1.63/1.73	MAYBE
1.63/1.73	
1.63/1.73	DP problem for innermost termination.
1.63/1.73	P =
1.63/1.73	  f4#(x1, x2, x3) -> f3#(x1, x2, x3) 
1.63/1.73	  f3#(I0, I1, I2) -> f1#(I0, I1, I2) 
1.63/1.73	  f2#(I3, I4, I5) -> f1#(I3, I4, I5) 
1.63/1.73	  f1#(I6, I7, I8) -> f2#(I6, I7, -1 * I6 + I8) [1 + I7 <= -1 * I6 + I8]
1.63/1.73	R = 
1.63/1.73	  f4(x1, x2, x3) -> f3(x1, x2, x3) 
1.63/1.73	  f3(I0, I1, I2) -> f1(I0, I1, I2) 
1.63/1.73	  f2(I3, I4, I5) -> f1(I3, I4, I5) 
1.63/1.73	  f1(I6, I7, I8) -> f2(I6, I7, -1 * I6 + I8) [1 + I7 <= -1 * I6 + I8]
1.63/1.73	
1.63/1.73	The dependency graph for this problem is:
1.63/1.73	  0 -> 1
1.63/1.73	  1 -> 3
1.63/1.73	  2 -> 3
1.63/1.73	  3 -> 2
1.63/1.73	Where:
1.63/1.73	  0) f4#(x1, x2, x3) -> f3#(x1, x2, x3) 
1.63/1.73	  1) f3#(I0, I1, I2) -> f1#(I0, I1, I2) 
1.63/1.73	  2) f2#(I3, I4, I5) -> f1#(I3, I4, I5) 
1.63/1.73	  3) f1#(I6, I7, I8) -> f2#(I6, I7, -1 * I6 + I8) [1 + I7 <= -1 * I6 + I8]
1.63/1.73	
1.63/1.73	We have the following SCCs.
1.63/1.73	  { 2, 3 }
1.63/1.73	
1.63/1.73	DP problem for innermost termination.
1.63/1.73	P =
1.63/1.73	  f2#(I3, I4, I5) -> f1#(I3, I4, I5) 
1.63/1.73	  f1#(I6, I7, I8) -> f2#(I6, I7, -1 * I6 + I8) [1 + I7 <= -1 * I6 + I8]
1.63/1.73	R = 
1.63/1.73	  f4(x1, x2, x3) -> f3(x1, x2, x3) 
1.63/1.73	  f3(I0, I1, I2) -> f1(I0, I1, I2) 
1.63/1.73	  f2(I3, I4, I5) -> f1(I3, I4, I5) 
1.63/1.73	  f1(I6, I7, I8) -> f2(I6, I7, -1 * I6 + I8) [1 + I7 <= -1 * I6 + I8]
1.63/1.73	
1.76/1.74	EOF