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


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


YES

DP problem for innermost termination.
P =
  f5#(x1, x2, x3, x4, x5, x6, x7, x8, x9, x10) -> f4#(x1, x2, x3, x4, x5, x6, x7, x8, x9, x10) 
  f4#(I0, I1, I2, I3, I4, I5, I6, I7, I8, I9) -> f3#(1, 0, I2, I2, I4, 2, I2, I7, I8, I9) 
  f3#(I10, I11, I12, I13, I14, I15, I16, I17, I18, I19) -> f1#(I10, I11, I12, I13, I14, I15, I16, I17, I18, I19) 
  f1#(I20, I21, I22, I23, I24, I25, I26, I27, I28, I29) -> f3#(I20 + I21, rnd2, I22, I23, I24, 1 + I25, I26, I27, I20, I29) [rnd2 = I20 /\ I25 <= I26]
R = 
  f5(x1, x2, x3, x4, x5, x6, x7, x8, x9, x10) -> f4(x1, x2, x3, x4, x5, x6, x7, x8, x9, x10) 
  f4(I0, I1, I2, I3, I4, I5, I6, I7, I8, I9) -> f3(1, 0, I2, I2, I4, 2, I2, I7, I8, I9) 
  f3(I10, I11, I12, I13, I14, I15, I16, I17, I18, I19) -> f1(I10, I11, I12, I13, I14, I15, I16, I17, I18, I19) 
  f1(I20, I21, I22, I23, I24, I25, I26, I27, I28, I29) -> f3(I20 + I21, rnd2, I22, I23, I24, 1 + I25, I26, I27, I20, I29) [rnd2 = I20 /\ I25 <= I26]
  f1(I30, I31, I32, I33, I34, I35, I36, I37, I38, I39) -> f2(I30, I31, I32, I33, I30, I35, I36, I30, I38, I30) [1 + I36 <= I35]

The dependency graph for this problem is:
  0 -> 1
  1 -> 2
  2 -> 3
  3 -> 2
Where:
  0) f5#(x1, x2, x3, x4, x5, x6, x7, x8, x9, x10) -> f4#(x1, x2, x3, x4, x5, x6, x7, x8, x9, x10) 
  1) f4#(I0, I1, I2, I3, I4, I5, I6, I7, I8, I9) -> f3#(1, 0, I2, I2, I4, 2, I2, I7, I8, I9) 
  2) f3#(I10, I11, I12, I13, I14, I15, I16, I17, I18, I19) -> f1#(I10, I11, I12, I13, I14, I15, I16, I17, I18, I19) 
  3) f1#(I20, I21, I22, I23, I24, I25, I26, I27, I28, I29) -> f3#(I20 + I21, rnd2, I22, I23, I24, 1 + I25, I26, I27, I20, I29) [rnd2 = I20 /\ I25 <= I26]

We have the following SCCs.
  { 2, 3 }

DP problem for innermost termination.
P =
  f3#(I10, I11, I12, I13, I14, I15, I16, I17, I18, I19) -> f1#(I10, I11, I12, I13, I14, I15, I16, I17, I18, I19) 
  f1#(I20, I21, I22, I23, I24, I25, I26, I27, I28, I29) -> f3#(I20 + I21, rnd2, I22, I23, I24, 1 + I25, I26, I27, I20, I29) [rnd2 = I20 /\ I25 <= I26]
R = 
  f5(x1, x2, x3, x4, x5, x6, x7, x8, x9, x10) -> f4(x1, x2, x3, x4, x5, x6, x7, x8, x9, x10) 
  f4(I0, I1, I2, I3, I4, I5, I6, I7, I8, I9) -> f3(1, 0, I2, I2, I4, 2, I2, I7, I8, I9) 
  f3(I10, I11, I12, I13, I14, I15, I16, I17, I18, I19) -> f1(I10, I11, I12, I13, I14, I15, I16, I17, I18, I19) 
  f1(I20, I21, I22, I23, I24, I25, I26, I27, I28, I29) -> f3(I20 + I21, rnd2, I22, I23, I24, 1 + I25, I26, I27, I20, I29) [rnd2 = I20 /\ I25 <= I26]
  f1(I30, I31, I32, I33, I34, I35, I36, I37, I38, I39) -> f2(I30, I31, I32, I33, I30, I35, I36, I30, I38, I30) [1 + I36 <= I35]

We use the reverse value criterion with the projection function NU:
  NU[f1#(z1,z2,z3,z4,z5,z6,z7,z8,z9,z10)] = z7 + -1 * z6
  NU[f3#(z1,z2,z3,z4,z5,z6,z7,z8,z9,z10)] = z7 + -1 * z6

This gives the following inequalities:
    ==>  I16 + -1 * I15 >= I16 + -1 * I15
  rnd2 = I20 /\ I25 <= I26  ==>  I26 + -1 * I25 > I26 + -1 * (1 + I25)  with  I26 + -1 * I25 >= 0

We remove all the strictly oriented dependency pairs.

DP problem for innermost termination.
P =
  f3#(I10, I11, I12, I13, I14, I15, I16, I17, I18, I19) -> f1#(I10, I11, I12, I13, I14, I15, I16, I17, I18, I19) 
R = 
  f5(x1, x2, x3, x4, x5, x6, x7, x8, x9, x10) -> f4(x1, x2, x3, x4, x5, x6, x7, x8, x9, x10) 
  f4(I0, I1, I2, I3, I4, I5, I6, I7, I8, I9) -> f3(1, 0, I2, I2, I4, 2, I2, I7, I8, I9) 
  f3(I10, I11, I12, I13, I14, I15, I16, I17, I18, I19) -> f1(I10, I11, I12, I13, I14, I15, I16, I17, I18, I19) 
  f1(I20, I21, I22, I23, I24, I25, I26, I27, I28, I29) -> f3(I20 + I21, rnd2, I22, I23, I24, 1 + I25, I26, I27, I20, I29) [rnd2 = I20 /\ I25 <= I26]
  f1(I30, I31, I32, I33, I34, I35, I36, I37, I38, I39) -> f2(I30, I31, I32, I33, I30, I35, I36, I30, I38, I30) [1 + I36 <= I35]

The dependency graph for this problem is:
  2 -> 
Where:
  2) f3#(I10, I11, I12, I13, I14, I15, I16, I17, I18, I19) -> f1#(I10, I11, I12, I13, I14, I15, I16, I17, I18, I19) 

We have the following SCCs.