/export/starexec/sandbox/solver/bin/starexec_run_Transition /export/starexec/sandbox/benchmark/theBenchmark.smt2 /export/starexec/sandbox/output/output_files


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YES

DP problem for innermost termination.
P =
  f5#(x1, x2, x3, x4) -> f4#(x1, x2, x3, x4) 
  f2#(I4, I5, I6, I7) -> f1#(I4, 1, 1 + I6, I7) [0 <= -1 * I5 /\ 0 <= -1 - I6 + I7]
  f4#(I8, I9, I10, I11) -> f2#(I8, 0, I10, I11) 
  f1#(I17, I18, I19, I20) -> f2#(I17, 0, I19, -1 + I20) [1 - I18 <= 0 /\ 0 <= -1 - I19 + I20]
R = 
  f5(x1, x2, x3, x4) -> f4(x1, x2, x3, x4) 
  f2(I0, I1, I2, I3) -> f3(rnd1, I1, I2, I3) [rnd1 = rnd1 /\ -1 * I2 + I3 <= 0]
  f2(I4, I5, I6, I7) -> f1(I4, 1, 1 + I6, I7) [0 <= -1 * I5 /\ 0 <= -1 - I6 + I7]
  f4(I8, I9, I10, I11) -> f2(I8, 0, I10, I11) 
  f1(I12, I13, I14, I15) -> f3(I16, I13, I14, I15) [I16 = I16 /\ -1 * I14 + I15 <= 0]
  f1(I17, I18, I19, I20) -> f2(I17, 0, I19, -1 + I20) [1 - I18 <= 0 /\ 0 <= -1 - I19 + I20]

The dependency graph for this problem is:
  0 -> 2
  1 -> 3
  2 -> 1
  3 -> 1
Where:
  0) f5#(x1, x2, x3, x4) -> f4#(x1, x2, x3, x4) 
  1) f2#(I4, I5, I6, I7) -> f1#(I4, 1, 1 + I6, I7) [0 <= -1 * I5 /\ 0 <= -1 - I6 + I7]
  2) f4#(I8, I9, I10, I11) -> f2#(I8, 0, I10, I11) 
  3) f1#(I17, I18, I19, I20) -> f2#(I17, 0, I19, -1 + I20) [1 - I18 <= 0 /\ 0 <= -1 - I19 + I20]

We have the following SCCs.
  { 1, 3 }

DP problem for innermost termination.
P =
  f2#(I4, I5, I6, I7) -> f1#(I4, 1, 1 + I6, I7) [0 <= -1 * I5 /\ 0 <= -1 - I6 + I7]
  f1#(I17, I18, I19, I20) -> f2#(I17, 0, I19, -1 + I20) [1 - I18 <= 0 /\ 0 <= -1 - I19 + I20]
R = 
  f5(x1, x2, x3, x4) -> f4(x1, x2, x3, x4) 
  f2(I0, I1, I2, I3) -> f3(rnd1, I1, I2, I3) [rnd1 = rnd1 /\ -1 * I2 + I3 <= 0]
  f2(I4, I5, I6, I7) -> f1(I4, 1, 1 + I6, I7) [0 <= -1 * I5 /\ 0 <= -1 - I6 + I7]
  f4(I8, I9, I10, I11) -> f2(I8, 0, I10, I11) 
  f1(I12, I13, I14, I15) -> f3(I16, I13, I14, I15) [I16 = I16 /\ -1 * I14 + I15 <= 0]
  f1(I17, I18, I19, I20) -> f2(I17, 0, I19, -1 + I20) [1 - I18 <= 0 /\ 0 <= -1 - I19 + I20]

We use the reverse value criterion with the projection function NU:
  NU[f1#(z1,z2,z3,z4)] = -1 - z3 + z4 + -1 * 0
  NU[f2#(z1,z2,z3,z4)] = -1 - z3 + z4 + -1 * 0

This gives the following inequalities:
  0 <= -1 * I5 /\ 0 <= -1 - I6 + I7  ==>  -1 - I6 + I7 + -1 * 0 > -1 - (1 + I6) + I7 + -1 * 0  with  -1 - I6 + I7 + -1 * 0 >= 0
  1 - I18 <= 0 /\ 0 <= -1 - I19 + I20  ==>  -1 - I19 + I20 + -1 * 0 > -1 - I19 + (-1 + I20) + -1 * 0  with  -1 - I19 + I20 + -1 * 0 >= 0

All dependency pairs are strictly oriented, so the entire dependency pair problem may be removed.