/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 =
  f7#(x1, x2, x3, x4, x5, x6, x7) -> f6#(x1, x2, x3, x4, x5, x6, x7) 
  f6#(I0, I1, I2, I3, I4, I5, I6) -> f4#(I0, I1, I2, 4, 0, 0, I6) 
  f2#(I7, I8, I9, I10, I11, I12, I13) -> f4#(I7, I8, I9, I10, 1 + I11, I12, I13) 
  f4#(I14, I15, I16, I17, I18, I19, I20) -> f2#(rnd1, I20, I16, I17, I18, rnd6, I20) [y1 = I20 /\ rnd6 = y1 /\ rnd1 = rnd1]
  f3#(I29, I30, I31, I32, I33, I34, I35) -> f2#(I36, I35, I31, I32, I33, I37, I35) [I38 = I35 /\ I37 = I38 /\ I36 = I36]
  f1#(I39, I40, I41, I42, I43, I44, I45) -> f2#(I46, I45, I41, I42, I43, I47, I45) [I48 = I45 /\ I47 = I48 /\ I46 = I46]
R = 
  f7(x1, x2, x3, x4, x5, x6, x7) -> f6(x1, x2, x3, x4, x5, x6, x7) 
  f6(I0, I1, I2, I3, I4, I5, I6) -> f4(I0, I1, I2, 4, 0, 0, I6) 
  f2(I7, I8, I9, I10, I11, I12, I13) -> f4(I7, I8, I9, I10, 1 + I11, I12, I13) 
  f4(I14, I15, I16, I17, I18, I19, I20) -> f2(rnd1, I20, I16, I17, I18, rnd6, I20) [y1 = I20 /\ rnd6 = y1 /\ rnd1 = rnd1]
  f4(I21, I22, I23, I24, I25, I26, I27) -> f5(I28, I22, I26, I24, I25, I26, I27) [I28 = I26]
  f3(I29, I30, I31, I32, I33, I34, I35) -> f2(I36, I35, I31, I32, I33, I37, I35) [I38 = I35 /\ I37 = I38 /\ I36 = I36]
  f1(I39, I40, I41, I42, I43, I44, I45) -> f2(I46, I45, I41, I42, I43, I47, I45) [I48 = I45 /\ I47 = I48 /\ I46 = I46]

The dependency graph for this problem is:
  0 -> 1
  1 -> 3
  2 -> 3
  3 -> 2
  4 -> 2
  5 -> 2
Where:
  0) f7#(x1, x2, x3, x4, x5, x6, x7) -> f6#(x1, x2, x3, x4, x5, x6, x7) 
  1) f6#(I0, I1, I2, I3, I4, I5, I6) -> f4#(I0, I1, I2, 4, 0, 0, I6) 
  2) f2#(I7, I8, I9, I10, I11, I12, I13) -> f4#(I7, I8, I9, I10, 1 + I11, I12, I13) 
  3) f4#(I14, I15, I16, I17, I18, I19, I20) -> f2#(rnd1, I20, I16, I17, I18, rnd6, I20) [y1 = I20 /\ rnd6 = y1 /\ rnd1 = rnd1]
  4) f3#(I29, I30, I31, I32, I33, I34, I35) -> f2#(I36, I35, I31, I32, I33, I37, I35) [I38 = I35 /\ I37 = I38 /\ I36 = I36]
  5) f1#(I39, I40, I41, I42, I43, I44, I45) -> f2#(I46, I45, I41, I42, I43, I47, I45) [I48 = I45 /\ I47 = I48 /\ I46 = I46]

We have the following SCCs.
  { 2, 3 }

DP problem for innermost termination.
P =
  f2#(I7, I8, I9, I10, I11, I12, I13) -> f4#(I7, I8, I9, I10, 1 + I11, I12, I13) 
  f4#(I14, I15, I16, I17, I18, I19, I20) -> f2#(rnd1, I20, I16, I17, I18, rnd6, I20) [y1 = I20 /\ rnd6 = y1 /\ rnd1 = rnd1]
R = 
  f7(x1, x2, x3, x4, x5, x6, x7) -> f6(x1, x2, x3, x4, x5, x6, x7) 
  f6(I0, I1, I2, I3, I4, I5, I6) -> f4(I0, I1, I2, 4, 0, 0, I6) 
  f2(I7, I8, I9, I10, I11, I12, I13) -> f4(I7, I8, I9, I10, 1 + I11, I12, I13) 
  f4(I14, I15, I16, I17, I18, I19, I20) -> f2(rnd1, I20, I16, I17, I18, rnd6, I20) [y1 = I20 /\ rnd6 = y1 /\ rnd1 = rnd1]
  f4(I21, I22, I23, I24, I25, I26, I27) -> f5(I28, I22, I26, I24, I25, I26, I27) [I28 = I26]
  f3(I29, I30, I31, I32, I33, I34, I35) -> f2(I36, I35, I31, I32, I33, I37, I35) [I38 = I35 /\ I37 = I38 /\ I36 = I36]
  f1(I39, I40, I41, I42, I43, I44, I45) -> f2(I46, I45, I41, I42, I43, I47, I45) [I48 = I45 /\ I47 = I48 /\ I46 = I46]