0.00/0.36	YES
0.00/0.36	
0.00/0.36	DP problem for innermost termination.
0.00/0.36	P =
0.00/0.36	  f5#(x1, x2) -> f1#(x1, x2) 
0.00/0.36	  f3#(I2, I3) -> f2#(I2, I3) 
0.00/0.36	  f2#(I4, I5) -> f3#(I4, -1 + I5) [-1 * (-1 + I5) <= 0]
0.00/0.36	  f1#(I6, I7) -> f2#(I6, I7) 
0.00/0.36	R = 
0.00/0.36	  f5(x1, x2) -> f1(x1, x2) 
0.00/0.36	  f2(I0, I1) -> f4(rnd1, -1 + I1) [rnd1 = rnd1 /\ 0 <= -1 - (-1 + I1)]
0.00/0.36	  f3(I2, I3) -> f2(I2, I3) 
0.00/0.36	  f2(I4, I5) -> f3(I4, -1 + I5) [-1 * (-1 + I5) <= 0]
0.00/0.36	  f1(I6, I7) -> f2(I6, I7) 
0.00/0.36	
0.00/0.36	The dependency graph for this problem is:
0.00/0.36	  0 -> 3
0.00/0.36	  1 -> 2
0.00/0.36	  2 -> 1
0.00/0.36	  3 -> 2
0.00/0.36	Where:
0.00/0.36	  0) f5#(x1, x2) -> f1#(x1, x2) 
0.00/0.36	  1) f3#(I2, I3) -> f2#(I2, I3) 
0.00/0.36	  2) f2#(I4, I5) -> f3#(I4, -1 + I5) [-1 * (-1 + I5) <= 0]
0.00/0.36	  3) f1#(I6, I7) -> f2#(I6, I7) 
0.00/0.36	
0.00/0.36	We have the following SCCs.
0.00/0.36	  { 1, 2 }
0.00/0.36	
0.00/0.36	DP problem for innermost termination.
0.00/0.36	P =
0.00/0.36	  f3#(I2, I3) -> f2#(I2, I3) 
0.00/0.36	  f2#(I4, I5) -> f3#(I4, -1 + I5) [-1 * (-1 + I5) <= 0]
0.00/0.36	R = 
0.00/0.36	  f5(x1, x2) -> f1(x1, x2) 
0.00/0.36	  f2(I0, I1) -> f4(rnd1, -1 + I1) [rnd1 = rnd1 /\ 0 <= -1 - (-1 + I1)]
0.00/0.36	  f3(I2, I3) -> f2(I2, I3) 
0.00/0.36	  f2(I4, I5) -> f3(I4, -1 + I5) [-1 * (-1 + I5) <= 0]
0.00/0.36	  f1(I6, I7) -> f2(I6, I7) 
0.00/0.36	
0.00/0.36	We use the basic value criterion with the projection function NU:
0.00/0.36	  NU[f2#(z1,z2)] = z2
0.00/0.36	  NU[f3#(z1,z2)] = z2
0.00/0.36	
0.00/0.36	This gives the following inequalities:
0.00/0.36	    ==>  I3 (>! \union =) I3
0.00/0.36	  -1 * (-1 + I5) <= 0  ==>  I5 >! -1 + I5
0.00/0.36	
0.00/0.36	We remove all the strictly oriented dependency pairs.
0.00/0.36	
0.00/0.36	DP problem for innermost termination.
0.00/0.36	P =
0.00/0.36	  f3#(I2, I3) -> f2#(I2, I3) 
0.00/0.36	R = 
0.00/0.36	  f5(x1, x2) -> f1(x1, x2) 
0.00/0.36	  f2(I0, I1) -> f4(rnd1, -1 + I1) [rnd1 = rnd1 /\ 0 <= -1 - (-1 + I1)]
0.00/0.36	  f3(I2, I3) -> f2(I2, I3) 
0.00/0.36	  f2(I4, I5) -> f3(I4, -1 + I5) [-1 * (-1 + I5) <= 0]
0.00/0.36	  f1(I6, I7) -> f2(I6, I7) 
0.00/0.36	
0.00/0.36	The dependency graph for this problem is:
0.00/0.36	  1 -> 
0.00/0.36	Where:
0.00/0.36	  1) f3#(I2, I3) -> f2#(I2, I3) 
0.00/0.36	
0.00/0.36	We have the following SCCs.
0.00/0.36	  
0.00/3.34	EOF