/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 =
  f10#(x1, x2, x3, x4) -> f9#(x1, x2, x3, x4) 
  f9#(I0, I1, I2, I3) -> f4#(0, 0, rnd3, I3) [rnd3 = rnd3]
  f5#(I4, I5, I6, I7) -> f4#(I4, 0, I8, I7) [I7 <= 0 /\ y1 = 1 /\ I8 = I8]
  f5#(I9, I10, I11, I12) -> f2#(I9, I10, I11, I12) [1 <= I12]
  f8#(I13, I14, I15, I16) -> f3#(I13, I14, I15, I16) 
  f3#(I17, I18, I19, I20) -> f8#(I17, I18, I19, I20) 
  f2#(I25, I26, I27, I28) -> f5#(I25, I26, I27, I28) 
  f4#(I29, I30, I31, I32) -> f1#(I29, I30, I31, I32) 
  f1#(I33, I34, I35, I36) -> f3#(I33, I34, I35, I36) [1 <= I35]
  f1#(I37, I38, I39, I40) -> f2#(0, I38, I39, rnd4) [I39 <= 0 /\ I41 = 1 /\ rnd4 = rnd4 /\ 1 <= rnd4]
R = 
  f10(x1, x2, x3, x4) -> f9(x1, x2, x3, x4) 
  f9(I0, I1, I2, I3) -> f4(0, 0, rnd3, I3) [rnd3 = rnd3]
  f5(I4, I5, I6, I7) -> f4(I4, 0, I8, I7) [I7 <= 0 /\ y1 = 1 /\ I8 = I8]
  f5(I9, I10, I11, I12) -> f2(I9, I10, I11, I12) [1 <= I12]
  f8(I13, I14, I15, I16) -> f3(I13, I14, I15, I16) 
  f3(I17, I18, I19, I20) -> f8(I17, I18, I19, I20) 
  f6(I21, I22, I23, I24) -> f7(I21, I22, I23, I24) 
  f2(I25, I26, I27, I28) -> f5(I25, I26, I27, I28) 
  f4(I29, I30, I31, I32) -> f1(I29, I30, I31, I32) 
  f1(I33, I34, I35, I36) -> f3(I33, I34, I35, I36) [1 <= I35]
  f1(I37, I38, I39, I40) -> f2(0, I38, I39, rnd4) [I39 <= 0 /\ I41 = 1 /\ rnd4 = rnd4 /\ 1 <= rnd4]

The dependency graph for this problem is:
  0 -> 1
  1 -> 7
  2 -> 7
  3 -> 6
  4 -> 5
  5 -> 4
  6 -> 2, 3
  7 -> 8, 9
  8 -> 5
  9 -> 6
Where:
  0) f10#(x1, x2, x3, x4) -> f9#(x1, x2, x3, x4) 
  1) f9#(I0, I1, I2, I3) -> f4#(0, 0, rnd3, I3) [rnd3 = rnd3]
  2) f5#(I4, I5, I6, I7) -> f4#(I4, 0, I8, I7) [I7 <= 0 /\ y1 = 1 /\ I8 = I8]
  3) f5#(I9, I10, I11, I12) -> f2#(I9, I10, I11, I12) [1 <= I12]
  4) f8#(I13, I14, I15, I16) -> f3#(I13, I14, I15, I16) 
  5) f3#(I17, I18, I19, I20) -> f8#(I17, I18, I19, I20) 
  6) f2#(I25, I26, I27, I28) -> f5#(I25, I26, I27, I28) 
  7) f4#(I29, I30, I31, I32) -> f1#(I29, I30, I31, I32) 
  8) f1#(I33, I34, I35, I36) -> f3#(I33, I34, I35, I36) [1 <= I35]
  9) f1#(I37, I38, I39, I40) -> f2#(0, I38, I39, rnd4) [I39 <= 0 /\ I41 = 1 /\ rnd4 = rnd4 /\ 1 <= rnd4]

We have the following SCCs.
  { 2, 3, 6, 7, 9 }
  { 4, 5 }

DP problem for innermost termination.
P =
  f8#(I13, I14, I15, I16) -> f3#(I13, I14, I15, I16) 
  f3#(I17, I18, I19, I20) -> f8#(I17, I18, I19, I20) 
R = 
  f10(x1, x2, x3, x4) -> f9(x1, x2, x3, x4) 
  f9(I0, I1, I2, I3) -> f4(0, 0, rnd3, I3) [rnd3 = rnd3]
  f5(I4, I5, I6, I7) -> f4(I4, 0, I8, I7) [I7 <= 0 /\ y1 = 1 /\ I8 = I8]
  f5(I9, I10, I11, I12) -> f2(I9, I10, I11, I12) [1 <= I12]
  f8(I13, I14, I15, I16) -> f3(I13, I14, I15, I16) 
  f3(I17, I18, I19, I20) -> f8(I17, I18, I19, I20) 
  f6(I21, I22, I23, I24) -> f7(I21, I22, I23, I24) 
  f2(I25, I26, I27, I28) -> f5(I25, I26, I27, I28) 
  f4(I29, I30, I31, I32) -> f1(I29, I30, I31, I32) 
  f1(I33, I34, I35, I36) -> f3(I33, I34, I35, I36) [1 <= I35]
  f1(I37, I38, I39, I40) -> f2(0, I38, I39, rnd4) [I39 <= 0 /\ I41 = 1 /\ rnd4 = rnd4 /\ 1 <= rnd4]