0.00/0.06	YES
0.00/0.06	
0.00/0.06	DP problem for innermost termination.
0.00/0.06	P =
0.00/0.06	  f5#(x1, x2) -> f3#(x1, x2) 
0.00/0.06	  f3#(I2, I3) -> f1#(I2, I3) 
0.00/0.06	  f1#(I4, I5) -> f2#(I4, I5) [0 <= -1 - I5]
0.00/0.06	  f1#(I6, I7) -> f2#(I6, 1 + I7) [-1 * I7 <= 0]
0.00/0.06	R = 
0.00/0.06	  f5(x1, x2) -> f3(x1, x2) 
0.00/0.06	  f2(I0, I1) -> f4(rnd1, I1) [rnd1 = rnd1]
0.00/0.06	  f3(I2, I3) -> f1(I2, I3) 
0.00/0.06	  f1(I4, I5) -> f2(I4, I5) [0 <= -1 - I5]
0.00/0.06	  f1(I6, I7) -> f2(I6, 1 + I7) [-1 * I7 <= 0]
0.00/0.06	
0.00/0.06	The dependency graph for this problem is:
0.00/0.06	  0 -> 1
0.00/0.06	  1 -> 2, 3
0.00/0.06	  2 -> 
0.00/0.06	  3 -> 
0.00/0.06	Where:
0.00/0.06	  0) f5#(x1, x2) -> f3#(x1, x2) 
0.00/0.06	  1) f3#(I2, I3) -> f1#(I2, I3) 
0.00/0.06	  2) f1#(I4, I5) -> f2#(I4, I5) [0 <= -1 - I5]
0.00/0.06	  3) f1#(I6, I7) -> f2#(I6, 1 + I7) [-1 * I7 <= 0]
0.00/0.06	
0.00/0.06	We have the following SCCs.
0.00/0.06	  
0.00/3.05	EOF