1.68/1.69	YES
1.68/1.69	
1.68/1.69	DP problem for innermost termination.
1.68/1.69	P =
1.68/1.69	  f#(true, x, y, z) -> f#(x > y + z, x, y, z + 1) 
1.68/1.69	  f#(true, I0, I1, I2) -> f#(I0 > I1 + I2, I0, I1 + 1, I2) 
1.68/1.69	R = 
1.68/1.69	  f(true, x, y, z) -> f(x > y + z, x, y, z + 1) 
1.68/1.69	  f(true, I0, I1, I2) -> f(I0 > I1 + I2, I0, I1 + 1, I2) 
1.68/1.69	
1.68/1.69	This problem is converted using chaining, where edges between chained DPs are removed.
1.68/1.69	
1.68/1.69	DP problem for innermost termination.
1.68/1.69	P =
1.68/1.69	  f#(true, x, y, z) -> f_2#(x, y, z) 
1.68/1.69	  f#(true, I0, I1, I2) -> f_1#(I0, I1, I2) 
1.68/1.69	  f_1#(I0, I1, I2) -> f#(I0 > I1 + I2, I0, I1 + 1, I2) 
1.68/1.69	  f_2#(x, y, z) -> f#(x > y + z, x, y, z + 1) 
1.68/1.69	  f_2#(x, y, z) -> f_1#(x, y, z + 1) [x > y + z]
1.68/1.69	  f_2#(x, y, z) -> f_2#(x, y, z + 1) [x > y + z]
1.68/1.69	  f_1#(I0, I1, I2) -> f_1#(I0, I1 + 1, I2) [I0 > I1 + I2]
1.68/1.69	  f_1#(I0, I1, I2) -> f_2#(I0, I1 + 1, I2) [I0 > I1 + I2]
1.68/1.69	R = 
1.68/1.69	  f(true, x, y, z) -> f(x > y + z, x, y, z + 1) 
1.68/1.69	  f(true, I0, I1, I2) -> f(I0 > I1 + I2, I0, I1 + 1, I2) 
1.68/1.69	
1.68/1.69	The dependency graph for this problem is:
1.68/1.69	  0 -> 5, 4, 3
1.68/1.69	  1 -> 7, 6, 2
1.68/1.69	  2 -> 
1.68/1.69	  3 -> 
1.68/1.69	  4 -> 7, 6, 2
1.68/1.69	  5 -> 5, 3, 4
1.68/1.69	  6 -> 7, 6, 2
1.68/1.69	  7 -> 3, 4, 5
1.68/1.69	Where:
1.68/1.69	  0) f#(true, x, y, z) -> f_2#(x, y, z) 
1.68/1.69	  1) f#(true, I0, I1, I2) -> f_1#(I0, I1, I2) 
1.68/1.69	  2) f_1#(I0, I1, I2) -> f#(I0 > I1 + I2, I0, I1 + 1, I2) 
1.68/1.69	  3) f_2#(x, y, z) -> f#(x > y + z, x, y, z + 1) 
1.68/1.69	  4) f_2#(x, y, z) -> f_1#(x, y, z + 1) [x > y + z]
1.68/1.69	  5) f_2#(x, y, z) -> f_2#(x, y, z + 1) [x > y + z]
1.68/1.69	  6) f_1#(I0, I1, I2) -> f_1#(I0, I1 + 1, I2) [I0 > I1 + I2]
1.68/1.69	  7) f_1#(I0, I1, I2) -> f_2#(I0, I1 + 1, I2) [I0 > I1 + I2]
1.68/1.69	
1.68/1.69	We have the following SCCs.
1.68/1.69	  { 4, 5, 6, 7 }
1.68/1.69	
1.68/1.69	DP problem for innermost termination.
1.68/1.69	P =
1.68/1.69	  f_2#(x, y, z) -> f_1#(x, y, z + 1) [x > y + z]
1.68/1.69	  f_2#(x, y, z) -> f_2#(x, y, z + 1) [x > y + z]
1.68/1.69	  f_1#(I0, I1, I2) -> f_1#(I0, I1 + 1, I2) [I0 > I1 + I2]
1.68/1.69	  f_1#(I0, I1, I2) -> f_2#(I0, I1 + 1, I2) [I0 > I1 + I2]
1.68/1.69	R = 
1.68/1.69	  f(true, x, y, z) -> f(x > y + z, x, y, z + 1) 
1.68/1.69	  f(true, I0, I1, I2) -> f(I0 > I1 + I2, I0, I1 + 1, I2) 
1.68/1.69	
1.68/1.69	We use the reverse value criterion with the projection function NU:
1.68/1.69	  NU[f_1#(z1,z2,z3)] = z1 + -1 * (z2 + 1 + z3)
1.68/1.69	  NU[f_2#(z1,z2,z3)] = z1 + -1 * (z2 + (z3 + 1))
1.68/1.69	
1.68/1.69	This gives the following inequalities:
1.68/1.69	  x > y + z  ==>  x + -1 * (y + (z + 1)) > x + -1 * (y + 1 + (z + 1))  with  x + -1 * (y + (z + 1)) >= 0
1.68/1.69	  x > y + z  ==>  x + -1 * (y + (z + 1)) > x + -1 * (y + (z + 1 + 1))  with  x + -1 * (y + (z + 1)) >= 0
1.68/1.69	  I0 > I1 + I2  ==>  I0 + -1 * (I1 + 1 + I2) > I0 + -1 * (I1 + 1 + 1 + I2)  with  I0 + -1 * (I1 + 1 + I2) >= 0
1.68/1.69	  I0 > I1 + I2  ==>  I0 + -1 * (I1 + 1 + I2) > I0 + -1 * (I1 + 1 + (I2 + 1))  with  I0 + -1 * (I1 + 1 + I2) >= 0
1.68/1.69	
1.68/1.69	All dependency pairs are strictly oriented, so the entire dependency pair problem may be removed.
1.68/4.67	EOF