Integer Transition Systems pair #487528278

details
property	value
status	complete
benchmark	IntListDupRec.jar-obl-9.smt2
ran by	Akihisa Yamada
cpu timeout	1200 seconds
wallclock timeout	300 seconds
memory limit	137438953472 bytes
execution host	n027.star.cs.uiowa.edu
space	From_AProVE_2014
run statistics
property	value
solver	Ctrl
configuration	Transition
runtime (wallclock)	1.08862185478 seconds
cpu usage	1.151667309
max memory	1.2161024E7
stage attributes
key	value
output-size	10558
starexec-result	YES
output
/export/starexec/sandbox/solver/bin/starexec_run_Transition /export/starexec/sandbox/benchmark/theBenchmark.smt2 /export/starexec/sandbox/output/output_files


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YES

DP problem for innermost termination.
P =
  init#(x1, x2, x3) -> f1#(rnd1, rnd2, rnd3) 
  f4#(I0, I1, I2) -> f4#(I3, I4, 1) [0 = I2 /\ 0 <= I4 - 1 /\ 2 <= I3 - 1 /\ 0 <= I1 - 1 /\ 2 <= I0 - 1 /\ I4 <= I1 /\ I4 + 2 <= I0 /\ I3 - 2 <= I1 /\ I3 <= I0]
  f4#(I5, I6, I7) -> f4#(I8, I9, I10) [I10 + 2 <= I6 /\ I7 + 2 <= I5 /\ I10 + 4 <= I5 /\ -1 <= I9 - 1 /\ 0 <= I8 - 1 /\ 0 <= I6 - 1 /\ 2 <= I5 - 1 /\ I9 + 1 <= I6 /\ I9 + 3 <= I5 /\ I8 <= I6 /\ 0 <= I7 - 1 /\ I8 + 2 <= I5]
  f4#(I11, I12, I13) -> f4#(I14, I15, I16) [I16 + 2 <= I12 /\ I13 + 2 <= I11 /\ I16 + 4 <= I11 /\ -1 <= I15 - 1 /\ 0 <= I14 - 1 /\ 0 <= I12 - 1 /\ 2 <= I11 - 1 /\ I15 + 1 <= I12 /\ I15 + 3 <= I11 /\ I14 <= I12 /\ I13 <= -1 /\ I14 + 2 <= I11]
  f3#(I17, I18, I19) -> f4#(I20, I21, 0) [I20 <= I18 /\ 0 <= y1 - 1 /\ I21 + 2 <= I18 /\ 0 <= I17 - 1 /\ 1 <= I18 - 1 /\ 1 <= I20 - 1 /\ -1 <= I21 - 1 /\ 0 = I19]
  f2#(I22, I23, I24) -> f2#(I22 - 1, I25, I26) [I22 - 1 <= I22 - 1 /\ -1 <= I22 - 1]
  f1#(I27, I28, I29) -> f3#(I30, I31, 0) [-1 <= I32 - 1 /\ 0 <= I28 - 1 /\ I30 <= I27 /\ I31 - 1 <= I27 /\ 0 <= I27 - 1 /\ 0 <= I30 - 1 /\ 1 <= I31 - 1]
  f1#(I33, I34, I35) -> f3#(I36, I37, 0) [-1 <= I38 - 1 /\ 0 <= I34 - 1 /\ I36 <= I33 /\ 0 <= I33 - 1 /\ 0 <= I36 - 1 /\ 3 <= I37 - 1]
  f1#(I39, I40, I41) -> f2#(I42, I43, I44) [0 <= I39 - 1 /\ 0 <= I40 - 1 /\ -1 <= I42 - 1]
R = 
  init(x1, x2, x3) -> f1(rnd1, rnd2, rnd3) 
  f4(I0, I1, I2) -> f4(I3, I4, 1) [0 = I2 /\ 0 <= I4 - 1 /\ 2 <= I3 - 1 /\ 0 <= I1 - 1 /\ 2 <= I0 - 1 /\ I4 <= I1 /\ I4 + 2 <= I0 /\ I3 - 2 <= I1 /\ I3 <= I0]
  f4(I5, I6, I7) -> f4(I8, I9, I10) [I10 + 2 <= I6 /\ I7 + 2 <= I5 /\ I10 + 4 <= I5 /\ -1 <= I9 - 1 /\ 0 <= I8 - 1 /\ 0 <= I6 - 1 /\ 2 <= I5 - 1 /\ I9 + 1 <= I6 /\ I9 + 3 <= I5 /\ I8 <= I6 /\ 0 <= I7 - 1 /\ I8 + 2 <= I5]
  f4(I11, I12, I13) -> f4(I14, I15, I16) [I16 + 2 <= I12 /\ I13 + 2 <= I11 /\ I16 + 4 <= I11 /\ -1 <= I15 - 1 /\ 0 <= I14 - 1 /\ 0 <= I12 - 1 /\ 2 <= I11 - 1 /\ I15 + 1 <= I12 /\ I15 + 3 <= I11 /\ I14 <= I12 /\ I13 <= -1 /\ I14 + 2 <= I11]
  f3(I17, I18, I19) -> f4(I20, I21, 0) [I20 <= I18 /\ 0 <= y1 - 1 /\ I21 + 2 <= I18 /\ 0 <= I17 - 1 /\ 1 <= I18 - 1 /\ 1 <= I20 - 1 /\ -1 <= I21 - 1 /\ 0 = I19]
  f2(I22, I23, I24) -> f2(I22 - 1, I25, I26) [I22 - 1 <= I22 - 1 /\ -1 <= I22 - 1]
  f1(I27, I28, I29) -> f3(I30, I31, 0) [-1 <= I32 - 1 /\ 0 <= I28 - 1 /\ I30 <= I27 /\ I31 - 1 <= I27 /\ 0 <= I27 - 1 /\ 0 <= I30 - 1 /\ 1 <= I31 - 1]
  f1(I33, I34, I35) -> f3(I36, I37, 0) [-1 <= I38 - 1 /\ 0 <= I34 - 1 /\ I36 <= I33 /\ 0 <= I33 - 1 /\ 0 <= I36 - 1 /\ 3 <= I37 - 1]
  f1(I39, I40, I41) -> f2(I42, I43, I44) [0 <= I39 - 1 /\ 0 <= I40 - 1 /\ -1 <= I42 - 1]

The dependency graph for this problem is:
  0 -> 6, 7, 8
  1 -> 2
  2 -> 1, 2, 3
  3 -> 1, 2, 3
  4 -> 1
  5 -> 5
  6 -> 4
  7 -> 4
  8 -> 5
Where:
  0) init#(x1, x2, x3) -> f1#(rnd1, rnd2, rnd3) 
  1) f4#(I0, I1, I2) -> f4#(I3, I4, 1) [0 = I2 /\ 0 <= I4 - 1 /\ 2 <= I3 - 1 /\ 0 <= I1 - 1 /\ 2 <= I0 - 1 /\ I4 <= I1 /\ I4 + 2 <= I0 /\ I3 - 2 <= I1 /\ I3 <= I0]
  2) f4#(I5, I6, I7) -> f4#(I8, I9, I10) [I10 + 2 <= I6 /\ I7 + 2 <= I5 /\ I10 + 4 <= I5 /\ -1 <= I9 - 1 /\ 0 <= I8 - 1 /\ 0 <= I6 - 1 /\ 2 <= I5 - 1 /\ I9 + 1 <= I6 /\ I9 + 3 <= I5 /\ I8 <= I6 /\ 0 <= I7 - 1 /\ I8 + 2 <= I5]
  3) f4#(I11, I12, I13) -> f4#(I14, I15, I16) [I16 + 2 <= I12 /\ I13 + 2 <= I11 /\ I16 + 4 <= I11 /\ -1 <= I15 - 1 /\ 0 <= I14 - 1 /\ 0 <= I12 - 1 /\ 2 <= I11 - 1 /\ I15 + 1 <= I12 /\ I15 + 3 <= I11 /\ I14 <= I12 /\ I13 <= -1 /\ I14 + 2 <= I11]
  4) f3#(I17, I18, I19) -> f4#(I20, I21, 0) [I20 <= I18 /\ 0 <= y1 - 1 /\ I21 + 2 <= I18 /\ 0 <= I17 - 1 /\ 1 <= I18 - 1 /\ 1 <= I20 - 1 /\ -1 <= I21 - 1 /\ 0 = I19]
  5) f2#(I22, I23, I24) -> f2#(I22 - 1, I25, I26) [I22 - 1 <= I22 - 1 /\ -1 <= I22 - 1]
  6) f1#(I27, I28, I29) -> f3#(I30, I31, 0) [-1 <= I32 - 1 /\ 0 <= I28 - 1 /\ I30 <= I27 /\ I31 - 1 <= I27 /\ 0 <= I27 - 1 /\ 0 <= I30 - 1 /\ 1 <= I31 - 1]
  7) f1#(I33, I34, I35) -> f3#(I36, I37, 0) [-1 <= I38 - 1 /\ 0 <= I34 - 1 /\ I36 <= I33 /\ 0 <= I33 - 1 /\ 0 <= I36 - 1 /\ 3 <= I37 - 1]
  8) f1#(I39, I40, I41) -> f2#(I42, I43, I44) [0 <= I39 - 1 /\ 0 <= I40 - 1 /\ -1 <= I42 - 1]

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

DP problem for innermost termination.
P =
  f2#(I22, I23, I24) -> f2#(I22 - 1, I25, I26) [I22 - 1 <= I22 - 1 /\ -1 <= I22 - 1]
R = 
  init(x1, x2, x3) -> f1(rnd1, rnd2, rnd3) 
  f4(I0, I1, I2) -> f4(I3, I4, 1) [0 = I2 /\ 0 <= I4 - 1 /\ 2 <= I3 - 1 /\ 0 <= I1 - 1 /\ 2 <= I0 - 1 /\ I4 <= I1 /\ I4 + 2 <= I0 /\ I3 - 2 <= I1 /\ I3 <= I0]
  f4(I5, I6, I7) -> f4(I8, I9, I10) [I10 + 2 <= I6 /\ I7 + 2 <= I5 /\ I10 + 4 <= I5 /\ -1 <= I9 - 1 /\ 0 <= I8 - 1 /\ 0 <= I6 - 1 /\ 2 <= I5 - 1 /\ I9 + 1 <= I6 /\ I9 + 3 <= I5 /\ I8 <= I6 /\ 0 <= I7 - 1 /\ I8 + 2 <= I5]
  f4(I11, I12, I13) -> f4(I14, I15, I16) [I16 + 2 <= I12 /\ I13 + 2 <= I11 /\ I16 + 4 <= I11 /\ -1 <= I15 - 1 /\ 0 <= I14 - 1 /\ 0 <= I12 - 1 /\ 2 <= I11 - 1 /\ I15 + 1 <= I12 /\ I15 + 3 <= I11 /\ I14 <= I12 /\ I13 <= -1 /\ I14 + 2 <= I11]
  f3(I17, I18, I19) -> f4(I20, I21, 0) [I20 <= I18 /\ 0 <= y1 - 1 /\ I21 + 2 <= I18 /\ 0 <= I17 - 1 /\ 1 <= I18 - 1 /\ 1 <= I20 - 1 /\ -1 <= I21 - 1 /\ 0 = I19]
  f2(I22, I23, I24) -> f2(I22 - 1, I25, I26) [I22 - 1 <= I22 - 1 /\ -1 <= I22 - 1]
  f1(I27, I28, I29) -> f3(I30, I31, 0) [-1 <= I32 - 1 /\ 0 <= I28 - 1 /\ I30 <= I27 /\ I31 - 1 <= I27 /\ 0 <= I27 - 1 /\ 0 <= I30 - 1 /\ 1 <= I31 - 1]
  f1(I33, I34, I35) -> f3(I36, I37, 0) [-1 <= I38 - 1 /\ 0 <= I34 - 1 /\ I36 <= I33 /\ 0 <= I33 - 1 /\ 0 <= I36 - 1 /\ 3 <= I37 - 1]
  f1(I39, I40, I41) -> f2(I42, I43, I44) [0 <= I39 - 1 /\ 0 <= I40 - 1 /\ -1 <= I42 - 1]

We use the basic value criterion with the projection function NU:
  NU[f2#(z1,z2,z3)] = z1

This gives the following inequalities:
  I22 - 1 <= I22 - 1 /\ -1 <= I22 - 1  ==>  I22 >! I22 - 1

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

DP problem for innermost termination.
P =
  f4#(I0, I1, I2) -> f4#(I3, I4, 1) [0 = I2 /\ 0 <= I4 - 1 /\ 2 <= I3 - 1 /\ 0 <= I1 - 1 /\ 2 <= I0 - 1 /\ I4 <= I1 /\ I4 + 2 <= I0 /\ I3 - 2 <= I1 /\ I3 <= I0]
  f4#(I5, I6, I7) -> f4#(I8, I9, I10) [I10 + 2 <= I6 /\ I7 + 2 <= I5 /\ I10 + 4 <= I5 /\ -1 <= I9 - 1 /\ 0 <= I8 - 1 /\ 0 <= I6 - 1 /\ 2 <= I5 - 1 /\ I9 + 1 <= I6 /\ I9 + 3 <= I5 /\ I8 <= I6 /\ 0 <= I7 - 1 /\ I8 + 2 <= I5]
  f4#(I11, I12, I13) -> f4#(I14, I15, I16) [I16 + 2 <= I12 /\ I13 + 2 <= I11 /\ I16 + 4 <= I11 /\ -1 <= I15 - 1 /\ 0 <= I14 - 1 /\ 0 <= I12 - 1 /\ 2 <= I11 - 1 /\ I15 + 1 <= I12 /\ I15 + 3 <= I11 /\ I14 <= I12 /\ I13 <= -1 /\ I14 + 2 <= I11]
R = 
  init(x1, x2, x3) -> f1(rnd1, rnd2, rnd3) 
  f4(I0, I1, I2) -> f4(I3, I4, 1) [0 = I2 /\ 0 <= I4 - 1 /\ 2 <= I3 - 1 /\ 0 <= I1 - 1 /\ 2 <= I0 - 1 /\ I4 <= I1 /\ I4 + 2 <= I0 /\ I3 - 2 <= I1 /\ I3 <= I0]
  f4(I5, I6, I7) -> f4(I8, I9, I10) [I10 + 2 <= I6 /\ I7 + 2 <= I5 /\ I10 + 4 <= I5 /\ -1 <= I9 - 1 /\ 0 <= I8 - 1 /\ 0 <= I6 - 1 /\ 2 <= I5 - 1 /\ I9 + 1 <= I6 /\ I9 + 3 <= I5 /\ I8 <= I6 /\ 0 <= I7 - 1 /\ I8 + 2 <= I5]
  f4(I11, I12, I13) -> f4(I14, I15, I16) [I16 + 2 <= I12 /\ I13 + 2 <= I11 /\ I16 + 4 <= I11 /\ -1 <= I15 - 1 /\ 0 <= I14 - 1 /\ 0 <= I12 - 1 /\ 2 <= I11 - 1 /\ I15 + 1 <= I12 /\ I15 + 3 <= I11 /\ I14 <= I12 /\ I13 <= -1 /\ I14 + 2 <= I11]
  f3(I17, I18, I19) -> f4(I20, I21, 0) [I20 <= I18 /\ 0 <= y1 - 1 /\ I21 + 2 <= I18 /\ 0 <= I17 - 1 /\ 1 <= I18 - 1 /\ 1 <= I20 - 1 /\ -1 <= I21 - 1 /\ 0 = I19]
  f2(I22, I23, I24) -> f2(I22 - 1, I25, I26) [I22 - 1 <= I22 - 1 /\ -1 <= I22 - 1]
  f1(I27, I28, I29) -> f3(I30, I31, 0) [-1 <= I32 - 1 /\ 0 <= I28 - 1 /\ I30 <= I27 /\ I31 - 1 <= I27 /\ 0 <= I27 - 1 /\ 0 <= I30 - 1 /\ 1 <= I31 - 1]
  f1(I33, I34, I35) -> f3(I36, I37, 0) [-1 <= I38 - 1 /\ 0 <= I34 - 1 /\ I36 <= I33 /\ 0 <= I33 - 1 /\ 0 <= I36 - 1 /\ 3 <= I37 - 1]
  f1(I39, I40, I41) -> f2(I42, I43, I44) [0 <= I39 - 1 /\ 0 <= I40 - 1 /\ -1 <= I42 - 1]

We use the basic value criterion with the projection function NU:
  NU[f4#(z1,z2,z3)] = z2

This gives the following inequalities:
  0 = I2 /\ 0 <= I4 - 1 /\ 2 <= I3 - 1 /\ 0 <= I1 - 1 /\ 2 <= I0 - 1 /\ I4 <= I1 /\ I4 + 2 <= I0 /\ I3 - 2 <= I1 /\ I3 <= I0  ==>  I1 (>! \union =) I4
  I10 + 2 <= I6 /\ I7 + 2 <= I5 /\ I10 + 4 <= I5 /\ -1 <= I9 - 1 /\ 0 <= I8 - 1 /\ 0 <= I6 - 1 /\ 2 <= I5 - 1 /\ I9 + 1 <= I6 /\ I9 + 3 <= I5 /\ I8 <= I6 /\ 0 <= I7 - 1 /\ I8 + 2 <= I5  ==>  I6 >! I9
  I16 + 2 <= I12 /\ I13 + 2 <= I11 /\ I16 + 4 <= I11 /\ -1 <= I15 - 1 /\ 0 <= I14 - 1 /\ 0 <= I12 - 1 /\ 2 <= I11 - 1 /\ I15 + 1 <= I12 /\ I15 + 3 <= I11 /\ I14 <= I12 /\ I13 <= -1 /\ I14 + 2 <= I11  ==>  I12 >! I15
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