4.25/4.69	MAYBE
4.25/4.69	
4.25/4.69	DP problem for innermost termination.
4.25/4.69	P =
4.25/4.69	  f4#(x1, x2, x3) -> f3#(x1, x2, x3) 
4.25/4.69	  f3#(I0, I1, I2) -> f2#(I0, I1, I2) 
4.25/4.69	  f2#(I3, I4, I5) -> f1#(I3, I4, I5) [I4 <= I5 /\ 0 <= I3]
4.25/4.69	  f1#(I6, I7, I8) -> f2#(-1 + I6, -1 + I7, I8) 
4.25/4.69	  f1#(I9, I10, I11) -> f2#(I9, -1 + I10, -1 + I10 + I11) 
4.25/4.69	R = 
4.25/4.69	  f4(x1, x2, x3) -> f3(x1, x2, x3) 
4.25/4.69	  f3(I0, I1, I2) -> f2(I0, I1, I2) 
4.25/4.69	  f2(I3, I4, I5) -> f1(I3, I4, I5) [I4 <= I5 /\ 0 <= I3]
4.25/4.69	  f1(I6, I7, I8) -> f2(-1 + I6, -1 + I7, I8) 
4.25/4.69	  f1(I9, I10, I11) -> f2(I9, -1 + I10, -1 + I10 + I11) 
4.25/4.69	
4.25/4.69	The dependency graph for this problem is:
4.25/4.69	  0 -> 1
4.25/4.69	  1 -> 2
4.25/4.69	  2 -> 3, 4
4.25/4.69	  3 -> 2
4.25/4.69	  4 -> 2
4.25/4.69	Where:
4.25/4.69	  0) f4#(x1, x2, x3) -> f3#(x1, x2, x3) 
4.25/4.69	  1) f3#(I0, I1, I2) -> f2#(I0, I1, I2) 
4.25/4.69	  2) f2#(I3, I4, I5) -> f1#(I3, I4, I5) [I4 <= I5 /\ 0 <= I3]
4.25/4.69	  3) f1#(I6, I7, I8) -> f2#(-1 + I6, -1 + I7, I8) 
4.25/4.69	  4) f1#(I9, I10, I11) -> f2#(I9, -1 + I10, -1 + I10 + I11) 
4.25/4.69	
4.25/4.69	We have the following SCCs.
4.25/4.69	  { 2, 3, 4 }
4.25/4.69	
4.25/4.69	DP problem for innermost termination.
4.25/4.69	P =
4.25/4.69	  f2#(I3, I4, I5) -> f1#(I3, I4, I5) [I4 <= I5 /\ 0 <= I3]
4.25/4.69	  f1#(I6, I7, I8) -> f2#(-1 + I6, -1 + I7, I8) 
4.25/4.69	  f1#(I9, I10, I11) -> f2#(I9, -1 + I10, -1 + I10 + I11) 
4.25/4.69	R = 
4.25/4.69	  f4(x1, x2, x3) -> f3(x1, x2, x3) 
4.25/4.69	  f3(I0, I1, I2) -> f2(I0, I1, I2) 
4.25/4.69	  f2(I3, I4, I5) -> f1(I3, I4, I5) [I4 <= I5 /\ 0 <= I3]
4.25/4.69	  f1(I6, I7, I8) -> f2(-1 + I6, -1 + I7, I8) 
4.25/4.69	  f1(I9, I10, I11) -> f2(I9, -1 + I10, -1 + I10 + I11) 
4.25/4.69	
4.25/7.67	EOF