5.81/5.87	MAYBE
5.81/5.87	
5.81/5.87	DP problem for innermost termination.
5.81/5.87	P =
5.81/5.87	  init#(x1, x2) -> f1#(rnd1, rnd2) 
5.81/5.87	  f3#(I0, I1) -> f2#(I0, I2) [30 <= I0 - 1]
5.81/5.87	  f3#(I3, I4) -> f2#(I3, I5) [I3 <= 29]
5.81/5.87	  f2#(I6, I7) -> f3#(I6 + 1, I8) [20 <= I6 - 1]
5.81/5.87	  f2#(I9, I10) -> f3#(I9 - 1, I11) [10 <= I9 - 1 /\ I9 <= 20]
5.81/5.87	  f1#(I12, I13) -> f2#(I13, I14) [-1 <= I13 - 1 /\ 0 <= I12 - 1]
5.81/5.87	R = 
5.81/5.87	  init(x1, x2) -> f1(rnd1, rnd2) 
5.81/5.87	  f3(I0, I1) -> f2(I0, I2) [30 <= I0 - 1]
5.81/5.87	  f3(I3, I4) -> f2(I3, I5) [I3 <= 29]
5.81/5.87	  f2(I6, I7) -> f3(I6 + 1, I8) [20 <= I6 - 1]
5.81/5.87	  f2(I9, I10) -> f3(I9 - 1, I11) [10 <= I9 - 1 /\ I9 <= 20]
5.81/5.87	  f1(I12, I13) -> f2(I13, I14) [-1 <= I13 - 1 /\ 0 <= I12 - 1]
5.81/5.87	
5.81/5.87	The dependency graph for this problem is:
5.81/5.87	  0 -> 5
5.81/5.87	  1 -> 3
5.81/5.87	  2 -> 3, 4
5.81/5.87	  3 -> 1, 2
5.81/5.87	  4 -> 2
5.81/5.87	  5 -> 3, 4
5.81/5.87	Where:
5.81/5.87	  0) init#(x1, x2) -> f1#(rnd1, rnd2) 
5.81/5.87	  1) f3#(I0, I1) -> f2#(I0, I2) [30 <= I0 - 1]
5.81/5.87	  2) f3#(I3, I4) -> f2#(I3, I5) [I3 <= 29]
5.81/5.87	  3) f2#(I6, I7) -> f3#(I6 + 1, I8) [20 <= I6 - 1]
5.81/5.87	  4) f2#(I9, I10) -> f3#(I9 - 1, I11) [10 <= I9 - 1 /\ I9 <= 20]
5.81/5.87	  5) f1#(I12, I13) -> f2#(I13, I14) [-1 <= I13 - 1 /\ 0 <= I12 - 1]
5.81/5.87	
5.81/5.87	We have the following SCCs.
5.81/5.87	  { 1, 2, 3, 4 }
5.81/5.87	
5.81/5.87	DP problem for innermost termination.
5.81/5.87	P =
5.81/5.87	  f3#(I0, I1) -> f2#(I0, I2) [30 <= I0 - 1]
5.81/5.87	  f3#(I3, I4) -> f2#(I3, I5) [I3 <= 29]
5.81/5.87	  f2#(I6, I7) -> f3#(I6 + 1, I8) [20 <= I6 - 1]
5.81/5.87	  f2#(I9, I10) -> f3#(I9 - 1, I11) [10 <= I9 - 1 /\ I9 <= 20]
5.81/5.87	R = 
5.81/5.87	  init(x1, x2) -> f1(rnd1, rnd2) 
5.81/5.87	  f3(I0, I1) -> f2(I0, I2) [30 <= I0 - 1]
5.81/5.87	  f3(I3, I4) -> f2(I3, I5) [I3 <= 29]
5.81/5.87	  f2(I6, I7) -> f3(I6 + 1, I8) [20 <= I6 - 1]
5.81/5.87	  f2(I9, I10) -> f3(I9 - 1, I11) [10 <= I9 - 1 /\ I9 <= 20]
5.81/5.87	  f1(I12, I13) -> f2(I13, I14) [-1 <= I13 - 1 /\ 0 <= I12 - 1]
5.81/5.87	
5.81/8.84	EOF