0.90/0.94	MAYBE
0.90/0.94	
0.90/0.94	DP problem for innermost termination.
0.90/0.94	P =
0.90/0.94	  f5#(x1, x2) -> f1#(x1, x2) 
0.90/0.94	  f3#(I2, I3) -> f2#(I2, I3) 
0.90/0.94	  f2#(I4, I5) -> f3#(I4, I5) [-1 * I5 <= 0]
0.90/0.94	  f1#(I6, I7) -> f2#(I6, I7) 
0.90/0.94	R = 
0.90/0.94	  f5(x1, x2) -> f1(x1, x2) 
0.90/0.94	  f2(I0, I1) -> f4(rnd1, I1) [rnd1 = rnd1 /\ 0 <= -1 - I1]
0.90/0.94	  f3(I2, I3) -> f2(I2, I3) 
0.90/0.94	  f2(I4, I5) -> f3(I4, I5) [-1 * I5 <= 0]
0.90/0.94	  f1(I6, I7) -> f2(I6, I7) 
0.90/0.94	
0.90/0.94	The dependency graph for this problem is:
0.90/0.94	  0 -> 3
0.90/0.94	  1 -> 2
0.90/0.94	  2 -> 1
0.90/0.94	  3 -> 2
0.90/0.94	Where:
0.90/0.94	  0) f5#(x1, x2) -> f1#(x1, x2) 
0.90/0.94	  1) f3#(I2, I3) -> f2#(I2, I3) 
0.90/0.94	  2) f2#(I4, I5) -> f3#(I4, I5) [-1 * I5 <= 0]
0.90/0.94	  3) f1#(I6, I7) -> f2#(I6, I7) 
0.90/0.94	
0.90/0.94	We have the following SCCs.
0.90/0.94	  { 1, 2 }
0.90/0.94	
0.90/0.94	DP problem for innermost termination.
0.90/0.94	P =
0.90/0.94	  f3#(I2, I3) -> f2#(I2, I3) 
0.90/0.94	  f2#(I4, I5) -> f3#(I4, I5) [-1 * I5 <= 0]
0.90/0.94	R = 
0.90/0.94	  f5(x1, x2) -> f1(x1, x2) 
0.90/0.94	  f2(I0, I1) -> f4(rnd1, I1) [rnd1 = rnd1 /\ 0 <= -1 - I1]
0.90/0.94	  f3(I2, I3) -> f2(I2, I3) 
0.90/0.94	  f2(I4, I5) -> f3(I4, I5) [-1 * I5 <= 0]
0.90/0.94	  f1(I6, I7) -> f2(I6, I7) 
0.90/0.94	
0.90/3.91	EOF