3.50/3.54	MAYBE
3.50/3.54	
3.50/3.54	DP problem for innermost termination.
3.50/3.54	P =
3.50/3.54	  f4#(x1, x2) -> f1#(x1, x2) 
3.50/3.54	  f3#(I0, I1) -> f1#(I0, I1) 
3.50/3.54	  f1#(I2, I3) -> f3#(I3, I3) [0 <= I2 /\ I2 <= 0 /\ 1 + I3 <= 0]
3.50/3.54	  f2#(I4, I5) -> f1#(I4, I5) 
3.50/3.54	  f1#(I6, I7) -> f2#(I7, I7) [0 <= I6 /\ I6 <= 0 /\ 1 <= I7]
3.50/3.54	R = 
3.50/3.54	  f4(x1, x2) -> f1(x1, x2) 
3.50/3.54	  f3(I0, I1) -> f1(I0, I1) 
3.50/3.54	  f1(I2, I3) -> f3(I3, I3) [0 <= I2 /\ I2 <= 0 /\ 1 + I3 <= 0]
3.50/3.54	  f2(I4, I5) -> f1(I4, I5) 
3.50/3.54	  f1(I6, I7) -> f2(I7, I7) [0 <= I6 /\ I6 <= 0 /\ 1 <= I7]
3.50/3.54	
3.50/3.54	The dependency graph for this problem is:
3.50/3.54	  0 -> 2, 4
3.50/3.54	  1 -> 2, 4
3.50/3.54	  2 -> 1
3.50/3.54	  3 -> 2, 4
3.50/3.54	  4 -> 3
3.50/3.54	Where:
3.50/3.54	  0) f4#(x1, x2) -> f1#(x1, x2) 
3.50/3.54	  1) f3#(I0, I1) -> f1#(I0, I1) 
3.50/3.54	  2) f1#(I2, I3) -> f3#(I3, I3) [0 <= I2 /\ I2 <= 0 /\ 1 + I3 <= 0]
3.50/3.54	  3) f2#(I4, I5) -> f1#(I4, I5) 
3.50/3.54	  4) f1#(I6, I7) -> f2#(I7, I7) [0 <= I6 /\ I6 <= 0 /\ 1 <= I7]
3.50/3.54	
3.50/3.54	We have the following SCCs.
3.50/3.54	  { 1, 2, 3, 4 }
3.50/3.54	
3.50/3.54	DP problem for innermost termination.
3.50/3.54	P =
3.50/3.54	  f3#(I0, I1) -> f1#(I0, I1) 
3.50/3.54	  f1#(I2, I3) -> f3#(I3, I3) [0 <= I2 /\ I2 <= 0 /\ 1 + I3 <= 0]
3.50/3.54	  f2#(I4, I5) -> f1#(I4, I5) 
3.50/3.54	  f1#(I6, I7) -> f2#(I7, I7) [0 <= I6 /\ I6 <= 0 /\ 1 <= I7]
3.50/3.54	R = 
3.50/3.54	  f4(x1, x2) -> f1(x1, x2) 
3.50/3.54	  f3(I0, I1) -> f1(I0, I1) 
3.50/3.54	  f1(I2, I3) -> f3(I3, I3) [0 <= I2 /\ I2 <= 0 /\ 1 + I3 <= 0]
3.50/3.54	  f2(I4, I5) -> f1(I4, I5) 
3.50/3.54	  f1(I6, I7) -> f2(I7, I7) [0 <= I6 /\ I6 <= 0 /\ 1 <= I7]
3.50/3.54	
3.59/6.52	EOF