6.40/6.74	MAYBE
6.40/6.74	
6.40/6.74	DP problem for innermost termination.
6.40/6.74	P =
6.40/6.74	  f4#(x1, x2, x3, x4, x5, x6) -> f3#(x1, x2, x3, x4, x5, x6) 
6.40/6.74	  f3#(I0, I1, I2, I3, I4, I5) -> f1#(I0, I1, I2, I3, I4, I3) 
6.40/6.74	  f2#(I6, I7, I8, I9, I10, I11) -> f1#(I6, I7, I8, I9, I10, I11) 
6.40/6.74	  f1#(I12, I13, I14, I15, I16, I17) -> f2#(I12, I13, I14, I15, I12 + I16, I17) [1 + I13 <= I12 + I16 /\ 1 + I17 <= I14]
6.40/6.74	R = 
6.40/6.74	  f4(x1, x2, x3, x4, x5, x6) -> f3(x1, x2, x3, x4, x5, x6) 
6.40/6.74	  f3(I0, I1, I2, I3, I4, I5) -> f1(I0, I1, I2, I3, I4, I3) 
6.40/6.74	  f2(I6, I7, I8, I9, I10, I11) -> f1(I6, I7, I8, I9, I10, I11) 
6.40/6.74	  f1(I12, I13, I14, I15, I16, I17) -> f2(I12, I13, I14, I15, I12 + I16, I17) [1 + I13 <= I12 + I16 /\ 1 + I17 <= I14]
6.40/6.74	
6.40/6.74	The dependency graph for this problem is:
6.40/6.74	  0 -> 1
6.40/6.74	  1 -> 3
6.40/6.74	  2 -> 3
6.40/6.74	  3 -> 2
6.40/6.74	Where:
6.40/6.74	  0) f4#(x1, x2, x3, x4, x5, x6) -> f3#(x1, x2, x3, x4, x5, x6) 
6.40/6.74	  1) f3#(I0, I1, I2, I3, I4, I5) -> f1#(I0, I1, I2, I3, I4, I3) 
6.40/6.74	  2) f2#(I6, I7, I8, I9, I10, I11) -> f1#(I6, I7, I8, I9, I10, I11) 
6.40/6.74	  3) f1#(I12, I13, I14, I15, I16, I17) -> f2#(I12, I13, I14, I15, I12 + I16, I17) [1 + I13 <= I12 + I16 /\ 1 + I17 <= I14]
6.40/6.74	
6.40/6.74	We have the following SCCs.
6.40/6.74	  { 2, 3 }
6.40/6.74	
6.40/6.74	DP problem for innermost termination.
6.40/6.74	P =
6.40/6.74	  f2#(I6, I7, I8, I9, I10, I11) -> f1#(I6, I7, I8, I9, I10, I11) 
6.40/6.74	  f1#(I12, I13, I14, I15, I16, I17) -> f2#(I12, I13, I14, I15, I12 + I16, I17) [1 + I13 <= I12 + I16 /\ 1 + I17 <= I14]
6.40/6.74	R = 
6.40/6.74	  f4(x1, x2, x3, x4, x5, x6) -> f3(x1, x2, x3, x4, x5, x6) 
6.40/6.74	  f3(I0, I1, I2, I3, I4, I5) -> f1(I0, I1, I2, I3, I4, I3) 
6.40/6.74	  f2(I6, I7, I8, I9, I10, I11) -> f1(I6, I7, I8, I9, I10, I11) 
6.40/6.74	  f1(I12, I13, I14, I15, I16, I17) -> f2(I12, I13, I14, I15, I12 + I16, I17) [1 + I13 <= I12 + I16 /\ 1 + I17 <= I14]
6.40/6.74	
6.40/9.72	EOF