0.00/0.09	YES
0.00/0.09	
0.00/0.09	DP problem for innermost termination.
0.00/0.09	P =
0.00/0.09	  f4#(x1) -> f3#(x1) 
0.00/0.09	  f3#(I0) -> f1#(I0) 
0.00/0.09	  f1#(I1) -> f3#(rnd1) [1 <= I1 /\ y1 = I1 /\ rnd1 = -1 + y1]
0.00/0.09	R = 
0.00/0.09	  f4(x1) -> f3(x1) 
0.00/0.09	  f3(I0) -> f1(I0) 
0.00/0.09	  f1(I1) -> f3(rnd1) [1 <= I1 /\ y1 = I1 /\ rnd1 = -1 + y1]
0.00/0.09	  f1(I2) -> f2(I3) [I2 <= 0 /\ I4 = I2 /\ y2 = I4 /\ y3 = y2 /\ I3 = y3]
0.00/0.09	
0.00/0.09	The dependency graph for this problem is:
0.00/0.09	  0 -> 1
0.00/0.09	  1 -> 2
0.00/0.09	  2 -> 1
0.00/0.09	Where:
0.00/0.09	  0) f4#(x1) -> f3#(x1) 
0.00/0.09	  1) f3#(I0) -> f1#(I0) 
0.00/0.09	  2) f1#(I1) -> f3#(rnd1) [1 <= I1 /\ y1 = I1 /\ rnd1 = -1 + y1]
0.00/0.09	
0.00/0.09	We have the following SCCs.
0.00/0.09	  { 1, 2 }
0.00/0.09	
0.00/0.09	DP problem for innermost termination.
0.00/0.09	P =
0.00/0.09	  f3#(I0) -> f1#(I0) 
0.00/0.09	  f1#(I1) -> f3#(rnd1) [1 <= I1 /\ y1 = I1 /\ rnd1 = -1 + y1]
0.00/0.09	R = 
0.00/0.09	  f4(x1) -> f3(x1) 
0.00/0.09	  f3(I0) -> f1(I0) 
0.00/0.09	  f1(I1) -> f3(rnd1) [1 <= I1 /\ y1 = I1 /\ rnd1 = -1 + y1]
0.00/0.09	  f1(I2) -> f2(I3) [I2 <= 0 /\ I4 = I2 /\ y2 = I4 /\ y3 = y2 /\ I3 = y3]
0.00/0.09	
0.00/0.09	We use the basic value criterion with the projection function NU:
0.00/0.09	  NU[f1#(z1)] = z1
0.00/0.09	  NU[f3#(z1)] = z1
0.00/0.09	
0.00/0.09	This gives the following inequalities:
0.00/0.09	    ==>  I0 (>! \union =) I0
0.00/0.09	  1 <= I1 /\ y1 = I1 /\ rnd1 = -1 + y1  ==>  I1 >! rnd1
0.00/0.09	
0.00/0.09	We remove all the strictly oriented dependency pairs.
0.00/0.09	
0.00/0.09	DP problem for innermost termination.
0.00/0.09	P =
0.00/0.09	  f3#(I0) -> f1#(I0) 
0.00/0.09	R = 
0.00/0.09	  f4(x1) -> f3(x1) 
0.00/0.09	  f3(I0) -> f1(I0) 
0.00/0.09	  f1(I1) -> f3(rnd1) [1 <= I1 /\ y1 = I1 /\ rnd1 = -1 + y1]
0.00/0.09	  f1(I2) -> f2(I3) [I2 <= 0 /\ I4 = I2 /\ y2 = I4 /\ y3 = y2 /\ I3 = y3]
0.00/0.09	
0.00/0.09	The dependency graph for this problem is:
0.00/0.09	  1 -> 
0.00/0.09	Where:
0.00/0.09	  1) f3#(I0) -> f1#(I0) 
0.00/0.09	
0.00/0.09	We have the following SCCs.
0.00/0.09	  
0.00/3.07	EOF