/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 =
  f8#(x1, x2, x3) -> f7#(x1, x2, x3) 
  f7#(I0, I1, I2) -> f3#(I0, I1, I2) 
  f7#(I3, I4, I5) -> f6#(I3, I4, I5) 
  f7#(I6, I7, I8) -> f5#(I6, I7, I8) 
  f7#(I9, I10, I11) -> f4#(I9, I10, I11) 
  f7#(I12, I13, I14) -> f1#(I12, I13, I14) 
  f3#(I18, I19, I20) -> f6#(I20, I19, I20) 
  f6#(I21, I22, I23) -> f5#(I23, I22, I23) [I23 <= 0]
  f6#(I24, I25, I26) -> f4#(I26, I25, I26) [1 <= I26]
  f4#(I30, I31, I32) -> f5#(I32, I31, I32) [1 <= I32 /\ I32 <= 1]
  f4#(I33, I34, I35) -> f1#(I35, I34, I35) [1 + I35 <= 1]
  f4#(I36, I37, I38) -> f1#(I38, I37, I38) [2 <= I38]
  f1#(I39, I40, I41) -> f3#(I41, I40, -1 + I41) 
  f1#(I42, I43, I44) -> f3#(I44, I43, -2 + I44) 
R = 
  f8(x1, x2, x3) -> f7(x1, x2, x3) 
  f7(I0, I1, I2) -> f3(I0, I1, I2) 
  f7(I3, I4, I5) -> f6(I3, I4, I5) 
  f7(I6, I7, I8) -> f5(I6, I7, I8) 
  f7(I9, I10, I11) -> f4(I9, I10, I11) 
  f7(I12, I13, I14) -> f1(I12, I13, I14) 
  f7(I15, I16, I17) -> f2(I15, I16, I17) 
  f3(I18, I19, I20) -> f6(I20, I19, I20) 
  f6(I21, I22, I23) -> f5(I23, I22, I23) [I23 <= 0]
  f6(I24, I25, I26) -> f4(I26, I25, I26) [1 <= I26]
  f5(I27, I28, I29) -> f2(I29, rnd2, rnd3) [rnd3 = rnd2 /\ rnd2 = rnd2]
  f4(I30, I31, I32) -> f5(I32, I31, I32) [1 <= I32 /\ I32 <= 1]
  f4(I33, I34, I35) -> f1(I35, I34, I35) [1 + I35 <= 1]
  f4(I36, I37, I38) -> f1(I38, I37, I38) [2 <= I38]
  f1(I39, I40, I41) -> f3(I41, I40, -1 + I41) 
  f1(I42, I43, I44) -> f3(I44, I43, -2 + I44) 
  f1(I45, I46, I47) -> f2(I47, I48, I49) [I49 = I48 /\ I48 = I48]

The dependency graph for this problem is:
  0 -> 1, 2, 3, 4, 5
  1 -> 6
  2 -> 7, 8
  3 -> 
  4 -> 9, 10, 11
  5 -> 12, 13
  6 -> 7, 8
  7 -> 
  8 -> 9, 11
  9 -> 
  10 -> 12, 13
  11 -> 12, 13
  12 -> 6
  13 -> 6
Where:
  0) f8#(x1, x2, x3) -> f7#(x1, x2, x3) 
  1) f7#(I0, I1, I2) -> f3#(I0, I1, I2) 
  2) f7#(I3, I4, I5) -> f6#(I3, I4, I5) 
  3) f7#(I6, I7, I8) -> f5#(I6, I7, I8) 
  4) f7#(I9, I10, I11) -> f4#(I9, I10, I11) 
  5) f7#(I12, I13, I14) -> f1#(I12, I13, I14) 
  6) f3#(I18, I19, I20) -> f6#(I20, I19, I20) 
  7) f6#(I21, I22, I23) -> f5#(I23, I22, I23) [I23 <= 0]
  8) f6#(I24, I25, I26) -> f4#(I26, I25, I26) [1 <= I26]
  9) f4#(I30, I31, I32) -> f5#(I32, I31, I32) [1 <= I32 /\ I32 <= 1]
  10) f4#(I33, I34, I35) -> f1#(I35, I34, I35) [1 + I35 <= 1]
  11) f4#(I36, I37, I38) -> f1#(I38, I37, I38) [2 <= I38]
  12) f1#(I39, I40, I41) -> f3#(I41, I40, -1 + I41) 
  13) f1#(I42, I43, I44) -> f3#(I44, I43, -2 + I44) 

We have the following SCCs.
  { 6, 8, 11, 12, 13 }

DP problem for innermost termination.
P =
  f3#(I18, I19, I20) -> f6#(I20, I19, I20) 
  f6#(I24, I25, I26) -> f4#(I26, I25, I26) [1 <= I26]
  f4#(I36, I37, I38) -> f1#(I38, I37, I38) [2 <= I38]
  f1#(I39, I40, I41) -> f3#(I41, I40, -1 + I41) 
  f1#(I42, I43, I44) -> f3#(I44, I43, -2 + I44) 
R = 
  f8(x1, x2, x3) -> f7(x1, x2, x3) 
  f7(I0, I1, I2) -> f3(I0, I1, I2) 
  f7(I3, I4, I5) -> f6(I3, I4, I5) 
  f7(I6, I7, I8) -> f5(I6, I7, I8) 
  f7(I9, I10, I11) -> f4(I9, I10, I11) 
  f7(I12, I13, I14) -> f1(I12, I13, I14) 
  f7(I15, I16, I17) -> f2(I15, I16, I17) 
  f3(I18, I19, I20) -> f6(I20, I19, I20) 
  f6(I21, I22, I23) -> f5(I23, I22, I23) [I23 <= 0]
  f6(I24, I25, I26) -> f4(I26, I25, I26) [1 <= I26]
  f5(I27, I28, I29) -> f2(I29, rnd2, rnd3) [rnd3 = rnd2 /\ rnd2 = rnd2]
  f4(I30, I31, I32) -> f5(I32, I31, I32) [1 <= I32 /\ I32 <= 1]
  f4(I33, I34, I35) -> f1(I35, I34, I35) [1 + I35 <= 1]
  f4(I36, I37, I38) -> f1(I38, I37, I38) [2 <= I38]
  f1(I39, I40, I41) -> f3(I41, I40, -1 + I41) 
  f1(I42, I43, I44) -> f3(I44, I43, -2 + I44) 
  f1(I45, I46, I47) -> f2(I47, I48, I49) [I49 = I48 /\ I48 = I48]

We use the extended value criterion with the projection function NU:
  NU[f1#(x0,x1,x2)] = x2 - 2
  NU[f4#(x0,x1,x2)] = x2 - 2
  NU[f6#(x0,x1,x2)] = x2 - 1
  NU[f3#(x0,x1,x2)] = x2 - 1

This gives the following inequalities:
    ==>  I20 - 1 >= I20 - 1
  1 <= I26  ==>  I26 - 1 > I26 - 2  with  I26 - 1 >= 0
  2 <= I38  ==>  I38 - 2 >= I38 - 2
    ==>  I41 - 2 >= (-1 + I41) - 1
    ==>  I44 - 2 >= (-2 + I44) - 1

We remove all the strictly oriented dependency pairs.

DP problem for innermost termination.
P =
  f3#(I18, I19, I20) -> f6#(I20, I19, I20) 
  f4#(I36, I37, I38) -> f1#(I38, I37, I38) [2 <= I38]
  f1#(I39, I40, I41) -> f3#(I41, I40, -1 + I41) 
  f1#(I42, I43, I44) -> f3#(I44, I43, -2 + I44) 
R = 
  f8(x1, x2, x3) -> f7(x1, x2, x3) 
  f7(I0, I1, I2) -> f3(I0, I1, I2) 
  f7(I3, I4, I5) -> f6(I3, I4, I5) 
  f7(I6, I7, I8) -> f5(I6, I7, I8) 
  f7(I9, I10, I11) -> f4(I9, I10, I11) 
  f7(I12, I13, I14) -> f1(I12, I13, I14) 
  f7(I15, I16, I17) -> f2(I15, I16, I17) 
  f3(I18, I19, I20) -> f6(I20, I19, I20) 
  f6(I21, I22, I23) -> f5(I23, I22, I23) [I23 <= 0]
  f6(I24, I25, I26) -> f4(I26, I25, I26) [1 <= I26]
  f5(I27, I28, I29) -> f2(I29, rnd2, rnd3) [rnd3 = rnd2 /\ rnd2 = rnd2]
  f4(I30, I31, I32) -> f5(I32, I31, I32) [1 <= I32 /\ I32 <= 1]
  f4(I33, I34, I35) -> f1(I35, I34, I35) [1 + I35 <= 1]
  f4(I36, I37, I38) -> f1(I38, I37, I38) [2 <= I38]
  f1(I39, I40, I41) -> f3(I41, I40, -1 + I41) 
  f1(I42, I43, I44) -> f3(I44, I43, -2 + I44) 
  f1(I45, I46, I47) -> f2(I47, I48, I49) [I49 = I48 /\ I48 = I48]

The dependency graph for this problem is:
  6 -> 
  11 -> 12, 13
  12 -> 6
  13 -> 6
Where:
  6) f3#(I18, I19, I20) -> f6#(I20, I19, I20) 
  11) f4#(I36, I37, I38) -> f1#(I38, I37, I38) [2 <= I38]
  12) f1#(I39, I40, I41) -> f3#(I41, I40, -1 + I41) 
  13) f1#(I42, I43, I44) -> f3#(I44, I43, -2 + I44) 

We have the following SCCs.