2.76/2.82	MAYBE
2.76/2.82	
2.76/2.82	DP problem for innermost termination.
2.76/2.82	P =
2.76/2.82	  f5#(x1, x2, x3, x4) -> f1#(x1, x2, x3, x4) 
2.76/2.82	  f4#(I0, I1, I2, I3) -> f2#(I0, I1, I2, I3) 
2.76/2.82	  f2#(I4, I5, I6, I7) -> f4#(I4, I5, I6 + I7, -1 + I7) [1 <= I6]
2.76/2.82	  f1#(I12, I13, I14, I15) -> f2#(I12, I13, I14, I15) 
2.76/2.82	R = 
2.76/2.82	  f5(x1, x2, x3, x4) -> f1(x1, x2, x3, x4) 
2.76/2.82	  f4(I0, I1, I2, I3) -> f2(I0, I1, I2, I3) 
2.76/2.82	  f2(I4, I5, I6, I7) -> f4(I4, I5, I6 + I7, -1 + I7) [1 <= I6]
2.76/2.82	  f2(I8, I9, I10, I11) -> f3(I9, I9, I10, I11) [I10 <= 0]
2.76/2.82	  f1(I12, I13, I14, I15) -> f2(I12, I13, I14, I15) 
2.76/2.82	
2.76/2.82	The dependency graph for this problem is:
2.76/2.82	  0 -> 3
2.76/2.82	  1 -> 2
2.76/2.82	  2 -> 1
2.76/2.82	  3 -> 2
2.76/2.82	Where:
2.76/2.82	  0) f5#(x1, x2, x3, x4) -> f1#(x1, x2, x3, x4) 
2.76/2.82	  1) f4#(I0, I1, I2, I3) -> f2#(I0, I1, I2, I3) 
2.76/2.82	  2) f2#(I4, I5, I6, I7) -> f4#(I4, I5, I6 + I7, -1 + I7) [1 <= I6]
2.76/2.82	  3) f1#(I12, I13, I14, I15) -> f2#(I12, I13, I14, I15) 
2.76/2.82	
2.76/2.82	We have the following SCCs.
2.76/2.82	  { 1, 2 }
2.76/2.82	
2.76/2.82	DP problem for innermost termination.
2.76/2.82	P =
2.76/2.82	  f4#(I0, I1, I2, I3) -> f2#(I0, I1, I2, I3) 
2.76/2.82	  f2#(I4, I5, I6, I7) -> f4#(I4, I5, I6 + I7, -1 + I7) [1 <= I6]
2.76/2.82	R = 
2.76/2.82	  f5(x1, x2, x3, x4) -> f1(x1, x2, x3, x4) 
2.76/2.82	  f4(I0, I1, I2, I3) -> f2(I0, I1, I2, I3) 
2.76/2.82	  f2(I4, I5, I6, I7) -> f4(I4, I5, I6 + I7, -1 + I7) [1 <= I6]
2.76/2.82	  f2(I8, I9, I10, I11) -> f3(I9, I9, I10, I11) [I10 <= 0]
2.76/2.82	  f1(I12, I13, I14, I15) -> f2(I12, I13, I14, I15) 
2.76/2.82	
2.76/5.80	EOF