/export/starexec/sandbox2/solver/bin/starexec_run_Transition /export/starexec/sandbox2/benchmark/theBenchmark.smt2 /export/starexec/sandbox2/output/output_files


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MAYBE

DP problem for innermost termination.
P =
  f7#(x1, x2, x3) -> f6#(x1, x2, x3) 
  f6#(I0, I1, I2) -> f1#(2, I1, I2) 
  f2#(I3, I4, I5) -> f3#(I3, I3, I5) [I3 <= 10]
  f4#(I9, I10, I11) -> f1#(1 + I9, I10, I11) 
  f4#(I12, I13, I14) -> f3#(I12, -1 + I13, rnd3) [rnd3 = rnd3]
  f3#(I15, I16, I17) -> f4#(I15, I16, I17) 
  f1#(I18, I19, I20) -> f2#(I18, I19, I20) 
R = 
  f7(x1, x2, x3) -> f6(x1, x2, x3) 
  f6(I0, I1, I2) -> f1(2, I1, I2) 
  f2(I3, I4, I5) -> f3(I3, I3, I5) [I3 <= 10]
  f2(I6, I7, I8) -> f5(I6, I7, I8) [11 <= I6]
  f4(I9, I10, I11) -> f1(1 + I9, I10, I11) 
  f4(I12, I13, I14) -> f3(I12, -1 + I13, rnd3) [rnd3 = rnd3]
  f3(I15, I16, I17) -> f4(I15, I16, I17) 
  f1(I18, I19, I20) -> f2(I18, I19, I20) 

The dependency graph for this problem is:
  0 -> 1
  1 -> 6
  2 -> 5
  3 -> 6
  4 -> 5
  5 -> 3, 4
  6 -> 2
Where:
  0) f7#(x1, x2, x3) -> f6#(x1, x2, x3) 
  1) f6#(I0, I1, I2) -> f1#(2, I1, I2) 
  2) f2#(I3, I4, I5) -> f3#(I3, I3, I5) [I3 <= 10]
  3) f4#(I9, I10, I11) -> f1#(1 + I9, I10, I11) 
  4) f4#(I12, I13, I14) -> f3#(I12, -1 + I13, rnd3) [rnd3 = rnd3]
  5) f3#(I15, I16, I17) -> f4#(I15, I16, I17) 
  6) f1#(I18, I19, I20) -> f2#(I18, I19, I20) 

We have the following SCCs.
  { 2, 3, 4, 5, 6 }

DP problem for innermost termination.
P =
  f2#(I3, I4, I5) -> f3#(I3, I3, I5) [I3 <= 10]
  f4#(I9, I10, I11) -> f1#(1 + I9, I10, I11) 
  f4#(I12, I13, I14) -> f3#(I12, -1 + I13, rnd3) [rnd3 = rnd3]
  f3#(I15, I16, I17) -> f4#(I15, I16, I17) 
  f1#(I18, I19, I20) -> f2#(I18, I19, I20) 
R = 
  f7(x1, x2, x3) -> f6(x1, x2, x3) 
  f6(I0, I1, I2) -> f1(2, I1, I2) 
  f2(I3, I4, I5) -> f3(I3, I3, I5) [I3 <= 10]
  f2(I6, I7, I8) -> f5(I6, I7, I8) [11 <= I6]
  f4(I9, I10, I11) -> f1(1 + I9, I10, I11) 
  f4(I12, I13, I14) -> f3(I12, -1 + I13, rnd3) [rnd3 = rnd3]
  f3(I15, I16, I17) -> f4(I15, I16, I17) 
  f1(I18, I19, I20) -> f2(I18, I19, I20) 

We use the extended value criterion with the projection function NU:
  NU[f1#(x0,x1,x2)] = -x0 + 10
  NU[f4#(x0,x1,x2)] = -x0 + 9
  NU[f3#(x0,x1,x2)] = -x0 + 9
  NU[f2#(x0,x1,x2)] = -x0 + 10

This gives the following inequalities:
  I3 <= 10  ==>  -I3 + 10 > -I3 + 9  with  -I3 + 10 >= 0
    ==>  -I9 + 9 >= -(1 + I9) + 10
  rnd3 = rnd3  ==>  -I12 + 9 >= -I12 + 9
    ==>  -I15 + 9 >= -I15 + 9
    ==>  -I18 + 10 >= -I18 + 10

We remove all the strictly oriented dependency pairs.

DP problem for innermost termination.
P =
  f4#(I9, I10, I11) -> f1#(1 + I9, I10, I11) 
  f4#(I12, I13, I14) -> f3#(I12, -1 + I13, rnd3) [rnd3 = rnd3]
  f3#(I15, I16, I17) -> f4#(I15, I16, I17) 
  f1#(I18, I19, I20) -> f2#(I18, I19, I20) 
R = 
  f7(x1, x2, x3) -> f6(x1, x2, x3) 
  f6(I0, I1, I2) -> f1(2, I1, I2) 
  f2(I3, I4, I5) -> f3(I3, I3, I5) [I3 <= 10]
  f2(I6, I7, I8) -> f5(I6, I7, I8) [11 <= I6]
  f4(I9, I10, I11) -> f1(1 + I9, I10, I11) 
  f4(I12, I13, I14) -> f3(I12, -1 + I13, rnd3) [rnd3 = rnd3]
  f3(I15, I16, I17) -> f4(I15, I16, I17) 
  f1(I18, I19, I20) -> f2(I18, I19, I20) 

The dependency graph for this problem is:
  3 -> 6
  4 -> 5
  5 -> 3, 4
  6 -> 
Where:
  3) f4#(I9, I10, I11) -> f1#(1 + I9, I10, I11) 
  4) f4#(I12, I13, I14) -> f3#(I12, -1 + I13, rnd3) [rnd3 = rnd3]
  5) f3#(I15, I16, I17) -> f4#(I15, I16, I17) 
  6) f1#(I18, I19, I20) -> f2#(I18, I19, I20) 

We have the following SCCs.
  { 4, 5 }

DP problem for innermost termination.
P =
  f4#(I12, I13, I14) -> f3#(I12, -1 + I13, rnd3) [rnd3 = rnd3]
  f3#(I15, I16, I17) -> f4#(I15, I16, I17) 
R = 
  f7(x1, x2, x3) -> f6(x1, x2, x3) 
  f6(I0, I1, I2) -> f1(2, I1, I2) 
  f2(I3, I4, I5) -> f3(I3, I3, I5) [I3 <= 10]
  f2(I6, I7, I8) -> f5(I6, I7, I8) [11 <= I6]
  f4(I9, I10, I11) -> f1(1 + I9, I10, I11) 
  f4(I12, I13, I14) -> f3(I12, -1 + I13, rnd3) [rnd3 = rnd3]
  f3(I15, I16, I17) -> f4(I15, I16, I17) 
  f1(I18, I19, I20) -> f2(I18, I19, I20)