1.67/1.68	MAYBE
1.67/1.68	
1.67/1.68	DP problem for innermost termination.
1.67/1.68	P =
1.67/1.68	  f5#(x1, x2, x3) -> f4#(x1, x2, x3) 
1.67/1.68	  f4#(I0, I1, I2) -> f1#(I0, I1, I0) 
1.67/1.68	  f3#(I3, I4, I5) -> f1#(I3, I4, I5) 
1.67/1.68	  f2#(I6, I7, I8) -> f3#(I6, I7, I8) [I7 = I7]
1.67/1.68	  f1#(I9, I10, I11) -> f2#(I9, I10, -1 + I11) 
1.67/1.68	R = 
1.67/1.68	  f5(x1, x2, x3) -> f4(x1, x2, x3) 
1.67/1.68	  f4(I0, I1, I2) -> f1(I0, I1, I0) 
1.67/1.68	  f3(I3, I4, I5) -> f1(I3, I4, I5) 
1.67/1.68	  f2(I6, I7, I8) -> f3(I6, I7, I8) [I7 = I7]
1.67/1.68	  f1(I9, I10, I11) -> f2(I9, I10, -1 + I11) 
1.67/1.68	
1.67/1.68	The dependency graph for this problem is:
1.67/1.68	  0 -> 1
1.67/1.68	  1 -> 4
1.67/1.68	  2 -> 4
1.67/1.68	  3 -> 2
1.67/1.68	  4 -> 3
1.67/1.68	Where:
1.67/1.68	  0) f5#(x1, x2, x3) -> f4#(x1, x2, x3) 
1.67/1.68	  1) f4#(I0, I1, I2) -> f1#(I0, I1, I0) 
1.67/1.68	  2) f3#(I3, I4, I5) -> f1#(I3, I4, I5) 
1.67/1.68	  3) f2#(I6, I7, I8) -> f3#(I6, I7, I8) [I7 = I7]
1.67/1.68	  4) f1#(I9, I10, I11) -> f2#(I9, I10, -1 + I11) 
1.67/1.68	
1.67/1.68	We have the following SCCs.
1.67/1.68	  { 2, 3, 4 }
1.67/1.68	
1.67/1.68	DP problem for innermost termination.
1.67/1.68	P =
1.67/1.68	  f3#(I3, I4, I5) -> f1#(I3, I4, I5) 
1.67/1.68	  f2#(I6, I7, I8) -> f3#(I6, I7, I8) [I7 = I7]
1.67/1.68	  f1#(I9, I10, I11) -> f2#(I9, I10, -1 + I11) 
1.67/1.68	R = 
1.67/1.68	  f5(x1, x2, x3) -> f4(x1, x2, x3) 
1.67/1.68	  f4(I0, I1, I2) -> f1(I0, I1, I0) 
1.67/1.68	  f3(I3, I4, I5) -> f1(I3, I4, I5) 
1.67/1.68	  f2(I6, I7, I8) -> f3(I6, I7, I8) [I7 = I7]
1.67/1.68	  f1(I9, I10, I11) -> f2(I9, I10, -1 + I11) 
1.67/1.68	
1.67/4.66	EOF