0.00/0.43	MAYBE
0.00/0.43	
0.00/0.43	DP problem for innermost termination.
0.00/0.43	P =
0.00/0.43	  round#(x) -> if#(x % 2 = 0, x, x + 1) 
0.00/0.43	  fNat#(true, I2, y) -> f#(I2 > y, I2, round(y + 1)) 
0.00/0.43	  fNat#(true, I2, y) -> round#(y + 1) 
0.00/0.43	  f#(true, I3, I4) -> fNat#(I3 >= 0 && I4 >= 0, I3, I4) 
0.00/0.43	R = 
0.00/0.43	  if(false, u, v) -> v 
0.00/0.43	  if(true, I0, I1) -> I0 
0.00/0.43	  round(x) -> if(x % 2 = 0, x, x + 1) 
0.00/0.43	  fNat(true, I2, y) -> f(I2 > y, I2, round(y + 1)) 
0.00/0.43	  f(true, I3, I4) -> fNat(I3 >= 0 && I4 >= 0, I3, I4) 
0.00/0.43	
0.00/0.43	This problem is converted using chaining, where edges between chained DPs are removed.
0.00/0.43	
0.00/0.43	DP problem for innermost termination.
0.00/0.43	P =
0.00/0.43	  round#(x) -> if#(x % 2 = 0, x, x + 1) 
0.00/0.43	  fNat#(true, I2, y) -> f#(I2 > y, I2, round(y + 1)) 
0.00/0.43	  fNat#(true, I2, y) -> round#(y + 1) 
0.00/0.43	  f#(true, I3, I4) -> f_1#(I3, I4) 
0.00/0.43	  f_1#(I3, I4) -> fNat#(I3 >= 0 && I4 >= 0, I3, I4) 
0.00/0.43	  f_1#(I3, I4) -> round#(I4 + 1) [I3 >= 0 && I4 >= 0]
0.00/0.43	  f_1#(I3, I4) -> f#(I3 > I4, I3, round(I4 + 1)) [I3 >= 0 && I4 >= 0]
0.00/0.43	  f_1#(I3, I4) -> f_1#(I3, round(I4 + 1)) [I3 >= 0 && I4 >= 0, I3 > I4]
0.00/0.43	  fNat#(true, I2, y) -> f_1#(I2, round(y + 1)) [I2 > y]
0.00/0.43	R = 
0.00/0.43	  if(false, u, v) -> v 
0.00/0.43	  if(true, I0, I1) -> I0 
0.00/0.43	  round(x) -> if(x % 2 = 0, x, x + 1) 
0.00/0.43	  fNat(true, I2, y) -> f(I2 > y, I2, round(y + 1)) 
0.00/0.43	  f(true, I3, I4) -> fNat(I3 >= 0 && I4 >= 0, I3, I4) 
0.00/0.43	
0.00/0.43	The dependency graph for this problem is:
0.00/0.43	  0 -> 
0.00/0.43	  1 -> 
0.00/0.43	  2 -> 0
0.00/0.43	  3 -> 7, 6, 5, 4
0.00/0.43	  4 -> 
0.00/0.43	  5 -> 0
0.00/0.43	  6 -> 
0.00/0.43	  7 -> 7, 4, 5, 6
0.00/0.43	  8 -> 4, 5, 6, 7
0.00/0.43	Where:
0.00/0.43	  0) round#(x) -> if#(x % 2 = 0, x, x + 1) 
0.00/0.43	  1) fNat#(true, I2, y) -> f#(I2 > y, I2, round(y + 1)) 
0.00/0.43	  2) fNat#(true, I2, y) -> round#(y + 1) 
0.00/0.43	  3) f#(true, I3, I4) -> f_1#(I3, I4) 
0.00/0.43	  4) f_1#(I3, I4) -> fNat#(I3 >= 0 && I4 >= 0, I3, I4) 
0.00/0.43	  5) f_1#(I3, I4) -> round#(I4 + 1) [I3 >= 0 && I4 >= 0]
0.00/0.43	  6) f_1#(I3, I4) -> f#(I3 > I4, I3, round(I4 + 1)) [I3 >= 0 && I4 >= 0]
0.00/0.43	  7) f_1#(I3, I4) -> f_1#(I3, round(I4 + 1)) [I3 >= 0 && I4 >= 0, I3 > I4]
0.00/0.43	  8) fNat#(true, I2, y) -> f_1#(I2, round(y + 1)) [I2 > y]
0.00/0.43	
0.00/0.43	We have the following SCCs.
0.00/0.43	  { 7 }
0.00/0.43	
0.00/0.43	DP problem for innermost termination.
0.00/0.43	P =
0.00/0.43	  f_1#(I3, I4) -> f_1#(I3, round(I4 + 1)) [I3 >= 0 && I4 >= 0, I3 > I4]
0.00/0.43	R = 
0.00/0.43	  if(false, u, v) -> v 
0.00/0.43	  if(true, I0, I1) -> I0 
0.00/0.43	  round(x) -> if(x % 2 = 0, x, x + 1) 
0.00/0.43	  fNat(true, I2, y) -> f(I2 > y, I2, round(y + 1)) 
0.00/0.43	  f(true, I3, I4) -> fNat(I3 >= 0 && I4 >= 0, I3, I4) 
0.00/0.43	
0.00/3.42	EOF