0.00/0.38	YES
0.00/0.38	
0.00/0.38	DP problem for innermost termination.
0.00/0.38	P =
0.00/0.38	  f5#(x1, x2) -> f4#(x1, x2) 
0.00/0.38	  f4#(I0, I1) -> f1#(I0, I1) 
0.00/0.38	  f2#(I4, I5) -> f1#(I4, I5) 
0.00/0.38	  f1#(I6, I7) -> f2#(I6, rnd2) [0 <= I7 /\ y1 = 1 + I7 /\ rnd2 = -2 + y1]
0.00/0.38	R = 
0.00/0.38	  f5(x1, x2) -> f4(x1, x2) 
0.00/0.38	  f4(I0, I1) -> f1(I0, I1) 
0.00/0.38	  f1(I2, I3) -> f3(rnd1, I3) [rnd1 = rnd1 /\ 1 + I3 <= 0]
0.00/0.38	  f2(I4, I5) -> f1(I4, I5) 
0.00/0.38	  f1(I6, I7) -> f2(I6, rnd2) [0 <= I7 /\ y1 = 1 + I7 /\ rnd2 = -2 + y1]
0.00/0.38	
0.00/0.38	The dependency graph for this problem is:
0.00/0.38	  0 -> 1
0.00/0.38	  1 -> 3
0.00/0.38	  2 -> 3
0.00/0.38	  3 -> 2
0.00/0.38	Where:
0.00/0.38	  0) f5#(x1, x2) -> f4#(x1, x2) 
0.00/0.38	  1) f4#(I0, I1) -> f1#(I0, I1) 
0.00/0.38	  2) f2#(I4, I5) -> f1#(I4, I5) 
0.00/0.38	  3) f1#(I6, I7) -> f2#(I6, rnd2) [0 <= I7 /\ y1 = 1 + I7 /\ rnd2 = -2 + y1]
0.00/0.38	
0.00/0.38	We have the following SCCs.
0.00/0.38	  { 2, 3 }
0.00/0.38	
0.00/0.38	DP problem for innermost termination.
0.00/0.38	P =
0.00/0.38	  f2#(I4, I5) -> f1#(I4, I5) 
0.00/0.38	  f1#(I6, I7) -> f2#(I6, rnd2) [0 <= I7 /\ y1 = 1 + I7 /\ rnd2 = -2 + y1]
0.00/0.38	R = 
0.00/0.38	  f5(x1, x2) -> f4(x1, x2) 
0.00/0.38	  f4(I0, I1) -> f1(I0, I1) 
0.00/0.38	  f1(I2, I3) -> f3(rnd1, I3) [rnd1 = rnd1 /\ 1 + I3 <= 0]
0.00/0.38	  f2(I4, I5) -> f1(I4, I5) 
0.00/0.38	  f1(I6, I7) -> f2(I6, rnd2) [0 <= I7 /\ y1 = 1 + I7 /\ rnd2 = -2 + y1]
0.00/0.38	
0.00/0.38	We use the basic value criterion with the projection function NU:
0.00/0.38	  NU[f1#(z1,z2)] = z2
0.00/0.38	  NU[f2#(z1,z2)] = z2
0.00/0.38	
0.00/0.38	This gives the following inequalities:
0.00/0.38	    ==>  I5 (>! \union =) I5
0.00/0.38	  0 <= I7 /\ y1 = 1 + I7 /\ rnd2 = -2 + y1  ==>  I7 >! rnd2
0.00/0.38	
0.00/0.38	We remove all the strictly oriented dependency pairs.
0.00/0.38	
0.00/0.38	DP problem for innermost termination.
0.00/0.38	P =
0.00/0.38	  f2#(I4, I5) -> f1#(I4, I5) 
0.00/0.38	R = 
0.00/0.38	  f5(x1, x2) -> f4(x1, x2) 
0.00/0.38	  f4(I0, I1) -> f1(I0, I1) 
0.00/0.38	  f1(I2, I3) -> f3(rnd1, I3) [rnd1 = rnd1 /\ 1 + I3 <= 0]
0.00/0.38	  f2(I4, I5) -> f1(I4, I5) 
0.00/0.38	  f1(I6, I7) -> f2(I6, rnd2) [0 <= I7 /\ y1 = 1 + I7 /\ rnd2 = -2 + y1]
0.00/0.38	
0.00/0.38	The dependency graph for this problem is:
0.00/0.38	  2 -> 
0.00/0.38	Where:
0.00/0.38	  2) f2#(I4, I5) -> f1#(I4, I5) 
0.00/0.38	
0.00/0.38	We have the following SCCs.
0.00/0.38	  
0.00/3.36	EOF