0.00/0.01	YES
0.00/0.01	
0.00/0.01	DP problem for innermost termination.
0.00/0.01	P =
0.00/0.01	  f5#(x1) -> f4#(x1) 
0.00/0.01	  f4#(I0) -> f3#(I0) 
0.00/0.01	  f4#(I1) -> f1#(rnd1) [rnd1 = rnd1]
0.00/0.01	  f4#(I2) -> f3#(I2) 
0.00/0.01	  f3#(I3) -> f1#(I3) 
0.00/0.01	R = 
0.00/0.01	  f5(x1) -> f4(x1) 
0.00/0.01	  f4(I0) -> f3(I0) 
0.00/0.01	  f4(I1) -> f1(rnd1) [rnd1 = rnd1]
0.00/0.01	  f4(I2) -> f3(I2) 
0.00/0.01	  f3(I3) -> f1(I3) 
0.00/0.01	  f1(I4) -> f2(I4) 
0.00/0.01	
0.00/0.01	The dependency graph for this problem is:
0.00/0.01	  0 -> 1, 2, 3
0.00/0.01	  1 -> 4
0.00/0.01	  2 -> 
0.00/0.01	  3 -> 4
0.00/0.01	  4 -> 
0.00/0.01	Where:
0.00/0.01	  0) f5#(x1) -> f4#(x1) 
0.00/0.01	  1) f4#(I0) -> f3#(I0) 
0.00/0.01	  2) f4#(I1) -> f1#(rnd1) [rnd1 = rnd1]
0.00/0.01	  3) f4#(I2) -> f3#(I2) 
0.00/0.01	  4) f3#(I3) -> f1#(I3) 
0.00/0.01	
0.00/0.01	We have the following SCCs.
0.00/0.01	  
0.00/0.01	EOF