2.61/2.61	YES
2.61/2.61	
2.61/2.61	DP problem for innermost termination.
2.61/2.61	P =
2.61/2.61	  f#(true, x, y, z) -> f#(x > y && x > z, x, y, z + 1) 
2.61/2.61	  f#(true, I0, I1, I2) -> f#(I0 > I1 && I0 > I2, I0, I1 + 1, I2) 
2.61/2.61	R = 
2.61/2.61	  f(true, x, y, z) -> f(x > y && x > z, x, y, z + 1) 
2.61/2.61	  f(true, I0, I1, I2) -> f(I0 > I1 && I0 > I2, I0, I1 + 1, I2) 
2.61/2.61	
2.61/2.61	This problem is converted using chaining, where edges between chained DPs are removed.
2.61/2.61	
2.61/2.61	DP problem for innermost termination.
2.61/2.61	P =
2.61/2.61	  f#(true, x, y, z) -> f_2#(x, y, z) 
2.61/2.61	  f#(true, I0, I1, I2) -> f_1#(I0, I1, I2) 
2.61/2.61	  f_1#(I0, I1, I2) -> f#(I0 > I1 && I0 > I2, I0, I1 + 1, I2) 
2.61/2.61	  f_2#(x, y, z) -> f#(x > y && x > z, x, y, z + 1) 
2.61/2.61	  f_2#(x, y, z) -> f_1#(x, y, z + 1) [x > y && x > z]
2.61/2.61	  f_2#(x, y, z) -> f_2#(x, y, z + 1) [x > y && x > z]
2.61/2.61	  f_1#(I0, I1, I2) -> f_1#(I0, I1 + 1, I2) [I0 > I1 && I0 > I2]
2.61/2.61	  f_1#(I0, I1, I2) -> f_2#(I0, I1 + 1, I2) [I0 > I1 && I0 > I2]
2.61/2.61	R = 
2.61/2.61	  f(true, x, y, z) -> f(x > y && x > z, x, y, z + 1) 
2.61/2.61	  f(true, I0, I1, I2) -> f(I0 > I1 && I0 > I2, I0, I1 + 1, I2) 
2.61/2.61	
2.61/2.61	The dependency graph for this problem is:
2.61/2.61	  0 -> 5, 4, 3
2.61/2.61	  1 -> 7, 6, 2
2.61/2.61	  2 -> 
2.61/2.61	  3 -> 
2.61/2.61	  4 -> 7, 6, 2
2.61/2.61	  5 -> 5, 3, 4
2.61/2.61	  6 -> 7, 6, 2
2.61/2.61	  7 -> 3, 4, 5
2.61/2.61	Where:
2.61/2.61	  0) f#(true, x, y, z) -> f_2#(x, y, z) 
2.61/2.61	  1) f#(true, I0, I1, I2) -> f_1#(I0, I1, I2) 
2.61/2.61	  2) f_1#(I0, I1, I2) -> f#(I0 > I1 && I0 > I2, I0, I1 + 1, I2) 
2.61/2.61	  3) f_2#(x, y, z) -> f#(x > y && x > z, x, y, z + 1) 
2.61/2.61	  4) f_2#(x, y, z) -> f_1#(x, y, z + 1) [x > y && x > z]
2.61/2.61	  5) f_2#(x, y, z) -> f_2#(x, y, z + 1) [x > y && x > z]
2.61/2.61	  6) f_1#(I0, I1, I2) -> f_1#(I0, I1 + 1, I2) [I0 > I1 && I0 > I2]
2.61/2.61	  7) f_1#(I0, I1, I2) -> f_2#(I0, I1 + 1, I2) [I0 > I1 && I0 > I2]
2.61/2.61	
2.61/2.61	We have the following SCCs.
2.61/2.61	  { 4, 5, 6, 7 }
2.61/2.61	
2.61/2.61	DP problem for innermost termination.
2.61/2.61	P =
2.61/2.61	  f_2#(x, y, z) -> f_1#(x, y, z + 1) [x > y && x > z]
2.61/2.61	  f_2#(x, y, z) -> f_2#(x, y, z + 1) [x > y && x > z]
2.61/2.61	  f_1#(I0, I1, I2) -> f_1#(I0, I1 + 1, I2) [I0 > I1 && I0 > I2]
2.61/2.61	  f_1#(I0, I1, I2) -> f_2#(I0, I1 + 1, I2) [I0 > I1 && I0 > I2]
2.61/2.61	R = 
2.61/2.61	  f(true, x, y, z) -> f(x > y && x > z, x, y, z + 1) 
2.61/2.61	  f(true, I0, I1, I2) -> f(I0 > I1 && I0 > I2, I0, I1 + 1, I2) 
2.61/2.61	
2.61/2.61	We use the reverse value criterion with the projection function NU:
2.61/2.62	  NU[f_1#(z1,z2,z3)] = z1 + -1 * z2
2.61/2.62	  NU[f_2#(z1,z2,z3)] = z1 + -1 * z2
2.61/2.62	
2.61/2.62	This gives the following inequalities:
2.61/2.62	  x > y && x > z  ==>  x + -1 * y >= x + -1 * y
2.61/2.62	  x > y && x > z  ==>  x + -1 * y >= x + -1 * y
2.61/2.62	  I0 > I1 && I0 > I2  ==>  I0 + -1 * I1 > I0 + -1 * (I1 + 1)  with  I0 + -1 * I1 >= 0
2.61/2.62	  I0 > I1 && I0 > I2  ==>  I0 + -1 * I1 > I0 + -1 * (I1 + 1)  with  I0 + -1 * I1 >= 0
2.61/2.62	
2.61/2.62	We remove all the strictly oriented dependency pairs.
2.61/2.62	
2.61/2.62	DP problem for innermost termination.
2.61/2.62	P =
2.61/2.62	  f_2#(x, y, z) -> f_1#(x, y, z + 1) [x > y && x > z]
2.61/2.62	  f_2#(x, y, z) -> f_2#(x, y, z + 1) [x > y && x > z]
2.61/2.62	R = 
2.61/2.62	  f(true, x, y, z) -> f(x > y && x > z, x, y, z + 1) 
2.61/2.62	  f(true, I0, I1, I2) -> f(I0 > I1 && I0 > I2, I0, I1 + 1, I2) 
2.61/2.62	
2.61/2.62	The dependency graph for this problem is:
2.61/2.62	  4 -> 
2.61/2.62	  5 -> 5, 4
2.61/2.62	Where:
2.61/2.62	  4) f_2#(x, y, z) -> f_1#(x, y, z + 1) [x > y && x > z]
2.61/2.62	  5) f_2#(x, y, z) -> f_2#(x, y, z + 1) [x > y && x > z]
2.61/2.62	
2.61/2.62	We have the following SCCs.
2.61/2.62	  { 5 }
2.61/2.62	
2.61/2.62	DP problem for innermost termination.
2.61/2.62	P =
2.61/2.62	  f_2#(x, y, z) -> f_2#(x, y, z + 1) [x > y && x > z]
2.61/2.62	R = 
2.61/2.62	  f(true, x, y, z) -> f(x > y && x > z, x, y, z + 1) 
2.61/2.62	  f(true, I0, I1, I2) -> f(I0 > I1 && I0 > I2, I0, I1 + 1, I2) 
2.61/2.62	
2.61/2.62	We use the reverse value criterion with the projection function NU:
2.61/2.62	  NU[f_2#(z1,z2,z3)] = z1 + -1 * z3
2.61/2.62	
2.61/2.62	This gives the following inequalities:
2.61/2.62	  x > y && x > z  ==>  x + -1 * z > x + -1 * (z + 1)  with  x + -1 * z >= 0
2.61/2.62	
2.61/2.62	All dependency pairs are strictly oriented, so the entire dependency pair problem may be removed.
2.61/5.60	EOF