/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 =
  init#(x1, x2, x3) -> f1#(rnd1, rnd2, rnd3) 
  f2#(I0, I1, I2) -> f2#(-1, 1, 0) [1 = I2 /\ I1 <= 1 /\ 5 <= I0 - 1]
  f2#(I3, I4, I5) -> f2#(I3 + 1, 0 - (I3 + 1), I5 - 1) [I5 <= I3 + 1 - 1 /\ 1 <= I5 - 1 /\ 5 <= I3 + I5 /\ 5 <= I3 - I5 /\ I4 <= I5 /\ I5 <= I3]
  f2#(I6, I7, I8) -> f2#(I6 + 1, 0 - (I6 + 1), I8 - 1) [I8 <= I6 + 1 - 1 /\ I8 <= 0 /\ 5 <= I6 + I8 /\ 5 <= I6 - I8 /\ I7 <= I8 /\ I8 <= I6]
  f3#(I9, I10, I11) -> f2#(I11, 0 - I11, I10) [I9 <= I11]
  f2#(I12, I13, I14) -> f3#(I14, -1 * I14, I12) [I12 + I14 <= 4 /\ 5 <= I12 - I14 /\ I13 <= I14 /\ I14 <= I12]
  f2#(I15, I16, I17) -> f3#(I17, -1 * I17, I15) [I16 <= I17 /\ I17 <= I15 /\ I15 - I17 <= 4]
  f1#(I18, I19, I20) -> f2#(20, -20, I19) [-1 <= I19 - 1 /\ 0 <= I18 - 1]
R = 
  init(x1, x2, x3) -> f1(rnd1, rnd2, rnd3) 
  f2(I0, I1, I2) -> f2(-1, 1, 0) [1 = I2 /\ I1 <= 1 /\ 5 <= I0 - 1]
  f2(I3, I4, I5) -> f2(I3 + 1, 0 - (I3 + 1), I5 - 1) [I5 <= I3 + 1 - 1 /\ 1 <= I5 - 1 /\ 5 <= I3 + I5 /\ 5 <= I3 - I5 /\ I4 <= I5 /\ I5 <= I3]
  f2(I6, I7, I8) -> f2(I6 + 1, 0 - (I6 + 1), I8 - 1) [I8 <= I6 + 1 - 1 /\ I8 <= 0 /\ 5 <= I6 + I8 /\ 5 <= I6 - I8 /\ I7 <= I8 /\ I8 <= I6]
  f3(I9, I10, I11) -> f2(I11, 0 - I11, I10) [I9 <= I11]
  f2(I12, I13, I14) -> f3(I14, -1 * I14, I12) [I12 + I14 <= 4 /\ 5 <= I12 - I14 /\ I13 <= I14 /\ I14 <= I12]
  f2(I15, I16, I17) -> f3(I17, -1 * I17, I15) [I16 <= I17 /\ I17 <= I15 /\ I15 - I17 <= 4]
  f1(I18, I19, I20) -> f2(20, -20, I19) [-1 <= I19 - 1 /\ 0 <= I18 - 1]

The dependency graph for this problem is:
  0 -> 7
  1 -> 
  2 -> 1, 2
  3 -> 3
  4 -> 1, 2, 3, 5, 6
  5 -> 4
  6 -> 4
  7 -> 1, 2, 3, 6
Where:
  0) init#(x1, x2, x3) -> f1#(rnd1, rnd2, rnd3) 
  1) f2#(I0, I1, I2) -> f2#(-1, 1, 0) [1 = I2 /\ I1 <= 1 /\ 5 <= I0 - 1]
  2) f2#(I3, I4, I5) -> f2#(I3 + 1, 0 - (I3 + 1), I5 - 1) [I5 <= I3 + 1 - 1 /\ 1 <= I5 - 1 /\ 5 <= I3 + I5 /\ 5 <= I3 - I5 /\ I4 <= I5 /\ I5 <= I3]
  3) f2#(I6, I7, I8) -> f2#(I6 + 1, 0 - (I6 + 1), I8 - 1) [I8 <= I6 + 1 - 1 /\ I8 <= 0 /\ 5 <= I6 + I8 /\ 5 <= I6 - I8 /\ I7 <= I8 /\ I8 <= I6]
  4) f3#(I9, I10, I11) -> f2#(I11, 0 - I11, I10) [I9 <= I11]
  5) f2#(I12, I13, I14) -> f3#(I14, -1 * I14, I12) [I12 + I14 <= 4 /\ 5 <= I12 - I14 /\ I13 <= I14 /\ I14 <= I12]
  6) f2#(I15, I16, I17) -> f3#(I17, -1 * I17, I15) [I16 <= I17 /\ I17 <= I15 /\ I15 - I17 <= 4]
  7) f1#(I18, I19, I20) -> f2#(20, -20, I19) [-1 <= I19 - 1 /\ 0 <= I18 - 1]

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

DP problem for innermost termination.
P =
  f2#(I3, I4, I5) -> f2#(I3 + 1, 0 - (I3 + 1), I5 - 1) [I5 <= I3 + 1 - 1 /\ 1 <= I5 - 1 /\ 5 <= I3 + I5 /\ 5 <= I3 - I5 /\ I4 <= I5 /\ I5 <= I3]
R = 
  init(x1, x2, x3) -> f1(rnd1, rnd2, rnd3) 
  f2(I0, I1, I2) -> f2(-1, 1, 0) [1 = I2 /\ I1 <= 1 /\ 5 <= I0 - 1]
  f2(I3, I4, I5) -> f2(I3 + 1, 0 - (I3 + 1), I5 - 1) [I5 <= I3 + 1 - 1 /\ 1 <= I5 - 1 /\ 5 <= I3 + I5 /\ 5 <= I3 - I5 /\ I4 <= I5 /\ I5 <= I3]
  f2(I6, I7, I8) -> f2(I6 + 1, 0 - (I6 + 1), I8 - 1) [I8 <= I6 + 1 - 1 /\ I8 <= 0 /\ 5 <= I6 + I8 /\ 5 <= I6 - I8 /\ I7 <= I8 /\ I8 <= I6]
  f3(I9, I10, I11) -> f2(I11, 0 - I11, I10) [I9 <= I11]
  f2(I12, I13, I14) -> f3(I14, -1 * I14, I12) [I12 + I14 <= 4 /\ 5 <= I12 - I14 /\ I13 <= I14 /\ I14 <= I12]
  f2(I15, I16, I17) -> f3(I17, -1 * I17, I15) [I16 <= I17 /\ I17 <= I15 /\ I15 - I17 <= 4]
  f1(I18, I19, I20) -> f2(20, -20, I19) [-1 <= I19 - 1 /\ 0 <= I18 - 1]

We use the basic value criterion with the projection function NU:
  NU[f2#(z1,z2,z3)] = z3

This gives the following inequalities:
  I5 <= I3 + 1 - 1 /\ 1 <= I5 - 1 /\ 5 <= I3 + I5 /\ 5 <= I3 - I5 /\ I4 <= I5 /\ I5 <= I3  ==>  I5 >! I5 - 1

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

DP problem for innermost termination.
P =
  f2#(I6, I7, I8) -> f2#(I6 + 1, 0 - (I6 + 1), I8 - 1) [I8 <= I6 + 1 - 1 /\ I8 <= 0 /\ 5 <= I6 + I8 /\ 5 <= I6 - I8 /\ I7 <= I8 /\ I8 <= I6]
R = 
  init(x1, x2, x3) -> f1(rnd1, rnd2, rnd3) 
  f2(I0, I1, I2) -> f2(-1, 1, 0) [1 = I2 /\ I1 <= 1 /\ 5 <= I0 - 1]
  f2(I3, I4, I5) -> f2(I3 + 1, 0 - (I3 + 1), I5 - 1) [I5 <= I3 + 1 - 1 /\ 1 <= I5 - 1 /\ 5 <= I3 + I5 /\ 5 <= I3 - I5 /\ I4 <= I5 /\ I5 <= I3]
  f2(I6, I7, I8) -> f2(I6 + 1, 0 - (I6 + 1), I8 - 1) [I8 <= I6 + 1 - 1 /\ I8 <= 0 /\ 5 <= I6 + I8 /\ 5 <= I6 - I8 /\ I7 <= I8 /\ I8 <= I6]
  f3(I9, I10, I11) -> f2(I11, 0 - I11, I10) [I9 <= I11]
  f2(I12, I13, I14) -> f3(I14, -1 * I14, I12) [I12 + I14 <= 4 /\ 5 <= I12 - I14 /\ I13 <= I14 /\ I14 <= I12]
  f2(I15, I16, I17) -> f3(I17, -1 * I17, I15) [I16 <= I17 /\ I17 <= I15 /\ I15 - I17 <= 4]
  f1(I18, I19, I20) -> f2(20, -20, I19) [-1 <= I19 - 1 /\ 0 <= I18 - 1]