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


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YES

DP problem for innermost termination.
P =
  f9#(x1, x2, x3, x4) -> f8#(x1, x2, x3, x4) 
  f8#(I0, I1, I2, I3) -> f7#(0, I1, I2, I3) 
  f2#(I4, I5, I6, I7) -> f7#(1 + I4, I5, I6, I7) 
  f4#(I8, I9, I10, I11) -> f5#(I8, I9, I10, I11) 
  f4#(I12, I13, I14, I15) -> f5#(I12, I13, I14, I15) 
  f7#(I16, I17, I18, I19) -> f3#(I16, I17, I18, I19) 
  f3#(I24, I25, I26, I27) -> f1#(I24, I25, I26, I27) [1 + I24 <= 4]
  f3#(I28, I29, I30, I31) -> f4#(I28, I29, I30, I31) [4 <= I28]
  f1#(I32, I33, I34, I35) -> f2#(I32, I33, I34, I35) 
  f1#(I36, I37, I38, I39) -> f2#(I36, rnd2, I36, 1 + I36) [rnd2 = rnd2]
R = 
  f9(x1, x2, x3, x4) -> f8(x1, x2, x3, x4) 
  f8(I0, I1, I2, I3) -> f7(0, I1, I2, I3) 
  f2(I4, I5, I6, I7) -> f7(1 + I4, I5, I6, I7) 
  f4(I8, I9, I10, I11) -> f5(I8, I9, I10, I11) 
  f4(I12, I13, I14, I15) -> f5(I12, I13, I14, I15) 
  f7(I16, I17, I18, I19) -> f3(I16, I17, I18, I19) 
  f5(I20, I21, I22, I23) -> f6(I20, I21, I22, I23) 
  f3(I24, I25, I26, I27) -> f1(I24, I25, I26, I27) [1 + I24 <= 4]
  f3(I28, I29, I30, I31) -> f4(I28, I29, I30, I31) [4 <= I28]
  f1(I32, I33, I34, I35) -> f2(I32, I33, I34, I35) 
  f1(I36, I37, I38, I39) -> f2(I36, rnd2, I36, 1 + I36) [rnd2 = rnd2]

The dependency graph for this problem is:
  0 -> 1
  1 -> 5
  2 -> 5
  3 -> 
  4 -> 
  5 -> 6, 7
  6 -> 8, 9
  7 -> 3, 4
  8 -> 2
  9 -> 2
Where:
  0) f9#(x1, x2, x3, x4) -> f8#(x1, x2, x3, x4) 
  1) f8#(I0, I1, I2, I3) -> f7#(0, I1, I2, I3) 
  2) f2#(I4, I5, I6, I7) -> f7#(1 + I4, I5, I6, I7) 
  3) f4#(I8, I9, I10, I11) -> f5#(I8, I9, I10, I11) 
  4) f4#(I12, I13, I14, I15) -> f5#(I12, I13, I14, I15) 
  5) f7#(I16, I17, I18, I19) -> f3#(I16, I17, I18, I19) 
  6) f3#(I24, I25, I26, I27) -> f1#(I24, I25, I26, I27) [1 + I24 <= 4]
  7) f3#(I28, I29, I30, I31) -> f4#(I28, I29, I30, I31) [4 <= I28]
  8) f1#(I32, I33, I34, I35) -> f2#(I32, I33, I34, I35) 
  9) f1#(I36, I37, I38, I39) -> f2#(I36, rnd2, I36, 1 + I36) [rnd2 = rnd2]

We have the following SCCs.
  { 2, 5, 6, 8, 9 }

DP problem for innermost termination.
P =
  f2#(I4, I5, I6, I7) -> f7#(1 + I4, I5, I6, I7) 
  f7#(I16, I17, I18, I19) -> f3#(I16, I17, I18, I19) 
  f3#(I24, I25, I26, I27) -> f1#(I24, I25, I26, I27) [1 + I24 <= 4]
  f1#(I32, I33, I34, I35) -> f2#(I32, I33, I34, I35) 
  f1#(I36, I37, I38, I39) -> f2#(I36, rnd2, I36, 1 + I36) [rnd2 = rnd2]
R = 
  f9(x1, x2, x3, x4) -> f8(x1, x2, x3, x4) 
  f8(I0, I1, I2, I3) -> f7(0, I1, I2, I3) 
  f2(I4, I5, I6, I7) -> f7(1 + I4, I5, I6, I7) 
  f4(I8, I9, I10, I11) -> f5(I8, I9, I10, I11) 
  f4(I12, I13, I14, I15) -> f5(I12, I13, I14, I15) 
  f7(I16, I17, I18, I19) -> f3(I16, I17, I18, I19) 
  f5(I20, I21, I22, I23) -> f6(I20, I21, I22, I23) 
  f3(I24, I25, I26, I27) -> f1(I24, I25, I26, I27) [1 + I24 <= 4]
  f3(I28, I29, I30, I31) -> f4(I28, I29, I30, I31) [4 <= I28]
  f1(I32, I33, I34, I35) -> f2(I32, I33, I34, I35) 
  f1(I36, I37, I38, I39) -> f2(I36, rnd2, I36, 1 + I36) [rnd2 = rnd2]

We use the extended value criterion with the projection function NU:
  NU[f1#(x0,x1,x2,x3)] = -x0 + 2
  NU[f3#(x0,x1,x2,x3)] = -x0 + 3
  NU[f7#(x0,x1,x2,x3)] = -x0 + 3
  NU[f2#(x0,x1,x2,x3)] = -x0 + 2

This gives the following inequalities:
    ==>  -I4 + 2 >= -(1 + I4) + 3
    ==>  -I16 + 3 >= -I16 + 3
  1 + I24 <= 4  ==>  -I24 + 3 > -I24 + 2  with  -I24 + 3 >= 0
    ==>  -I32 + 2 >= -I32 + 2
  rnd2 = rnd2  ==>  -I36 + 2 >= -I36 + 2

We remove all the strictly oriented dependency pairs.

DP problem for innermost termination.
P =
  f2#(I4, I5, I6, I7) -> f7#(1 + I4, I5, I6, I7) 
  f7#(I16, I17, I18, I19) -> f3#(I16, I17, I18, I19) 
  f1#(I32, I33, I34, I35) -> f2#(I32, I33, I34, I35) 
  f1#(I36, I37, I38, I39) -> f2#(I36, rnd2, I36, 1 + I36) [rnd2 = rnd2]
R = 
  f9(x1, x2, x3, x4) -> f8(x1, x2, x3, x4) 
  f8(I0, I1, I2, I3) -> f7(0, I1, I2, I3) 
  f2(I4, I5, I6, I7) -> f7(1 + I4, I5, I6, I7) 
  f4(I8, I9, I10, I11) -> f5(I8, I9, I10, I11) 
  f4(I12, I13, I14, I15) -> f5(I12, I13, I14, I15) 
  f7(I16, I17, I18, I19) -> f3(I16, I17, I18, I19) 
  f5(I20, I21, I22, I23) -> f6(I20, I21, I22, I23) 
  f3(I24, I25, I26, I27) -> f1(I24, I25, I26, I27) [1 + I24 <= 4]
  f3(I28, I29, I30, I31) -> f4(I28, I29, I30, I31) [4 <= I28]
  f1(I32, I33, I34, I35) -> f2(I32, I33, I34, I35) 
  f1(I36, I37, I38, I39) -> f2(I36, rnd2, I36, 1 + I36) [rnd2 = rnd2]

The dependency graph for this problem is:
  2 -> 5
  5 -> 
  8 -> 2
  9 -> 2
Where:
  2) f2#(I4, I5, I6, I7) -> f7#(1 + I4, I5, I6, I7) 
  5) f7#(I16, I17, I18, I19) -> f3#(I16, I17, I18, I19) 
  8) f1#(I32, I33, I34, I35) -> f2#(I32, I33, I34, I35) 
  9) f1#(I36, I37, I38, I39) -> f2#(I36, rnd2, I36, 1 + I36) [rnd2 = rnd2]

We have the following SCCs.