0.00/0.49	MAYBE
0.00/0.49	
0.00/0.49	DP problem for innermost termination.
0.00/0.49	P =
0.00/0.49	  f5#(x1, x2) -> f4#(x1, x2) 
0.00/0.49	  f4#(I0, I1) -> f1#(0, 0) 
0.00/0.49	  f3#(I2, I3) -> f1#(I2, I3) 
0.00/0.49	  f1#(I4, I5) -> f3#(1 + I4, 1 + I5) 
0.00/0.49	R = 
0.00/0.49	  f5(x1, x2) -> f4(x1, x2) 
0.00/0.49	  f4(I0, I1) -> f1(0, 0) 
0.00/0.49	  f3(I2, I3) -> f1(I2, I3) 
0.00/0.49	  f1(I4, I5) -> f3(1 + I4, 1 + I5) 
0.00/0.49	  f1(I6, I7) -> f2(I6, I7) [1 + I7 <= I6]
0.00/0.49	
0.00/0.49	The dependency graph for this problem is:
0.00/0.49	  0 -> 1
0.00/0.49	  1 -> 3
0.00/0.49	  2 -> 3
0.00/0.49	  3 -> 2
0.00/0.49	Where:
0.00/0.49	  0) f5#(x1, x2) -> f4#(x1, x2) 
0.00/0.49	  1) f4#(I0, I1) -> f1#(0, 0) 
0.00/0.49	  2) f3#(I2, I3) -> f1#(I2, I3) 
0.00/0.49	  3) f1#(I4, I5) -> f3#(1 + I4, 1 + I5) 
0.00/0.49	
0.00/0.49	We have the following SCCs.
0.00/0.49	  { 2, 3 }
0.00/0.49	
0.00/0.49	DP problem for innermost termination.
0.00/0.49	P =
0.00/0.49	  f3#(I2, I3) -> f1#(I2, I3) 
0.00/0.49	  f1#(I4, I5) -> f3#(1 + I4, 1 + I5) 
0.00/0.49	R = 
0.00/0.49	  f5(x1, x2) -> f4(x1, x2) 
0.00/0.49	  f4(I0, I1) -> f1(0, 0) 
0.00/0.49	  f3(I2, I3) -> f1(I2, I3) 
0.00/0.49	  f1(I4, I5) -> f3(1 + I4, 1 + I5) 
0.00/0.49	  f1(I6, I7) -> f2(I6, I7) [1 + I7 <= I6]
0.00/0.49	
0.00/3.47	EOF