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, x4, x5) -> f4#(x1, x2, x3, x4, x5) 
0.00/0.07	  f4#(I0, I1, I2, I3, I4) -> f3#(0, 1, rnd3, rnd4, rnd5) [rnd5 = rnd5 /\ rnd4 = rnd4 /\ rnd3 = rnd3]
0.00/0.07	  f3#(I5, I6, I7, I8, I9) -> f1#(I5, I6, I7, I8, I9) 
0.00/0.07	  f3#(I10, I11, I12, I13, I14) -> f1#(I10, I11, I12, I13, I14) 
0.00/0.07	R = 
0.00/0.07	  f5(x1, x2, x3, x4, x5) -> f4(x1, x2, x3, x4, x5) 
0.00/0.07	  f4(I0, I1, I2, I3, I4) -> f3(0, 1, rnd3, rnd4, rnd5) [rnd5 = rnd5 /\ rnd4 = rnd4 /\ rnd3 = rnd3]
0.00/0.07	  f3(I5, I6, I7, I8, I9) -> f1(I5, I6, I7, I8, I9) 
0.00/0.07	  f3(I10, I11, I12, I13, I14) -> f1(I10, I11, I12, I13, I14) 
0.00/0.07	  f1(I15, I16, I17, I18, I19) -> f2(I15, I16, I17, I18, I19) 
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, x4, x5) -> f4#(x1, x2, x3, x4, x5) 
0.00/0.07	  1) f4#(I0, I1, I2, I3, I4) -> f3#(0, 1, rnd3, rnd4, rnd5) [rnd5 = rnd5 /\ rnd4 = rnd4 /\ rnd3 = rnd3]
0.00/0.07	  2) f3#(I5, I6, I7, I8, I9) -> f1#(I5, I6, I7, I8, I9) 
0.00/0.07	  3) f3#(I10, I11, I12, I13, I14) -> f1#(I10, I11, I12, I13, I14) 
0.00/0.07	
0.00/0.07	We have the following SCCs.
0.00/0.07	  
0.00/3.05	EOF