0.00/0.38	MAYBE
0.00/0.38	
0.00/0.38	DP problem for innermost termination.
0.00/0.38	P =
0.00/0.38	  init#(x1, x2, x3) -> f1#(rnd1, rnd2, rnd3) 
0.00/0.38	  f2#(I0, I1, I2) -> f2#(I0 - I1 + I1 - (I0 - I1), I0 - I1 + I1, I0 - I1 + I1 - (I0 - I1) + I0 - I1 + I1) [I2 <= 4]
0.00/0.38	  f1#(I3, I4, I5) -> f2#(1, 2, 3) 
0.00/0.38	R = 
0.00/0.38	  init(x1, x2, x3) -> f1(rnd1, rnd2, rnd3) 
0.00/0.38	  f2(I0, I1, I2) -> f2(I0 - I1 + I1 - (I0 - I1), I0 - I1 + I1, I0 - I1 + I1 - (I0 - I1) + I0 - I1 + I1) [I2 <= 4]
0.00/0.38	  f1(I3, I4, I5) -> f2(1, 2, 3) 
0.00/0.38	
0.00/0.38	The dependency graph for this problem is:
0.00/0.38	  0 -> 2
0.00/0.38	  1 -> 1
0.00/0.38	  2 -> 1
0.00/0.38	Where:
0.00/0.38	  0) init#(x1, x2, x3) -> f1#(rnd1, rnd2, rnd3) 
0.00/0.38	  1) f2#(I0, I1, I2) -> f2#(I0 - I1 + I1 - (I0 - I1), I0 - I1 + I1, I0 - I1 + I1 - (I0 - I1) + I0 - I1 + I1) [I2 <= 4]
0.00/0.38	  2) f1#(I3, I4, I5) -> f2#(1, 2, 3) 
0.00/0.38	
0.00/0.38	We have the following SCCs.
0.00/0.38	  { 1 }
0.00/0.38	
0.00/0.38	DP problem for innermost termination.
0.00/0.38	P =
0.00/0.38	  f2#(I0, I1, I2) -> f2#(I0 - I1 + I1 - (I0 - I1), I0 - I1 + I1, I0 - I1 + I1 - (I0 - I1) + I0 - I1 + I1) [I2 <= 4]
0.00/0.38	R = 
0.00/0.38	  init(x1, x2, x3) -> f1(rnd1, rnd2, rnd3) 
0.00/0.38	  f2(I0, I1, I2) -> f2(I0 - I1 + I1 - (I0 - I1), I0 - I1 + I1, I0 - I1 + I1 - (I0 - I1) + I0 - I1 + I1) [I2 <= 4]
0.00/0.38	  f1(I3, I4, I5) -> f2(1, 2, 3) 
0.00/0.38	
0.00/3.36	EOF