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


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MAYBE

DP problem for innermost termination.
P =
  f5#(x1, x2, x3, x4) -> f4#(x1, x2, x3, x4) 
  f4#(I0, I1, I2, I3) -> f1#(I0, I1, I2, I3) 
  f3#(I4, I5, I6, I7) -> f1#(I4, I5, I6, I7) 
  f1#(I8, I9, I10, I11) -> f3#(-1 + I8, I9 + I10, I10 + I11, I8 + I11) [1 <= I9]
  f2#(I12, I13, I14, I15) -> f1#(I12, I13, I14, I15) 
  f1#(I16, I17, I18, I19) -> f2#(I16, -1 + I17, I18, I19) [1 <= I17]
R = 
  f5(x1, x2, x3, x4) -> f4(x1, x2, x3, x4) 
  f4(I0, I1, I2, I3) -> f1(I0, I1, I2, I3) 
  f3(I4, I5, I6, I7) -> f1(I4, I5, I6, I7) 
  f1(I8, I9, I10, I11) -> f3(-1 + I8, I9 + I10, I10 + I11, I8 + I11) [1 <= I9]
  f2(I12, I13, I14, I15) -> f1(I12, I13, I14, I15) 
  f1(I16, I17, I18, I19) -> f2(I16, -1 + I17, I18, I19) [1 <= I17]

The dependency graph for this problem is:
  0 -> 1
  1 -> 3, 5
  2 -> 3, 5
  3 -> 2
  4 -> 3, 5
  5 -> 4
Where:
  0) f5#(x1, x2, x3, x4) -> f4#(x1, x2, x3, x4) 
  1) f4#(I0, I1, I2, I3) -> f1#(I0, I1, I2, I3) 
  2) f3#(I4, I5, I6, I7) -> f1#(I4, I5, I6, I7) 
  3) f1#(I8, I9, I10, I11) -> f3#(-1 + I8, I9 + I10, I10 + I11, I8 + I11) [1 <= I9]
  4) f2#(I12, I13, I14, I15) -> f1#(I12, I13, I14, I15) 
  5) f1#(I16, I17, I18, I19) -> f2#(I16, -1 + I17, I18, I19) [1 <= I17]

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

DP problem for innermost termination.
P =
  f3#(I4, I5, I6, I7) -> f1#(I4, I5, I6, I7) 
  f1#(I8, I9, I10, I11) -> f3#(-1 + I8, I9 + I10, I10 + I11, I8 + I11) [1 <= I9]
  f2#(I12, I13, I14, I15) -> f1#(I12, I13, I14, I15) 
  f1#(I16, I17, I18, I19) -> f2#(I16, -1 + I17, I18, I19) [1 <= I17]
R = 
  f5(x1, x2, x3, x4) -> f4(x1, x2, x3, x4) 
  f4(I0, I1, I2, I3) -> f1(I0, I1, I2, I3) 
  f3(I4, I5, I6, I7) -> f1(I4, I5, I6, I7) 
  f1(I8, I9, I10, I11) -> f3(-1 + I8, I9 + I10, I10 + I11, I8 + I11) [1 <= I9]
  f2(I12, I13, I14, I15) -> f1(I12, I13, I14, I15) 
  f1(I16, I17, I18, I19) -> f2(I16, -1 + I17, I18, I19) [1 <= I17]