0.00/0.27	MAYBE
0.00/0.27	
0.00/0.27	DP problem for innermost termination.
0.00/0.27	P =
0.00/0.27	  init#(x1) -> f1#(rnd1) 
0.00/0.27	  f2#(I0) -> f2#(49) [49 <= I0 - 1]
0.00/0.27	  f2#(I1) -> f2#(51) [I1 <= 49]
0.00/0.27	  f1#(I2) -> f2#(0) 
0.00/0.27	R = 
0.00/0.27	  init(x1) -> f1(rnd1) 
0.00/0.27	  f2(I0) -> f2(49) [49 <= I0 - 1]
0.00/0.27	  f2(I1) -> f2(51) [I1 <= 49]
0.00/0.27	  f1(I2) -> f2(0) 
0.00/0.27	
0.00/0.27	The dependency graph for this problem is:
0.00/0.27	  0 -> 3
0.00/0.27	  1 -> 2
0.00/0.27	  2 -> 1
0.00/0.27	  3 -> 2
0.00/0.27	Where:
0.00/0.27	  0) init#(x1) -> f1#(rnd1) 
0.00/0.27	  1) f2#(I0) -> f2#(49) [49 <= I0 - 1]
0.00/0.27	  2) f2#(I1) -> f2#(51) [I1 <= 49]
0.00/0.27	  3) f1#(I2) -> f2#(0) 
0.00/0.27	
0.00/0.27	We have the following SCCs.
0.00/0.27	  { 1, 2 }
0.00/0.27	
0.00/0.27	DP problem for innermost termination.
0.00/0.27	P =
0.00/0.27	  f2#(I0) -> f2#(49) [49 <= I0 - 1]
0.00/0.27	  f2#(I1) -> f2#(51) [I1 <= 49]
0.00/0.27	R = 
0.00/0.27	  init(x1) -> f1(rnd1) 
0.00/0.27	  f2(I0) -> f2(49) [49 <= I0 - 1]
0.00/0.27	  f2(I1) -> f2(51) [I1 <= 49]
0.00/0.27	  f1(I2) -> f2(0) 
0.00/0.27	
0.00/3.25	EOF