0.81/0.90	MAYBE
0.81/0.90	
0.81/0.90	DP problem for innermost termination.
0.81/0.90	P =
0.81/0.90	  f4#(x1, x2) -> f3#(x1, x2) 
0.81/0.90	  f3#(I0, I1) -> f1#(I0, I1) 
0.81/0.90	  f2#(I2, I3) -> f1#(I2, I3) 
0.81/0.90	  f1#(I4, I5) -> f2#(I4, -1 * I4 + I5) [1 <= -1 * I4 + I5]
0.81/0.90	R = 
0.81/0.90	  f4(x1, x2) -> f3(x1, x2) 
0.81/0.90	  f3(I0, I1) -> f1(I0, I1) 
0.81/0.90	  f2(I2, I3) -> f1(I2, I3) 
0.81/0.90	  f1(I4, I5) -> f2(I4, -1 * I4 + I5) [1 <= -1 * I4 + I5]
0.81/0.90	
0.81/0.90	The dependency graph for this problem is:
0.81/0.90	  0 -> 1
0.81/0.90	  1 -> 3
0.81/0.90	  2 -> 3
0.81/0.90	  3 -> 2
0.81/0.90	Where:
0.81/0.90	  0) f4#(x1, x2) -> f3#(x1, x2) 
0.81/0.90	  1) f3#(I0, I1) -> f1#(I0, I1) 
0.81/0.90	  2) f2#(I2, I3) -> f1#(I2, I3) 
0.81/0.90	  3) f1#(I4, I5) -> f2#(I4, -1 * I4 + I5) [1 <= -1 * I4 + I5]
0.81/0.90	
0.81/0.90	We have the following SCCs.
0.81/0.90	  { 2, 3 }
0.81/0.90	
0.81/0.90	DP problem for innermost termination.
0.81/0.90	P =
0.81/0.90	  f2#(I2, I3) -> f1#(I2, I3) 
0.81/0.90	  f1#(I4, I5) -> f2#(I4, -1 * I4 + I5) [1 <= -1 * I4 + I5]
0.81/0.90	R = 
0.81/0.90	  f4(x1, x2) -> f3(x1, x2) 
0.81/0.90	  f3(I0, I1) -> f1(I0, I1) 
0.81/0.90	  f2(I2, I3) -> f1(I2, I3) 
0.81/0.90	  f1(I4, I5) -> f2(I4, -1 * I4 + I5) [1 <= -1 * I4 + I5]
0.81/0.90	
0.81/3.87	EOF