6.30/6.32	YES
6.30/6.32	
6.30/6.32	DP problem for innermost termination.
6.30/6.32	P =
6.30/6.32	  f5#(x1, x2, x3, x4, x5, x6) -> f4#(x1, x2, x3, x4, x5, x6) 
6.30/6.32	  f4#(I0, I1, I2, I3, I4, I5) -> f3#(I0, 0, I0, rnd4, rnd5, I5) [rnd4 = rnd5 /\ rnd5 = rnd5]
6.30/6.32	  f3#(I6, I7, I8, I9, I10, I11) -> f1#(I6, I7, I8, I9, I10, I11) 
6.30/6.32	  f1#(I12, I13, I14, I15, I16, I17) -> f3#(I12, 1 + I13, I14, I15, I16, rnd6) [rnd6 = rnd6 /\ 1 + I13 <= I14]
6.30/6.32	R = 
6.30/6.32	  f5(x1, x2, x3, x4, x5, x6) -> f4(x1, x2, x3, x4, x5, x6) 
6.30/6.32	  f4(I0, I1, I2, I3, I4, I5) -> f3(I0, 0, I0, rnd4, rnd5, I5) [rnd4 = rnd5 /\ rnd5 = rnd5]
6.30/6.32	  f3(I6, I7, I8, I9, I10, I11) -> f1(I6, I7, I8, I9, I10, I11) 
6.30/6.32	  f1(I12, I13, I14, I15, I16, I17) -> f3(I12, 1 + I13, I14, I15, I16, rnd6) [rnd6 = rnd6 /\ 1 + I13 <= I14]
6.30/6.32	  f1(I18, I19, I20, I21, I22, I23) -> f2(I18, I19, I20, I21, I22, I23) [I20 <= I19]
6.30/6.32	
6.30/6.32	The dependency graph for this problem is:
6.30/6.32	  0 -> 1
6.30/6.32	  1 -> 2
6.30/6.32	  2 -> 3
6.30/6.32	  3 -> 2
6.30/6.32	Where:
6.30/6.32	  0) f5#(x1, x2, x3, x4, x5, x6) -> f4#(x1, x2, x3, x4, x5, x6) 
6.30/6.32	  1) f4#(I0, I1, I2, I3, I4, I5) -> f3#(I0, 0, I0, rnd4, rnd5, I5) [rnd4 = rnd5 /\ rnd5 = rnd5]
6.30/6.32	  2) f3#(I6, I7, I8, I9, I10, I11) -> f1#(I6, I7, I8, I9, I10, I11) 
6.30/6.32	  3) f1#(I12, I13, I14, I15, I16, I17) -> f3#(I12, 1 + I13, I14, I15, I16, rnd6) [rnd6 = rnd6 /\ 1 + I13 <= I14]
6.30/6.32	
6.30/6.32	We have the following SCCs.
6.30/6.32	  { 2, 3 }
6.30/6.32	
6.30/6.32	DP problem for innermost termination.
6.30/6.32	P =
6.30/6.32	  f3#(I6, I7, I8, I9, I10, I11) -> f1#(I6, I7, I8, I9, I10, I11) 
6.30/6.32	  f1#(I12, I13, I14, I15, I16, I17) -> f3#(I12, 1 + I13, I14, I15, I16, rnd6) [rnd6 = rnd6 /\ 1 + I13 <= I14]
6.30/6.32	R = 
6.30/6.32	  f5(x1, x2, x3, x4, x5, x6) -> f4(x1, x2, x3, x4, x5, x6) 
6.30/6.32	  f4(I0, I1, I2, I3, I4, I5) -> f3(I0, 0, I0, rnd4, rnd5, I5) [rnd4 = rnd5 /\ rnd5 = rnd5]
6.30/6.32	  f3(I6, I7, I8, I9, I10, I11) -> f1(I6, I7, I8, I9, I10, I11) 
6.30/6.32	  f1(I12, I13, I14, I15, I16, I17) -> f3(I12, 1 + I13, I14, I15, I16, rnd6) [rnd6 = rnd6 /\ 1 + I13 <= I14]
6.30/6.32	  f1(I18, I19, I20, I21, I22, I23) -> f2(I18, I19, I20, I21, I22, I23) [I20 <= I19]
6.30/6.32	
6.30/6.32	We use the reverse value criterion with the projection function NU:
6.30/6.32	  NU[f1#(z1,z2,z3,z4,z5,z6)] = z3 + -1 * (1 + z2)
6.30/6.32	  NU[f3#(z1,z2,z3,z4,z5,z6)] = z3 + -1 * (1 + z2)
6.30/6.32	
6.30/6.32	This gives the following inequalities:
6.30/6.32	    ==>  I8 + -1 * (1 + I7) >= I8 + -1 * (1 + I7)
6.30/6.32	  rnd6 = rnd6 /\ 1 + I13 <= I14  ==>  I14 + -1 * (1 + I13) > I14 + -1 * (1 + (1 + I13))  with  I14 + -1 * (1 + I13) >= 0
6.30/6.32	
6.30/6.32	We remove all the strictly oriented dependency pairs.
6.30/6.32	
6.30/6.32	DP problem for innermost termination.
6.30/6.32	P =
6.30/6.32	  f3#(I6, I7, I8, I9, I10, I11) -> f1#(I6, I7, I8, I9, I10, I11) 
6.30/6.32	R = 
6.30/6.32	  f5(x1, x2, x3, x4, x5, x6) -> f4(x1, x2, x3, x4, x5, x6) 
6.30/6.32	  f4(I0, I1, I2, I3, I4, I5) -> f3(I0, 0, I0, rnd4, rnd5, I5) [rnd4 = rnd5 /\ rnd5 = rnd5]
6.30/6.32	  f3(I6, I7, I8, I9, I10, I11) -> f1(I6, I7, I8, I9, I10, I11) 
6.30/6.32	  f1(I12, I13, I14, I15, I16, I17) -> f3(I12, 1 + I13, I14, I15, I16, rnd6) [rnd6 = rnd6 /\ 1 + I13 <= I14]
6.30/6.32	  f1(I18, I19, I20, I21, I22, I23) -> f2(I18, I19, I20, I21, I22, I23) [I20 <= I19]
6.30/6.32	
6.30/6.32	The dependency graph for this problem is:
6.30/6.32	  2 -> 
6.30/6.32	Where:
6.30/6.32	  2) f3#(I6, I7, I8, I9, I10, I11) -> f1#(I6, I7, I8, I9, I10, I11) 
6.30/6.32	
6.30/6.32	We have the following SCCs.
6.30/6.32	  
6.40/9.29	EOF