6.49/6.40	MAYBE
6.49/6.40	
6.49/6.40	DP problem for innermost termination.
6.49/6.40	P =
6.49/6.40	  f4#(x1, x2, x3, x4, x5, x6) -> f3#(x1, x2, x3, x4, x5, x6) 
6.49/6.40	  f3#(I0, I1, I2, I3, I4, I5) -> f1#(I0, I1, I2, I3, I4, I2) 
6.49/6.40	  f2#(I6, I7, I8, I9, I10, I11) -> f1#(I6, I7, I8, I9, I10, I11) 
6.49/6.40	  f1#(I12, I13, I14, I15, I16, I17) -> f2#(I12, I13, I14, I15, I12 + I16, I17) [1 + I13 <= I12 + I16 /\ 1 + I17 <= I15]
6.49/6.40	R = 
6.49/6.40	  f4(x1, x2, x3, x4, x5, x6) -> f3(x1, x2, x3, x4, x5, x6) 
6.49/6.40	  f3(I0, I1, I2, I3, I4, I5) -> f1(I0, I1, I2, I3, I4, I2) 
6.49/6.40	  f2(I6, I7, I8, I9, I10, I11) -> f1(I6, I7, I8, I9, I10, I11) 
6.49/6.40	  f1(I12, I13, I14, I15, I16, I17) -> f2(I12, I13, I14, I15, I12 + I16, I17) [1 + I13 <= I12 + I16 /\ 1 + I17 <= I15]
6.49/6.40	
6.49/6.40	The dependency graph for this problem is:
6.49/6.40	  0 -> 1
6.49/6.40	  1 -> 3
6.49/6.40	  2 -> 3
6.49/6.40	  3 -> 2
6.49/6.40	Where:
6.49/6.40	  0) f4#(x1, x2, x3, x4, x5, x6) -> f3#(x1, x2, x3, x4, x5, x6) 
6.49/6.40	  1) f3#(I0, I1, I2, I3, I4, I5) -> f1#(I0, I1, I2, I3, I4, I2) 
6.49/6.40	  2) f2#(I6, I7, I8, I9, I10, I11) -> f1#(I6, I7, I8, I9, I10, I11) 
6.49/6.40	  3) f1#(I12, I13, I14, I15, I16, I17) -> f2#(I12, I13, I14, I15, I12 + I16, I17) [1 + I13 <= I12 + I16 /\ 1 + I17 <= I15]
6.49/6.41	
6.49/6.41	We have the following SCCs.
6.49/6.41	  { 2, 3 }
6.49/6.41	
6.49/6.41	DP problem for innermost termination.
6.49/6.41	P =
6.49/6.41	  f2#(I6, I7, I8, I9, I10, I11) -> f1#(I6, I7, I8, I9, I10, I11) 
6.49/6.41	  f1#(I12, I13, I14, I15, I16, I17) -> f2#(I12, I13, I14, I15, I12 + I16, I17) [1 + I13 <= I12 + I16 /\ 1 + I17 <= I15]
6.49/6.41	R = 
6.49/6.41	  f4(x1, x2, x3, x4, x5, x6) -> f3(x1, x2, x3, x4, x5, x6) 
6.49/6.41	  f3(I0, I1, I2, I3, I4, I5) -> f1(I0, I1, I2, I3, I4, I2) 
6.49/6.41	  f2(I6, I7, I8, I9, I10, I11) -> f1(I6, I7, I8, I9, I10, I11) 
6.49/6.41	  f1(I12, I13, I14, I15, I16, I17) -> f2(I12, I13, I14, I15, I12 + I16, I17) [1 + I13 <= I12 + I16 /\ 1 + I17 <= I15]
6.49/6.41	
6.49/9.38	EOF