0.00/0.21	MAYBE
0.00/0.21	
0.00/0.21	DP problem for innermost termination.
0.00/0.21	P =
0.00/0.21	  min#(I2, I3) -> if#(I2 < I3, I2, I3) 
0.00/0.21	  cond#(true, x, y) -> minus#(x, y + 1) 
0.00/0.21	  minus#(I4, I5) -> cond#(min(I4, I5) = I5, I4, I5) 
0.00/0.21	  minus#(I4, I5) -> min#(I4, I5) 
0.00/0.21	R = 
0.00/0.21	  if(false, u, v) -> v 
0.00/0.21	  if(true, I0, I1) -> I0 
0.00/0.21	  min(I2, I3) -> if(I2 < I3, I2, I3) 
0.00/0.21	  cond(true, x, y) -> 1 + minus(x, y + 1) 
0.00/0.21	  minus(I4, I5) -> cond(min(I4, I5) = I5, I4, I5) 
0.00/0.21	  minus(I6, I6) -> 0 
0.00/0.21	
0.00/0.21	The dependency graph for this problem is:
0.00/0.21	  0 -> 
0.00/0.21	  1 -> 2, 3
0.00/0.21	  2 -> 1
0.00/0.21	  3 -> 0
0.00/0.21	Where:
0.00/0.21	  0) min#(I2, I3) -> if#(I2 < I3, I2, I3) 
0.00/0.21	  1) cond#(true, x, y) -> minus#(x, y + 1) 
0.00/0.21	  2) minus#(I4, I5) -> cond#(min(I4, I5) = I5, I4, I5) 
0.00/0.21	  3) minus#(I4, I5) -> min#(I4, I5) 
0.00/0.21	
0.00/0.21	We have the following SCCs.
0.00/0.21	  { 1, 2 }
0.00/0.21	
0.00/0.21	DP problem for innermost termination.
0.00/0.21	P =
0.00/0.21	  cond#(true, x, y) -> minus#(x, y + 1) 
0.00/0.21	  minus#(I4, I5) -> cond#(min(I4, I5) = I5, I4, I5) 
0.00/0.21	R = 
0.00/0.21	  if(false, u, v) -> v 
0.00/0.21	  if(true, I0, I1) -> I0 
0.00/0.21	  min(I2, I3) -> if(I2 < I3, I2, I3) 
0.00/0.21	  cond(true, x, y) -> 1 + minus(x, y + 1) 
0.00/0.21	  minus(I4, I5) -> cond(min(I4, I5) = I5, I4, I5) 
0.00/0.21	  minus(I6, I6) -> 0 
0.00/0.21	
0.00/3.20	EOF