8.49/8.43	YES
8.49/8.43	
8.49/8.43	DP problem for innermost termination.
8.49/8.43	P =
8.49/8.43	  eval_2#(x, y, z) -> eval_1#(x - 1, y, z) [z >= x]
8.49/8.43	  eval_2#(I0, I1, I2) -> eval_1#(I0, I1, I2 + 1) [I0 > I2]
8.49/8.43	  eval_2#(I3, I4, I5) -> eval_1#(I3, I4 + 1, I5) [I3 > I5]
8.49/8.43	  eval_1#(I6, I7, I8) -> eval_2#(I6, I7, I8) [I6 > I7]
8.49/8.43	R = 
8.49/8.43	  eval_2(x, y, z) -> eval_1(x - 1, y, z) [z >= x]
8.49/8.43	  eval_2(I0, I1, I2) -> eval_1(I0, I1, I2 + 1) [I0 > I2]
8.49/8.43	  eval_2(I3, I4, I5) -> eval_1(I3, I4 + 1, I5) [I3 > I5]
8.49/8.43	  eval_1(I6, I7, I8) -> eval_2(I6, I7, I8) [I6 > I7]
8.49/8.43	
8.49/8.43	We use the reverse value criterion with the projection function NU:
8.49/8.43	  NU[eval_1#(z1,z2,z3)] = z1 + -1 * z3
8.49/8.43	  NU[eval_2#(z1,z2,z3)] = z1 + -1 * z3
8.49/8.43	
8.49/8.43	This gives the following inequalities:
8.49/8.43	  z >= x  ==>  x + -1 * z >= x - 1 + -1 * z
8.49/8.43	  I0 > I2  ==>  I0 + -1 * I2 > I0 + -1 * (I2 + 1)  with  I0 + -1 * I2 >= 0
8.49/8.43	  I3 > I5  ==>  I3 + -1 * I5 >= I3 + -1 * I5
8.49/8.43	  I6 > I7  ==>  I6 + -1 * I8 >= I6 + -1 * I8
8.49/8.43	
8.49/8.43	We remove all the strictly oriented dependency pairs.
8.49/8.43	
8.49/8.43	DP problem for innermost termination.
8.49/8.43	P =
8.49/8.43	  eval_2#(x, y, z) -> eval_1#(x - 1, y, z) [z >= x]
8.49/8.43	  eval_2#(I3, I4, I5) -> eval_1#(I3, I4 + 1, I5) [I3 > I5]
8.49/8.43	  eval_1#(I6, I7, I8) -> eval_2#(I6, I7, I8) [I6 > I7]
8.49/8.43	R = 
8.49/8.43	  eval_2(x, y, z) -> eval_1(x - 1, y, z) [z >= x]
8.49/8.43	  eval_2(I0, I1, I2) -> eval_1(I0, I1, I2 + 1) [I0 > I2]
8.49/8.43	  eval_2(I3, I4, I5) -> eval_1(I3, I4 + 1, I5) [I3 > I5]
8.49/8.43	  eval_1(I6, I7, I8) -> eval_2(I6, I7, I8) [I6 > I7]
8.49/8.43	
8.49/8.43	We use the reverse value criterion with the projection function NU:
8.49/8.43	  NU[eval_1#(z1,z2,z3)] = z1 + -1 * z2
8.49/8.43	  NU[eval_2#(z1,z2,z3)] = z1 - 1 + -1 * z2
8.49/8.43	
8.49/8.43	This gives the following inequalities:
8.49/8.43	  z >= x  ==>  x - 1 + -1 * y >= x - 1 + -1 * y
8.49/8.43	  I3 > I5  ==>  I3 - 1 + -1 * I4 >= I3 + -1 * (I4 + 1)
8.49/8.43	  I6 > I7  ==>  I6 + -1 * I7 > I6 - 1 + -1 * I7  with  I6 + -1 * I7 >= 0
8.49/8.43	
8.49/8.43	We remove all the strictly oriented dependency pairs.
8.49/8.43	
8.49/8.43	DP problem for innermost termination.
8.49/8.43	P =
8.49/8.43	  eval_2#(x, y, z) -> eval_1#(x - 1, y, z) [z >= x]
8.49/8.43	  eval_2#(I3, I4, I5) -> eval_1#(I3, I4 + 1, I5) [I3 > I5]
8.49/8.43	R = 
8.49/8.43	  eval_2(x, y, z) -> eval_1(x - 1, y, z) [z >= x]
8.49/8.43	  eval_2(I0, I1, I2) -> eval_1(I0, I1, I2 + 1) [I0 > I2]
8.49/8.43	  eval_2(I3, I4, I5) -> eval_1(I3, I4 + 1, I5) [I3 > I5]
8.49/8.43	  eval_1(I6, I7, I8) -> eval_2(I6, I7, I8) [I6 > I7]
8.49/8.43	
8.49/8.43	The dependency graph for this problem is:
8.49/8.43	  0 -> 
8.49/8.43	  2 -> 
8.49/8.43	Where:
8.49/8.43	  0) eval_2#(x, y, z) -> eval_1#(x - 1, y, z) [z >= x]
8.49/8.43	  2) eval_2#(I3, I4, I5) -> eval_1#(I3, I4 + 1, I5) [I3 > I5]
8.49/8.43	
8.49/8.43	We have the following SCCs.
8.49/8.43	  
8.49/11.40	EOF