0.76/0.79	MAYBE
0.76/0.79	
0.76/0.79	DP problem for innermost termination.
0.76/0.79	P =
0.76/0.79	  cond2#(false, x) -> f#(3 * x + 1) 
0.76/0.79	  cond2#(true, I0) -> f#(I0 / 2) 
0.76/0.79	  cond1#(true, I2) -> cond2#(I2 % 2 = 0, I2) 
0.76/0.79	  f#(I3) -> cond1#(I3 > 1, I3) 
0.76/0.79	R = 
0.76/0.79	  cond2(false, x) -> f(3 * x + 1) 
0.76/0.79	  cond2(true, I0) -> f(I0 / 2) 
0.76/0.79	  cond1(false, I1) -> I1 
0.76/0.79	  cond1(true, I2) -> cond2(I2 % 2 = 0, I2) 
0.76/0.79	  f(I3) -> cond1(I3 > 1, I3) 
0.76/0.79	
0.76/0.79	This problem is converted using chaining, where edges between chained DPs are removed.
0.76/0.79	
0.76/0.79	DP problem for innermost termination.
0.76/0.79	P =
0.76/0.79	  cond2#(false, x) -> f#(3 * x + 1) 
0.76/0.79	  cond2#(true, I0) -> f#(I0 / 2) 
0.76/0.79	  cond1#(true, I2) -> cond2#(I2 % 2 = 0, I2) 
0.76/0.79	  f#(I3) -> cond1#(I3 > 1, I3) 
0.76/0.79	  f#(I3) -> cond2#(I3 % 2 = 0, I3) [I3 > 1]
0.76/0.79	  f#(I3) -> f#(I3 / 2) [I3 > 1, I3 % 2 = 0]
0.76/0.79	  f#(I3) -> f#(3 * I3 + 1) [I3 > 1, not(I3 % 2 = 0)]
0.76/0.79	  cond1#(true, I2) -> f#(3 * I2 + 1) [not(I2 % 2 = 0)]
0.76/0.79	  cond1#(true, I2) -> f#(I2 / 2) [I2 % 2 = 0]
0.76/0.79	R = 
0.76/0.79	  cond2(false, x) -> f(3 * x + 1) 
0.76/0.79	  cond2(true, I0) -> f(I0 / 2) 
0.76/0.79	  cond1(false, I1) -> I1 
0.76/0.79	  cond1(true, I2) -> cond2(I2 % 2 = 0, I2) 
0.76/0.80	  f(I3) -> cond1(I3 > 1, I3) 
0.76/0.80	
0.76/0.80	The dependency graph for this problem is:
0.76/0.80	  0 -> 6, 5, 4, 3
0.76/0.80	  1 -> 6, 5, 4, 3
0.76/0.80	  2 -> 
0.76/0.80	  3 -> 
0.76/0.80	  4 -> 
0.76/0.80	  5 -> 6, 5, 3, 4
0.76/0.80	  6 -> 3, 4, 5
0.76/0.80	  7 -> 3, 4, 5
0.76/0.80	  8 -> 3, 4, 5, 6
0.76/0.80	Where:
0.76/0.80	  0) cond2#(false, x) -> f#(3 * x + 1) 
0.76/0.80	  1) cond2#(true, I0) -> f#(I0 / 2) 
0.76/0.80	  2) cond1#(true, I2) -> cond2#(I2 % 2 = 0, I2) 
0.76/0.80	  3) f#(I3) -> cond1#(I3 > 1, I3) 
0.76/0.80	  4) f#(I3) -> cond2#(I3 % 2 = 0, I3) [I3 > 1]
0.76/0.80	  5) f#(I3) -> f#(I3 / 2) [I3 > 1, I3 % 2 = 0]
0.76/0.80	  6) f#(I3) -> f#(3 * I3 + 1) [I3 > 1, not(I3 % 2 = 0)]
0.76/0.80	  7) cond1#(true, I2) -> f#(3 * I2 + 1) [not(I2 % 2 = 0)]
0.76/0.80	  8) cond1#(true, I2) -> f#(I2 / 2) [I2 % 2 = 0]
0.76/0.80	
0.76/0.80	We have the following SCCs.
0.76/0.80	  { 5, 6 }
0.76/0.80	
0.76/0.80	DP problem for innermost termination.
0.76/0.80	P =
0.76/0.80	  f#(I3) -> f#(I3 / 2) [I3 > 1, I3 % 2 = 0]
0.76/0.80	  f#(I3) -> f#(3 * I3 + 1) [I3 > 1, not(I3 % 2 = 0)]
0.76/0.80	R = 
0.76/0.80	  cond2(false, x) -> f(3 * x + 1) 
0.76/0.80	  cond2(true, I0) -> f(I0 / 2) 
0.76/0.80	  cond1(false, I1) -> I1 
0.76/0.80	  cond1(true, I2) -> cond2(I2 % 2 = 0, I2) 
0.76/0.80	  f(I3) -> cond1(I3 > 1, I3) 
0.76/0.80	
0.76/3.78	EOF