0.00/0.50	YES
0.00/0.50	
0.00/0.50	DP problem for innermost termination.
0.00/0.50	P =
0.00/0.50	  f#(x, y, z) -> f#(x, y, z + 1) [x > y && x > z]
0.00/0.50	  f#(I0, I1, I2) -> f#(I0, I1 + 1, I2) [I0 > I1 && I0 > I2]
0.00/0.50	R = 
0.00/0.50	  f(x, y, z) -> f(x, y, z + 1) [x > y && x > z]
0.00/0.50	  f(I0, I1, I2) -> f(I0, I1 + 1, I2) [I0 > I1 && I0 > I2]
0.00/0.50	
0.00/0.50	We use the reverse value criterion with the projection function NU:
0.00/0.50	  NU[f#(z1,z2,z3)] = z1 + -1 * z3
0.00/0.50	
0.00/0.50	This gives the following inequalities:
0.00/0.50	  x > y && x > z  ==>  x + -1 * z > x + -1 * (z + 1)  with  x + -1 * z >= 0
0.00/0.50	  I0 > I1 && I0 > I2  ==>  I0 + -1 * I2 >= I0 + -1 * I2
0.00/0.50	
0.00/0.50	We remove all the strictly oriented dependency pairs.
0.00/0.50	
0.00/0.50	DP problem for innermost termination.
0.00/0.50	P =
0.00/0.50	  f#(I0, I1, I2) -> f#(I0, I1 + 1, I2) [I0 > I1 && I0 > I2]
0.00/0.50	R = 
0.00/0.50	  f(x, y, z) -> f(x, y, z + 1) [x > y && x > z]
0.00/0.50	  f(I0, I1, I2) -> f(I0, I1 + 1, I2) [I0 > I1 && I0 > I2]
0.00/0.50	
0.00/0.50	We use the reverse value criterion with the projection function NU:
0.00/0.50	  NU[f#(z1,z2,z3)] = z1 + -1 * z2
0.00/0.50	
0.00/0.50	This gives the following inequalities:
0.00/0.50	  I0 > I1 && I0 > I2  ==>  I0 + -1 * I1 > I0 + -1 * (I1 + 1)  with  I0 + -1 * I1 >= 0
0.00/0.50	
0.00/0.50	All dependency pairs are strictly oriented, so the entire dependency pair problem may be removed.
0.00/3.48	EOF