4.31/4.29	MAYBE
4.31/4.29	
4.31/4.29	DP problem for innermost termination.
4.31/4.29	P =
4.31/4.29	  f5#(x1, x2, x3) -> f4#(x1, x2, x3) 
4.31/4.29	  f4#(I0, I1, I2) -> f1#(2, I1, I2) [I1 <= 3 /\ 0 <= I1 /\ I2 <= 3 /\ 0 <= I2]
4.31/4.29	  f3#(I3, I4, I5) -> f1#(I3, I4, I5) 
4.31/4.29	  f1#(I6, I7, I8) -> f3#(I6, 1 + I7, I8) [1 + 2 * I7 <= I6 + I8]
4.31/4.29	  f2#(I9, I10, I11) -> f1#(I9, I10, I11) 
4.31/4.29	  f1#(I12, I13, I14) -> f2#(I12, -1 + I13, I14) [1 + I12 + I14 <= -1 + 2 * I13]
4.31/4.29	R = 
4.31/4.29	  f5(x1, x2, x3) -> f4(x1, x2, x3) 
4.31/4.29	  f4(I0, I1, I2) -> f1(2, I1, I2) [I1 <= 3 /\ 0 <= I1 /\ I2 <= 3 /\ 0 <= I2]
4.31/4.29	  f3(I3, I4, I5) -> f1(I3, I4, I5) 
4.31/4.29	  f1(I6, I7, I8) -> f3(I6, 1 + I7, I8) [1 + 2 * I7 <= I6 + I8]
4.31/4.29	  f2(I9, I10, I11) -> f1(I9, I10, I11) 
4.31/4.29	  f1(I12, I13, I14) -> f2(I12, -1 + I13, I14) [1 + I12 + I14 <= -1 + 2 * I13]
4.31/4.29	
4.31/4.29	The dependency graph for this problem is:
4.31/4.29	  0 -> 1
4.31/4.29	  1 -> 3, 5
4.31/4.29	  2 -> 3, 5
4.31/4.29	  3 -> 2
4.31/4.29	  4 -> 3, 5
4.31/4.29	  5 -> 4
4.31/4.29	Where:
4.31/4.29	  0) f5#(x1, x2, x3) -> f4#(x1, x2, x3) 
4.31/4.29	  1) f4#(I0, I1, I2) -> f1#(2, I1, I2) [I1 <= 3 /\ 0 <= I1 /\ I2 <= 3 /\ 0 <= I2]
4.31/4.29	  2) f3#(I3, I4, I5) -> f1#(I3, I4, I5) 
4.31/4.29	  3) f1#(I6, I7, I8) -> f3#(I6, 1 + I7, I8) [1 + 2 * I7 <= I6 + I8]
4.31/4.30	  4) f2#(I9, I10, I11) -> f1#(I9, I10, I11) 
4.31/4.30	  5) f1#(I12, I13, I14) -> f2#(I12, -1 + I13, I14) [1 + I12 + I14 <= -1 + 2 * I13]
4.31/4.30	
4.31/4.30	We have the following SCCs.
4.31/4.30	  { 2, 3, 4, 5 }
4.31/4.30	
4.31/4.30	DP problem for innermost termination.
4.31/4.30	P =
4.31/4.30	  f3#(I3, I4, I5) -> f1#(I3, I4, I5) 
4.31/4.30	  f1#(I6, I7, I8) -> f3#(I6, 1 + I7, I8) [1 + 2 * I7 <= I6 + I8]
4.31/4.30	  f2#(I9, I10, I11) -> f1#(I9, I10, I11) 
4.31/4.30	  f1#(I12, I13, I14) -> f2#(I12, -1 + I13, I14) [1 + I12 + I14 <= -1 + 2 * I13]
4.31/4.30	R = 
4.31/4.30	  f5(x1, x2, x3) -> f4(x1, x2, x3) 
4.31/4.30	  f4(I0, I1, I2) -> f1(2, I1, I2) [I1 <= 3 /\ 0 <= I1 /\ I2 <= 3 /\ 0 <= I2]
4.31/4.30	  f3(I3, I4, I5) -> f1(I3, I4, I5) 
4.31/4.30	  f1(I6, I7, I8) -> f3(I6, 1 + I7, I8) [1 + 2 * I7 <= I6 + I8]
4.31/4.30	  f2(I9, I10, I11) -> f1(I9, I10, I11) 
4.31/4.30	  f1(I12, I13, I14) -> f2(I12, -1 + I13, I14) [1 + I12 + I14 <= -1 + 2 * I13]
4.31/4.30	
4.31/7.27	EOF