0.00/0.07	YES
0.00/0.07	
0.00/0.07	DP problem for innermost termination.
0.00/0.07	P =
0.00/0.07	  f5#(x1, x2, x3) -> f4#(x1, x2, x3) 
0.00/0.07	  f4#(I0, I1, I2) -> f1#(0, 1, rnd3) [rnd3 = rnd3]
0.00/0.07	  f1#(I6, I7, I8) -> f2#(I6, I7, I8) 
0.00/0.07	  f1#(I9, I10, I11) -> f2#(I9, I10, I11) 
0.00/0.07	R = 
0.00/0.07	  f5(x1, x2, x3) -> f4(x1, x2, x3) 
0.00/0.07	  f4(I0, I1, I2) -> f1(0, 1, rnd3) [rnd3 = rnd3]
0.00/0.07	  f2(I3, I4, I5) -> f3(I3, I4, I5) 
0.00/0.07	  f1(I6, I7, I8) -> f2(I6, I7, I8) 
0.00/0.07	  f1(I9, I10, I11) -> f2(I9, I10, I11) 
0.00/0.07	
0.00/0.07	The dependency graph for this problem is:
0.00/0.07	  0 -> 1
0.00/0.07	  1 -> 2, 3
0.00/0.07	  2 -> 
0.00/0.07	  3 -> 
0.00/0.07	Where:
0.00/0.07	  0) f5#(x1, x2, x3) -> f4#(x1, x2, x3) 
0.00/0.07	  1) f4#(I0, I1, I2) -> f1#(0, 1, rnd3) [rnd3 = rnd3]
0.00/0.07	  2) f1#(I6, I7, I8) -> f2#(I6, I7, I8) 
0.00/0.07	  3) f1#(I9, I10, I11) -> f2#(I9, I10, I11) 
0.00/0.07	
0.00/0.07	We have the following SCCs.
0.00/0.07	  
0.00/3.05	EOF