3.77/4.11	MAYBE
3.77/4.11	
3.77/4.11	DP problem for innermost termination.
3.77/4.11	P =
3.77/4.11	  f5#(x1, x2, x3) -> f4#(x1, x2, x3) 
3.77/4.11	  f4#(I0, I1, I2) -> f2#(I0, I1, I2) 
3.77/4.11	  f2#(I3, I4, I5) -> f1#(I3, I4, I5) [2 <= I3]
3.77/4.11	  f3#(I6, I7, I8) -> f1#(I6, I7, I8) 
3.77/4.11	  f1#(I9, I10, I11) -> f3#(I9, 1 + I10, -1 + I11) [2 <= I11]
3.77/4.11	  f1#(I12, I13, I14) -> f2#(I12 - I13, I13, rnd3) [rnd3 = rnd3]
3.77/4.11	R = 
3.77/4.11	  f5(x1, x2, x3) -> f4(x1, x2, x3) 
3.77/4.11	  f4(I0, I1, I2) -> f2(I0, I1, I2) 
3.77/4.11	  f2(I3, I4, I5) -> f1(I3, I4, I5) [2 <= I3]
3.77/4.11	  f3(I6, I7, I8) -> f1(I6, I7, I8) 
3.77/4.11	  f1(I9, I10, I11) -> f3(I9, 1 + I10, -1 + I11) [2 <= I11]
3.77/4.11	  f1(I12, I13, I14) -> f2(I12 - I13, I13, rnd3) [rnd3 = rnd3]
3.77/4.11	
3.77/4.11	The dependency graph for this problem is:
3.77/4.11	  0 -> 1
3.77/4.11	  1 -> 2
3.77/4.11	  2 -> 4, 5
3.77/4.11	  3 -> 4, 5
3.77/4.11	  4 -> 3
3.77/4.11	  5 -> 2
3.77/4.11	Where:
3.77/4.11	  0) f5#(x1, x2, x3) -> f4#(x1, x2, x3) 
3.77/4.11	  1) f4#(I0, I1, I2) -> f2#(I0, I1, I2) 
3.77/4.11	  2) f2#(I3, I4, I5) -> f1#(I3, I4, I5) [2 <= I3]
3.77/4.11	  3) f3#(I6, I7, I8) -> f1#(I6, I7, I8) 
3.77/4.11	  4) f1#(I9, I10, I11) -> f3#(I9, 1 + I10, -1 + I11) [2 <= I11]
3.77/4.11	  5) f1#(I12, I13, I14) -> f2#(I12 - I13, I13, rnd3) [rnd3 = rnd3]
3.77/4.11	
3.77/4.11	We have the following SCCs.
3.77/4.11	  { 2, 3, 4, 5 }
3.77/4.11	
3.77/4.11	DP problem for innermost termination.
3.77/4.11	P =
3.77/4.11	  f2#(I3, I4, I5) -> f1#(I3, I4, I5) [2 <= I3]
3.77/4.11	  f3#(I6, I7, I8) -> f1#(I6, I7, I8) 
3.77/4.11	  f1#(I9, I10, I11) -> f3#(I9, 1 + I10, -1 + I11) [2 <= I11]
3.77/4.11	  f1#(I12, I13, I14) -> f2#(I12 - I13, I13, rnd3) [rnd3 = rnd3]
3.77/4.11	R = 
3.77/4.11	  f5(x1, x2, x3) -> f4(x1, x2, x3) 
3.77/4.11	  f4(I0, I1, I2) -> f2(I0, I1, I2) 
3.77/4.11	  f2(I3, I4, I5) -> f1(I3, I4, I5) [2 <= I3]
3.77/4.11	  f3(I6, I7, I8) -> f1(I6, I7, I8) 
3.77/4.11	  f1(I9, I10, I11) -> f3(I9, 1 + I10, -1 + I11) [2 <= I11]
3.77/4.11	  f1(I12, I13, I14) -> f2(I12 - I13, I13, rnd3) [rnd3 = rnd3]
3.77/4.11	
3.77/7.09	EOF