2.69/2.96	MAYBE
2.69/2.96	
2.69/2.96	DP problem for innermost termination.
2.69/2.96	P =
2.69/2.96	  f5#(x1, x2, x3, x4) -> f1#(x1, x2, x3, x4) 
2.69/2.96	  f4#(I0, I1, I2, I3) -> f3#(I0, I1, I2, I3) 
2.69/2.96	  f3#(I4, I5, I6, I7) -> f4#(I4, I5, I6 - I7, 1 + I7) [1 <= I6]
2.69/2.96	  f1#(I12, I13, I14, I15) -> f3#(I12, I13, I14, I15) [1 <= I15]
2.69/2.96	R = 
2.69/2.96	  f5(x1, x2, x3, x4) -> f1(x1, x2, x3, x4) 
2.69/2.96	  f4(I0, I1, I2, I3) -> f3(I0, I1, I2, I3) 
2.69/2.96	  f3(I4, I5, I6, I7) -> f4(I4, I5, I6 - I7, 1 + I7) [1 <= I6]
2.69/2.96	  f3(I8, I9, I10, I11) -> f2(I9, I9, I10, I11) [I10 <= 0]
2.69/2.96	  f1(I12, I13, I14, I15) -> f3(I12, I13, I14, I15) [1 <= I15]
2.69/2.96	  f1(I16, I17, I18, I19) -> f2(I17, I17, I18, I19) [1 + I19 <= 1]
2.69/2.96	
2.69/2.96	The dependency graph for this problem is:
2.69/2.96	  0 -> 3
2.69/2.96	  1 -> 2
2.69/2.96	  2 -> 1
2.69/2.96	  3 -> 2
2.69/2.96	Where:
2.69/2.96	  0) f5#(x1, x2, x3, x4) -> f1#(x1, x2, x3, x4) 
2.69/2.96	  1) f4#(I0, I1, I2, I3) -> f3#(I0, I1, I2, I3) 
2.69/2.96	  2) f3#(I4, I5, I6, I7) -> f4#(I4, I5, I6 - I7, 1 + I7) [1 <= I6]
2.69/2.96	  3) f1#(I12, I13, I14, I15) -> f3#(I12, I13, I14, I15) [1 <= I15]
2.69/2.96	
2.69/2.96	We have the following SCCs.
2.69/2.96	  { 1, 2 }
2.69/2.96	
2.69/2.96	DP problem for innermost termination.
2.69/2.96	P =
2.69/2.96	  f4#(I0, I1, I2, I3) -> f3#(I0, I1, I2, I3) 
2.69/2.96	  f3#(I4, I5, I6, I7) -> f4#(I4, I5, I6 - I7, 1 + I7) [1 <= I6]
2.69/2.96	R = 
2.69/2.96	  f5(x1, x2, x3, x4) -> f1(x1, x2, x3, x4) 
2.69/2.96	  f4(I0, I1, I2, I3) -> f3(I0, I1, I2, I3) 
2.69/2.96	  f3(I4, I5, I6, I7) -> f4(I4, I5, I6 - I7, 1 + I7) [1 <= I6]
2.69/2.96	  f3(I8, I9, I10, I11) -> f2(I9, I9, I10, I11) [I10 <= 0]
2.69/2.96	  f1(I12, I13, I14, I15) -> f3(I12, I13, I14, I15) [1 <= I15]
2.69/2.96	  f1(I16, I17, I18, I19) -> f2(I17, I17, I18, I19) [1 + I19 <= 1]
2.69/2.96	
2.69/5.94	EOF