1.43/1.46	MAYBE
1.43/1.46	
1.43/1.46	DP problem for innermost termination.
1.43/1.46	P =
1.43/1.46	  f5#(x1, x2, x3) -> f4#(x1, x2, x3) 
1.43/1.46	  f4#(I0, I1, I2) -> f3#(I0, I1, I2) 
1.43/1.46	  f3#(I3, I4, I5) -> f1#(I4, rnd2, rnd3) [rnd3 = rnd3 /\ y1 = rnd3 /\ rnd2 = 1 + y1]
1.43/1.46	  f1#(I6, I7, I8) -> f3#(I6, I7, I8) 
1.43/1.46	R = 
1.43/1.46	  f5(x1, x2, x3) -> f4(x1, x2, x3) 
1.43/1.46	  f4(I0, I1, I2) -> f3(I0, I1, I2) 
1.43/1.46	  f3(I3, I4, I5) -> f1(I4, rnd2, rnd3) [rnd3 = rnd3 /\ y1 = rnd3 /\ rnd2 = 1 + y1]
1.43/1.46	  f1(I6, I7, I8) -> f3(I6, I7, I8) 
1.43/1.46	  f1(I9, I10, I11) -> f2(I9, I10, I11) [1 + I10 <= I11]
1.43/1.46	
1.43/1.46	The dependency graph for this problem is:
1.43/1.46	  0 -> 1
1.43/1.46	  1 -> 2
1.43/1.46	  2 -> 3
1.43/1.46	  3 -> 2
1.43/1.46	Where:
1.43/1.46	  0) f5#(x1, x2, x3) -> f4#(x1, x2, x3) 
1.43/1.46	  1) f4#(I0, I1, I2) -> f3#(I0, I1, I2) 
1.43/1.46	  2) f3#(I3, I4, I5) -> f1#(I4, rnd2, rnd3) [rnd3 = rnd3 /\ y1 = rnd3 /\ rnd2 = 1 + y1]
1.43/1.46	  3) f1#(I6, I7, I8) -> f3#(I6, I7, I8) 
1.43/1.46	
1.43/1.46	We have the following SCCs.
1.43/1.46	  { 2, 3 }
1.43/1.46	
1.43/1.46	DP problem for innermost termination.
1.43/1.46	P =
1.43/1.46	  f3#(I3, I4, I5) -> f1#(I4, rnd2, rnd3) [rnd3 = rnd3 /\ y1 = rnd3 /\ rnd2 = 1 + y1]
1.43/1.46	  f1#(I6, I7, I8) -> f3#(I6, I7, I8) 
1.43/1.46	R = 
1.43/1.46	  f5(x1, x2, x3) -> f4(x1, x2, x3) 
1.43/1.46	  f4(I0, I1, I2) -> f3(I0, I1, I2) 
1.43/1.46	  f3(I3, I4, I5) -> f1(I4, rnd2, rnd3) [rnd3 = rnd3 /\ y1 = rnd3 /\ rnd2 = 1 + y1]
1.43/1.46	  f1(I6, I7, I8) -> f3(I6, I7, I8) 
1.43/1.46	  f1(I9, I10, I11) -> f2(I9, I10, I11) [1 + I10 <= I11]
1.43/1.46	
1.43/4.44	EOF