/export/starexec/sandbox2/solver/bin/starexec_run_Transition /export/starexec/sandbox2/benchmark/theBenchmark.smt2 /export/starexec/sandbox2/output/output_files


--------------------------------------------------------------------------------


YES

DP problem for innermost termination.
P =
  f6#(x1, x2, x3, x4, x5) -> f5#(x1, x2, x3, x4, x5) 
  f5#(I5, I6, I7, I8, I9) -> f1#(I5, I6, I7, I8, I9) [0 <= -2 + I7 /\ 0 <= -2 + I6]
  f3#(I16, I17, I18, I19, I20) -> f1#(I16, I17, I18, I19, I20) 
  f2#(I21, I22, I23, I24, I25) -> f3#(I21, I22, I23, I24, I25) [1 <= I25]
  f2#(I26, I27, I28, I29, I30) -> f3#(I26, I27, I28, I29, I30) [1 + I30 <= 0]
  f1#(I31, I32, I33, I34, I35) -> f2#(I31, I32, I33, 1 + I34, 1 + I35) [0 <= -2 - I34]
R = 
  f6(x1, x2, x3, x4, x5) -> f5(x1, x2, x3, x4, x5) 
  f5(I0, I1, I2, I3, I4) -> f4(rnd1, I1, I2, I3, I4) [rnd1 = rnd1 /\ -1 + I1 <= 0]
  f5(I5, I6, I7, I8, I9) -> f1(I5, I6, I7, I8, I9) [0 <= -2 + I7 /\ 0 <= -2 + I6]
  f1(I10, I11, I12, I13, I14) -> f4(I15, I11, I12, I13, I14) [I15 = I15 /\ -1 - I13 <= 0]
  f3(I16, I17, I18, I19, I20) -> f1(I16, I17, I18, I19, I20) 
  f2(I21, I22, I23, I24, I25) -> f3(I21, I22, I23, I24, I25) [1 <= I25]
  f2(I26, I27, I28, I29, I30) -> f3(I26, I27, I28, I29, I30) [1 + I30 <= 0]
  f1(I31, I32, I33, I34, I35) -> f2(I31, I32, I33, 1 + I34, 1 + I35) [0 <= -2 - I34]

The dependency graph for this problem is:
  0 -> 1
  1 -> 5
  2 -> 5
  3 -> 2
  4 -> 2
  5 -> 3, 4
Where:
  0) f6#(x1, x2, x3, x4, x5) -> f5#(x1, x2, x3, x4, x5) 
  1) f5#(I5, I6, I7, I8, I9) -> f1#(I5, I6, I7, I8, I9) [0 <= -2 + I7 /\ 0 <= -2 + I6]
  2) f3#(I16, I17, I18, I19, I20) -> f1#(I16, I17, I18, I19, I20) 
  3) f2#(I21, I22, I23, I24, I25) -> f3#(I21, I22, I23, I24, I25) [1 <= I25]
  4) f2#(I26, I27, I28, I29, I30) -> f3#(I26, I27, I28, I29, I30) [1 + I30 <= 0]
  5) f1#(I31, I32, I33, I34, I35) -> f2#(I31, I32, I33, 1 + I34, 1 + I35) [0 <= -2 - I34]

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

DP problem for innermost termination.
P =
  f3#(I16, I17, I18, I19, I20) -> f1#(I16, I17, I18, I19, I20) 
  f2#(I21, I22, I23, I24, I25) -> f3#(I21, I22, I23, I24, I25) [1 <= I25]
  f2#(I26, I27, I28, I29, I30) -> f3#(I26, I27, I28, I29, I30) [1 + I30 <= 0]
  f1#(I31, I32, I33, I34, I35) -> f2#(I31, I32, I33, 1 + I34, 1 + I35) [0 <= -2 - I34]
R = 
  f6(x1, x2, x3, x4, x5) -> f5(x1, x2, x3, x4, x5) 
  f5(I0, I1, I2, I3, I4) -> f4(rnd1, I1, I2, I3, I4) [rnd1 = rnd1 /\ -1 + I1 <= 0]
  f5(I5, I6, I7, I8, I9) -> f1(I5, I6, I7, I8, I9) [0 <= -2 + I7 /\ 0 <= -2 + I6]
  f1(I10, I11, I12, I13, I14) -> f4(I15, I11, I12, I13, I14) [I15 = I15 /\ -1 - I13 <= 0]
  f3(I16, I17, I18, I19, I20) -> f1(I16, I17, I18, I19, I20) 
  f2(I21, I22, I23, I24, I25) -> f3(I21, I22, I23, I24, I25) [1 <= I25]
  f2(I26, I27, I28, I29, I30) -> f3(I26, I27, I28, I29, I30) [1 + I30 <= 0]
  f1(I31, I32, I33, I34, I35) -> f2(I31, I32, I33, 1 + I34, 1 + I35) [0 <= -2 - I34]

We use the extended value criterion with the projection function NU:
  NU[f2#(x0,x1,x2,x3,x4)] = -x3 - 2
  NU[f1#(x0,x1,x2,x3,x4)] = -x3 - 2
  NU[f3#(x0,x1,x2,x3,x4)] = -x3 - 2

This gives the following inequalities:
    ==>  -I19 - 2 >= -I19 - 2
  1 <= I25  ==>  -I24 - 2 >= -I24 - 2
  1 + I30 <= 0  ==>  -I29 - 2 >= -I29 - 2
  0 <= -2 - I34  ==>  -I34 - 2 > -(1 + I34) - 2  with  -I34 - 2 >= 0

We remove all the strictly oriented dependency pairs.

DP problem for innermost termination.
P =
  f3#(I16, I17, I18, I19, I20) -> f1#(I16, I17, I18, I19, I20) 
  f2#(I21, I22, I23, I24, I25) -> f3#(I21, I22, I23, I24, I25) [1 <= I25]
  f2#(I26, I27, I28, I29, I30) -> f3#(I26, I27, I28, I29, I30) [1 + I30 <= 0]
R = 
  f6(x1, x2, x3, x4, x5) -> f5(x1, x2, x3, x4, x5) 
  f5(I0, I1, I2, I3, I4) -> f4(rnd1, I1, I2, I3, I4) [rnd1 = rnd1 /\ -1 + I1 <= 0]
  f5(I5, I6, I7, I8, I9) -> f1(I5, I6, I7, I8, I9) [0 <= -2 + I7 /\ 0 <= -2 + I6]
  f1(I10, I11, I12, I13, I14) -> f4(I15, I11, I12, I13, I14) [I15 = I15 /\ -1 - I13 <= 0]
  f3(I16, I17, I18, I19, I20) -> f1(I16, I17, I18, I19, I20) 
  f2(I21, I22, I23, I24, I25) -> f3(I21, I22, I23, I24, I25) [1 <= I25]
  f2(I26, I27, I28, I29, I30) -> f3(I26, I27, I28, I29, I30) [1 + I30 <= 0]
  f1(I31, I32, I33, I34, I35) -> f2(I31, I32, I33, 1 + I34, 1 + I35) [0 <= -2 - I34]

The dependency graph for this problem is:
  2 -> 
  3 -> 2
  4 -> 2
Where:
  2) f3#(I16, I17, I18, I19, I20) -> f1#(I16, I17, I18, I19, I20) 
  3) f2#(I21, I22, I23, I24, I25) -> f3#(I21, I22, I23, I24, I25) [1 <= I25]
  4) f2#(I26, I27, I28, I29, I30) -> f3#(I26, I27, I28, I29, I30) [1 + I30 <= 0]

We have the following SCCs.