12.16/11.99	YES
12.16/11.99	
12.16/11.99	DP problem for innermost termination.
12.16/11.99	P =
12.16/11.99	  f5#(x1, x2, x3, x4, x5, x6, x7, x8) -> f4#(x1, x2, x3, x4, x5, x6, x7, x8) 
12.16/11.99	  f4#(I0, I1, I2, I3, I4, I5, I6, I7) -> f1#(I0, I1, I2, I3, I4, I5, I2, I7) 
12.16/11.99	  f3#(I8, I9, I10, I11, I12, I13, I14, I15) -> f1#(I8, I9, I10, I11, I12, I13, I14, I15) 
12.16/11.99	  f1#(I16, I17, I18, I19, I20, I21, I22, I23) -> f3#(I16, I17, I18, I19, I20, I21, -1 + I22, I23) [0 <= -1 * I19 + I22 /\ 0 <= I20 - I23]
12.16/11.99	R = 
12.16/11.99	  f5(x1, x2, x3, x4, x5, x6, x7, x8) -> f4(x1, x2, x3, x4, x5, x6, x7, x8) 
12.16/11.99	  f4(I0, I1, I2, I3, I4, I5, I6, I7) -> f1(I0, I1, I2, I3, I4, I5, I2, I7) 
12.16/11.99	  f3(I8, I9, I10, I11, I12, I13, I14, I15) -> f1(I8, I9, I10, I11, I12, I13, I14, I15) 
12.16/11.99	  f1(I16, I17, I18, I19, I20, I21, I22, I23) -> f3(I16, I17, I18, I19, I20, I21, -1 + I22, I23) [0 <= -1 * I19 + I22 /\ 0 <= I20 - I23]
12.16/11.99	  f1(I24, I25, I26, I27, I28, I29, I30, I31) -> f2(rnd1, rnd2, I26, I27, I28, 0, I30, I31) [rnd1 = rnd2 /\ rnd2 = 0 /\ 1 - I27 + I30 <= 0 /\ 0 <= I28 - I31]
12.16/11.99	  f1(I32, I33, I34, I35, I36, I37, I38, I39) -> f2(I40, I41, I34, I35, I36, 0, I38, I39) [I40 = I41 /\ I41 = 0 /\ 1 + I36 - I39 <= 0]
12.16/11.99	
12.16/11.99	The dependency graph for this problem is:
12.16/11.99	  0 -> 1
12.16/11.99	  1 -> 3
12.16/11.99	  2 -> 3
12.16/11.99	  3 -> 2
12.16/11.99	Where:
12.16/11.99	  0) f5#(x1, x2, x3, x4, x5, x6, x7, x8) -> f4#(x1, x2, x3, x4, x5, x6, x7, x8) 
12.16/11.99	  1) f4#(I0, I1, I2, I3, I4, I5, I6, I7) -> f1#(I0, I1, I2, I3, I4, I5, I2, I7) 
12.16/11.99	  2) f3#(I8, I9, I10, I11, I12, I13, I14, I15) -> f1#(I8, I9, I10, I11, I12, I13, I14, I15) 
12.16/11.99	  3) f1#(I16, I17, I18, I19, I20, I21, I22, I23) -> f3#(I16, I17, I18, I19, I20, I21, -1 + I22, I23) [0 <= -1 * I19 + I22 /\ 0 <= I20 - I23]
12.16/11.99	
12.16/11.99	We have the following SCCs.
12.16/11.99	  { 2, 3 }
12.16/11.99	
12.16/11.99	DP problem for innermost termination.
12.16/11.99	P =
12.16/11.99	  f3#(I8, I9, I10, I11, I12, I13, I14, I15) -> f1#(I8, I9, I10, I11, I12, I13, I14, I15) 
12.16/11.99	  f1#(I16, I17, I18, I19, I20, I21, I22, I23) -> f3#(I16, I17, I18, I19, I20, I21, -1 + I22, I23) [0 <= -1 * I19 + I22 /\ 0 <= I20 - I23]
12.16/11.99	R = 
12.16/11.99	  f5(x1, x2, x3, x4, x5, x6, x7, x8) -> f4(x1, x2, x3, x4, x5, x6, x7, x8) 
12.16/11.99	  f4(I0, I1, I2, I3, I4, I5, I6, I7) -> f1(I0, I1, I2, I3, I4, I5, I2, I7) 
12.16/11.99	  f3(I8, I9, I10, I11, I12, I13, I14, I15) -> f1(I8, I9, I10, I11, I12, I13, I14, I15) 
12.16/11.99	  f1(I16, I17, I18, I19, I20, I21, I22, I23) -> f3(I16, I17, I18, I19, I20, I21, -1 + I22, I23) [0 <= -1 * I19 + I22 /\ 0 <= I20 - I23]
12.16/11.99	  f1(I24, I25, I26, I27, I28, I29, I30, I31) -> f2(rnd1, rnd2, I26, I27, I28, 0, I30, I31) [rnd1 = rnd2 /\ rnd2 = 0 /\ 1 - I27 + I30 <= 0 /\ 0 <= I28 - I31]
12.16/11.99	  f1(I32, I33, I34, I35, I36, I37, I38, I39) -> f2(I40, I41, I34, I35, I36, 0, I38, I39) [I40 = I41 /\ I41 = 0 /\ 1 + I36 - I39 <= 0]
12.16/11.99	
12.16/11.99	We use the reverse value criterion with the projection function NU:
12.16/11.99	  NU[f1#(z1,z2,z3,z4,z5,z6,z7,z8)] = -1 * z4 + z7 + -1 * 0
12.16/11.99	  NU[f3#(z1,z2,z3,z4,z5,z6,z7,z8)] = -1 * z4 + z7 + -1 * 0
12.16/11.99	
12.16/11.99	This gives the following inequalities:
12.16/11.99	    ==>  -1 * I11 + I14 + -1 * 0 >= -1 * I11 + I14 + -1 * 0
12.16/11.99	  0 <= -1 * I19 + I22 /\ 0 <= I20 - I23  ==>  -1 * I19 + I22 + -1 * 0 > -1 * I19 + (-1 + I22) + -1 * 0  with  -1 * I19 + I22 + -1 * 0 >= 0
12.16/11.99	
12.16/11.99	We remove all the strictly oriented dependency pairs.
12.16/11.99	
12.16/11.99	DP problem for innermost termination.
12.16/11.99	P =
12.16/11.99	  f3#(I8, I9, I10, I11, I12, I13, I14, I15) -> f1#(I8, I9, I10, I11, I12, I13, I14, I15) 
12.16/11.99	R = 
12.16/11.99	  f5(x1, x2, x3, x4, x5, x6, x7, x8) -> f4(x1, x2, x3, x4, x5, x6, x7, x8) 
12.16/11.99	  f4(I0, I1, I2, I3, I4, I5, I6, I7) -> f1(I0, I1, I2, I3, I4, I5, I2, I7) 
12.16/11.99	  f3(I8, I9, I10, I11, I12, I13, I14, I15) -> f1(I8, I9, I10, I11, I12, I13, I14, I15) 
12.16/11.99	  f1(I16, I17, I18, I19, I20, I21, I22, I23) -> f3(I16, I17, I18, I19, I20, I21, -1 + I22, I23) [0 <= -1 * I19 + I22 /\ 0 <= I20 - I23]
12.16/11.99	  f1(I24, I25, I26, I27, I28, I29, I30, I31) -> f2(rnd1, rnd2, I26, I27, I28, 0, I30, I31) [rnd1 = rnd2 /\ rnd2 = 0 /\ 1 - I27 + I30 <= 0 /\ 0 <= I28 - I31]
12.16/11.99	  f1(I32, I33, I34, I35, I36, I37, I38, I39) -> f2(I40, I41, I34, I35, I36, 0, I38, I39) [I40 = I41 /\ I41 = 0 /\ 1 + I36 - I39 <= 0]
12.16/11.99	
12.16/11.99	The dependency graph for this problem is:
12.16/11.99	  2 -> 
12.16/11.99	Where:
12.16/11.99	  2) f3#(I8, I9, I10, I11, I12, I13, I14, I15) -> f1#(I8, I9, I10, I11, I12, I13, I14, I15) 
12.16/11.99	
12.16/11.99	We have the following SCCs.
12.16/11.99	  
12.16/14.96	EOF