3.23/3.25	MAYBE
3.23/3.25	
3.23/3.25	DP problem for innermost termination.
3.23/3.25	P =
3.23/3.25	  f5#(x1, x2, x3) -> f4#(x1, x2, x3) 
3.23/3.25	  f4#(I0, I1, I2) -> f1#(I0, I1, I2) 
3.23/3.25	  f3#(I3, I4, I5) -> f1#(I3, I4, I5) 
3.23/3.25	  f1#(I6, I7, I8) -> f3#(1 + I6, 1 + I6 + I7, I8) [I7 <= I6]
3.23/3.25	  f2#(I9, I10, I11) -> f1#(I9, I10, I11) 
3.23/3.25	  f1#(I12, I13, I14) -> f2#(I12 - I14, I13 - 2 * I14, -1 + I14) [I13 <= I12]
3.23/3.25	R = 
3.23/3.25	  f5(x1, x2, x3) -> f4(x1, x2, x3) 
3.23/3.25	  f4(I0, I1, I2) -> f1(I0, I1, I2) 
3.23/3.25	  f3(I3, I4, I5) -> f1(I3, I4, I5) 
3.23/3.25	  f1(I6, I7, I8) -> f3(1 + I6, 1 + I6 + I7, I8) [I7 <= I6]
3.23/3.25	  f2(I9, I10, I11) -> f1(I9, I10, I11) 
3.23/3.25	  f1(I12, I13, I14) -> f2(I12 - I14, I13 - 2 * I14, -1 + I14) [I13 <= I12]
3.23/3.25	
3.23/3.25	The dependency graph for this problem is:
3.23/3.25	  0 -> 1
3.23/3.25	  1 -> 3, 5
3.23/3.25	  2 -> 3, 5
3.23/3.25	  3 -> 2
3.23/3.25	  4 -> 3, 5
3.23/3.25	  5 -> 4
3.23/3.25	Where:
3.23/3.25	  0) f5#(x1, x2, x3) -> f4#(x1, x2, x3) 
3.23/3.25	  1) f4#(I0, I1, I2) -> f1#(I0, I1, I2) 
3.23/3.25	  2) f3#(I3, I4, I5) -> f1#(I3, I4, I5) 
3.23/3.25	  3) f1#(I6, I7, I8) -> f3#(1 + I6, 1 + I6 + I7, I8) [I7 <= I6]
3.23/3.25	  4) f2#(I9, I10, I11) -> f1#(I9, I10, I11) 
3.23/3.25	  5) f1#(I12, I13, I14) -> f2#(I12 - I14, I13 - 2 * I14, -1 + I14) [I13 <= I12]
3.23/3.25	
3.23/3.25	We have the following SCCs.
3.23/3.25	  { 2, 3, 4, 5 }
3.23/3.25	
3.23/3.25	DP problem for innermost termination.
3.23/3.25	P =
3.23/3.25	  f3#(I3, I4, I5) -> f1#(I3, I4, I5) 
3.23/3.25	  f1#(I6, I7, I8) -> f3#(1 + I6, 1 + I6 + I7, I8) [I7 <= I6]
3.23/3.25	  f2#(I9, I10, I11) -> f1#(I9, I10, I11) 
3.23/3.25	  f1#(I12, I13, I14) -> f2#(I12 - I14, I13 - 2 * I14, -1 + I14) [I13 <= I12]
3.23/3.25	R = 
3.23/3.25	  f5(x1, x2, x3) -> f4(x1, x2, x3) 
3.23/3.25	  f4(I0, I1, I2) -> f1(I0, I1, I2) 
3.23/3.25	  f3(I3, I4, I5) -> f1(I3, I4, I5) 
3.23/3.25	  f1(I6, I7, I8) -> f3(1 + I6, 1 + I6 + I7, I8) [I7 <= I6]
3.23/3.25	  f2(I9, I10, I11) -> f1(I9, I10, I11) 
3.23/3.25	  f1(I12, I13, I14) -> f2(I12 - I14, I13 - 2 * I14, -1 + I14) [I13 <= I12]
3.23/3.25	
3.23/6.23	EOF