0.00/0.38	MAYBE
0.00/0.38	
0.00/0.38	DP problem for innermost termination.
0.00/0.38	P =
0.00/0.38	  f4#(x1) -> f3#(x1) 
0.00/0.38	  f3#(I0) -> f1#(I0) 
0.00/0.38	  f2#(I1) -> f1#(I1) 
0.00/0.38	  f1#(I2) -> f2#(1000 + I2) [1 <= 1000 + I2]
0.00/0.38	R = 
0.00/0.38	  f4(x1) -> f3(x1) 
0.00/0.38	  f3(I0) -> f1(I0) 
0.00/0.38	  f2(I1) -> f1(I1) 
0.00/0.38	  f1(I2) -> f2(1000 + I2) [1 <= 1000 + I2]
0.00/0.38	
0.00/0.38	The dependency graph for this problem is:
0.00/0.38	  0 -> 1
0.00/0.38	  1 -> 3
0.00/0.38	  2 -> 3
0.00/0.38	  3 -> 2
0.00/0.38	Where:
0.00/0.38	  0) f4#(x1) -> f3#(x1) 
0.00/0.38	  1) f3#(I0) -> f1#(I0) 
0.00/0.38	  2) f2#(I1) -> f1#(I1) 
0.00/0.38	  3) f1#(I2) -> f2#(1000 + I2) [1 <= 1000 + I2]
0.00/0.38	
0.00/0.38	We have the following SCCs.
0.00/0.38	  { 2, 3 }
0.00/0.38	
0.00/0.38	DP problem for innermost termination.
0.00/0.38	P =
0.00/0.38	  f2#(I1) -> f1#(I1) 
0.00/0.38	  f1#(I2) -> f2#(1000 + I2) [1 <= 1000 + I2]
0.00/0.38	R = 
0.00/0.38	  f4(x1) -> f3(x1) 
0.00/0.38	  f3(I0) -> f1(I0) 
0.00/0.38	  f2(I1) -> f1(I1) 
0.00/0.38	  f1(I2) -> f2(1000 + I2) [1 <= 1000 + I2]
0.00/0.38	
0.00/3.36	EOF