1.10/1.16	YES
1.10/1.16	
1.10/1.16	DP problem for innermost termination.
1.10/1.16	P =
1.10/1.16	  g#(true, x, y, z) -> f#(x > z, x, y, z + 1) 
1.10/1.16	  g#(true, I0, I1, I2) -> f#(I0 > I2, I0, I1 + 1, I2) 
1.10/1.16	  f#(true, I3, I4, I5) -> g#(I3 > I4, I3, I4, I5) 
1.10/1.16	R = 
1.10/1.16	  g(true, x, y, z) -> f(x > z, x, y, z + 1) 
1.10/1.16	  g(true, I0, I1, I2) -> f(I0 > I2, I0, I1 + 1, I2) 
1.10/1.16	  f(true, I3, I4, I5) -> g(I3 > I4, I3, I4, I5) 
1.10/1.16	
1.10/1.16	This problem is converted using chaining, where edges between chained DPs are removed.
1.10/1.16	
1.10/1.16	DP problem for innermost termination.
1.10/1.16	P =
1.10/1.16	  g#(true, x, y, z) -> f#(x > z, x, y, z + 1) 
1.10/1.16	  g#(true, I0, I1, I2) -> f#(I0 > I2, I0, I1 + 1, I2) 
1.10/1.16	  f#(true, I3, I4, I5) -> f_1#(I3, I4, I5) 
1.10/1.16	  f_1#(I3, I4, I5) -> g#(I3 > I4, I3, I4, I5) 
1.10/1.16	  f_1#(I3, I4, I5) -> f#(I3 > I5, I3, I4 + 1, I5) [I3 > I4]
1.10/1.16	  f_1#(I3, I4, I5) -> f_1#(I3, I4 + 1, I5) [I3 > I4, I3 > I5]
1.10/1.16	  f_1#(I3, I4, I5) -> f#(I3 > I5, I3, I4, I5 + 1) [I3 > I4]
1.10/1.16	  f_1#(I3, I4, I5) -> f_1#(I3, I4, I5 + 1) [I3 > I4, I3 > I5]
1.10/1.16	  g#(true, I0, I1, I2) -> f_1#(I0, I1 + 1, I2) [I0 > I2]
1.10/1.16	  g#(true, x, y, z) -> f_1#(x, y, z + 1) [x > z]
1.10/1.16	R = 
1.10/1.16	  g(true, x, y, z) -> f(x > z, x, y, z + 1) 
1.10/1.16	  g(true, I0, I1, I2) -> f(I0 > I2, I0, I1 + 1, I2) 
1.10/1.16	  f(true, I3, I4, I5) -> g(I3 > I4, I3, I4, I5) 
1.10/1.16	
1.10/1.16	The dependency graph for this problem is:
1.10/1.16	  0 -> 
1.10/1.16	  1 -> 
1.10/1.16	  2 -> 7, 6, 5, 4, 3
1.10/1.16	  3 -> 
1.10/1.16	  4 -> 
1.10/1.16	  5 -> 7, 6, 5, 3, 4
1.10/1.16	  6 -> 
1.10/1.16	  7 -> 7, 3, 4, 5, 6
1.10/1.16	  8 -> 3, 4, 5, 6, 7
1.10/1.16	  9 -> 3, 4, 5, 6, 7
1.10/1.16	Where:
1.10/1.16	  0) g#(true, x, y, z) -> f#(x > z, x, y, z + 1) 
1.10/1.16	  1) g#(true, I0, I1, I2) -> f#(I0 > I2, I0, I1 + 1, I2) 
1.10/1.16	  2) f#(true, I3, I4, I5) -> f_1#(I3, I4, I5) 
1.10/1.16	  3) f_1#(I3, I4, I5) -> g#(I3 > I4, I3, I4, I5) 
1.10/1.16	  4) f_1#(I3, I4, I5) -> f#(I3 > I5, I3, I4 + 1, I5) [I3 > I4]
1.10/1.16	  5) f_1#(I3, I4, I5) -> f_1#(I3, I4 + 1, I5) [I3 > I4, I3 > I5]
1.10/1.16	  6) f_1#(I3, I4, I5) -> f#(I3 > I5, I3, I4, I5 + 1) [I3 > I4]
1.10/1.16	  7) f_1#(I3, I4, I5) -> f_1#(I3, I4, I5 + 1) [I3 > I4, I3 > I5]
1.10/1.16	  8) g#(true, I0, I1, I2) -> f_1#(I0, I1 + 1, I2) [I0 > I2]
1.10/1.16	  9) g#(true, x, y, z) -> f_1#(x, y, z + 1) [x > z]
1.10/1.16	
1.10/1.16	We have the following SCCs.
1.10/1.16	  { 5, 7 }
1.10/1.16	
1.10/1.16	DP problem for innermost termination.
1.10/1.16	P =
1.10/1.16	  f_1#(I3, I4, I5) -> f_1#(I3, I4 + 1, I5) [I3 > I4, I3 > I5]
1.10/1.16	  f_1#(I3, I4, I5) -> f_1#(I3, I4, I5 + 1) [I3 > I4, I3 > I5]
1.10/1.16	R = 
1.10/1.16	  g(true, x, y, z) -> f(x > z, x, y, z + 1) 
1.10/1.16	  g(true, I0, I1, I2) -> f(I0 > I2, I0, I1 + 1, I2) 
1.10/1.16	  f(true, I3, I4, I5) -> g(I3 > I4, I3, I4, I5) 
1.10/1.16	
1.10/1.16	We use the reverse value criterion with the projection function NU:
1.10/1.16	  NU[f_1#(z1,z2,z3)] = z1 + -1 * z2
1.10/1.16	
1.10/1.16	This gives the following inequalities:
1.10/1.16	  I3 > I4/\I3 > I5  ==>  I3 + -1 * I4 > I3 + -1 * (I4 + 1)  with  I3 + -1 * I4 >= 0
1.10/1.16	  I3 > I4/\I3 > I5  ==>  I3 + -1 * I4 >= I3 + -1 * I4
1.10/1.16	
1.10/1.16	We remove all the strictly oriented dependency pairs.
1.10/1.16	
1.10/1.16	DP problem for innermost termination.
1.10/1.16	P =
1.10/1.16	  f_1#(I3, I4, I5) -> f_1#(I3, I4, I5 + 1) [I3 > I4, I3 > I5]
1.10/1.16	R = 
1.10/1.16	  g(true, x, y, z) -> f(x > z, x, y, z + 1) 
1.10/1.16	  g(true, I0, I1, I2) -> f(I0 > I2, I0, I1 + 1, I2) 
1.10/1.16	  f(true, I3, I4, I5) -> g(I3 > I4, I3, I4, I5) 
1.10/1.16	
1.10/1.16	We use the reverse value criterion with the projection function NU:
1.10/1.16	  NU[f_1#(z1,z2,z3)] = z1 + -1 * z3
1.10/1.16	
1.10/1.16	This gives the following inequalities:
1.10/1.16	  I3 > I4/\I3 > I5  ==>  I3 + -1 * I5 > I3 + -1 * (I5 + 1)  with  I3 + -1 * I5 >= 0
1.10/1.16	
1.10/1.16	All dependency pairs are strictly oriented, so the entire dependency pair problem may be removed.
1.10/4.14	EOF