4.04/4.02	YES
4.04/4.02	
4.04/4.02	DP problem for innermost termination.
4.04/4.02	P =
4.04/4.02	  f5#(x1, x2, x3, x4, x5) -> f4#(x1, x2, x3, x4, x5) 
4.04/4.02	  f4#(I0, I1, I2, I3, I4) -> f3#(0, 10, rnd3, rnd4, I4) [rnd3 = rnd4 /\ rnd4 = rnd4]
4.04/4.02	  f3#(I5, I6, I7, I8, I9) -> f1#(I5, I6, I7, I8, I9) 
4.04/4.02	  f1#(I10, I11, I12, I13, I14) -> f3#(1 + I10, I11, I12, I13, rnd5) [rnd5 = rnd5 /\ 1 + I10 <= I11]
4.04/4.02	R = 
4.04/4.02	  f5(x1, x2, x3, x4, x5) -> f4(x1, x2, x3, x4, x5) 
4.04/4.02	  f4(I0, I1, I2, I3, I4) -> f3(0, 10, rnd3, rnd4, I4) [rnd3 = rnd4 /\ rnd4 = rnd4]
4.04/4.02	  f3(I5, I6, I7, I8, I9) -> f1(I5, I6, I7, I8, I9) 
4.04/4.02	  f1(I10, I11, I12, I13, I14) -> f3(1 + I10, I11, I12, I13, rnd5) [rnd5 = rnd5 /\ 1 + I10 <= I11]
4.04/4.02	  f1(I15, I16, I17, I18, I19) -> f2(I15, I16, I17, I18, I19) [I16 <= I15]
4.04/4.02	
4.04/4.02	The dependency graph for this problem is:
4.04/4.02	  0 -> 1
4.04/4.02	  1 -> 2
4.04/4.02	  2 -> 3
4.04/4.02	  3 -> 2
4.04/4.02	Where:
4.04/4.02	  0) f5#(x1, x2, x3, x4, x5) -> f4#(x1, x2, x3, x4, x5) 
4.04/4.02	  1) f4#(I0, I1, I2, I3, I4) -> f3#(0, 10, rnd3, rnd4, I4) [rnd3 = rnd4 /\ rnd4 = rnd4]
4.04/4.02	  2) f3#(I5, I6, I7, I8, I9) -> f1#(I5, I6, I7, I8, I9) 
4.04/4.02	  3) f1#(I10, I11, I12, I13, I14) -> f3#(1 + I10, I11, I12, I13, rnd5) [rnd5 = rnd5 /\ 1 + I10 <= I11]
4.04/4.02	
4.04/4.02	We have the following SCCs.
4.04/4.02	  { 2, 3 }
4.04/4.02	
4.04/4.02	DP problem for innermost termination.
4.04/4.02	P =
4.04/4.02	  f3#(I5, I6, I7, I8, I9) -> f1#(I5, I6, I7, I8, I9) 
4.04/4.02	  f1#(I10, I11, I12, I13, I14) -> f3#(1 + I10, I11, I12, I13, rnd5) [rnd5 = rnd5 /\ 1 + I10 <= I11]
4.04/4.02	R = 
4.04/4.02	  f5(x1, x2, x3, x4, x5) -> f4(x1, x2, x3, x4, x5) 
4.04/4.02	  f4(I0, I1, I2, I3, I4) -> f3(0, 10, rnd3, rnd4, I4) [rnd3 = rnd4 /\ rnd4 = rnd4]
4.04/4.02	  f3(I5, I6, I7, I8, I9) -> f1(I5, I6, I7, I8, I9) 
4.04/4.02	  f1(I10, I11, I12, I13, I14) -> f3(1 + I10, I11, I12, I13, rnd5) [rnd5 = rnd5 /\ 1 + I10 <= I11]
4.04/4.02	  f1(I15, I16, I17, I18, I19) -> f2(I15, I16, I17, I18, I19) [I16 <= I15]
4.04/4.02	
4.04/4.02	We use the reverse value criterion with the projection function NU:
4.04/4.02	  NU[f1#(z1,z2,z3,z4,z5)] = z2 + -1 * (1 + z1)
4.04/4.02	  NU[f3#(z1,z2,z3,z4,z5)] = z2 + -1 * (1 + z1)
4.04/4.02	
4.04/4.02	This gives the following inequalities:
4.04/4.02	    ==>  I6 + -1 * (1 + I5) >= I6 + -1 * (1 + I5)
4.04/4.02	  rnd5 = rnd5 /\ 1 + I10 <= I11  ==>  I11 + -1 * (1 + I10) > I11 + -1 * (1 + (1 + I10))  with  I11 + -1 * (1 + I10) >= 0
4.04/4.02	
4.04/4.02	We remove all the strictly oriented dependency pairs.
4.04/4.02	
4.04/4.02	DP problem for innermost termination.
4.04/4.02	P =
4.04/4.02	  f3#(I5, I6, I7, I8, I9) -> f1#(I5, I6, I7, I8, I9) 
4.04/4.02	R = 
4.04/4.02	  f5(x1, x2, x3, x4, x5) -> f4(x1, x2, x3, x4, x5) 
4.04/4.02	  f4(I0, I1, I2, I3, I4) -> f3(0, 10, rnd3, rnd4, I4) [rnd3 = rnd4 /\ rnd4 = rnd4]
4.04/4.02	  f3(I5, I6, I7, I8, I9) -> f1(I5, I6, I7, I8, I9) 
4.04/4.02	  f1(I10, I11, I12, I13, I14) -> f3(1 + I10, I11, I12, I13, rnd5) [rnd5 = rnd5 /\ 1 + I10 <= I11]
4.04/4.02	  f1(I15, I16, I17, I18, I19) -> f2(I15, I16, I17, I18, I19) [I16 <= I15]
4.04/4.02	
4.04/4.02	The dependency graph for this problem is:
4.04/4.02	  2 -> 
4.04/4.02	Where:
4.04/4.02	  2) f3#(I5, I6, I7, I8, I9) -> f1#(I5, I6, I7, I8, I9) 
4.04/4.02	
4.04/4.02	We have the following SCCs.
4.04/4.02	  
4.04/7.00	EOF