0.00/0.31	MAYBE
0.00/0.31	
0.00/0.31	DP problem for innermost termination.
0.00/0.31	P =
0.00/0.31	  init#(x1) -> f1#(rnd1) 
0.00/0.31	  f2#(I0) -> f2#(I0) [I0 <= 19 /\ 11 <= I0 - 1]
0.00/0.31	  f2#(I1) -> f2#(I1 + 1) [I1 <= 19 /\ I1 <= 11]
0.00/0.31	  f1#(I2) -> f2#(0) 
0.00/0.31	R = 
0.00/0.31	  init(x1) -> f1(rnd1) 
0.00/0.31	  f2(I0) -> f2(I0) [I0 <= 19 /\ 11 <= I0 - 1]
0.00/0.31	  f2(I1) -> f2(I1 + 1) [I1 <= 19 /\ I1 <= 11]
0.00/0.31	  f1(I2) -> f2(0) 
0.00/0.31	
0.00/0.31	The dependency graph for this problem is:
0.00/0.31	  0 -> 3
0.00/0.31	  1 -> 1
0.00/0.31	  2 -> 1, 2
0.00/0.31	  3 -> 2
0.00/0.31	Where:
0.00/0.31	  0) init#(x1) -> f1#(rnd1) 
0.00/0.31	  1) f2#(I0) -> f2#(I0) [I0 <= 19 /\ 11 <= I0 - 1]
0.00/0.31	  2) f2#(I1) -> f2#(I1 + 1) [I1 <= 19 /\ I1 <= 11]
0.00/0.31	  3) f1#(I2) -> f2#(0) 
0.00/0.31	
0.00/0.31	We have the following SCCs.
0.00/0.31	  { 2 }
0.00/0.31	  { 1 }
0.00/0.31	
0.00/0.31	DP problem for innermost termination.
0.00/0.31	P =
0.00/0.31	  f2#(I0) -> f2#(I0) [I0 <= 19 /\ 11 <= I0 - 1]
0.00/0.31	R = 
0.00/0.31	  init(x1) -> f1(rnd1) 
0.00/0.31	  f2(I0) -> f2(I0) [I0 <= 19 /\ 11 <= I0 - 1]
0.00/0.31	  f2(I1) -> f2(I1 + 1) [I1 <= 19 /\ I1 <= 11]
0.00/0.31	  f1(I2) -> f2(0) 
0.00/0.31	
0.00/3.29	EOF