0.00/0.22	YES
0.00/0.22	
0.00/0.22	DP problem for innermost termination.
0.00/0.22	P =
0.00/0.22	  f#(I0, I1, I2, I3) -> f#(I0, I1 + 1, I2 + 2, I3 + I2 + 2) [I0 >= I3 && I2 >= 0]
0.00/0.22	  sqrt#(I4) -> f#(I4, 0, 1, 1) 
0.00/0.22	R = 
0.00/0.22	  f(x, z, u, w) -> z [w > x]
0.00/0.22	  f(I0, I1, I2, I3) -> f(I0, I1 + 1, I2 + 2, I3 + I2 + 2) [I0 >= I3 && I2 >= 0]
0.00/0.22	  sqrt(I4) -> f(I4, 0, 1, 1) 
0.00/0.22	
0.00/0.22	The dependency graph for this problem is:
0.00/0.22	  0 -> 0
0.00/0.22	  1 -> 0
0.00/0.22	Where:
0.00/0.22	  0) f#(I0, I1, I2, I3) -> f#(I0, I1 + 1, I2 + 2, I3 + I2 + 2) [I0 >= I3 && I2 >= 0]
0.00/0.22	  1) sqrt#(I4) -> f#(I4, 0, 1, 1) 
0.00/0.22	
0.00/0.22	We have the following SCCs.
0.00/0.22	  { 0 }
0.00/0.22	
0.00/0.22	DP problem for innermost termination.
0.00/0.22	P =
0.00/0.22	  f#(I0, I1, I2, I3) -> f#(I0, I1 + 1, I2 + 2, I3 + I2 + 2) [I0 >= I3 && I2 >= 0]
0.00/0.22	R = 
0.00/0.22	  f(x, z, u, w) -> z [w > x]
0.00/0.22	  f(I0, I1, I2, I3) -> f(I0, I1 + 1, I2 + 2, I3 + I2 + 2) [I0 >= I3 && I2 >= 0]
0.00/0.22	  sqrt(I4) -> f(I4, 0, 1, 1) 
0.00/0.22	
0.00/0.22	We use the reverse value criterion with the projection function NU:
0.00/0.22	  NU[f#(z1,z2,z3,z4)] = z1 + -1 * z4
0.00/0.22	
0.00/0.22	This gives the following inequalities:
0.00/0.22	  I0 >= I3 && I2 >= 0  ==>  I0 + -1 * I3 > I0 + -1 * (I3 + I2 + 2)  with  I0 + -1 * I3 >= 0
0.00/0.22	
0.00/0.22	All dependency pairs are strictly oriented, so the entire dependency pair problem may be removed.
0.00/3.20	EOF