0.00/0.50	MAYBE
0.00/0.50	
0.00/0.50	DP problem for innermost termination.
0.00/0.50	P =
0.00/0.50	  f5#(x1, x2) -> f1#(x1, x2) 
0.00/0.50	  f4#(I0, I1) -> f2#(I0, I1) 
0.00/0.50	  f2#(I2, I3) -> f4#(1 + I2, 1 + I3) 
0.00/0.50	  f1#(I6, I7) -> f2#(0, 0) 
0.00/0.50	R = 
0.00/0.50	  f5(x1, x2) -> f1(x1, x2) 
0.00/0.50	  f4(I0, I1) -> f2(I0, I1) 
0.00/0.50	  f2(I2, I3) -> f4(1 + I2, 1 + I3) 
0.00/0.50	  f2(I4, I5) -> f3(I4, I5) [1 + I5 <= I4]
0.00/0.50	  f1(I6, I7) -> f2(0, 0) 
0.00/0.50	
0.00/0.50	The dependency graph for this problem is:
0.00/0.50	  0 -> 3
0.00/0.50	  1 -> 2
0.00/0.50	  2 -> 1
0.00/0.50	  3 -> 2
0.00/0.50	Where:
0.00/0.50	  0) f5#(x1, x2) -> f1#(x1, x2) 
0.00/0.50	  1) f4#(I0, I1) -> f2#(I0, I1) 
0.00/0.50	  2) f2#(I2, I3) -> f4#(1 + I2, 1 + I3) 
0.00/0.50	  3) f1#(I6, I7) -> f2#(0, 0) 
0.00/0.50	
0.00/0.50	We have the following SCCs.
0.00/0.50	  { 1, 2 }
0.00/0.50	
0.00/0.50	DP problem for innermost termination.
0.00/0.50	P =
0.00/0.50	  f4#(I0, I1) -> f2#(I0, I1) 
0.00/0.50	  f2#(I2, I3) -> f4#(1 + I2, 1 + I3) 
0.00/0.50	R = 
0.00/0.50	  f5(x1, x2) -> f1(x1, x2) 
0.00/0.50	  f4(I0, I1) -> f2(I0, I1) 
0.00/0.50	  f2(I2, I3) -> f4(1 + I2, 1 + I3) 
0.00/0.50	  f2(I4, I5) -> f3(I4, I5) [1 + I5 <= I4]
0.00/0.50	  f1(I6, I7) -> f2(0, 0) 
0.00/0.50	
0.00/3.48	EOF