1.09/1.18	MAYBE
1.09/1.18	
1.09/1.18	DP problem for innermost termination.
1.09/1.18	P =
1.09/1.18	  f4#(x1, x2) -> f3#(x1, x2) 
1.09/1.18	  f3#(I0, I1) -> f1#(I0, I1) 
1.09/1.18	  f2#(I2, I3) -> f1#(I2, I3) 
1.09/1.18	  f1#(I4, I5) -> f2#(I4 - I5, I5) [1 + I5 <= 0 /\ 1 <= I4]
1.09/1.18	R = 
1.09/1.18	  f4(x1, x2) -> f3(x1, x2) 
1.09/1.18	  f3(I0, I1) -> f1(I0, I1) 
1.09/1.18	  f2(I2, I3) -> f1(I2, I3) 
1.09/1.18	  f1(I4, I5) -> f2(I4 - I5, I5) [1 + I5 <= 0 /\ 1 <= I4]
1.09/1.18	
1.09/1.18	The dependency graph for this problem is:
1.09/1.18	  0 -> 1
1.09/1.18	  1 -> 3
1.09/1.18	  2 -> 3
1.09/1.18	  3 -> 2
1.09/1.18	Where:
1.09/1.18	  0) f4#(x1, x2) -> f3#(x1, x2) 
1.09/1.18	  1) f3#(I0, I1) -> f1#(I0, I1) 
1.09/1.18	  2) f2#(I2, I3) -> f1#(I2, I3) 
1.09/1.18	  3) f1#(I4, I5) -> f2#(I4 - I5, I5) [1 + I5 <= 0 /\ 1 <= I4]
1.09/1.18	
1.09/1.18	We have the following SCCs.
1.09/1.18	  { 2, 3 }
1.09/1.18	
1.09/1.18	DP problem for innermost termination.
1.09/1.18	P =
1.09/1.18	  f2#(I2, I3) -> f1#(I2, I3) 
1.09/1.18	  f1#(I4, I5) -> f2#(I4 - I5, I5) [1 + I5 <= 0 /\ 1 <= I4]
1.09/1.18	R = 
1.09/1.18	  f4(x1, x2) -> f3(x1, x2) 
1.09/1.18	  f3(I0, I1) -> f1(I0, I1) 
1.09/1.18	  f2(I2, I3) -> f1(I2, I3) 
1.09/1.18	  f1(I4, I5) -> f2(I4 - I5, I5) [1 + I5 <= 0 /\ 1 <= I4]
1.09/1.18	
1.19/1.18	EOF