1.08/1.47	MAYBE
1.08/1.47	
1.08/1.47	DP problem for innermost termination.
1.08/1.47	P =
1.08/1.47	  f5#(x1, x2, x3) -> f4#(x1, x2, x3) 
1.08/1.47	  f4#(I0, I1, I2) -> f2#(0, 0, I2) 
1.08/1.47	  f2#(I3, I4, I5) -> f1#(I3, I4, I5) 
1.08/1.47	  f1#(I9, I10, I11) -> f2#(1 + I9, 2 + I10, rnd3) [rnd3 = rnd3]
1.08/1.47	R = 
1.08/1.47	  f5(x1, x2, x3) -> f4(x1, x2, x3) 
1.08/1.47	  f4(I0, I1, I2) -> f2(0, 0, I2) 
1.08/1.47	  f2(I3, I4, I5) -> f1(I3, I4, I5) 
1.08/1.47	  f1(I6, I7, I8) -> f3(I6, I7, I8) 
1.08/1.47	  f1(I9, I10, I11) -> f2(1 + I9, 2 + I10, rnd3) [rnd3 = rnd3]
1.08/1.47	
1.08/1.47	The dependency graph for this problem is:
1.08/1.47	  0 -> 1
1.08/1.47	  1 -> 2
1.08/1.47	  2 -> 3
1.08/1.47	  3 -> 2
1.08/1.47	Where:
1.08/1.47	  0) f5#(x1, x2, x3) -> f4#(x1, x2, x3) 
1.08/1.47	  1) f4#(I0, I1, I2) -> f2#(0, 0, I2) 
1.08/1.47	  2) f2#(I3, I4, I5) -> f1#(I3, I4, I5) 
1.08/1.47	  3) f1#(I9, I10, I11) -> f2#(1 + I9, 2 + I10, rnd3) [rnd3 = rnd3]
1.08/1.47	
1.08/1.47	We have the following SCCs.
1.08/1.47	  { 2, 3 }
1.08/1.47	
1.08/1.47	DP problem for innermost termination.
1.08/1.47	P =
1.08/1.47	  f2#(I3, I4, I5) -> f1#(I3, I4, I5) 
1.08/1.47	  f1#(I9, I10, I11) -> f2#(1 + I9, 2 + I10, rnd3) [rnd3 = rnd3]
1.08/1.47	R = 
1.08/1.47	  f5(x1, x2, x3) -> f4(x1, x2, x3) 
1.08/1.47	  f4(I0, I1, I2) -> f2(0, 0, I2) 
1.08/1.47	  f2(I3, I4, I5) -> f1(I3, I4, I5) 
1.08/1.47	  f1(I6, I7, I8) -> f3(I6, I7, I8) 
1.08/1.47	  f1(I9, I10, I11) -> f2(1 + I9, 2 + I10, rnd3) [rnd3 = rnd3]
1.08/1.47	
1.08/4.45	EOF