/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 =
  f14#(x1, x2, x3) -> f13#(x1, x2, x3) 
  f13#(I0, I1, I2) -> f1#(rnd1, rnd2, I2) [y1 = y1 /\ rnd2 = rnd2 /\ rnd1 = rnd2]
  f3#(I3, I4, I5) -> f12#(I3, I4, I5) [I3 <= 5]
  f3#(I6, I7, I8) -> f11#(I6, I7, I8) [6 <= I6]
  f12#(I9, I10, I11) -> f6#(I9, I10, I11) [1 <= I11]
  f12#(I12, I13, I14) -> f11#(I12, I13, I14) [I14 <= 0]
  f11#(I15, I16, I17) -> f2#(1 + I15, I16, I17) [I15 <= 5]
  f11#(I18, I19, I20) -> f2#(1 + I18, I19, I20) [6 <= I18]
  f10#(I21, I22, I23) -> f9#(I21, I22, I23) 
  f9#(I24, I25, I26) -> f10#(I24, I25, I26) 
  f7#(I27, I28, I29) -> f1#(1, I28, I29) [I27 <= 2]
  f7#(I30, I31, I32) -> f6#(-1 + I30, I31, I32) [3 <= I30]
  f8#(I33, I34, I35) -> f9#(I33, I34, I35) 
  f6#(I36, I37, I38) -> f7#(I36, I37, I38) 
  f2#(I42, I43, I44) -> f3#(I42, I43, rnd3) [rnd3 = rnd3]
  f1#(I45, I46, I47) -> f2#(I45, I46, I47) 
R = 
  f14(x1, x2, x3) -> f13(x1, x2, x3) 
  f13(I0, I1, I2) -> f1(rnd1, rnd2, I2) [y1 = y1 /\ rnd2 = rnd2 /\ rnd1 = rnd2]
  f3(I3, I4, I5) -> f12(I3, I4, I5) [I3 <= 5]
  f3(I6, I7, I8) -> f11(I6, I7, I8) [6 <= I6]
  f12(I9, I10, I11) -> f6(I9, I10, I11) [1 <= I11]
  f12(I12, I13, I14) -> f11(I12, I13, I14) [I14 <= 0]
  f11(I15, I16, I17) -> f2(1 + I15, I16, I17) [I15 <= 5]
  f11(I18, I19, I20) -> f2(1 + I18, I19, I20) [6 <= I18]
  f10(I21, I22, I23) -> f9(I21, I22, I23) 
  f9(I24, I25, I26) -> f10(I24, I25, I26) 
  f7(I27, I28, I29) -> f1(1, I28, I29) [I27 <= 2]
  f7(I30, I31, I32) -> f6(-1 + I30, I31, I32) [3 <= I30]
  f8(I33, I34, I35) -> f9(I33, I34, I35) 
  f6(I36, I37, I38) -> f7(I36, I37, I38) 
  f4(I39, I40, I41) -> f5(I39, I40, I41) 
  f2(I42, I43, I44) -> f3(I42, I43, rnd3) [rnd3 = rnd3]
  f1(I45, I46, I47) -> f2(I45, I46, I47) 

The dependency graph for this problem is:
  0 -> 1
  1 -> 15
  2 -> 4, 5
  3 -> 7
  4 -> 13
  5 -> 6, 7
  6 -> 14
  7 -> 14
  8 -> 9
  9 -> 8
  10 -> 15
  11 -> 13
  12 -> 9
  13 -> 10, 11
  14 -> 2, 3
  15 -> 14
Where:
  0) f14#(x1, x2, x3) -> f13#(x1, x2, x3) 
  1) f13#(I0, I1, I2) -> f1#(rnd1, rnd2, I2) [y1 = y1 /\ rnd2 = rnd2 /\ rnd1 = rnd2]
  2) f3#(I3, I4, I5) -> f12#(I3, I4, I5) [I3 <= 5]
  3) f3#(I6, I7, I8) -> f11#(I6, I7, I8) [6 <= I6]
  4) f12#(I9, I10, I11) -> f6#(I9, I10, I11) [1 <= I11]
  5) f12#(I12, I13, I14) -> f11#(I12, I13, I14) [I14 <= 0]
  6) f11#(I15, I16, I17) -> f2#(1 + I15, I16, I17) [I15 <= 5]
  7) f11#(I18, I19, I20) -> f2#(1 + I18, I19, I20) [6 <= I18]
  8) f10#(I21, I22, I23) -> f9#(I21, I22, I23) 
  9) f9#(I24, I25, I26) -> f10#(I24, I25, I26) 
  10) f7#(I27, I28, I29) -> f1#(1, I28, I29) [I27 <= 2]
  11) f7#(I30, I31, I32) -> f6#(-1 + I30, I31, I32) [3 <= I30]
  12) f8#(I33, I34, I35) -> f9#(I33, I34, I35) 
  13) f6#(I36, I37, I38) -> f7#(I36, I37, I38) 
  14) f2#(I42, I43, I44) -> f3#(I42, I43, rnd3) [rnd3 = rnd3]
  15) f1#(I45, I46, I47) -> f2#(I45, I46, I47) 

We have the following SCCs.
  { 8, 9 }
  { 2, 3, 4, 5, 6, 7, 10, 11, 13, 14, 15 }

DP problem for innermost termination.
P =
  f3#(I3, I4, I5) -> f12#(I3, I4, I5) [I3 <= 5]
  f3#(I6, I7, I8) -> f11#(I6, I7, I8) [6 <= I6]
  f12#(I9, I10, I11) -> f6#(I9, I10, I11) [1 <= I11]
  f12#(I12, I13, I14) -> f11#(I12, I13, I14) [I14 <= 0]
  f11#(I15, I16, I17) -> f2#(1 + I15, I16, I17) [I15 <= 5]
  f11#(I18, I19, I20) -> f2#(1 + I18, I19, I20) [6 <= I18]
  f7#(I27, I28, I29) -> f1#(1, I28, I29) [I27 <= 2]
  f7#(I30, I31, I32) -> f6#(-1 + I30, I31, I32) [3 <= I30]
  f6#(I36, I37, I38) -> f7#(I36, I37, I38) 
  f2#(I42, I43, I44) -> f3#(I42, I43, rnd3) [rnd3 = rnd3]
  f1#(I45, I46, I47) -> f2#(I45, I46, I47) 
R = 
  f14(x1, x2, x3) -> f13(x1, x2, x3) 
  f13(I0, I1, I2) -> f1(rnd1, rnd2, I2) [y1 = y1 /\ rnd2 = rnd2 /\ rnd1 = rnd2]
  f3(I3, I4, I5) -> f12(I3, I4, I5) [I3 <= 5]
  f3(I6, I7, I8) -> f11(I6, I7, I8) [6 <= I6]
  f12(I9, I10, I11) -> f6(I9, I10, I11) [1 <= I11]
  f12(I12, I13, I14) -> f11(I12, I13, I14) [I14 <= 0]
  f11(I15, I16, I17) -> f2(1 + I15, I16, I17) [I15 <= 5]
  f11(I18, I19, I20) -> f2(1 + I18, I19, I20) [6 <= I18]
  f10(I21, I22, I23) -> f9(I21, I22, I23) 
  f9(I24, I25, I26) -> f10(I24, I25, I26) 
  f7(I27, I28, I29) -> f1(1, I28, I29) [I27 <= 2]
  f7(I30, I31, I32) -> f6(-1 + I30, I31, I32) [3 <= I30]
  f8(I33, I34, I35) -> f9(I33, I34, I35) 
  f6(I36, I37, I38) -> f7(I36, I37, I38) 
  f4(I39, I40, I41) -> f5(I39, I40, I41) 
  f2(I42, I43, I44) -> f3(I42, I43, rnd3) [rnd3 = rnd3]
  f1(I45, I46, I47) -> f2(I45, I46, I47)