1.56/1.66	MAYBE
1.56/1.66	
1.56/1.66	DP problem for innermost termination.
1.56/1.66	P =
1.56/1.66	  f4#(x1, x2, x3) -> f3#(x1, x2, x3) 
1.56/1.66	  f3#(I0, I1, I2) -> f1#(I0, 1, 1) 
1.56/1.66	  f2#(I3, I4, I5) -> f1#(I3, I4, I5) 
1.56/1.66	  f1#(I6, I7, I8) -> f2#(I6, I8, 1 + I8) [1 + I7 <= I6]
1.56/1.66	R = 
1.56/1.66	  f4(x1, x2, x3) -> f3(x1, x2, x3) 
1.56/1.66	  f3(I0, I1, I2) -> f1(I0, 1, 1) 
1.56/1.66	  f2(I3, I4, I5) -> f1(I3, I4, I5) 
1.56/1.66	  f1(I6, I7, I8) -> f2(I6, I8, 1 + I8) [1 + I7 <= I6]
1.56/1.66	
1.56/1.66	The dependency graph for this problem is:
1.56/1.66	  0 -> 1
1.56/1.66	  1 -> 3
1.56/1.66	  2 -> 3
1.56/1.66	  3 -> 2
1.56/1.66	Where:
1.56/1.66	  0) f4#(x1, x2, x3) -> f3#(x1, x2, x3) 
1.56/1.66	  1) f3#(I0, I1, I2) -> f1#(I0, 1, 1) 
1.56/1.66	  2) f2#(I3, I4, I5) -> f1#(I3, I4, I5) 
1.56/1.66	  3) f1#(I6, I7, I8) -> f2#(I6, I8, 1 + I8) [1 + I7 <= I6]
1.56/1.66	
1.56/1.66	We have the following SCCs.
1.56/1.66	  { 2, 3 }
1.56/1.66	
1.56/1.66	DP problem for innermost termination.
1.56/1.66	P =
1.56/1.66	  f2#(I3, I4, I5) -> f1#(I3, I4, I5) 
1.56/1.66	  f1#(I6, I7, I8) -> f2#(I6, I8, 1 + I8) [1 + I7 <= I6]
1.56/1.66	R = 
1.56/1.66	  f4(x1, x2, x3) -> f3(x1, x2, x3) 
1.56/1.66	  f3(I0, I1, I2) -> f1(I0, 1, 1) 
1.56/1.66	  f2(I3, I4, I5) -> f1(I3, I4, I5) 
1.56/1.66	  f1(I6, I7, I8) -> f2(I6, I8, 1 + I8) [1 + I7 <= I6]
1.56/1.66	
1.66/1.67	EOF