0.00/0.16	YES
0.00/0.16	
0.00/0.16	DP problem for innermost termination.
0.00/0.16	P =
0.00/0.16	  cond#(true, x, y) -> minus#(x, y + 1) 
0.00/0.16	  minus#(I2, I3) -> cond#(I2 >= I3 + 1, I2, I3) 
0.00/0.16	R = 
0.00/0.16	  cond(true, x, y) -> 1 + minus(x, y + 1) 
0.00/0.16	  cond(false, I0, I1) -> 0 
0.00/0.16	  minus(I2, I3) -> cond(I2 >= I3 + 1, I2, I3) 
0.00/0.16	
0.00/0.16	This problem is converted using chaining, where edges between chained DPs are removed.
0.00/0.16	
0.00/0.16	DP problem for innermost termination.
0.00/0.16	P =
0.00/0.16	  cond#(true, x, y) -> minus#(x, y + 1) 
0.00/0.16	  minus#(I2, I3) -> cond#(I2 >= I3 + 1, I2, I3) 
0.00/0.16	  minus#(I2, I3) -> minus#(I2, I3 + 1) [I2 >= I3 + 1]
0.00/0.16	R = 
0.00/0.16	  cond(true, x, y) -> 1 + minus(x, y + 1) 
0.00/0.16	  cond(false, I0, I1) -> 0 
0.00/0.16	  minus(I2, I3) -> cond(I2 >= I3 + 1, I2, I3) 
0.00/0.16	
0.00/0.16	The dependency graph for this problem is:
0.00/0.16	  0 -> 2, 1
0.00/0.16	  1 -> 
0.00/0.16	  2 -> 2, 1
0.00/0.16	Where:
0.00/0.16	  0) cond#(true, x, y) -> minus#(x, y + 1) 
0.00/0.16	  1) minus#(I2, I3) -> cond#(I2 >= I3 + 1, I2, I3) 
0.00/0.16	  2) minus#(I2, I3) -> minus#(I2, I3 + 1) [I2 >= I3 + 1]
0.00/0.16	
0.00/0.16	We have the following SCCs.
0.00/0.16	  { 2 }
0.00/0.16	
0.00/0.16	DP problem for innermost termination.
0.00/0.16	P =
0.00/0.16	  minus#(I2, I3) -> minus#(I2, I3 + 1) [I2 >= I3 + 1]
0.00/0.16	R = 
0.00/0.16	  cond(true, x, y) -> 1 + minus(x, y + 1) 
0.00/0.16	  cond(false, I0, I1) -> 0 
0.00/0.16	  minus(I2, I3) -> cond(I2 >= I3 + 1, I2, I3) 
0.00/0.16	
0.00/0.16	We use the reverse value criterion with the projection function NU:
0.00/0.16	  NU[minus#(z1,z2)] = z1 + -1 * (z2 + 1)
0.00/0.16	
0.00/0.16	This gives the following inequalities:
0.00/0.16	  I2 >= I3 + 1  ==>  I2 + -1 * (I3 + 1) > I2 + -1 * (I3 + 1 + 1)  with  I2 + -1 * (I3 + 1) >= 0
0.00/0.16	
0.00/0.16	All dependency pairs are strictly oriented, so the entire dependency pair problem may be removed.
0.00/3.14	EOF