1.14/1.19	MAYBE
1.14/1.19	
1.14/1.19	DP problem for innermost termination.
1.14/1.19	P =
1.14/1.19	  f5#(x1, x2, x3) -> f4#(x1, x2, x3) 
1.14/1.19	  f4#(I0, I1, I2) -> f1#(I1, I1, I2) 
1.14/1.19	  f3#(I3, I4, I5) -> f1#(I3, I4, I5) 
1.14/1.19	  f1#(I6, I7, I8) -> f3#(-1 + I6, -1 + I7, 1 + I8) 
1.14/1.19	R = 
1.14/1.19	  f5(x1, x2, x3) -> f4(x1, x2, x3) 
1.14/1.19	  f4(I0, I1, I2) -> f1(I1, I1, I2) 
1.14/1.19	  f3(I3, I4, I5) -> f1(I3, I4, I5) 
1.14/1.19	  f1(I6, I7, I8) -> f3(-1 + I6, -1 + I7, 1 + I8) 
1.14/1.19	  f1(I9, I10, I11) -> f2(I9, I10, I11) [1 + I9 <= I10]
1.14/1.19	
1.14/1.19	The dependency graph for this problem is:
1.14/1.19	  0 -> 1
1.14/1.19	  1 -> 3
1.14/1.19	  2 -> 3
1.14/1.19	  3 -> 2
1.14/1.19	Where:
1.14/1.19	  0) f5#(x1, x2, x3) -> f4#(x1, x2, x3) 
1.14/1.19	  1) f4#(I0, I1, I2) -> f1#(I1, I1, I2) 
1.14/1.19	  2) f3#(I3, I4, I5) -> f1#(I3, I4, I5) 
1.14/1.19	  3) f1#(I6, I7, I8) -> f3#(-1 + I6, -1 + I7, 1 + I8) 
1.14/1.19	
1.14/1.19	We have the following SCCs.
1.14/1.19	  { 2, 3 }
1.14/1.19	
1.14/1.19	DP problem for innermost termination.
1.14/1.19	P =
1.14/1.19	  f3#(I3, I4, I5) -> f1#(I3, I4, I5) 
1.14/1.19	  f1#(I6, I7, I8) -> f3#(-1 + I6, -1 + I7, 1 + I8) 
1.14/1.19	R = 
1.14/1.19	  f5(x1, x2, x3) -> f4(x1, x2, x3) 
1.14/1.19	  f4(I0, I1, I2) -> f1(I1, I1, I2) 
1.14/1.19	  f3(I3, I4, I5) -> f1(I3, I4, I5) 
1.14/1.19	  f1(I6, I7, I8) -> f3(-1 + I6, -1 + I7, 1 + I8) 
1.14/1.19	  f1(I9, I10, I11) -> f2(I9, I10, I11) [1 + I9 <= I10]
1.14/1.19	
1.14/4.17	EOF