/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 =
  f4#(x1, x2, x3) -> f3#(x1, x2, x3) 
  f3#(I0, I1, I2) -> f1#(rnd1, I1, I2) [y1 = -1 /\ y2 = -1 + y1 /\ y3 = -1 + y2 /\ y4 = -1 + y3 /\ y5 = -1 + y4 /\ y6 = -1 + y5 /\ y7 = -1 + y6 /\ y8 = -1 + y7 /\ y9 = -1 + y8 /\ rnd1 = -1 + y9 /\ -1 <= I1 /\ rnd1 <= I2]
  f2#(I3, I4, I5) -> f1#(I3, I4, I5) 
  f1#(I6, I7, I8) -> f2#(I6, I7 + I8, I8) [I6 <= I8 /\ 0 <= I7]
R = 
  f4(x1, x2, x3) -> f3(x1, x2, x3) 
  f3(I0, I1, I2) -> f1(rnd1, I1, I2) [y1 = -1 /\ y2 = -1 + y1 /\ y3 = -1 + y2 /\ y4 = -1 + y3 /\ y5 = -1 + y4 /\ y6 = -1 + y5 /\ y7 = -1 + y6 /\ y8 = -1 + y7 /\ y9 = -1 + y8 /\ rnd1 = -1 + y9 /\ -1 <= I1 /\ rnd1 <= I2]
  f2(I3, I4, I5) -> f1(I3, I4, I5) 
  f1(I6, I7, I8) -> f2(I6, I7 + I8, I8) [I6 <= I8 /\ 0 <= I7]

The dependency graph for this problem is:
  0 -> 1
  1 -> 3
  2 -> 3
  3 -> 2
Where:
  0) f4#(x1, x2, x3) -> f3#(x1, x2, x3) 
  1) f3#(I0, I1, I2) -> f1#(rnd1, I1, I2) [y1 = -1 /\ y2 = -1 + y1 /\ y3 = -1 + y2 /\ y4 = -1 + y3 /\ y5 = -1 + y4 /\ y6 = -1 + y5 /\ y7 = -1 + y6 /\ y8 = -1 + y7 /\ y9 = -1 + y8 /\ rnd1 = -1 + y9 /\ -1 <= I1 /\ rnd1 <= I2]
  2) f2#(I3, I4, I5) -> f1#(I3, I4, I5) 
  3) f1#(I6, I7, I8) -> f2#(I6, I7 + I8, I8) [I6 <= I8 /\ 0 <= I7]

We have the following SCCs.
  { 2, 3 }

DP problem for innermost termination.
P =
  f2#(I3, I4, I5) -> f1#(I3, I4, I5) 
  f1#(I6, I7, I8) -> f2#(I6, I7 + I8, I8) [I6 <= I8 /\ 0 <= I7]
R = 
  f4(x1, x2, x3) -> f3(x1, x2, x3) 
  f3(I0, I1, I2) -> f1(rnd1, I1, I2) [y1 = -1 /\ y2 = -1 + y1 /\ y3 = -1 + y2 /\ y4 = -1 + y3 /\ y5 = -1 + y4 /\ y6 = -1 + y5 /\ y7 = -1 + y6 /\ y8 = -1 + y7 /\ y9 = -1 + y8 /\ rnd1 = -1 + y9 /\ -1 <= I1 /\ rnd1 <= I2]
  f2(I3, I4, I5) -> f1(I3, I4, I5) 
  f1(I6, I7, I8) -> f2(I6, I7 + I8, I8) [I6 <= I8 /\ 0 <= I7]