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	  round#(x) -> if#(x % 2 = 0, x, x + 1) 
0.00/0.21	  minusNat#(true, I2, y) -> minus#(I2, round(y)) 
0.00/0.21	  minusNat#(true, I2, y) -> round#(y) 
0.00/0.21	  minus#(I3, I4) -> minusNat#(I4 >= 0 && I3 = I4 + 1, I3, I4) 
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	  round(x) -> if(x % 2 = 0, x, x + 1) 
0.00/0.21	  minusNat(true, I2, y) -> minus(I2, round(y)) 
0.00/0.21	  minus(I3, I4) -> minusNat(I4 >= 0 && I3 = I4 + 1, I3, I4) 
0.00/0.21	
0.00/0.21	This problem is converted using chaining, where edges between chained DPs are removed.
0.00/0.21	
0.00/0.21	DP problem for innermost termination.
0.00/0.21	P =
0.00/0.21	  round#(x) -> if#(x % 2 = 0, x, x + 1) 
0.00/0.21	  minusNat#(true, I2, y) -> minus#(I2, round(y)) 
0.00/0.21	  minusNat#(true, I2, y) -> round#(y) 
0.00/0.21	  minus#(I3, I4) -> minusNat#(I4 >= 0 && I3 = I4 + 1, I3, I4) 
0.00/0.21	  minus#(I3, I4) -> minus#(I3, round(I4)) [I4 >= 0 && I3 = I4 + 1]
0.00/0.21	  minus#(I3, I4) -> round#(I4) [I4 >= 0 && I3 = I4 + 1]
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	  round(x) -> if(x % 2 = 0, x, x + 1) 
0.00/0.21	  minusNat(true, I2, y) -> minus(I2, round(y)) 
0.00/0.21	  minus(I3, I4) -> minusNat(I4 >= 0 && I3 = I4 + 1, I3, I4) 
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 -> 5, 4, 3
0.00/0.21	  2 -> 0
0.00/0.21	  3 -> 
0.00/0.21	  4 -> 5, 4, 3
0.00/0.21	  5 -> 0
0.00/0.21	Where:
0.00/0.21	  0) round#(x) -> if#(x % 2 = 0, x, x + 1) 
0.00/0.21	  1) minusNat#(true, I2, y) -> minus#(I2, round(y)) 
0.00/0.21	  2) minusNat#(true, I2, y) -> round#(y) 
0.00/0.21	  3) minus#(I3, I4) -> minusNat#(I4 >= 0 && I3 = I4 + 1, I3, I4) 
0.00/0.21	  4) minus#(I3, I4) -> minus#(I3, round(I4)) [I4 >= 0 && I3 = I4 + 1]
0.00/0.21	  5) minus#(I3, I4) -> round#(I4) [I4 >= 0 && I3 = I4 + 1]
0.00/0.21	
0.00/0.21	We have the following SCCs.
0.00/0.21	  { 4 }
0.00/0.21	
0.00/0.21	DP problem for innermost termination.
0.00/0.21	P =
0.00/0.21	  minus#(I3, I4) -> minus#(I3, round(I4)) [I4 >= 0 && I3 = I4 + 1]
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	  round(x) -> if(x % 2 = 0, x, x + 1) 
0.00/0.21	  minusNat(true, I2, y) -> minus(I2, round(y)) 
0.00/0.21	  minus(I3, I4) -> minusNat(I4 >= 0 && I3 = I4 + 1, I3, I4) 
0.00/0.21	
0.00/3.19	EOF