4.75/4.71	MAYBE
4.75/4.71	
4.75/4.71	DP problem for innermost termination.
4.75/4.71	P =
4.75/4.71	  f4#(x1, x2, x3, x4) -> f3#(x1, x2, x3, x4) 
4.75/4.71	  f3#(I0, I1, I2, I3) -> f1#(I0, I1, I2, I3) 
4.75/4.71	  f2#(I4, I5, I6, I7) -> f1#(I4, I5, I6, I7) 
4.75/4.71	  f1#(I8, I9, I10, I11) -> f2#(I9, I11, rnd3, I11) [1 <= 1 + I10 /\ 1 <= I10 /\ y1 = y1 /\ 1 <= 1 + y1 /\ 1 <= y1 /\ rnd3 = rnd3 /\ 0 <= -1 * I9 /\ -1 * I9 <= 0]
4.75/4.71	R = 
4.75/4.71	  f4(x1, x2, x3, x4) -> f3(x1, x2, x3, x4) 
4.75/4.71	  f3(I0, I1, I2, I3) -> f1(I0, I1, I2, I3) 
4.75/4.71	  f2(I4, I5, I6, I7) -> f1(I4, I5, I6, I7) 
4.75/4.71	  f1(I8, I9, I10, I11) -> f2(I9, I11, rnd3, I11) [1 <= 1 + I10 /\ 1 <= I10 /\ y1 = y1 /\ 1 <= 1 + y1 /\ 1 <= y1 /\ rnd3 = rnd3 /\ 0 <= -1 * I9 /\ -1 * I9 <= 0]
4.75/4.71	
4.75/4.71	The dependency graph for this problem is:
4.75/4.71	  0 -> 1
4.75/4.71	  1 -> 3
4.75/4.71	  2 -> 3
4.75/4.71	  3 -> 2
4.75/4.71	Where:
4.75/4.71	  0) f4#(x1, x2, x3, x4) -> f3#(x1, x2, x3, x4) 
4.75/4.71	  1) f3#(I0, I1, I2, I3) -> f1#(I0, I1, I2, I3) 
4.75/4.71	  2) f2#(I4, I5, I6, I7) -> f1#(I4, I5, I6, I7) 
4.75/4.71	  3) f1#(I8, I9, I10, I11) -> f2#(I9, I11, rnd3, I11) [1 <= 1 + I10 /\ 1 <= I10 /\ y1 = y1 /\ 1 <= 1 + y1 /\ 1 <= y1 /\ rnd3 = rnd3 /\ 0 <= -1 * I9 /\ -1 * I9 <= 0]
4.75/4.71	
4.75/4.71	We have the following SCCs.
4.75/4.71	  { 2, 3 }
4.75/4.71	
4.75/4.71	DP problem for innermost termination.
4.75/4.71	P =
4.75/4.71	  f2#(I4, I5, I6, I7) -> f1#(I4, I5, I6, I7) 
4.75/4.71	  f1#(I8, I9, I10, I11) -> f2#(I9, I11, rnd3, I11) [1 <= 1 + I10 /\ 1 <= I10 /\ y1 = y1 /\ 1 <= 1 + y1 /\ 1 <= y1 /\ rnd3 = rnd3 /\ 0 <= -1 * I9 /\ -1 * I9 <= 0]
4.75/4.71	R = 
4.75/4.71	  f4(x1, x2, x3, x4) -> f3(x1, x2, x3, x4) 
4.75/4.71	  f3(I0, I1, I2, I3) -> f1(I0, I1, I2, I3) 
4.75/4.71	  f2(I4, I5, I6, I7) -> f1(I4, I5, I6, I7) 
4.75/4.71	  f1(I8, I9, I10, I11) -> f2(I9, I11, rnd3, I11) [1 <= 1 + I10 /\ 1 <= I10 /\ y1 = y1 /\ 1 <= 1 + y1 /\ 1 <= y1 /\ rnd3 = rnd3 /\ 0 <= -1 * I9 /\ -1 * I9 <= 0]
4.75/4.71	
4.75/7.69	EOF