5.35/5.31	MAYBE
5.35/5.31	
5.35/5.31	DP problem for innermost termination.
5.35/5.31	P =
5.35/5.31	  f5#(x1, x2, x3, x4) -> f4#(x1, x2, x3, x4) 
5.35/5.31	  f4#(I0, I1, I2, I3) -> f1#(I0, I1, I2, I3) 
5.35/5.31	  f3#(I4, I5, I6, I7) -> f1#(I4, I5, I6, I7) 
5.35/5.31	  f1#(I8, I9, I10, I11) -> f3#(-1 + I8, I9 + I10, I10 + I11, I8 + I11) [1 <= I9]
5.35/5.31	  f2#(I12, I13, I14, I15) -> f1#(I12, I13, I14, I15) 
5.35/5.31	  f1#(I16, I17, I18, I19) -> f2#(I16, -1 + I17, I18, I19) [1 <= I17]
5.35/5.31	R = 
5.35/5.31	  f5(x1, x2, x3, x4) -> f4(x1, x2, x3, x4) 
5.35/5.31	  f4(I0, I1, I2, I3) -> f1(I0, I1, I2, I3) 
5.35/5.31	  f3(I4, I5, I6, I7) -> f1(I4, I5, I6, I7) 
5.35/5.31	  f1(I8, I9, I10, I11) -> f3(-1 + I8, I9 + I10, I10 + I11, I8 + I11) [1 <= I9]
5.35/5.31	  f2(I12, I13, I14, I15) -> f1(I12, I13, I14, I15) 
5.35/5.31	  f1(I16, I17, I18, I19) -> f2(I16, -1 + I17, I18, I19) [1 <= I17]
5.35/5.31	
5.35/5.31	The dependency graph for this problem is:
5.35/5.31	  0 -> 1
5.35/5.31	  1 -> 3, 5
5.35/5.31	  2 -> 3, 5
5.35/5.31	  3 -> 2
5.35/5.31	  4 -> 3, 5
5.35/5.31	  5 -> 4
5.35/5.31	Where:
5.35/5.31	  0) f5#(x1, x2, x3, x4) -> f4#(x1, x2, x3, x4) 
5.35/5.31	  1) f4#(I0, I1, I2, I3) -> f1#(I0, I1, I2, I3) 
5.35/5.31	  2) f3#(I4, I5, I6, I7) -> f1#(I4, I5, I6, I7) 
5.35/5.31	  3) f1#(I8, I9, I10, I11) -> f3#(-1 + I8, I9 + I10, I10 + I11, I8 + I11) [1 <= I9]
5.35/5.31	  4) f2#(I12, I13, I14, I15) -> f1#(I12, I13, I14, I15) 
5.35/5.31	  5) f1#(I16, I17, I18, I19) -> f2#(I16, -1 + I17, I18, I19) [1 <= I17]
5.35/5.31	
5.35/5.31	We have the following SCCs.
5.35/5.31	  { 2, 3, 4, 5 }
5.35/5.31	
5.35/5.31	DP problem for innermost termination.
5.35/5.31	P =
5.35/5.31	  f3#(I4, I5, I6, I7) -> f1#(I4, I5, I6, I7) 
5.35/5.31	  f1#(I8, I9, I10, I11) -> f3#(-1 + I8, I9 + I10, I10 + I11, I8 + I11) [1 <= I9]
5.35/5.31	  f2#(I12, I13, I14, I15) -> f1#(I12, I13, I14, I15) 
5.35/5.31	  f1#(I16, I17, I18, I19) -> f2#(I16, -1 + I17, I18, I19) [1 <= I17]
5.35/5.31	R = 
5.35/5.31	  f5(x1, x2, x3, x4) -> f4(x1, x2, x3, x4) 
5.35/5.31	  f4(I0, I1, I2, I3) -> f1(I0, I1, I2, I3) 
5.35/5.31	  f3(I4, I5, I6, I7) -> f1(I4, I5, I6, I7) 
5.35/5.31	  f1(I8, I9, I10, I11) -> f3(-1 + I8, I9 + I10, I10 + I11, I8 + I11) [1 <= I9]
5.35/5.31	  f2(I12, I13, I14, I15) -> f1(I12, I13, I14, I15) 
5.35/5.31	  f1(I16, I17, I18, I19) -> f2(I16, -1 + I17, I18, I19) [1 <= I17]
5.35/5.31	
5.35/8.29	EOF