3.12/3.20	MAYBE
3.12/3.20	
3.12/3.20	DP problem for innermost termination.
3.12/3.20	P =
3.12/3.20	  f6#(x1, x2, x3) -> f5#(x1, x2, x3) 
3.12/3.20	  f5#(I0, I1, I2) -> f2#(I0, 0, I0) 
3.12/3.20	  f2#(I3, I4, I5) -> f3#(I3, I4, I5) 
3.12/3.20	  f3#(I6, I7, I8) -> f1#(I6, I7, I8) [1 + I7 <= I8]
3.12/3.20	  f1#(I12, I13, I14) -> f2#(I12, I13, rnd3) [rnd3 = rnd3]
3.12/3.20	  f1#(I15, I16, I17) -> f2#(I15, rnd2, I17) [rnd2 = rnd2]
3.12/3.20	R = 
3.12/3.20	  f6(x1, x2, x3) -> f5(x1, x2, x3) 
3.12/3.20	  f5(I0, I1, I2) -> f2(I0, 0, I0) 
3.12/3.20	  f2(I3, I4, I5) -> f3(I3, I4, I5) 
3.12/3.20	  f3(I6, I7, I8) -> f1(I6, I7, I8) [1 + I7 <= I8]
3.12/3.20	  f3(I9, I10, I11) -> f4(I9, I10, I11) [I11 <= I10]
3.12/3.20	  f1(I12, I13, I14) -> f2(I12, I13, rnd3) [rnd3 = rnd3]
3.12/3.20	  f1(I15, I16, I17) -> f2(I15, rnd2, I17) [rnd2 = rnd2]
3.12/3.20	
3.12/3.20	The dependency graph for this problem is:
3.12/3.20	  0 -> 1
3.12/3.20	  1 -> 2
3.12/3.20	  2 -> 3
3.12/3.20	  3 -> 4, 5
3.12/3.20	  4 -> 2
3.12/3.20	  5 -> 2
3.12/3.20	Where:
3.12/3.20	  0) f6#(x1, x2, x3) -> f5#(x1, x2, x3) 
3.12/3.20	  1) f5#(I0, I1, I2) -> f2#(I0, 0, I0) 
3.12/3.20	  2) f2#(I3, I4, I5) -> f3#(I3, I4, I5) 
3.12/3.20	  3) f3#(I6, I7, I8) -> f1#(I6, I7, I8) [1 + I7 <= I8]
3.12/3.20	  4) f1#(I12, I13, I14) -> f2#(I12, I13, rnd3) [rnd3 = rnd3]
3.12/3.20	  5) f1#(I15, I16, I17) -> f2#(I15, rnd2, I17) [rnd2 = rnd2]
3.12/3.20	
3.12/3.20	We have the following SCCs.
3.12/3.20	  { 2, 3, 4, 5 }
3.12/3.20	
3.12/3.20	DP problem for innermost termination.
3.12/3.20	P =
3.12/3.20	  f2#(I3, I4, I5) -> f3#(I3, I4, I5) 
3.12/3.20	  f3#(I6, I7, I8) -> f1#(I6, I7, I8) [1 + I7 <= I8]
3.12/3.20	  f1#(I12, I13, I14) -> f2#(I12, I13, rnd3) [rnd3 = rnd3]
3.12/3.20	  f1#(I15, I16, I17) -> f2#(I15, rnd2, I17) [rnd2 = rnd2]
3.12/3.20	R = 
3.12/3.20	  f6(x1, x2, x3) -> f5(x1, x2, x3) 
3.12/3.20	  f5(I0, I1, I2) -> f2(I0, 0, I0) 
3.12/3.20	  f2(I3, I4, I5) -> f3(I3, I4, I5) 
3.12/3.20	  f3(I6, I7, I8) -> f1(I6, I7, I8) [1 + I7 <= I8]
3.12/3.20	  f3(I9, I10, I11) -> f4(I9, I10, I11) [I11 <= I10]
3.12/3.20	  f1(I12, I13, I14) -> f2(I12, I13, rnd3) [rnd3 = rnd3]
3.12/3.20	  f1(I15, I16, I17) -> f2(I15, rnd2, I17) [rnd2 = rnd2]
3.12/3.20	
3.12/6.18	EOF