4.82/4.78	MAYBE
4.82/4.78	
4.82/4.78	DP problem for innermost termination.
4.82/4.78	P =
4.82/4.78	  f6#(x1, x2, x3) -> f4#(x1, x2, x3) 
4.82/4.78	  f5#(I0, I1, I2) -> f3#(I0, I1, I2) 
4.82/4.78	  f3#(I3, I4, I5) -> f5#(I3, -1 + I4, I5) [0 <= 29 - I5]
4.82/4.78	  f3#(I6, I7, I8) -> f1#(I6, -1 + I7, I8) [30 - I8 <= 0]
4.82/4.78	  f4#(I9, I10, I11) -> f1#(I9, I10, I11) 
4.82/4.78	  f1#(I12, I13, I14) -> f3#(I12, I13, I14) [0 <= 19 - I13]
4.82/4.78	R = 
4.82/4.78	  f6(x1, x2, x3) -> f4(x1, x2, x3) 
4.82/4.78	  f5(I0, I1, I2) -> f3(I0, I1, I2) 
4.82/4.78	  f3(I3, I4, I5) -> f5(I3, -1 + I4, I5) [0 <= 29 - I5]
4.82/4.78	  f3(I6, I7, I8) -> f1(I6, -1 + I7, I8) [30 - I8 <= 0]
4.82/4.78	  f4(I9, I10, I11) -> f1(I9, I10, I11) 
4.82/4.78	  f1(I12, I13, I14) -> f3(I12, I13, I14) [0 <= 19 - I13]
4.82/4.78	  f1(I15, I16, I17) -> f2(rnd1, I16, I17) [rnd1 = rnd1 /\ 20 - I16 <= 0]
4.82/4.78	
4.82/4.78	The dependency graph for this problem is:
4.82/4.78	  0 -> 4
4.82/4.78	  1 -> 2, 3
4.82/4.78	  2 -> 1
4.82/4.78	  3 -> 5
4.82/4.78	  4 -> 5
4.82/4.78	  5 -> 2, 3
4.82/4.78	Where:
4.82/4.78	  0) f6#(x1, x2, x3) -> f4#(x1, x2, x3) 
4.82/4.78	  1) f5#(I0, I1, I2) -> f3#(I0, I1, I2) 
4.82/4.78	  2) f3#(I3, I4, I5) -> f5#(I3, -1 + I4, I5) [0 <= 29 - I5]
4.82/4.78	  3) f3#(I6, I7, I8) -> f1#(I6, -1 + I7, I8) [30 - I8 <= 0]
4.82/4.78	  4) f4#(I9, I10, I11) -> f1#(I9, I10, I11) 
4.82/4.78	  5) f1#(I12, I13, I14) -> f3#(I12, I13, I14) [0 <= 19 - I13]
4.82/4.78	
4.82/4.78	We have the following SCCs.
4.82/4.78	  { 1, 2, 3, 5 }
4.82/4.78	
4.82/4.78	DP problem for innermost termination.
4.82/4.78	P =
4.82/4.78	  f5#(I0, I1, I2) -> f3#(I0, I1, I2) 
4.82/4.78	  f3#(I3, I4, I5) -> f5#(I3, -1 + I4, I5) [0 <= 29 - I5]
4.82/4.78	  f3#(I6, I7, I8) -> f1#(I6, -1 + I7, I8) [30 - I8 <= 0]
4.82/4.78	  f1#(I12, I13, I14) -> f3#(I12, I13, I14) [0 <= 19 - I13]
4.82/4.78	R = 
4.82/4.78	  f6(x1, x2, x3) -> f4(x1, x2, x3) 
4.82/4.78	  f5(I0, I1, I2) -> f3(I0, I1, I2) 
4.82/4.78	  f3(I3, I4, I5) -> f5(I3, -1 + I4, I5) [0 <= 29 - I5]
4.82/4.78	  f3(I6, I7, I8) -> f1(I6, -1 + I7, I8) [30 - I8 <= 0]
4.82/4.78	  f4(I9, I10, I11) -> f1(I9, I10, I11) 
4.82/4.78	  f1(I12, I13, I14) -> f3(I12, I13, I14) [0 <= 19 - I13]
4.82/4.78	  f1(I15, I16, I17) -> f2(rnd1, I16, I17) [rnd1 = rnd1 /\ 20 - I16 <= 0]
4.82/4.78	
4.82/7.76	EOF