/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, x5, x6, x7) -> f1#(x1, x2, x3, x4, x5, x6, x7) 
  f4#(I0, I1, I2, I3, I4, I5, I6) -> f2#(I0, I1, I2, I3, I4, I5, I6) 
  f2#(I7, I8, I9, I10, I11, I12, I13) -> f4#(I7, rnd2, rnd3, I10, rnd5, rnd6, rnd7) [rnd7 = rnd7 /\ rnd3 = rnd3 /\ 1 <= I11 /\ 1 <= I12 /\ y1 = I11 /\ y2 = -2 + I12 /\ y3 = 1 + y1 /\ rnd2 = y2 /\ rnd5 = -2 + y3 /\ rnd6 = 1 + rnd2 /\ rnd5 <= -1 + rnd3 /\ -1 + rnd3 <= rnd5 /\ rnd6 <= 1 + rnd2 /\ 1 + rnd2 <= rnd6 /\ rnd2 <= -2 + rnd7 /\ -2 + rnd7 <= rnd2 /\ 1 <= rnd7 /\ 1 <= rnd3]
  f1#(I29, I30, I31, I32, I33, I34, I35) -> f2#(I29, I30, I31, I32, I33, I34, I35) 
R = 
  f5(x1, x2, x3, x4, x5, x6, x7) -> f1(x1, x2, x3, x4, x5, x6, x7) 
  f4(I0, I1, I2, I3, I4, I5, I6) -> f2(I0, I1, I2, I3, I4, I5, I6) 
  f2(I7, I8, I9, I10, I11, I12, I13) -> f4(I7, rnd2, rnd3, I10, rnd5, rnd6, rnd7) [rnd7 = rnd7 /\ rnd3 = rnd3 /\ 1 <= I11 /\ 1 <= I12 /\ y1 = I11 /\ y2 = -2 + I12 /\ y3 = 1 + y1 /\ rnd2 = y2 /\ rnd5 = -2 + y3 /\ rnd6 = 1 + rnd2 /\ rnd5 <= -1 + rnd3 /\ -1 + rnd3 <= rnd5 /\ rnd6 <= 1 + rnd2 /\ 1 + rnd2 <= rnd6 /\ rnd2 <= -2 + rnd7 /\ -2 + rnd7 <= rnd2 /\ 1 <= rnd7 /\ 1 <= rnd3]
  f2(I14, I15, I16, I17, I18, I19, I20) -> f3(I17, I15, I16, I17, I21, I19, I20) [I21 <= 0 /\ I21 = I21 /\ I18 <= 0]
  f2(I22, I23, I24, I25, I26, I27, I28) -> f3(I25, I23, I24, I25, I26, I27, I28) [I27 <= 0 /\ 1 <= I26]
  f1(I29, I30, I31, I32, I33, I34, I35) -> f2(I29, I30, I31, I32, I33, I34, I35) 

The dependency graph for this problem is:
  0 -> 3
  1 -> 2
  2 -> 1
  3 -> 2
Where:
  0) f5#(x1, x2, x3, x4, x5, x6, x7) -> f1#(x1, x2, x3, x4, x5, x6, x7) 
  1) f4#(I0, I1, I2, I3, I4, I5, I6) -> f2#(I0, I1, I2, I3, I4, I5, I6) 
  2) f2#(I7, I8, I9, I10, I11, I12, I13) -> f4#(I7, rnd2, rnd3, I10, rnd5, rnd6, rnd7) [rnd7 = rnd7 /\ rnd3 = rnd3 /\ 1 <= I11 /\ 1 <= I12 /\ y1 = I11 /\ y2 = -2 + I12 /\ y3 = 1 + y1 /\ rnd2 = y2 /\ rnd5 = -2 + y3 /\ rnd6 = 1 + rnd2 /\ rnd5 <= -1 + rnd3 /\ -1 + rnd3 <= rnd5 /\ rnd6 <= 1 + rnd2 /\ 1 + rnd2 <= rnd6 /\ rnd2 <= -2 + rnd7 /\ -2 + rnd7 <= rnd2 /\ 1 <= rnd7 /\ 1 <= rnd3]
  3) f1#(I29, I30, I31, I32, I33, I34, I35) -> f2#(I29, I30, I31, I32, I33, I34, I35) 

We have the following SCCs.
  { 1, 2 }

DP problem for innermost termination.
P =
  f4#(I0, I1, I2, I3, I4, I5, I6) -> f2#(I0, I1, I2, I3, I4, I5, I6) 
  f2#(I7, I8, I9, I10, I11, I12, I13) -> f4#(I7, rnd2, rnd3, I10, rnd5, rnd6, rnd7) [rnd7 = rnd7 /\ rnd3 = rnd3 /\ 1 <= I11 /\ 1 <= I12 /\ y1 = I11 /\ y2 = -2 + I12 /\ y3 = 1 + y1 /\ rnd2 = y2 /\ rnd5 = -2 + y3 /\ rnd6 = 1 + rnd2 /\ rnd5 <= -1 + rnd3 /\ -1 + rnd3 <= rnd5 /\ rnd6 <= 1 + rnd2 /\ 1 + rnd2 <= rnd6 /\ rnd2 <= -2 + rnd7 /\ -2 + rnd7 <= rnd2 /\ 1 <= rnd7 /\ 1 <= rnd3]
R = 
  f5(x1, x2, x3, x4, x5, x6, x7) -> f1(x1, x2, x3, x4, x5, x6, x7) 
  f4(I0, I1, I2, I3, I4, I5, I6) -> f2(I0, I1, I2, I3, I4, I5, I6) 
  f2(I7, I8, I9, I10, I11, I12, I13) -> f4(I7, rnd2, rnd3, I10, rnd5, rnd6, rnd7) [rnd7 = rnd7 /\ rnd3 = rnd3 /\ 1 <= I11 /\ 1 <= I12 /\ y1 = I11 /\ y2 = -2 + I12 /\ y3 = 1 + y1 /\ rnd2 = y2 /\ rnd5 = -2 + y3 /\ rnd6 = 1 + rnd2 /\ rnd5 <= -1 + rnd3 /\ -1 + rnd3 <= rnd5 /\ rnd6 <= 1 + rnd2 /\ 1 + rnd2 <= rnd6 /\ rnd2 <= -2 + rnd7 /\ -2 + rnd7 <= rnd2 /\ 1 <= rnd7 /\ 1 <= rnd3]
  f2(I14, I15, I16, I17, I18, I19, I20) -> f3(I17, I15, I16, I17, I21, I19, I20) [I21 <= 0 /\ I21 = I21 /\ I18 <= 0]
  f2(I22, I23, I24, I25, I26, I27, I28) -> f3(I25, I23, I24, I25, I26, I27, I28) [I27 <= 0 /\ 1 <= I26]
  f1(I29, I30, I31, I32, I33, I34, I35) -> f2(I29, I30, I31, I32, I33, I34, I35) 

We use the basic value criterion with the projection function NU:
  NU[f2#(z1,z2,z3,z4,z5,z6,z7)] = z6
  NU[f4#(z1,z2,z3,z4,z5,z6,z7)] = z6

This gives the following inequalities:
    ==>  I5 (>! \union =) I5
  rnd7 = rnd7 /\ rnd3 = rnd3 /\ 1 <= I11 /\ 1 <= I12 /\ y1 = I11 /\ y2 = -2 + I12 /\ y3 = 1 + y1 /\ rnd2 = y2 /\ rnd5 = -2 + y3 /\ rnd6 = 1 + rnd2 /\ rnd5 <= -1 + rnd3 /\ -1 + rnd3 <= rnd5 /\ rnd6 <= 1 + rnd2 /\ 1 + rnd2 <= rnd6 /\ rnd2 <= -2 + rnd7 /\ -2 + rnd7 <= rnd2 /\ 1 <= rnd7 /\ 1 <= rnd3  ==>  I12 >! rnd6

We remove all the strictly oriented dependency pairs.

DP problem for innermost termination.
P =
  f4#(I0, I1, I2, I3, I4, I5, I6) -> f2#(I0, I1, I2, I3, I4, I5, I6) 
R = 
  f5(x1, x2, x3, x4, x5, x6, x7) -> f1(x1, x2, x3, x4, x5, x6, x7) 
  f4(I0, I1, I2, I3, I4, I5, I6) -> f2(I0, I1, I2, I3, I4, I5, I6) 
  f2(I7, I8, I9, I10, I11, I12, I13) -> f4(I7, rnd2, rnd3, I10, rnd5, rnd6, rnd7) [rnd7 = rnd7 /\ rnd3 = rnd3 /\ 1 <= I11 /\ 1 <= I12 /\ y1 = I11 /\ y2 = -2 + I12 /\ y3 = 1 + y1 /\ rnd2 = y2 /\ rnd5 = -2 + y3 /\ rnd6 = 1 + rnd2 /\ rnd5 <= -1 + rnd3 /\ -1 + rnd3 <= rnd5 /\ rnd6 <= 1 + rnd2 /\ 1 + rnd2 <= rnd6 /\ rnd2 <= -2 + rnd7 /\ -2 + rnd7 <= rnd2 /\ 1 <= rnd7 /\ 1 <= rnd3]
  f2(I14, I15, I16, I17, I18, I19, I20) -> f3(I17, I15, I16, I17, I21, I19, I20) [I21 <= 0 /\ I21 = I21 /\ I18 <= 0]
  f2(I22, I23, I24, I25, I26, I27, I28) -> f3(I25, I23, I24, I25, I26, I27, I28) [I27 <= 0 /\ 1 <= I26]
  f1(I29, I30, I31, I32, I33, I34, I35) -> f2(I29, I30, I31, I32, I33, I34, I35) 

The dependency graph for this problem is:
  1 -> 
Where:
  1) f4#(I0, I1, I2, I3, I4, I5, I6) -> f2#(I0, I1, I2, I3, I4, I5, I6) 

We have the following SCCs.