3.01/3.03	MAYBE
3.01/3.03	
3.01/3.03	DP problem for innermost termination.
3.01/3.03	P =
3.01/3.03	  f4#(x1, x2, x3, x4) -> f3#(x1, x2, x3, x4) 
3.01/3.03	  f3#(I0, I1, I2, I3) -> f1#(0, I1, rnd3, I3) [rnd3 = rnd3]
3.01/3.03	  f2#(I4, I5, I6, I7) -> f1#(I4, I5, I6, I7) 
3.01/3.03	  f1#(I8, I9, I10, I11) -> f2#(1 + I8, rnd2, I10, I8 + I10) [rnd2 = I9 - (I8 + I10) /\ 1 <= I9]
3.01/3.03	R = 
3.01/3.03	  f4(x1, x2, x3, x4) -> f3(x1, x2, x3, x4) 
3.01/3.03	  f3(I0, I1, I2, I3) -> f1(0, I1, rnd3, I3) [rnd3 = rnd3]
3.01/3.03	  f2(I4, I5, I6, I7) -> f1(I4, I5, I6, I7) 
3.01/3.03	  f1(I8, I9, I10, I11) -> f2(1 + I8, rnd2, I10, I8 + I10) [rnd2 = I9 - (I8 + I10) /\ 1 <= I9]
3.01/3.03	
3.01/3.03	The dependency graph for this problem is:
3.01/3.03	  0 -> 1
3.01/3.03	  1 -> 3
3.01/3.03	  2 -> 3
3.01/3.03	  3 -> 2
3.01/3.03	Where:
3.01/3.03	  0) f4#(x1, x2, x3, x4) -> f3#(x1, x2, x3, x4) 
3.01/3.03	  1) f3#(I0, I1, I2, I3) -> f1#(0, I1, rnd3, I3) [rnd3 = rnd3]
3.01/3.03	  2) f2#(I4, I5, I6, I7) -> f1#(I4, I5, I6, I7) 
3.01/3.03	  3) f1#(I8, I9, I10, I11) -> f2#(1 + I8, rnd2, I10, I8 + I10) [rnd2 = I9 - (I8 + I10) /\ 1 <= I9]
3.01/3.03	
3.01/3.03	We have the following SCCs.
3.01/3.03	  { 2, 3 }
3.01/3.03	
3.01/3.03	DP problem for innermost termination.
3.01/3.03	P =
3.01/3.03	  f2#(I4, I5, I6, I7) -> f1#(I4, I5, I6, I7) 
3.01/3.03	  f1#(I8, I9, I10, I11) -> f2#(1 + I8, rnd2, I10, I8 + I10) [rnd2 = I9 - (I8 + I10) /\ 1 <= I9]
3.01/3.03	R = 
3.01/3.03	  f4(x1, x2, x3, x4) -> f3(x1, x2, x3, x4) 
3.01/3.03	  f3(I0, I1, I2, I3) -> f1(0, I1, rnd3, I3) [rnd3 = rnd3]
3.01/3.03	  f2(I4, I5, I6, I7) -> f1(I4, I5, I6, I7) 
3.01/3.03	  f1(I8, I9, I10, I11) -> f2(1 + I8, rnd2, I10, I8 + I10) [rnd2 = I9 - (I8 + I10) /\ 1 <= I9]
3.01/3.03	
3.01/6.01	EOF