/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, x5) -> f7#(x1, x2, x3, x4, x5) 
  f7#(I0, I1, I2, I3, I4) -> f3#(I0, I1, I2, I3, I4) [I0 <= 0]
  f3#(I5, I6, I7, I8, I9) -> f6#(I5, I6, I7, I8, I9) [1 <= I9 /\ 1 <= I8]
  f6#(I10, I11, I12, I13, I14) -> f5#(I10, I11, I12, -1 + I13, I14) 
  f6#(I15, I16, I17, I18, I19) -> f5#(I15, I16, I17, I18, -1 + I19) 
  f5#(I20, I21, I22, I23, I24) -> f1#(I20, I21, I22, I23, I24) [1 <= I20]
  f5#(I25, I26, I27, I28, I29) -> f4#(I25, I26, I27, I28, I29) [I25 <= 0]
  f4#(I30, I31, I32, I33, I34) -> f3#(1, I33, I34, I33, I34) 
  f4#(I35, I36, I37, I38, I39) -> f3#(I35, I36, I37, I38, I39) [I35 <= 0]
  f1#(I40, I41, I42, I43, I44) -> f3#(I40, I41, I42, I43, I44) 
R = 
  f8(x1, x2, x3, x4, x5) -> f7(x1, x2, x3, x4, x5) 
  f7(I0, I1, I2, I3, I4) -> f3(I0, I1, I2, I3, I4) [I0 <= 0]
  f3(I5, I6, I7, I8, I9) -> f6(I5, I6, I7, I8, I9) [1 <= I9 /\ 1 <= I8]
  f6(I10, I11, I12, I13, I14) -> f5(I10, I11, I12, -1 + I13, I14) 
  f6(I15, I16, I17, I18, I19) -> f5(I15, I16, I17, I18, -1 + I19) 
  f5(I20, I21, I22, I23, I24) -> f1(I20, I21, I22, I23, I24) [1 <= I20]
  f5(I25, I26, I27, I28, I29) -> f4(I25, I26, I27, I28, I29) [I25 <= 0]
  f4(I30, I31, I32, I33, I34) -> f3(1, I33, I34, I33, I34) 
  f4(I35, I36, I37, I38, I39) -> f3(I35, I36, I37, I38, I39) [I35 <= 0]
  f1(I40, I41, I42, I43, I44) -> f3(I40, I41, I42, I43, I44) 
  f1(I45, I46, I47, I48, I49) -> f2(I45, I46, I47, I48, I49) [I47 <= I49 /\ I46 <= I48]

The dependency graph for this problem is:
  0 -> 1
  1 -> 2
  2 -> 3, 4
  3 -> 5, 6
  4 -> 5, 6
  5 -> 9
  6 -> 7, 8
  7 -> 2
  8 -> 2
  9 -> 2
Where:
  0) f8#(x1, x2, x3, x4, x5) -> f7#(x1, x2, x3, x4, x5) 
  1) f7#(I0, I1, I2, I3, I4) -> f3#(I0, I1, I2, I3, I4) [I0 <= 0]
  2) f3#(I5, I6, I7, I8, I9) -> f6#(I5, I6, I7, I8, I9) [1 <= I9 /\ 1 <= I8]
  3) f6#(I10, I11, I12, I13, I14) -> f5#(I10, I11, I12, -1 + I13, I14) 
  4) f6#(I15, I16, I17, I18, I19) -> f5#(I15, I16, I17, I18, -1 + I19) 
  5) f5#(I20, I21, I22, I23, I24) -> f1#(I20, I21, I22, I23, I24) [1 <= I20]
  6) f5#(I25, I26, I27, I28, I29) -> f4#(I25, I26, I27, I28, I29) [I25 <= 0]
  7) f4#(I30, I31, I32, I33, I34) -> f3#(1, I33, I34, I33, I34) 
  8) f4#(I35, I36, I37, I38, I39) -> f3#(I35, I36, I37, I38, I39) [I35 <= 0]
  9) f1#(I40, I41, I42, I43, I44) -> f3#(I40, I41, I42, I43, I44) 

We have the following SCCs.
  { 2, 3, 4, 5, 6, 7, 8, 9 }

DP problem for innermost termination.
P =
  f3#(I5, I6, I7, I8, I9) -> f6#(I5, I6, I7, I8, I9) [1 <= I9 /\ 1 <= I8]
  f6#(I10, I11, I12, I13, I14) -> f5#(I10, I11, I12, -1 + I13, I14) 
  f6#(I15, I16, I17, I18, I19) -> f5#(I15, I16, I17, I18, -1 + I19) 
  f5#(I20, I21, I22, I23, I24) -> f1#(I20, I21, I22, I23, I24) [1 <= I20]
  f5#(I25, I26, I27, I28, I29) -> f4#(I25, I26, I27, I28, I29) [I25 <= 0]
  f4#(I30, I31, I32, I33, I34) -> f3#(1, I33, I34, I33, I34) 
  f4#(I35, I36, I37, I38, I39) -> f3#(I35, I36, I37, I38, I39) [I35 <= 0]
  f1#(I40, I41, I42, I43, I44) -> f3#(I40, I41, I42, I43, I44) 
R = 
  f8(x1, x2, x3, x4, x5) -> f7(x1, x2, x3, x4, x5) 
  f7(I0, I1, I2, I3, I4) -> f3(I0, I1, I2, I3, I4) [I0 <= 0]
  f3(I5, I6, I7, I8, I9) -> f6(I5, I6, I7, I8, I9) [1 <= I9 /\ 1 <= I8]
  f6(I10, I11, I12, I13, I14) -> f5(I10, I11, I12, -1 + I13, I14) 
  f6(I15, I16, I17, I18, I19) -> f5(I15, I16, I17, I18, -1 + I19) 
  f5(I20, I21, I22, I23, I24) -> f1(I20, I21, I22, I23, I24) [1 <= I20]
  f5(I25, I26, I27, I28, I29) -> f4(I25, I26, I27, I28, I29) [I25 <= 0]
  f4(I30, I31, I32, I33, I34) -> f3(1, I33, I34, I33, I34) 
  f4(I35, I36, I37, I38, I39) -> f3(I35, I36, I37, I38, I39) [I35 <= 0]
  f1(I40, I41, I42, I43, I44) -> f3(I40, I41, I42, I43, I44) 
  f1(I45, I46, I47, I48, I49) -> f2(I45, I46, I47, I48, I49) [I47 <= I49 /\ I46 <= I48]