1.67/1.96	YES
1.67/1.96	
1.67/1.96	DP problem for innermost termination.
1.67/1.96	P =
1.67/1.96	  f5#(x1, x2, x3) -> f4#(x1, x2, x3) 
1.67/1.96	  f4#(I0, I1, I2) -> f3#(I0, 0, rnd3) [rnd3 = rnd3]
1.67/1.96	  f3#(I3, I4, I5) -> f1#(I3, I4, I5) 
1.67/1.96	  f1#(I6, I7, I8) -> f3#(I6, 1 + I7, I8) [1 + I7 <= I6]
1.67/1.96	R = 
1.67/1.96	  f5(x1, x2, x3) -> f4(x1, x2, x3) 
1.67/1.96	  f4(I0, I1, I2) -> f3(I0, 0, rnd3) [rnd3 = rnd3]
1.67/1.96	  f3(I3, I4, I5) -> f1(I3, I4, I5) 
1.67/1.96	  f1(I6, I7, I8) -> f3(I6, 1 + I7, I8) [1 + I7 <= I6]
1.67/1.96	  f1(I9, I10, I11) -> f2(I9, I10, I11) [I9 <= I10]
1.67/1.96	
1.67/1.96	The dependency graph for this problem is:
1.67/1.96	  0 -> 1
1.67/1.96	  1 -> 2
1.67/1.96	  2 -> 3
1.67/1.96	  3 -> 2
1.67/1.96	Where:
1.67/1.96	  0) f5#(x1, x2, x3) -> f4#(x1, x2, x3) 
1.67/1.96	  1) f4#(I0, I1, I2) -> f3#(I0, 0, rnd3) [rnd3 = rnd3]
1.67/1.96	  2) f3#(I3, I4, I5) -> f1#(I3, I4, I5) 
1.67/1.96	  3) f1#(I6, I7, I8) -> f3#(I6, 1 + I7, I8) [1 + I7 <= I6]
1.67/1.96	
1.67/1.96	We have the following SCCs.
1.67/1.96	  { 2, 3 }
1.67/1.96	
1.67/1.96	DP problem for innermost termination.
1.67/1.96	P =
1.67/1.96	  f3#(I3, I4, I5) -> f1#(I3, I4, I5) 
1.67/1.96	  f1#(I6, I7, I8) -> f3#(I6, 1 + I7, I8) [1 + I7 <= I6]
1.67/1.96	R = 
1.67/1.96	  f5(x1, x2, x3) -> f4(x1, x2, x3) 
1.67/1.96	  f4(I0, I1, I2) -> f3(I0, 0, rnd3) [rnd3 = rnd3]
1.67/1.96	  f3(I3, I4, I5) -> f1(I3, I4, I5) 
1.67/1.96	  f1(I6, I7, I8) -> f3(I6, 1 + I7, I8) [1 + I7 <= I6]
1.67/1.96	  f1(I9, I10, I11) -> f2(I9, I10, I11) [I9 <= I10]
1.67/1.96	
1.67/1.96	We use the reverse value criterion with the projection function NU:
1.67/1.96	  NU[f1#(z1,z2,z3)] = z1 + -1 * (1 + z2)
1.67/1.96	  NU[f3#(z1,z2,z3)] = z1 + -1 * (1 + z2)
1.67/1.96	
1.67/1.96	This gives the following inequalities:
1.67/1.96	    ==>  I3 + -1 * (1 + I4) >= I3 + -1 * (1 + I4)
1.67/1.96	  1 + I7 <= I6  ==>  I6 + -1 * (1 + I7) > I6 + -1 * (1 + (1 + I7))  with  I6 + -1 * (1 + I7) >= 0
1.67/1.96	
1.67/1.96	We remove all the strictly oriented dependency pairs.
1.67/1.96	
1.67/1.96	DP problem for innermost termination.
1.67/1.96	P =
1.67/1.96	  f3#(I3, I4, I5) -> f1#(I3, I4, I5) 
1.67/1.96	R = 
1.67/1.96	  f5(x1, x2, x3) -> f4(x1, x2, x3) 
1.67/1.96	  f4(I0, I1, I2) -> f3(I0, 0, rnd3) [rnd3 = rnd3]
1.67/1.96	  f3(I3, I4, I5) -> f1(I3, I4, I5) 
1.67/1.96	  f1(I6, I7, I8) -> f3(I6, 1 + I7, I8) [1 + I7 <= I6]
1.67/1.96	  f1(I9, I10, I11) -> f2(I9, I10, I11) [I9 <= I10]
1.67/1.96	
1.67/1.96	The dependency graph for this problem is:
1.67/1.96	  2 -> 
1.67/1.96	Where:
1.67/1.96	  2) f3#(I3, I4, I5) -> f1#(I3, I4, I5) 
1.67/1.96	
1.67/1.96	We have the following SCCs.
1.67/1.96	  
1.67/4.94	EOF