/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) -> f2#(0, I1, I2) 
  f2#(I3, I4, I5) -> f4#(1, I4, I4) 
  f4#(I9, I10, I11) -> f1#(0, 1 + I10, I11) 
  f4#(I12, I13, I14) -> f1#(I12, I13, I14) [I13 <= I13]
  f1#(I18, I19, I20) -> f2#(I18, I19, I20) [1 + I20 <= I19]
R = 
  f7(x1, x2, x3) -> f6(x1, x2, x3) 
  f6(I0, I1, I2) -> f2(0, I1, I2) 
  f2(I3, I4, I5) -> f4(1, I4, I4) 
  f2(I6, I7, I8) -> f5(I6, I7, I8) [1 <= I6 /\ I6 <= 1]
  f4(I9, I10, I11) -> f1(0, 1 + I10, I11) 
  f4(I12, I13, I14) -> f1(I12, I13, I14) [I13 <= I13]
  f1(I15, I16, I17) -> f3(I15, I16, I17) [I16 <= I17]
  f1(I18, I19, I20) -> f2(I18, I19, I20) [1 + I20 <= I19]

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

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

DP problem for innermost termination.
P =
  f2#(I3, I4, I5) -> f4#(1, I4, I4) 
  f4#(I9, I10, I11) -> f1#(0, 1 + I10, I11) 
  f4#(I12, I13, I14) -> f1#(I12, I13, I14) [I13 <= I13]
  f1#(I18, I19, I20) -> f2#(I18, I19, I20) [1 + I20 <= I19]
R = 
  f7(x1, x2, x3) -> f6(x1, x2, x3) 
  f6(I0, I1, I2) -> f2(0, I1, I2) 
  f2(I3, I4, I5) -> f4(1, I4, I4) 
  f2(I6, I7, I8) -> f5(I6, I7, I8) [1 <= I6 /\ I6 <= 1]
  f4(I9, I10, I11) -> f1(0, 1 + I10, I11) 
  f4(I12, I13, I14) -> f1(I12, I13, I14) [I13 <= I13]
  f1(I15, I16, I17) -> f3(I15, I16, I17) [I16 <= I17]
  f1(I18, I19, I20) -> f2(I18, I19, I20) [1 + I20 <= I19]