/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 =
  f8#(x1, x2, x3, x4) -> f7#(x1, x2, x3, x4) 
  f7#(I0, I1, I2, I3) -> f4#(I0, rnd2, rnd3, I3) [rnd2 = rnd3 /\ rnd3 = rnd3]
  f5#(I4, I5, I6, I7) -> f3#(1 + I4, I5, I6, I7) 
  f3#(I12, I13, I14, I15) -> f5#(I12, I13, I14, I15) 
  f4#(I16, I17, I18, I19) -> f2#(I16, I17, I18, I19) [I17 <= 0]
  f4#(I20, I21, I22, I23) -> f1#(I20, I21, I22, I23) [1 <= I21]
  f2#(I24, I25, I26, I27) -> f3#(0, I25, I26, I27) 
  f1#(I28, I29, I30, I31) -> f2#(I28, I29, I30, I31) [1024 <= I29]
  f1#(I32, I33, I34, I35) -> f2#(I32, I33, I34, rnd4) [rnd4 = rnd4 /\ 1 + I33 <= 1024]
R = 
  f8(x1, x2, x3, x4) -> f7(x1, x2, x3, x4) 
  f7(I0, I1, I2, I3) -> f4(I0, rnd2, rnd3, I3) [rnd2 = rnd3 /\ rnd3 = rnd3]
  f5(I4, I5, I6, I7) -> f3(1 + I4, I5, I6, I7) 
  f5(I8, I9, I10, I11) -> f6(I8, I9, I10, I11) 
  f3(I12, I13, I14, I15) -> f5(I12, I13, I14, I15) 
  f4(I16, I17, I18, I19) -> f2(I16, I17, I18, I19) [I17 <= 0]
  f4(I20, I21, I22, I23) -> f1(I20, I21, I22, I23) [1 <= I21]
  f2(I24, I25, I26, I27) -> f3(0, I25, I26, I27) 
  f1(I28, I29, I30, I31) -> f2(I28, I29, I30, I31) [1024 <= I29]
  f1(I32, I33, I34, I35) -> f2(I32, I33, I34, rnd4) [rnd4 = rnd4 /\ 1 + I33 <= 1024]

The dependency graph for this problem is:
  0 -> 1
  1 -> 4, 5
  2 -> 3
  3 -> 2
  4 -> 6
  5 -> 7, 8
  6 -> 3
  7 -> 6
  8 -> 6
Where:
  0) f8#(x1, x2, x3, x4) -> f7#(x1, x2, x3, x4) 
  1) f7#(I0, I1, I2, I3) -> f4#(I0, rnd2, rnd3, I3) [rnd2 = rnd3 /\ rnd3 = rnd3]
  2) f5#(I4, I5, I6, I7) -> f3#(1 + I4, I5, I6, I7) 
  3) f3#(I12, I13, I14, I15) -> f5#(I12, I13, I14, I15) 
  4) f4#(I16, I17, I18, I19) -> f2#(I16, I17, I18, I19) [I17 <= 0]
  5) f4#(I20, I21, I22, I23) -> f1#(I20, I21, I22, I23) [1 <= I21]
  6) f2#(I24, I25, I26, I27) -> f3#(0, I25, I26, I27) 
  7) f1#(I28, I29, I30, I31) -> f2#(I28, I29, I30, I31) [1024 <= I29]
  8) f1#(I32, I33, I34, I35) -> f2#(I32, I33, I34, rnd4) [rnd4 = rnd4 /\ 1 + I33 <= 1024]

We have the following SCCs.
  { 2, 3 }

DP problem for innermost termination.
P =
  f5#(I4, I5, I6, I7) -> f3#(1 + I4, I5, I6, I7) 
  f3#(I12, I13, I14, I15) -> f5#(I12, I13, I14, I15) 
R = 
  f8(x1, x2, x3, x4) -> f7(x1, x2, x3, x4) 
  f7(I0, I1, I2, I3) -> f4(I0, rnd2, rnd3, I3) [rnd2 = rnd3 /\ rnd3 = rnd3]
  f5(I4, I5, I6, I7) -> f3(1 + I4, I5, I6, I7) 
  f5(I8, I9, I10, I11) -> f6(I8, I9, I10, I11) 
  f3(I12, I13, I14, I15) -> f5(I12, I13, I14, I15) 
  f4(I16, I17, I18, I19) -> f2(I16, I17, I18, I19) [I17 <= 0]
  f4(I20, I21, I22, I23) -> f1(I20, I21, I22, I23) [1 <= I21]
  f2(I24, I25, I26, I27) -> f3(0, I25, I26, I27) 
  f1(I28, I29, I30, I31) -> f2(I28, I29, I30, I31) [1024 <= I29]
  f1(I32, I33, I34, I35) -> f2(I32, I33, I34, rnd4) [rnd4 = rnd4 /\ 1 + I33 <= 1024]