1.20/1.25	MAYBE
1.20/1.25	
1.20/1.25	DP problem for innermost termination.
1.20/1.25	P =
1.20/1.25	  f4#(x1, x2) -> f3#(x1, x2) 
1.20/1.25	  f3#(I0, I1) -> f1#(I0, 3000) 
1.20/1.25	  f2#(I2, I3) -> f1#(I2, I3) 
1.20/1.25	  f1#(I4, I5) -> f2#(1000 + I4, I5) [111 <= 1000 + I4 /\ 1 + I5 <= 2000]
1.20/1.25	R = 
1.20/1.25	  f4(x1, x2) -> f3(x1, x2) 
1.20/1.25	  f3(I0, I1) -> f1(I0, 3000) 
1.20/1.25	  f2(I2, I3) -> f1(I2, I3) 
1.20/1.25	  f1(I4, I5) -> f2(1000 + I4, I5) [111 <= 1000 + I4 /\ 1 + I5 <= 2000]
1.20/1.25	
1.20/1.25	The dependency graph for this problem is:
1.20/1.25	  0 -> 1
1.20/1.25	  1 -> 
1.20/1.25	  2 -> 3
1.20/1.25	  3 -> 2
1.20/1.25	Where:
1.20/1.25	  0) f4#(x1, x2) -> f3#(x1, x2) 
1.20/1.25	  1) f3#(I0, I1) -> f1#(I0, 3000) 
1.20/1.25	  2) f2#(I2, I3) -> f1#(I2, I3) 
1.20/1.25	  3) f1#(I4, I5) -> f2#(1000 + I4, I5) [111 <= 1000 + I4 /\ 1 + I5 <= 2000]
1.20/1.25	
1.20/1.25	We have the following SCCs.
1.20/1.25	  { 2, 3 }
1.20/1.25	
1.20/1.25	DP problem for innermost termination.
1.20/1.25	P =
1.20/1.25	  f2#(I2, I3) -> f1#(I2, I3) 
1.20/1.25	  f1#(I4, I5) -> f2#(1000 + I4, I5) [111 <= 1000 + I4 /\ 1 + I5 <= 2000]
1.20/1.25	R = 
1.20/1.25	  f4(x1, x2) -> f3(x1, x2) 
1.20/1.25	  f3(I0, I1) -> f1(I0, 3000) 
1.20/1.25	  f2(I2, I3) -> f1(I2, I3) 
1.20/1.25	  f1(I4, I5) -> f2(1000 + I4, I5) [111 <= 1000 + I4 /\ 1 + I5 <= 2000]
1.20/1.25	
1.20/4.23	EOF