0.00/0.25	MAYBE
0.00/0.25	
0.00/0.25	DP problem for innermost termination.
0.00/0.25	P =
0.00/0.25	  f4#(x1) -> f3#(x1) 
0.00/0.25	  f3#(I0) -> f1#(rnd1) [2 <= rnd1 /\ rnd1 = rnd1]
0.00/0.25	  f2#(I1) -> f1#(I1) 
0.00/0.25	  f1#(I2) -> f2#(I3) [y1 = y1 /\ 1 <= y1 /\ 2 <= y1 /\ I3 = I3 /\ -1 <= I3 /\ 1 + I3 <= 2]
0.00/0.25	R = 
0.00/0.25	  f4(x1) -> f3(x1) 
0.00/0.25	  f3(I0) -> f1(rnd1) [2 <= rnd1 /\ rnd1 = rnd1]
0.00/0.25	  f2(I1) -> f1(I1) 
0.00/0.25	  f1(I2) -> f2(I3) [y1 = y1 /\ 1 <= y1 /\ 2 <= y1 /\ I3 = I3 /\ -1 <= I3 /\ 1 + I3 <= 2]
0.00/0.25	
0.00/0.25	The dependency graph for this problem is:
0.00/0.25	  0 -> 1
0.00/0.25	  1 -> 3
0.00/0.25	  2 -> 3
0.00/0.25	  3 -> 2
0.00/0.25	Where:
0.00/0.25	  0) f4#(x1) -> f3#(x1) 
0.00/0.25	  1) f3#(I0) -> f1#(rnd1) [2 <= rnd1 /\ rnd1 = rnd1]
0.00/0.25	  2) f2#(I1) -> f1#(I1) 
0.00/0.25	  3) f1#(I2) -> f2#(I3) [y1 = y1 /\ 1 <= y1 /\ 2 <= y1 /\ I3 = I3 /\ -1 <= I3 /\ 1 + I3 <= 2]
0.00/0.25	
0.00/0.25	We have the following SCCs.
0.00/0.25	  { 2, 3 }
0.00/0.25	
0.00/0.25	DP problem for innermost termination.
0.00/0.25	P =
0.00/0.25	  f2#(I1) -> f1#(I1) 
0.00/0.25	  f1#(I2) -> f2#(I3) [y1 = y1 /\ 1 <= y1 /\ 2 <= y1 /\ I3 = I3 /\ -1 <= I3 /\ 1 + I3 <= 2]
0.00/0.25	R = 
0.00/0.25	  f4(x1) -> f3(x1) 
0.00/0.25	  f3(I0) -> f1(rnd1) [2 <= rnd1 /\ rnd1 = rnd1]
0.00/0.25	  f2(I1) -> f1(I1) 
0.00/0.25	  f1(I2) -> f2(I3) [y1 = y1 /\ 1 <= y1 /\ 2 <= y1 /\ I3 = I3 /\ -1 <= I3 /\ 1 + I3 <= 2]
0.00/0.25	
0.00/3.23	EOF