0.00/0.23	MAYBE
0.00/0.23	
0.00/0.23	DP problem for innermost termination.
0.00/0.23	P =
0.00/0.23	  round#(x) -> if#(x % 2 = 0, x, x + 1) 
0.00/0.23	  minusNat#(true, I2, y) -> minus#(I2, round(I2)) 
0.00/0.23	  minusNat#(true, I2, y) -> round#(I2) 
0.00/0.23	  minus#(I3, I4) -> minusNat#(I4 >= 0 && I3 = I4 + 1, I3, I4) 
0.00/0.23	R = 
0.00/0.23	  if(false, u, v) -> v 
0.00/0.23	  if(true, I0, I1) -> I0 
0.00/0.23	  round(x) -> if(x % 2 = 0, x, x + 1) 
0.00/0.23	  minusNat(true, I2, y) -> minus(I2, round(I2)) 
0.00/0.23	  minus(I3, I4) -> minusNat(I4 >= 0 && I3 = I4 + 1, I3, I4) 
0.00/0.23	
0.00/0.23	This problem is converted using chaining, where edges between chained DPs are removed.
0.00/0.23	
0.00/0.23	DP problem for innermost termination.
0.00/0.23	P =
0.00/0.23	  round#(x) -> if#(x % 2 = 0, x, x + 1) 
0.00/0.23	  minusNat#(true, I2, y) -> minus#(I2, round(I2)) 
0.00/0.23	  minusNat#(true, I2, y) -> round#(I2) 
0.00/0.23	  minus#(I3, I4) -> minusNat#(I4 >= 0 && I3 = I4 + 1, I3, I4) 
0.00/0.23	  minus#(I3, I4) -> minus#(I3, round(I3)) [I4 >= 0 && I3 = I4 + 1]
0.00/0.23	  minus#(I3, I4) -> round#(I3) [I4 >= 0 && I3 = I4 + 1]
0.00/0.23	R = 
0.00/0.23	  if(false, u, v) -> v 
0.00/0.23	  if(true, I0, I1) -> I0 
0.00/0.23	  round(x) -> if(x % 2 = 0, x, x + 1) 
0.00/0.23	  minusNat(true, I2, y) -> minus(I2, round(I2)) 
0.00/0.23	  minus(I3, I4) -> minusNat(I4 >= 0 && I3 = I4 + 1, I3, I4) 
0.00/0.23	
0.00/0.23	The dependency graph for this problem is:
0.00/0.23	  0 -> 
0.00/0.23	  1 -> 5, 4, 3
0.00/0.23	  2 -> 0
0.00/0.23	  3 -> 
0.00/0.23	  4 -> 5, 4, 3
0.00/0.23	  5 -> 0
0.00/0.23	Where:
0.00/0.23	  0) round#(x) -> if#(x % 2 = 0, x, x + 1) 
0.00/0.23	  1) minusNat#(true, I2, y) -> minus#(I2, round(I2)) 
0.00/0.23	  2) minusNat#(true, I2, y) -> round#(I2) 
0.00/0.23	  3) minus#(I3, I4) -> minusNat#(I4 >= 0 && I3 = I4 + 1, I3, I4) 
0.00/0.23	  4) minus#(I3, I4) -> minus#(I3, round(I3)) [I4 >= 0 && I3 = I4 + 1]
0.00/0.23	  5) minus#(I3, I4) -> round#(I3) [I4 >= 0 && I3 = I4 + 1]
0.00/0.23	
0.00/0.23	We have the following SCCs.
0.00/0.23	  { 4 }
0.00/0.23	
0.00/0.23	DP problem for innermost termination.
0.00/0.23	P =
0.00/0.23	  minus#(I3, I4) -> minus#(I3, round(I3)) [I4 >= 0 && I3 = I4 + 1]
0.00/0.23	R = 
0.00/0.23	  if(false, u, v) -> v 
0.00/0.23	  if(true, I0, I1) -> I0 
0.00/0.23	  round(x) -> if(x % 2 = 0, x, x + 1) 
0.00/0.23	  minusNat(true, I2, y) -> minus(I2, round(I2)) 
0.00/0.23	  minus(I3, I4) -> minusNat(I4 >= 0 && I3 = I4 + 1, I3, I4) 
0.00/0.23	
0.00/3.21	EOF