ITS pair #487099777

details
property	value
status	complete
benchmark	Diff.jar-obl-8.smt2
ran by	Akihisa Yamada
cpu timeout	1200 seconds
wallclock timeout	300 seconds
memory limit	137438953472 bytes
execution host	n150.star.cs.uiowa.edu
space	From_AProVE_2014
run statistics
property	value
solver	Ctrl
configuration	Transition
runtime (wallclock)	21.9964 seconds
cpu usage	22.2435
user time	12.1185
system time	10.125
max virtual memory	733868.0
max residence set size	11528.0
stage attributes
key	value
starexec-result	YES
output
YES

DP problem for innermost termination.
P =
  init#(x1, x2, x3, x4, x5) -> f1#(rnd1, rnd2, rnd3, rnd4, rnd5) 
  f3#(I0, I1, I2, I3, I4) -> f3#(I0, I1, 1, I3, I3) [I3 = I4 /\ 0 = I2 /\ I1 <= 19 /\ I3 <= 19]
  f3#(I5, I6, I7, I8, I9) -> f2#(I5 + 1, I6 + 1, I6 + 1, I10, I11) [I8 = I9 /\ 0 = I7 /\ 19 <= I8 - 1 /\ I6 <= 19 /\ I5 <= 19]
  f3#(I12, I13, I14, I15, I16) -> f3#(I12, I13, 0, I15 + 1, I15 + 1) [I13 <= 19 /\ y1 <= y2 - 1 /\ I15 <= 19 /\ 0 = I14 /\ I15 = I16]
  f3#(I17, I18, I19, I20, I21) -> f3#(I17, I18, 0, I20 + 1, I20 + 1) [I18 <= 19 /\ I22 <= I23 - 1 /\ I20 <= 19 /\ 0 = I19 /\ I20 = I21]
  f3#(I24, I25, I26, I27, I28) -> f2#(I24, I25 + 1, I25 + 1, I29, I30) [I27 = I28 /\ 1 = I26 /\ I27 <= 19]
  f3#(I31, I32, I33, I34, I35) -> f2#(I31, I32 + 1, I32 + 1, I36, I37) [I34 = I35 /\ 1 = I33 /\ 19 <= I34 - 1]
  f2#(I38, I39, I40, I41, I42) -> f3#(I38, I39, 0, 0, 0) [I39 = I40 /\ I39 <= 19]
  f1#(I43, I44, I45, I46, I47) -> f2#(0, 0, 0, I48, I49) 
R = 
  init(x1, x2, x3, x4, x5) -> f1(rnd1, rnd2, rnd3, rnd4, rnd5) 
  f3(I0, I1, I2, I3, I4) -> f3(I0, I1, 1, I3, I3) [I3 = I4 /\ 0 = I2 /\ I1 <= 19 /\ I3 <= 19]
  f3(I5, I6, I7, I8, I9) -> f2(I5 + 1, I6 + 1, I6 + 1, I10, I11) [I8 = I9 /\ 0 = I7 /\ 19 <= I8 - 1 /\ I6 <= 19 /\ I5 <= 19]
  f3(I12, I13, I14, I15, I16) -> f3(I12, I13, 0, I15 + 1, I15 + 1) [I13 <= 19 /\ y1 <= y2 - 1 /\ I15 <= 19 /\ 0 = I14 /\ I15 = I16]
  f3(I17, I18, I19, I20, I21) -> f3(I17, I18, 0, I20 + 1, I20 + 1) [I18 <= 19 /\ I22 <= I23 - 1 /\ I20 <= 19 /\ 0 = I19 /\ I20 = I21]
  f3(I24, I25, I26, I27, I28) -> f2(I24, I25 + 1, I25 + 1, I29, I30) [I27 = I28 /\ 1 = I26 /\ I27 <= 19]
  f3(I31, I32, I33, I34, I35) -> f2(I31, I32 + 1, I32 + 1, I36, I37) [I34 = I35 /\ 1 = I33 /\ 19 <= I34 - 1]
  f2(I38, I39, I40, I41, I42) -> f3(I38, I39, 0, 0, 0) [I39 = I40 /\ I39 <= 19]
  f1(I43, I44, I45, I46, I47) -> f2(0, 0, 0, I48, I49) 

The dependency graph for this problem is:
  0 -> 8
  1 -> 5
  2 -> 7
  3 -> 1, 2, 3, 4
  4 -> 1, 2, 3, 4
  5 -> 7
  6 -> 7
  7 -> 1, 3, 4
  8 -> 7
Where:
  0) init#(x1, x2, x3, x4, x5) -> f1#(rnd1, rnd2, rnd3, rnd4, rnd5) 
  1) f3#(I0, I1, I2, I3, I4) -> f3#(I0, I1, 1, I3, I3) [I3 = I4 /\ 0 = I2 /\ I1 <= 19 /\ I3 <= 19]
  2) f3#(I5, I6, I7, I8, I9) -> f2#(I5 + 1, I6 + 1, I6 + 1, I10, I11) [I8 = I9 /\ 0 = I7 /\ 19 <= I8 - 1 /\ I6 <= 19 /\ I5 <= 19]
  3) f3#(I12, I13, I14, I15, I16) -> f3#(I12, I13, 0, I15 + 1, I15 + 1) [I13 <= 19 /\ y1 <= y2 - 1 /\ I15 <= 19 /\ 0 = I14 /\ I15 = I16]
  4) f3#(I17, I18, I19, I20, I21) -> f3#(I17, I18, 0, I20 + 1, I20 + 1) [I18 <= 19 /\ I22 <= I23 - 1 /\ I20 <= 19 /\ 0 = I19 /\ I20 = I21]
  5) f3#(I24, I25, I26, I27, I28) -> f2#(I24, I25 + 1, I25 + 1, I29, I30) [I27 = I28 /\ 1 = I26 /\ I27 <= 19]
  6) f3#(I31, I32, I33, I34, I35) -> f2#(I31, I32 + 1, I32 + 1, I36, I37) [I34 = I35 /\ 1 = I33 /\ 19 <= I34 - 1]
  7) f2#(I38, I39, I40, I41, I42) -> f3#(I38, I39, 0, 0, 0) [I39 = I40 /\ I39 <= 19]
  8) f1#(I43, I44, I45, I46, I47) -> f2#(0, 0, 0, I48, I49) 

We have the following SCCs.
  { 1, 2, 3, 4, 5, 7 }

DP problem for innermost termination.
P =
  f3#(I0, I1, I2, I3, I4) -> f3#(I0, I1, 1, I3, I3) [I3 = I4 /\ 0 = I2 /\ I1 <= 19 /\ I3 <= 19]
  f3#(I5, I6, I7, I8, I9) -> f2#(I5 + 1, I6 + 1, I6 + 1, I10, I11) [I8 = I9 /\ 0 = I7 /\ 19 <= I8 - 1 /\ I6 <= 19 /\ I5 <= 19]
  f3#(I12, I13, I14, I15, I16) -> f3#(I12, I13, 0, I15 + 1, I15 + 1) [I13 <= 19 /\ y1 <= y2 - 1 /\ I15 <= 19 /\ 0 = I14 /\ I15 = I16]
  f3#(I17, I18, I19, I20, I21) -> f3#(I17, I18, 0, I20 + 1, I20 + 1) [I18 <= 19 /\ I22 <= I23 - 1 /\ I20 <= 19 /\ 0 = I19 /\ I20 = I21]
  f3#(I24, I25, I26, I27, I28) -> f2#(I24, I25 + 1, I25 + 1, I29, I30) [I27 = I28 /\ 1 = I26 /\ I27 <= 19]
  f2#(I38, I39, I40, I41, I42) -> f3#(I38, I39, 0, 0, 0) [I39 = I40 /\ I39 <= 19]
R = 
  init(x1, x2, x3, x4, x5) -> f1(rnd1, rnd2, rnd3, rnd4, rnd5) 
  f3(I0, I1, I2, I3, I4) -> f3(I0, I1, 1, I3, I3) [I3 = I4 /\ 0 = I2 /\ I1 <= 19 /\ I3 <= 19]
  f3(I5, I6, I7, I8, I9) -> f2(I5 + 1, I6 + 1, I6 + 1, I10, I11) [I8 = I9 /\ 0 = I7 /\ 19 <= I8 - 1 /\ I6 <= 19 /\ I5 <= 19]
  f3(I12, I13, I14, I15, I16) -> f3(I12, I13, 0, I15 + 1, I15 + 1) [I13 <= 19 /\ y1 <= y2 - 1 /\ I15 <= 19 /\ 0 = I14 /\ I15 = I16]
  f3(I17, I18, I19, I20, I21) -> f3(I17, I18, 0, I20 + 1, I20 + 1) [I18 <= 19 /\ I22 <= I23 - 1 /\ I20 <= 19 /\ 0 = I19 /\ I20 = I21]
  f3(I24, I25, I26, I27, I28) -> f2(I24, I25 + 1, I25 + 1, I29, I30) [I27 = I28 /\ 1 = I26 /\ I27 <= 19]
  f3(I31, I32, I33, I34, I35) -> f2(I31, I32 + 1, I32 + 1, I36, I37) [I34 = I35 /\ 1 = I33 /\ 19 <= I34 - 1]
  f2(I38, I39, I40, I41, I42) -> f3(I38, I39, 0, 0, 0) [I39 = I40 /\ I39 <= 19]
  f1(I43, I44, I45, I46, I47) -> f2(0, 0, 0, I48, I49) 

We use the reverse value criterion with the projection function NU:
  NU[f2#(z1,z2,z3,z4,z5)] = 19 + -1 * z2
  NU[f3#(z1,z2,z3,z4,z5)] = 19 + -1 * (z2 + 1)

This gives the following inequalities:
  I3 = I4 /\ 0 = I2 /\ I1 <= 19 /\ I3 <= 19  ==>  19 + -1 * (I1 + 1) >= 19 + -1 * (I1 + 1)
  I8 = I9 /\ 0 = I7 /\ 19 <= I8 - 1 /\ I6 <= 19 /\ I5 <= 19  ==>  19 + -1 * (I6 + 1) >= 19 + -1 * (I6 + 1)
  I13 <= 19 /\ y1 <= y2 - 1 /\ I15 <= 19 /\ 0 = I14 /\ I15 = I16  ==>  19 + -1 * (I13 + 1) >= 19 + -1 * (I13 + 1)
  I18 <= 19 /\ I22 <= I23 - 1 /\ I20 <= 19 /\ 0 = I19 /\ I20 = I21  ==>  19 + -1 * (I18 + 1) >= 19 + -1 * (I18 + 1)
  I27 = I28 /\ 1 = I26 /\ I27 <= 19  ==>  19 + -1 * (I25 + 1) >= 19 + -1 * (I25 + 1)
  I39 = I40 /\ I39 <= 19  ==>  19 + -1 * I39 > 19 + -1 * (I39 + 1)  with  19 + -1 * I39 >= 0

We remove all the strictly oriented dependency pairs.

DP problem for innermost termination.
P =
  f3#(I0, I1, I2, I3, I4) -> f3#(I0, I1, 1, I3, I3) [I3 = I4 /\ 0 = I2 /\ I1 <= 19 /\ I3 <= 19]
  f3#(I5, I6, I7, I8, I9) -> f2#(I5 + 1, I6 + 1, I6 + 1, I10, I11) [I8 = I9 /\ 0 = I7 /\ 19 <= I8 - 1 /\ I6 <= 19 /\ I5 <= 19]
  f3#(I12, I13, I14, I15, I16) -> f3#(I12, I13, 0, I15 + 1, I15 + 1) [I13 <= 19 /\ y1 <= y2 - 1 /\ I15 <= 19 /\ 0 = I14 /\ I15 = I16]
  f3#(I17, I18, I19, I20, I21) -> f3#(I17, I18, 0, I20 + 1, I20 + 1) [I18 <= 19 /\ I22 <= I23 - 1 /\ I20 <= 19 /\ 0 = I19 /\ I20 = I21]
  f3#(I24, I25, I26, I27, I28) -> f2#(I24, I25 + 1, I25 + 1, I29, I30) [I27 = I28 /\ 1 = I26 /\ I27 <= 19]
R = 
  init(x1, x2, x3, x4, x5) -> f1(rnd1, rnd2, rnd3, rnd4, rnd5) 
  f3(I0, I1, I2, I3, I4) -> f3(I0, I1, 1, I3, I3) [I3 = I4 /\ 0 = I2 /\ I1 <= 19 /\ I3 <= 19]
  f3(I5, I6, I7, I8, I9) -> f2(I5 + 1, I6 + 1, I6 + 1, I10, I11) [I8 = I9 /\ 0 = I7 /\ 19 <= I8 - 1 /\ I6 <= 19 /\ I5 <= 19]
  f3(I12, I13, I14, I15, I16) -> f3(I12, I13, 0, I15 + 1, I15 + 1) [I13 <= 19 /\ y1 <= y2 - 1 /\ I15 <= 19 /\ 0 = I14 /\ I15 = I16]
  f3(I17, I18, I19, I20, I21) -> f3(I17, I18, 0, I20 + 1, I20 + 1) [I18 <= 19 /\ I22 <= I23 - 1 /\ I20 <= 19 /\ 0 = I19 /\ I20 = I21]
  f3(I24, I25, I26, I27, I28) -> f2(I24, I25 + 1, I25 + 1, I29, I30) [I27 = I28 /\ 1 = I26 /\ I27 <= 19]
  f3(I31, I32, I33, I34, I35) -> f2(I31, I32 + 1, I32 + 1, I36, I37) [I34 = I35 /\ 1 = I33 /\ 19 <= I34 - 1]
  f2(I38, I39, I40, I41, I42) -> f3(I38, I39, 0, 0, 0) [I39 = I40 /\ I39 <= 19]
  f1(I43, I44, I45, I46, I47) -> f2(0, 0, 0, I48, I49) 

The dependency graph for this problem is:
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