <= I9 - 1 /\ -1 <= I8 - 1 /\ 0 <= I7 - 1 /\ 0 <= I5 - 1 /\ I9 + 1 <= I7 /\ I9 + 1 <= I5 /\ I8 + 1 <= I7 /\ I8 + 1 <= I5]
  f2(I10, I11, I12) -> f5(I13, 1, I14) [I13 <= I11 /\ 0 <= y1 - 1 /\ I14 <= I11 /\ 0 <= I10 - 1 /\ -1 <= I11 - 1 /\ -1 <= I13 - 1 /\ -1 <= I14 - 1]
  f4(I15, I16, I17) -> f4(I15 - 1, I18, I19) [0 <= I15 - 1]
  f3(I20, I21, I22) -> f4(I23, I24, I25) [0 <= I20 - 1 /\ -1 <= I23 - 1 /\ -1 <= I21 - 1]
  f3(I26, I27, I28) -> f2(I29, I30, I31) [-1 <= I30 - 1 /\ 0 <= I29 - 1 /\ 0 <= I26 - 1 /\ I29 <= I26]
  f1(I32, I33, I34) -> f2(I35, I36, I37) [-1 <= I36 - 1 /\ 0 <= I35 - 1 /\ -1 <= I33 - 1 /\ 0 <= I32 - 1 /\ I36 <= I33 /\ I35 - 1 <= I33 /\ I35 <= I32]

The dependency graph for this problem is:
  0 -> 5, 6
  1 -> 2
  2 -> 1
  3 -> 1
  4 -> 4
  5 -> 4
  6 -> 3
  7 -> 3
Where:
  0) init#(x1, x2, x3) -> f3#(rnd1, rnd2, rnd3) 
  1) f5#(I0, I1, I2) -> f5#(I3, 0, I4) [1 = I1 /\ 0 <= I4 - 1 /\ 0 <= I3 - 1 /\ 0 <= I2 - 1 /\ 0 <= I0 - 1 /\ I4 <= I2 /\ I4 <= I0 /\ I3 <= I2 /\ I3 <= I0]
  2) f5#(I5, I6, I7) -> f5#(I8, 1, I9) [0 = I6 /\ -1 <= I9 - 1 /\ -1 <= I8 - 1 /\ 0 <= I7 - 1 /\ 0 <= I5 - 1 /\ I9 + 1 <= I7 /\ I9 + 1 <= I5 /\ I8 + 1 <= I7 /\ I8 + 1 <= I5]
  3) f2#(I10, I11, I12) -> f5#(I13, 1, I14) [I13 <= I11 /\ 0 <= y1 - 1 /\ I14 <= I11 /\ 0 <= I10 - 1 /\ -1 <= I11 - 1 /\ -1 <= I13 - 1 /\ -1 <= I14 - 1]
  4) f4#(I15, I16, I17) -> f4#(I15 - 1, I18, I19) [0 <= I15 - 1]
  5) f3#(I20, I21, I22) -> f4#(I23, I24, I25) [0 <= I20 - 1 /\ -1 <= I23 - 1 /\ -1 <= I21 - 1]
  6) f3#(I26, I27, I28) -> f2#(I29, I30, I31) [-1 <= I30 - 1 /\ 0 <= I29 - 1 /\ 0 <= I26 - 1 /\ I29 <= I26]
  7) f1#(I32, I33, I34) -> f2#(I35, I36, I37) [-1 <= I36 - 1 /\ 0 <= I35 - 1 /\ -1 <= I33 - 1 /\ 0 <= I32 - 1 /\ I36 <= I33 /\ I35 - 1 <= I33 /\ I35 <= I32]

We have the following SCCs.
  { 4 }
  { 1, 2 }

DP problem for innermost termination.
P =
  f5#(I0, I1, I2) -> f5#(I3, 0, I4) [1 = I1 /\ 0 <= I4 - 1 /\ 0 <= I3 - 1 /\ 0 <= I2 - 1 /\ 0 <= I0 - 1 /\ I4 <= I2 /\ I4 <= I0 /\ I3 <= I2 /\ I3 <= I0]
  f5#(I5, I6, I7) -> f5#(I8, 1, I9) [0 = I6 /\ -1 <= I9 - 1 /\ -1 <= I8 - 1 /\ 0 <= I7 - 1 /\ 0 <= I5 - 1 /\ I9 + 1 <= I7 /\ I9 + 1 <= I5 /\ I8 + 1 <= I7 /\ I8 + 1 <= I5]
R = 
  init(x1, x2, x3) -> f3(rnd1, rnd2, rnd3) 
  f5(I0, I1, I2) -> f5(I3, 0, I4) [1 = I1 /\ 0 <= I4 - 1 /\ 0 <= I3 - 1 /\ 0 <= I2 - 1 /\ 0 <= I0 - 1 /\ I4 <= I2 /\ I4 <= I0 /\ I3 <= I2 /\ I3 <= I0]
  f5(I5, I6, I7) -> f5(I8, 1, I9) [0 = I6 /\ -1 <= I9 - 1 /\ -1 <= I8 - 1 /\ 0 <= I7 - 1 /\ 0 <= I5 - 1 /\ I9 + 1 <= I7 /\ I9 + 1 <= I5 /\ I8 + 1 <= I7 /\ I8 + 1 <= I5]
  f2(I10, I11, I12) -> f5(I13, 1, I14) [I13 <= I11 /\ 0 <= y1 - 1 /\ I14 <= I11 /\ 0 <= I10 - 1 /\ -1 <= I11 - 1 /\ -1 <= I13 - 1 /\ -1 <= I14 - 1]
  f4(I15, I16, I17) -> f4(I15 - 1, I18, I19) [0 <= I15 - 1]
  f3(I20, I21, I22) -> f4(I23, I24, I25) [0 <= I20 - 1 /\ -1 <= I23 - 1 /\ -1 <= I21 - 1]
  f3(I26, I27, I28) -> f2(I29, I30, I31) [-1 <= I30 - 1 /\ 0 <= I29 - 1 /\ 0 <= I26 - 1 /\ I29 <= I26]
  f1(I32, I33, I34) -> f2(I35, I36, I37) [-1 <= I36 - 1 /\ 0 <= I35 - 1 /\ -1 <= I33 - 1 /\ 0 <= I32 - 1 /\ I36 <= I33 /\ I35 - 1 <= I33 /\ I35 <= I32]

We use the basic value criterion with the projection function NU:
  NU[f5#(z1,z2,z3)] = z3

This gives the following inequalities:
  1 = I1 /\ 0 <= I4 - 1 /\ 0 <= I3 - 1 /\ 0 <= I2 - 1 /\ 0 <= I0 - 1 /\ I4 <= I2 /\ I4 <= I0 /\ I3 <= I2 /\ I3 <= I0  ==>  I2 (>! \union =) I4
  0 = I6 /\ -1 <= I9 - 1 /\ -1 <= I8 - 1 /\ 0 <= I7 - 1 /\ 0 <= I5 - 1 /\ I9 + 1 <= I7 /\ I9 + 1 <= I5 /\ I8 + 1 <= I7 /\ I8 + 1 <= I5  ==>  I7 >! I9

We remove all the strictly oriented dependency pairs.

DP problem for innermost termination.
P =
  f5#(I0, I1, I2) -> f5#(I3, 0, I4) [1 = I1 /\ 0 <= I4 - 1 /\ 0 <= I3 - 1 /\ 0 <= I2 - 1 /\ 0 <= I0 - 1 /\ I4 <= I2 /\ I4 <= I0 /\ I3 <= I2 /\ I3 <= I0]
R = 
  init(x1, x2, x3) -> f3(rnd1, rnd2, rnd3) 
  f5(I0, I1, I2) -> f5(I3, 0, I4) [1 = I1 /\ 0 <= I4 - 1 /\ 0 <= I3 - 1 /\ 0 <= I2 - 1 /\ 0 <= I0 - 1 /\ I4 <= I2 /\ I4 <= I0 /\ I3 <= I2 /\ I3 <= I0]
  f5(I5, I6, I7) -> f5(I8, 1, I9) [0 = I6 /\ -1 <= I9 - 1 /\ -1 <= I8 - 1 /\ 0 <= I7 - 1 /\ 0 <= I5 - 1 /\ I9 + 1 <= I7 /\ I9 + 1 <= I5 /\ I8 + 1 <= I7 /\ I8 + 1 <= I5]
  f2(I10, I11, I12) -> f5(I13, 1, I14) [I13 <= I11 /\ 0 <= y1 - 1 /\ I14 <= I11 /\ 0 <= I10 - 1 /\ -1 <= I11 - 1 /\ -1 <= I13 - 1 /\ -1 <= I14 - 1]
  f4(I15, I16, I17) -> f4(I15 - 1, I18, I19) [0 <= I15 - 1]
  f3(I20, I21, I22) -> f4(I23, I24, I25) [0 <= I20 - 1 /\ -1 <= I23 - 1 /\ -1 <= I21 - 1]
  f3(I26, I27, I28) -> f2(I29, I30, I31) [-1 <= I30 - 1 /\ 0 <= I29 - 1 /\ 0 <= I26 - 1 /\ I29 <= I26]
  f1(I32, I33, I34) -> f2(I35, I36, I37) [-1 <= I36 - 1 /\ 0 <= I35 - 1 /\ -1 <= I33 - 1 /\ 0 <= I32 - 1 /\ I36 <= I33 /\ I35 - 1 <= I33 /\ I35 <= I32]

The dependency graph for this problem is:
  1 -> 
Where:
  1) f5#(I0, I1, I2) -> f5#(I3, 0, I4) [1 = I1 /\ 0 <= I4 - 1 /\ 0 <= I3 - 1 /\ 0 <= I2 - 1 /\ 0 <= I0 - 1 /\ I4 <= I2 /\ I4 <= I0 /\ I3 <= I2 /\ I3 <= I0]

We have the following SCCs.
  

DP problem for innermost termination.
P =
  f4#(I15, I16, I17) -> f4#(I15 - 1, I18, I19) [0 <= I15 - 1]
R = 
  init(x1, x2, x3) -> f3(rnd1, rnd2, rnd3) 
  f5(I0, I1, I2) -> f5(I3, 0, I4) [1 = I1 /\ 0 <= I4 - 1 /\ 0 <= I3 - 1 /\ 0 <= I2 - 1 /\ 0 <= I0 - 1 /\ I4 <= I2 /\ I4 <= I0 /\ I3 <= I2 /\ I3 <= I0]
  f5(I5, I6, I7) -> f5(I8, 1, I9) [0 = I6 /\ -1 <= I9 - 1 /\ -1 <= I8 - 1 /\ 0 <= I7 - 1 /\ 0 <= I5 - 1 /\ I9 + 1 <= I7 /\ I9 + 1 <= I5 /\ I8 + 1 <= I7 /\ I8 + 1 <= I5]
  f2(I10, I11, I12) -> f5(I13, 1, I14) [I13 <= I11 /\ 0 <= y1 - 1 /\ I14 <= I11 /\ 0 <= I10 - 1 /\ -1 <= I11 - 1 /\ -1 <= I13 - 1 /\ -1 <= I14 - 1]
  f4(I15, I16, I17) -> f4(I15 - 1, I18, I19) [0 <= I15 - 1]
  f3(I20, I21, I22) -> f4(I23, I24, I25) [0 <= I20 - 1 /\ -1 <= I23 - 1 /\ -1 <= I21 - 1]
  f3(I26, I27, I28) -> f2(I29, I30, I31) [-1 <= I30 - 1 /\ 0 <= I29 - 1 /\ 0 <= I26 - 1 /\ I29 <= I26]
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